1.291 problem 292

Internal problem ID [9273]
Book : Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section : Chapter 1, linear first order
Problem number : 292
Date solved : Thursday, October 17, 2024 at 01:42:47 PM
CAS classification : [[_homogeneous, `class C`], _rational]

Solve

\begin{align*} \left (a y+b x +c \right )^{2} y^{\prime }+\left (\alpha y+\beta x +\gamma \right )^{2}&=0 \end{align*}

1.291.1 Solved as first order homogeneous class Maple C ode

Time used: 1.307 (sec)

Let \(Y = y -y_{0}\) and \(X = x -x_{0}\) then the above is transformed to new ode in \(Y(X)\)

\[ \frac {d}{d X}Y \left (X \right ) = -\frac {\left (\alpha \left (Y \left (X \right )+y_{0} \right )+\beta \left (x_{0} +X \right )+\gamma \right )^{2}}{\left (a \left (Y \left (X \right )+y_{0} \right )+b \left (x_{0} +X \right )+c \right )^{2}} \]

Solving for possible values of \(x_{0}\) and \(y_{0}\) which makes the above ode a homogeneous ode results in

\begin{align*} x_{0}&=\frac {-a \gamma +\alpha c}{a \beta -\alpha b}\\ y_{0}&=\frac {b \gamma -\beta c}{a \beta -\alpha b} \end{align*}

Using these values now it is possible to easily solve for \(Y \left (X \right )\). The above ode now becomes

\begin{align*} \frac {d}{d X}Y \left (X \right ) = -\frac {\beta ^{2} X^{2}+2 \alpha Y \left (X \right ) \beta X +\frac {2 \beta X \alpha \left (b \gamma -\beta c \right )}{a \beta -\alpha b}+\frac {2 \beta ^{2} X \left (-a \gamma +\alpha c \right )}{a \beta -\alpha b}+\alpha ^{2} Y \left (X \right )^{2}+\frac {2 \alpha ^{2} Y \left (X \right ) \left (b \gamma -\beta c \right )}{a \beta -\alpha b}+\frac {2 \alpha Y \left (X \right ) \beta \left (-a \gamma +\alpha c \right )}{a \beta -\alpha b}+\frac {\alpha ^{2} \left (b \gamma -\beta c \right )^{2}}{\left (a \beta -\alpha b \right )^{2}}+\frac {2 \alpha \beta \left (-a \gamma +\alpha c \right ) \left (b \gamma -\beta c \right )}{\left (a \beta -\alpha b \right )^{2}}+\frac {\beta ^{2} \left (-a \gamma +\alpha c \right )^{2}}{\left (a \beta -\alpha b \right )^{2}}+2 \beta X \gamma +2 \alpha Y \left (X \right ) \gamma +\frac {2 \gamma \alpha \left (b \gamma -\beta c \right )}{a \beta -\alpha b}+\frac {2 \gamma \beta \left (-a \gamma +\alpha c \right )}{a \beta -\alpha b}+\gamma ^{2}}{b^{2} X^{2}+2 a Y \left (X \right ) b X +\frac {2 b X a \left (b \gamma -\beta c \right )}{a \beta -\alpha b}+\frac {2 b^{2} X \left (-a \gamma +\alpha c \right )}{a \beta -\alpha b}+a^{2} Y \left (X \right )^{2}+\frac {2 a^{2} Y \left (X \right ) \left (b \gamma -\beta c \right )}{a \beta -\alpha b}+\frac {2 a Y \left (X \right ) b \left (-a \gamma +\alpha c \right )}{a \beta -\alpha b}+\frac {a^{2} \left (b \gamma -\beta c \right )^{2}}{\left (a \beta -\alpha b \right )^{2}}+\frac {2 a b \left (-a \gamma +\alpha c \right ) \left (b \gamma -\beta c \right )}{\left (a \beta -\alpha b \right )^{2}}+\frac {b^{2} \left (-a \gamma +\alpha c \right )^{2}}{\left (a \beta -\alpha b \right )^{2}}+2 b X c +2 a Y \left (X \right ) c +\frac {2 c a \left (b \gamma -\beta c \right )}{a \beta -\alpha b}+\frac {2 c b \left (-a \gamma +\alpha c \right )}{a \beta -\alpha b}+c^{2}} \end{align*}

In canonical form, the ODE is

\begin{align*} Y' &= F(X,Y)\\ &= -\frac {\beta ^{2} X^{2}+2 \alpha Y \beta X +\alpha ^{2} Y^{2}}{b^{2} X^{2}+2 a Y b X +a^{2} Y^{2}}\tag {1} \end{align*}

An ode of the form \(Y' = \frac {M(X,Y)}{N(X,Y)}\) is called homogeneous if the functions \(M(X,Y)\) and \(N(X,Y)\) are both homogeneous functions and of the same order. Recall that a function \(f(X,Y)\) is homogeneous of order \(n\) if

\[ f(t^n X, t^n Y)= t^n f(X,Y) \]

In this case, it can be seen that both \(M=-\beta ^{2} X^{2}-2 \alpha Y \beta X -\alpha ^{2} Y^{2}\) and \(N=b^{2} X^{2}+2 a Y b X +a^{2} Y^{2}\) are both homogeneous and of the same order \(n=2\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {Y}{X}\), or \(Y=uX\). Hence

\[ \frac { \mathop {\mathrm {d}Y}}{\mathop {\mathrm {d}X}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}}X + u \]

Applying the transformation \(Y=uX\) to the above ODE in (1) gives

\begin{align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}}X + u &= -\frac {\left (\alpha u +\beta \right )^{2}}{\left (a u +b \right )^{2}}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {-\frac {\left (\alpha u \left (X \right )+\beta \right )^{2}}{\left (a u \left (X \right )+b \right )^{2}}-u \left (X \right )}{X} \end{align*}

Or

\[ \frac {d}{d X}u \left (X \right )-\frac {-\frac {\left (\alpha u \left (X \right )+\beta \right )^{2}}{\left (a u \left (X \right )+b \right )^{2}}-u \left (X \right )}{X} = 0 \]

Or

\[ \left (\frac {d}{d X}u \left (X \right )\right ) u \left (X \right )^{2} X \,a^{2}+2 \left (\frac {d}{d X}u \left (X \right )\right ) u \left (X \right ) X a b +u \left (X \right )^{3} a^{2}+\left (\frac {d}{d X}u \left (X \right )\right ) X \,b^{2}+2 u \left (X \right )^{2} a b +u \left (X \right )^{2} \alpha ^{2}+2 u \left (X \right ) \alpha \beta +u \left (X \right ) b^{2}+\beta ^{2} = 0 \]

Or

\[ X \left (a u \left (X \right )+b \right )^{2} \left (\frac {d}{d X}u \left (X \right )\right )+u \left (X \right )^{3} a^{2}+\left (2 a b +\alpha ^{2}\right ) u \left (X \right )^{2}+\left (2 \alpha \beta +b^{2}\right ) u \left (X \right )+\beta ^{2} = 0 \]

Which is now solved as separable in \(u \left (X \right )\).

The ode \(\frac {d}{d X}u \left (X \right ) = -\frac {u \left (X \right )^{3} a^{2}+2 u \left (X \right )^{2} a b +u \left (X \right )^{2} \alpha ^{2}+2 u \left (X \right ) \alpha \beta +u \left (X \right ) b^{2}+\beta ^{2}}{X \left (a^{2} u \left (X \right )^{2}+2 u \left (X \right ) a b +b^{2}\right )}\) is separable as it can be written as

\begin{align*} \frac {d}{d X}u \left (X \right )&= -\frac {u \left (X \right )^{3} a^{2}+2 u \left (X \right )^{2} a b +u \left (X \right )^{2} \alpha ^{2}+2 u \left (X \right ) \alpha \beta +u \left (X \right ) b^{2}+\beta ^{2}}{X \left (a^{2} u \left (X \right )^{2}+2 u \left (X \right ) a b +b^{2}\right )}\\ &= f(X) g(u) \end{align*}

Where

\begin{align*} f(X) &= -\frac {1}{X}\\ g(u) &= \frac {u^{3} a^{2}+2 u^{2} a b +\alpha ^{2} u^{2}+2 \alpha \beta u +u \,b^{2}+\beta ^{2}}{a^{2} u^{2}+2 a b u +b^{2}} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(X) \,dX}\\ \int { \frac {a^{2} u^{2}+2 a b u +b^{2}}{u^{3} a^{2}+2 u^{2} a b +\alpha ^{2} u^{2}+2 \alpha \beta u +u \,b^{2}+\beta ^{2}}\,du} &= \int { -\frac {1}{X} \,dX}\\ \int _{}^{u \left (X \right )}\frac {a^{2} \tau ^{2}+2 a b \tau +b^{2}}{a^{2} \tau ^{3}+2 a b \,\tau ^{2}+\alpha ^{2} \tau ^{2}+2 \alpha \beta \tau +b^{2} \tau +\beta ^{2}}d \tau = \ln \left (\frac {1}{X}\right )+c_1 \end{align*}

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or \(\frac {u^{3} a^{2}+2 u^{2} a b +\alpha ^{2} u^{2}+2 \alpha \beta u +u \,b^{2}+\beta ^{2}}{a^{2} u^{2}+2 a b u +b^{2}}=0\) for \(u \left (X \right )\) gives

\begin{align*} u \left (X \right )&=\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}-\frac {2 \left (6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}\right )}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}\\ u \left (X \right )&=-\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{12 a^{2}}+\frac {6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}-\frac {i \sqrt {3}\, \left (\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}+\frac {4 a^{2} \alpha \beta -\frac {2}{3} a^{2} b^{2}-\frac {8}{3} a \,\alpha ^{2} b -\frac {2}{3} \alpha ^{4}}{a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}\right )}{2}\\ u \left (X \right )&=-\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{12 a^{2}}+\frac {6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}+\frac {i \sqrt {3}\, \left (\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}+\frac {4 a^{2} \alpha \beta -\frac {2}{3} a^{2} b^{2}-\frac {8}{3} a \,\alpha ^{2} b -\frac {2}{3} \alpha ^{4}}{a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}\right )}{2} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \int _{}^{u \left (X \right )}\frac {a^{2} \tau ^{2}+2 a b \tau +b^{2}}{a^{2} \tau ^{3}+2 a b \,\tau ^{2}+\alpha ^{2} \tau ^{2}+2 \alpha \beta \tau +b^{2} \tau +\beta ^{2}}d \tau = \ln \left (\frac {1}{X}\right )+c_1\\ u \left (X \right ) = \frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}-\frac {2 \left (6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}\right )}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}\\ u \left (X \right ) = -\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{12 a^{2}}+\frac {6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}-\frac {i \sqrt {3}\, \left (\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}+\frac {4 a^{2} \alpha \beta -\frac {2}{3} a^{2} b^{2}-\frac {8}{3} a \,\alpha ^{2} b -\frac {2}{3} \alpha ^{4}}{a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}\right )}{2}\\ u \left (X \right ) = -\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{12 a^{2}}+\frac {6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}+\frac {i \sqrt {3}\, \left (\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}+\frac {4 a^{2} \alpha \beta -\frac {2}{3} a^{2} b^{2}-\frac {8}{3} a \,\alpha ^{2} b -\frac {2}{3} \alpha ^{4}}{a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}\right )}{2} \end{align*}

Converting \(\int _{}^{u \left (X \right )}\frac {a^{2} \tau ^{2}+2 a b \tau +b^{2}}{a^{2} \tau ^{3}+2 a b \,\tau ^{2}+\alpha ^{2} \tau ^{2}+2 \alpha \beta \tau +b^{2} \tau +\beta ^{2}}d \tau = \ln \left (\frac {1}{X}\right )+c_1\) back to \(Y \left (X \right )\) gives

\begin{align*} \int _{}^{\frac {Y \left (X \right )}{X}}\frac {\left (a \tau +b \right )^{2}}{a^{2} \tau ^{3}+\left (2 a b +\alpha ^{2}\right ) \tau ^{2}+\left (2 \alpha \beta +b^{2}\right ) \tau +\beta ^{2}}d \tau = \ln \left (\frac {1}{X}\right )+c_1 \end{align*}

Converting \(u \left (X \right ) = \frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}-\frac {2 \left (6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}\right )}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}\) back to \(Y \left (X \right )\) gives

\begin{align*} Y \left (X \right ) = X \left (\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}-\frac {2 \left (6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}\right )}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}\right ) \end{align*}

Converting \(u \left (X \right ) = -\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{12 a^{2}}+\frac {6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}-\frac {i \sqrt {3}\, \left (\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}+\frac {4 a^{2} \alpha \beta -\frac {2}{3} a^{2} b^{2}-\frac {8}{3} a \,\alpha ^{2} b -\frac {2}{3} \alpha ^{4}}{a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}\right )}{2}\) back to \(Y \left (X \right )\) gives

\begin{align*} Y \left (X \right ) = X \left (-\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{12 a^{2}}+\frac {6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}-\frac {i \sqrt {3}\, \left (\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}+\frac {4 a^{2} \alpha \beta -\frac {2}{3} a^{2} b^{2}-\frac {8}{3} a \,\alpha ^{2} b -\frac {2}{3} \alpha ^{4}}{a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}\right )}{2}\right ) \end{align*}

Converting \(u \left (X \right ) = -\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{12 a^{2}}+\frac {6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}+\frac {i \sqrt {3}\, \left (\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}+\frac {4 a^{2} \alpha \beta -\frac {2}{3} a^{2} b^{2}-\frac {8}{3} a \,\alpha ^{2} b -\frac {2}{3} \alpha ^{4}}{a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}\right )}{2}\) back to \(Y \left (X \right )\) gives

\begin{align*} Y \left (X \right ) = X \left (-\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{12 a^{2}}+\frac {6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}+\frac {i \sqrt {3}\, \left (\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}+\frac {4 a^{2} \alpha \beta -\frac {2}{3} a^{2} b^{2}-\frac {8}{3} a \,\alpha ^{2} b -\frac {2}{3} \alpha ^{4}}{a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}\right )}{2}\right ) \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} \int _{}^{\frac {Y \left (X \right )}{X}}\frac {\left (a \tau +b \right )^{2}}{a^{2} \tau ^{3}+\left (2 a b +\alpha ^{2}\right ) \tau ^{2}+\left (2 \alpha \beta +b^{2}\right ) \tau +\beta ^{2}}d \tau = \ln \left (\frac {1}{X}\right )+c_1\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= y +y_{0}\\ X &= x +x_{0} \end{align*}

Or

\begin{align*} Y &= y +\frac {b \gamma -\beta c}{a \beta -\alpha b}\\ X &= x +\frac {-a \gamma +\alpha c}{a \beta -\alpha b} \end{align*}

Then the solution in \(y\) becomes using EQ (A)

\begin{align*} \int _{}^{\frac {y-\frac {b \gamma -\beta c}{a \beta -\alpha b}}{x -\frac {-a \gamma +\alpha c}{a \beta -\alpha b}}}\frac {\left (a \tau +b \right )^{2}}{a^{2} \tau ^{3}+\left (2 a b +\alpha ^{2}\right ) \tau ^{2}+\left (2 \alpha \beta +b^{2}\right ) \tau +\beta ^{2}}d \tau = \ln \left (\frac {1}{x -\frac {-a \gamma +\alpha c}{a \beta -\alpha b}}\right )+c_1 \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} Y \left (X \right ) = X \left (\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}-\frac {2 \left (6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}\right )}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}\right )\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= y +y_{0}\\ X &= x +x_{0} \end{align*}

Or

\begin{align*} Y &= y +\frac {b \gamma -\beta c}{a \beta -\alpha b}\\ X &= x +\frac {-a \gamma +\alpha c}{a \beta -\alpha b} \end{align*}

Then the solution in \(y\) becomes using EQ (A)

\begin{align*} y-\frac {b \gamma -\beta c}{a \beta -\alpha b} = \left (x -\frac {-a \gamma +\alpha c}{a \beta -\alpha b}\right ) \left (\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}-\frac {2 \left (6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}\right )}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}\right ) \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} Y \left (X \right ) = X \left (-\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{12 a^{2}}+\frac {6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}-\frac {i \sqrt {3}\, \left (\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}+\frac {4 a^{2} \alpha \beta -\frac {2}{3} a^{2} b^{2}-\frac {8}{3} a \,\alpha ^{2} b -\frac {2}{3} \alpha ^{4}}{a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}\right )}{2}\right )\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= y +y_{0}\\ X &= x +x_{0} \end{align*}

Or

\begin{align*} Y &= y +\frac {b \gamma -\beta c}{a \beta -\alpha b}\\ X &= x +\frac {-a \gamma +\alpha c}{a \beta -\alpha b} \end{align*}

Then the solution in \(y\) becomes using EQ (A)

\begin{align*} y-\frac {b \gamma -\beta c}{a \beta -\alpha b} = \left (x -\frac {-a \gamma +\alpha c}{a \beta -\alpha b}\right ) \left (-\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{12 a^{2}}+\frac {6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}-\frac {i \sqrt {3}\, \left (\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}+\frac {4 a^{2} \alpha \beta -\frac {2}{3} a^{2} b^{2}-\frac {8}{3} a \,\alpha ^{2} b -\frac {2}{3} \alpha ^{4}}{a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}\right )}{2}\right ) \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} Y \left (X \right ) = X \left (-\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{12 a^{2}}+\frac {6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}+\frac {i \sqrt {3}\, \left (\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}+\frac {4 a^{2} \alpha \beta -\frac {2}{3} a^{2} b^{2}-\frac {8}{3} a \,\alpha ^{2} b -\frac {2}{3} \alpha ^{4}}{a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}\right )}{2}\right )\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= y +y_{0}\\ X &= x +x_{0} \end{align*}

Or

\begin{align*} Y &= y +\frac {b \gamma -\beta c}{a \beta -\alpha b}\\ X &= x +\frac {-a \gamma +\alpha c}{a \beta -\alpha b} \end{align*}

Then the solution in \(y\) becomes using EQ (A)

\begin{align*} y-\frac {b \gamma -\beta c}{a \beta -\alpha b} = \left (x -\frac {-a \gamma +\alpha c}{a \beta -\alpha b}\right ) \left (-\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{12 a^{2}}+\frac {6 a^{2} \alpha \beta -a^{2} b^{2}-4 a \,\alpha ^{2} b -\alpha ^{4}}{3 a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}-\frac {2 a b +\alpha ^{2}}{3 a^{2}}+\frac {i \sqrt {3}\, \left (\frac {\left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}{6 a^{2}}+\frac {4 a^{2} \alpha \beta -\frac {2}{3} a^{2} b^{2}-\frac {8}{3} a \,\alpha ^{2} b -\frac {2}{3} \alpha ^{4}}{a^{2} \left (-108 \beta ^{2} a^{4}+144 a^{3} \alpha b \beta +8 a^{3} b^{3}+72 a^{2} \alpha ^{3} \beta -60 a^{2} \alpha ^{2} b^{2}-48 a \,\alpha ^{4} b -8 \alpha ^{6}+12 \sqrt {3}\, \sqrt {27 a^{4} \beta ^{4}-72 a^{3} \alpha b \,\beta ^{3}-4 a^{3} b^{3} \beta ^{2}-4 a^{2} \alpha ^{3} \beta ^{3}+62 a^{2} \alpha ^{2} b^{2} \beta ^{2}+8 a^{2} \alpha \,b^{4} \beta +8 a \,\alpha ^{4} b \,\beta ^{2}-16 a \,\alpha ^{3} b^{3} \beta -4 a \,\alpha ^{2} b^{5}-4 \alpha ^{5} b^{2} \beta -\alpha ^{4} b^{4}}\, a^{2}\right )^{{1}/{3}}}\right )}{2}\right ) \end{align*}

1.291.2 Solved using Lie symmetry for first order ode

Time used: 1.224 (sec)

Writing the ode as

\begin{align*} y^{\prime }&=-\frac {y^{2} \alpha ^{2}+2 y \alpha \beta x +\beta ^{2} x^{2}+2 y \alpha \gamma +2 \beta \gamma x +\gamma ^{2}}{y^{2} a^{2}+2 y x b a +b^{2} x^{2}+2 y c a +2 b c x +c^{2}}\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end{align*}

To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*}

Where the unknown coefficients are

\[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \]

Substituting equations (1E,2E) and \(\omega \) into (A) gives

\begin{equation} \tag{5E} b_{2}-\frac {\left (y^{2} \alpha ^{2}+2 y \alpha \beta x +\beta ^{2} x^{2}+2 y \alpha \gamma +2 \beta \gamma x +\gamma ^{2}\right ) \left (b_{3}-a_{2}\right )}{y^{2} a^{2}+2 y x b a +b^{2} x^{2}+2 y c a +2 b c x +c^{2}}-\frac {\left (y^{2} \alpha ^{2}+2 y \alpha \beta x +\beta ^{2} x^{2}+2 y \alpha \gamma +2 \beta \gamma x +\gamma ^{2}\right )^{2} a_{3}}{\left (y^{2} a^{2}+2 y x b a +b^{2} x^{2}+2 y c a +2 b c x +c^{2}\right )^{2}}-\left (-\frac {2 y \beta \alpha +2 x \,\beta ^{2}+2 \gamma \beta }{y^{2} a^{2}+2 y x b a +b^{2} x^{2}+2 y c a +2 b c x +c^{2}}+\frac {\left (y^{2} \alpha ^{2}+2 y \alpha \beta x +\beta ^{2} x^{2}+2 y \alpha \gamma +2 \beta \gamma x +\gamma ^{2}\right ) \left (2 y b a +2 x \,b^{2}+2 b c \right )}{\left (y^{2} a^{2}+2 y x b a +b^{2} x^{2}+2 y c a +2 b c x +c^{2}\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (-\frac {2 \alpha ^{2} y +2 \alpha \beta x +2 \alpha \gamma }{y^{2} a^{2}+2 y x b a +b^{2} x^{2}+2 y c a +2 b c x +c^{2}}+\frac {\left (y^{2} \alpha ^{2}+2 y \alpha \beta x +\beta ^{2} x^{2}+2 y \alpha \gamma +2 \beta \gamma x +\gamma ^{2}\right ) \left (2 y \,a^{2}+2 x b a +2 a c \right )}{\left (y^{2} a^{2}+2 y x b a +b^{2} x^{2}+2 y c a +2 b c x +c^{2}\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation}

Putting the above in normal form gives

\[ \text {Expression too large to display} \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} \text {Expression too large to display} \end{equation}

Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them.

\[ \{x, y\} \]

The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them

\[ \{x = v_{1}, y = v_{2}\} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} \text {Expression too large to display} \end{equation}

Collecting the above on the terms \(v_i\) introduced, and these are

\[ \{v_{1}, v_{2}\} \]

Equation (7E) now becomes

\begin{equation} \tag{8E} \text {Expression too large to display} \end{equation}

Setting each coefficients in (8E) to zero gives the following equations to solve

\begin{align*} -2 a b \,\beta ^{2} b_{2}+2 \alpha \,b^{2} \beta b_{2}+b^{4} b_{2}+b^{2} \beta ^{2} a_{2}-b^{2} \beta ^{2} b_{3}-\beta ^{4} a_{3}&=0\\ -2 a^{2} \beta ^{2} b_{2}+4 a \,b^{3} b_{2}+4 a b \,\beta ^{2} a_{2}-4 a b \,\beta ^{2} b_{3}+2 \alpha ^{2} b^{2} b_{2}-4 \alpha \,\beta ^{3} a_{3}&=0\\ 4 a^{3} b b_{2}+4 a^{2} \alpha \beta a_{2}-4 a^{2} \alpha \beta b_{3}+2 a^{2} \beta ^{2} a_{3}-4 \alpha ^{3} \beta a_{3}-2 \alpha ^{2} b^{2} a_{3}&=0\\ a^{4} b_{2}+a^{2} \alpha ^{2} a_{2}-a^{2} \alpha ^{2} b_{3}+2 a^{2} \alpha \beta a_{3}-2 a \,\alpha ^{2} b a_{3}-\alpha ^{4} a_{3}&=0\\ -2 a c \,\gamma ^{2} b_{1}+2 \alpha \,c^{2} \gamma b_{1}-2 b c \,\gamma ^{2} a_{1}+2 \beta \,c^{2} \gamma a_{1}+c^{4} b_{2}+c^{2} \gamma ^{2} a_{2}-c^{2} \gamma ^{2} b_{3}-\gamma ^{4} a_{3}&=0\\ 4 a^{3} c b_{2}+2 a^{2} \alpha \beta a_{1}+2 a^{2} \alpha \gamma a_{2}-4 a^{2} \alpha \gamma b_{3}+2 a^{2} \beta \gamma a_{3}-2 a \,\alpha ^{2} b a_{1}+2 a \,\alpha ^{2} c a_{2}-4 a \alpha b \gamma a_{3}+4 a \alpha \beta c a_{3}-4 \alpha ^{3} \gamma a_{3}-2 \alpha ^{2} b c a_{3}&=0\\ -2 a b \,\beta ^{2} b_{1}-4 a b \beta \gamma b_{2}-2 a \,\beta ^{2} c b_{2}+2 \alpha \,b^{2} \beta b_{1}+2 \alpha \,b^{2} \gamma b_{2}+4 \alpha b \beta c b_{2}+4 b^{3} c b_{2}-2 b^{2} \beta \gamma b_{3}+4 b \,\beta ^{2} c a_{2}-2 b \,\beta ^{2} c b_{3}-4 \beta ^{3} \gamma a_{3}&=0\\ -2 a^{2} \alpha \beta b_{2}+6 a^{2} b^{2} b_{2}+3 a^{2} \beta ^{2} a_{2}-3 a^{2} \beta ^{2} b_{3}+2 a \,\alpha ^{2} b b_{2}+4 a \alpha b \beta a_{2}-4 a \alpha b \beta b_{3}+2 a b \,\beta ^{2} a_{3}-\alpha ^{2} b^{2} a_{2}+\alpha ^{2} b^{2} b_{3}-6 \alpha ^{2} \beta ^{2} a_{3}-2 \alpha \,b^{2} \beta a_{3}&=0\\ -2 a b \,\gamma ^{2} b_{1}-4 a \beta c \gamma b_{1}-2 a c \,\gamma ^{2} b_{2}+4 \alpha b c \gamma b_{1}+2 \alpha \beta \,c^{2} b_{1}+2 \alpha \,c^{2} \gamma b_{2}-2 b^{2} \gamma ^{2} a_{1}+4 b \,c^{3} b_{2}-2 b c \,\gamma ^{2} b_{3}+2 \beta ^{2} c^{2} a_{1}+4 \beta \,c^{2} \gamma a_{2}-2 \beta \,c^{2} \gamma b_{3}-4 \beta \,\gamma ^{3} a_{3}&=0\\ -2 a^{2} \gamma ^{2} b_{1}-2 a b \,\gamma ^{2} a_{1}+4 a \beta c \gamma a_{1}+4 a \,c^{3} b_{2}+2 a c \,\gamma ^{2} a_{2}-4 a c \,\gamma ^{2} b_{3}+2 \alpha ^{2} c^{2} b_{1}-4 \alpha b c \gamma a_{1}+2 \alpha \beta \,c^{2} a_{1}+2 \alpha \,c^{2} \gamma a_{2}-4 \alpha \,\gamma ^{3} a_{3}-2 b c \,\gamma ^{2} a_{3}+2 \beta \,c^{2} \gamma a_{3}&=0\\ -4 a^{2} \beta \gamma b_{1}-2 a^{2} \gamma ^{2} b_{2}+12 a b \,c^{2} b_{2}-4 a b \,\gamma ^{2} b_{3}+4 a \,\beta ^{2} c a_{1}+8 a \beta c \gamma a_{2}-8 a \beta c \gamma b_{3}+4 \alpha ^{2} b c b_{1}+2 \alpha ^{2} c^{2} b_{2}-4 \alpha \,b^{2} \gamma a_{1}+4 \alpha \beta \,c^{2} a_{2}-12 \alpha \beta \,\gamma ^{2} a_{3}-2 b^{2} \gamma ^{2} a_{3}+2 \beta ^{2} c^{2} a_{3}&=0\\ -2 a^{2} \beta ^{2} b_{1}-4 a^{2} \beta \gamma b_{2}+12 a \,b^{2} c b_{2}+2 a b \,\beta ^{2} a_{1}+4 a b \beta \gamma a_{2}-8 a b \beta \gamma b_{3}+6 a \,\beta ^{2} c a_{2}-4 a \,\beta ^{2} c b_{3}+2 \alpha ^{2} b^{2} b_{1}+4 \alpha ^{2} b c b_{2}-2 \alpha \,b^{2} \beta a_{1}-2 \alpha \,b^{2} \gamma a_{2}+4 \alpha b \beta c a_{2}-12 \alpha \,\beta ^{2} \gamma a_{3}-2 b^{2} \beta \gamma a_{3}+2 b \,\beta ^{2} c a_{3}&=0\\ -2 a^{2} \alpha \beta b_{1}-2 a^{2} \alpha \gamma b_{2}+12 a^{2} b c b_{2}+2 a^{2} \beta ^{2} a_{1}+4 a^{2} \beta \gamma a_{2}-6 a^{2} \beta \gamma b_{3}+2 a \,\alpha ^{2} b b_{1}+2 a \,\alpha ^{2} c b_{2}-4 a \alpha b \gamma b_{3}+8 a \alpha \beta c a_{2}-4 a \alpha \beta c b_{3}+4 a \,\beta ^{2} c a_{3}-2 \alpha ^{2} b^{2} a_{1}+2 \alpha ^{2} b c b_{3}-12 \alpha ^{2} \beta \gamma a_{3}-4 \alpha \,b^{2} \gamma a_{3}&=0\\ -4 a b \beta \gamma b_{1}-2 a b \,\gamma ^{2} b_{2}-2 a \,\beta ^{2} c b_{1}-4 a \beta c \gamma b_{2}+2 \alpha \,b^{2} \gamma b_{1}+4 \alpha b \beta c b_{1}+4 \alpha b c \gamma b_{2}+2 \alpha \beta \,c^{2} b_{2}-2 b^{2} \beta \gamma a_{1}+6 b^{2} c^{2} b_{2}-b^{2} \gamma ^{2} a_{2}-b^{2} \gamma ^{2} b_{3}+2 b \,\beta ^{2} c a_{1}+4 b \beta c \gamma a_{2}-4 b \beta c \gamma b_{3}+3 \beta ^{2} c^{2} a_{2}-\beta ^{2} c^{2} b_{3}-6 \beta ^{2} \gamma ^{2} a_{3}&=0\\ -2 a^{2} \alpha \gamma b_{1}+2 a^{2} \beta \gamma a_{1}+6 a^{2} c^{2} b_{2}+a^{2} \gamma ^{2} a_{2}-3 a^{2} \gamma ^{2} b_{3}+2 a \,\alpha ^{2} c b_{1}-4 a \alpha b \gamma a_{1}+4 a \alpha \beta c a_{1}+4 a \alpha c \gamma a_{2}-4 a \alpha c \gamma b_{3}-2 a b \,\gamma ^{2} a_{3}+4 a \beta c \gamma a_{3}-2 \alpha ^{2} b c a_{1}+\alpha ^{2} c^{2} a_{2}+\alpha ^{2} c^{2} b_{3}-6 \alpha ^{2} \gamma ^{2} a_{3}-4 \alpha b c \gamma a_{3}+2 \alpha \beta \,c^{2} a_{3}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=\frac {\left (a \gamma -\alpha c \right ) b_{3}}{a \beta -\alpha b}\\ a_{2}&=b_{3}\\ a_{3}&=0\\ b_{1}&=-\frac {\left (b \gamma -\beta c \right ) b_{3}}{a \beta -\alpha b}\\ b_{2}&=0\\ b_{3}&=b_{3} \end{align*}

Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives

\begin{align*} \xi &= \frac {a \beta x -\alpha b x +a \gamma -\alpha c}{a \beta -\alpha b} \\ \eta &= \frac {a \beta y -\alpha b y -b \gamma +\beta c}{a \beta -\alpha b} \\ \end{align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.

The characteristic pde which is used to find the canonical coordinates is

\begin{align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end{align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Therefore

\begin{align*} \frac {dy}{dx} &= \frac {\eta }{\xi }\\ &= \frac {\frac {a \beta y -\alpha b y -b \gamma +\beta c}{a \beta -\alpha b}}{\frac {a \beta x -\alpha b x +a \gamma -\alpha c}{a \beta -\alpha b}}\\ &= \frac {-\alpha b y +\left (a y +c \right ) \beta -b \gamma }{\left (\beta x +\gamma \right ) a +\left (-b x -c \right ) \alpha } \end{align*}

This is easily solved to give

\begin{align*} y = \left (a \beta x -\alpha b x +a \gamma -\alpha c \right ) c_1 +\frac {b \gamma -\beta c}{a \beta -\alpha b} \end{align*}

Where now the coordinate \(R\) is taken as the constant of integration. Hence

\begin{align*} R &= \frac {a \beta y -\alpha b y -b \gamma +\beta c}{\left (a \beta -\alpha b \right ) \left (a \beta x -\alpha b x +a \gamma -\alpha c \right )} \end{align*}

And \(S\) is found from

\begin{align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{\frac {a \beta x -\alpha b x +a \gamma -\alpha c}{a \beta -\alpha b}} \end{align*}

Integrating gives

\begin{align*} S &= \int { \frac {dx}{T}}\\ &= \ln \left (\left (a \beta -\alpha b \right ) x +a \gamma -\alpha c \right ) \end{align*}

Where the constant of integration is set to zero as we just need one solution. Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating

\begin{align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end{align*}

Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode given by

\begin{align*} \omega (x,y) &= -\frac {y^{2} \alpha ^{2}+2 y \alpha \beta x +\beta ^{2} x^{2}+2 y \alpha \gamma +2 \beta \gamma x +\gamma ^{2}}{y^{2} a^{2}+2 y x b a +b^{2} x^{2}+2 y c a +2 b c x +c^{2}} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= \frac {\alpha b y +\left (-a y -c \right ) \beta +b \gamma }{\left (\left (\beta x +\gamma \right ) a +\left (-b x -c \right ) \alpha \right )^{2}}\\ R_{y} &= \frac {1}{\left (\beta x +\gamma \right ) a +\left (-b x -c \right ) \alpha }\\ S_{x} &= \frac {a \beta -\alpha b}{\left (\beta x +\gamma \right ) a +\left (-b x -c \right ) \alpha }\\ S_{y} &= 0 \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= \frac {a \beta -\alpha b}{\left (\left (\beta x +\gamma \right ) a +\left (-b x -c \right ) \alpha \right ) \left (\frac {\alpha b y +\left (-a y -c \right ) \beta +b \gamma }{\left (\left (\beta x +\gamma \right ) a +\left (-b x -c \right ) \alpha \right )^{2}}-\frac {y^{2} \alpha ^{2}+2 y \alpha \beta x +\beta ^{2} x^{2}+2 y \alpha \gamma +2 \beta \gamma x +\gamma ^{2}}{\left (\left (\beta x +\gamma \right ) a +\left (-b x -c \right ) \alpha \right ) \left (y^{2} a^{2}+2 y x b a +b^{2} x^{2}+2 y c a +2 b c x +c^{2}\right )}\right )}\tag {2A} \end{align*}

We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives

\begin{align*} \frac {dS}{dR} &= -\frac {\left (a \beta -\alpha b \right ) {\left (\left (a^{2} \beta -a \alpha b \right ) R +b \right )}^{2}}{a^{2} \left (a \beta -\alpha b \right )^{3} R^{3}+2 \left (a b +\frac {\alpha ^{2}}{2}\right ) \left (a \beta -\alpha b \right )^{2} R^{2}+\left (2 \alpha \beta +b^{2}\right ) \left (a \beta -\alpha b \right ) R +\beta ^{2}} \end{align*}

The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\).

Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {-\frac {\left (a \beta -\alpha b \right ) \left (R \,a^{2} \beta -R a \alpha b +b \right )^{2}}{R^{3} a^{5} \beta ^{3}-3 R^{3} a^{4} \alpha b \,\beta ^{2}+3 R^{3} a^{3} \alpha ^{2} b^{2} \beta -R^{3} a^{2} \alpha ^{3} b^{3}+2 R^{2} a^{3} b \,\beta ^{2}+R^{2} a^{2} \alpha ^{2} \beta ^{2}-4 R^{2} a^{2} \alpha \,b^{2} \beta -2 R^{2} a \,\alpha ^{3} b \beta +2 R^{2} a \,\alpha ^{2} b^{3}+R^{2} \alpha ^{4} b^{2}+2 R a \alpha \,\beta ^{2}+R a \,b^{2} \beta -2 R \,\alpha ^{2} b \beta -R \alpha \,b^{3}+\beta ^{2}}\, dR}\\ S \left (R \right ) &= \left (-a \beta +\alpha b \right ) \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (a^{5} \beta ^{3}-3 a^{4} \alpha b \,\beta ^{2}+3 a^{3} \alpha ^{2} b^{2} \beta -a^{2} \alpha ^{3} b^{3}\right ) \textit {\_Z}^{3}+\left (2 a^{3} b \,\beta ^{2}+a^{2} \alpha ^{2} \beta ^{2}-4 a^{2} \alpha \,b^{2} \beta -2 a \,\alpha ^{3} b \beta +2 a \,\alpha ^{2} b^{3}+\alpha ^{4} b^{2}\right ) \textit {\_Z}^{2}+\left (2 a \alpha \,\beta ^{2}+a \,b^{2} \beta -2 \alpha ^{2} b \beta -\alpha \,b^{3}\right ) \textit {\_Z} +\beta ^{2}\right )}{\sum }\frac {\left (\textit {\_R}^{2} a^{4} \beta ^{2}-2 \textit {\_R}^{2} a^{3} \alpha b \beta +\textit {\_R}^{2} a^{2} \alpha ^{2} b^{2}+2 \textit {\_R} \,a^{2} b \beta -2 \textit {\_R} a \alpha \,b^{2}+b^{2}\right ) \ln \left (R -\textit {\_R} \right )}{3 \textit {\_R}^{2} a^{5} \beta ^{3}-9 \textit {\_R}^{2} a^{4} \alpha b \,\beta ^{2}+9 \textit {\_R}^{2} a^{3} \alpha ^{2} b^{2} \beta -3 \textit {\_R}^{2} a^{2} \alpha ^{3} b^{3}+4 \textit {\_R} \,a^{3} b \,\beta ^{2}+2 \textit {\_R} \,a^{2} \alpha ^{2} \beta ^{2}-8 \textit {\_R} \,a^{2} \alpha \,b^{2} \beta -4 \textit {\_R} a \,\alpha ^{3} b \beta +4 \textit {\_R} a \,\alpha ^{2} b^{3}+2 \textit {\_R} \,\alpha ^{4} b^{2}+2 a \alpha \,\beta ^{2}+a \,b^{2} \beta -2 \alpha ^{2} b \beta -\alpha \,b^{3}}\right ) + c_2 \end{align*}

\begin{align*} S \left (R \right )&= \int -\frac {\left (a \beta -\alpha b \right ) \left (R \,a^{2} \beta -R a \alpha b +b \right )^{2}}{R^{3} a^{5} \beta ^{3}-3 R^{3} a^{4} \alpha b \,\beta ^{2}+3 R^{3} a^{3} \alpha ^{2} b^{2} \beta -R^{3} a^{2} \alpha ^{3} b^{3}+2 R^{2} a^{3} b \,\beta ^{2}+R^{2} a^{2} \alpha ^{2} \beta ^{2}-4 R^{2} a^{2} \alpha \,b^{2} \beta -2 R^{2} a \,\alpha ^{3} b \beta +2 R^{2} a \,\alpha ^{2} b^{3}+R^{2} \alpha ^{4} b^{2}+2 R a \alpha \,\beta ^{2}+R a \,b^{2} \beta -2 R \,\alpha ^{2} b \beta -R \alpha \,b^{3}+\beta ^{2}}d R +c_2 \end{align*}

To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This results in

\begin{align*} \ln \left (\left (\beta x +\gamma \right ) a +\left (-b x -c \right ) \alpha \right ) = \int _{}^{\frac {-\alpha b y+\left (a y+c \right ) \beta -b \gamma }{\left (a \beta -\alpha b \right ) \left (\left (\beta x +\gamma \right ) a +\left (-b x -c \right ) \alpha \right )}}-\frac {\left (a \beta -\alpha b \right ) \left (\textit {\_a} \,a^{2} \beta -\textit {\_a} a \alpha b +b \right )^{2}}{\textit {\_a}^{3} a^{5} \beta ^{3}-3 \textit {\_a}^{3} a^{4} \alpha b \,\beta ^{2}+3 \textit {\_a}^{3} a^{3} \alpha ^{2} b^{2} \beta -\textit {\_a}^{3} a^{2} \alpha ^{3} b^{3}+2 \textit {\_a}^{2} a^{3} b \,\beta ^{2}+\textit {\_a}^{2} a^{2} \alpha ^{2} \beta ^{2}-4 \textit {\_a}^{2} a^{2} \alpha \,b^{2} \beta -2 \textit {\_a}^{2} a \,\alpha ^{3} b \beta +2 \textit {\_a}^{2} a \,\alpha ^{2} b^{3}+\textit {\_a}^{2} \alpha ^{4} b^{2}+2 \textit {\_a} a \alpha \,\beta ^{2}+\textit {\_a} a \,b^{2} \beta -2 \textit {\_a} \,\alpha ^{2} b \beta -\textit {\_a} \alpha \,b^{3}+\beta ^{2}}d \textit {\_a} +c_2 \end{align*}

1.291.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a y \left (x \right )+b x +c \right )^{2} \left (\frac {d}{d x}y \left (x \right )\right )+\left (\alpha y \left (x \right )+\beta x +\gamma \right )^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\frac {\left (\alpha y \left (x \right )+\beta x +\gamma \right )^{2}}{\left (a y \left (x \right )+b x +c \right )^{2}} \end {array} \]

1.291.4 Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
trying inverse linear 
trying homogeneous types: 
trying homogeneous C 
trying homogeneous types: 
trying homogeneous D 
<- homogeneous successful 
<- homogeneous successful`
 

1.291.5 Maple dsolve solution

Solving time : 0.014 (sec)
Leaf size : 115

dsolve((a*y(x)+b*x+c)^2*diff(y(x),x)+(alpha*y(x)+beta*x+gamma)^2 = 0, 
       y(x),singsol=all)
 

\[ y = \frac {\left (\left (b x +c \right ) \alpha -\left (\beta x +\gamma \right ) a \right ) \operatorname {RootOf}\left (\int _{}^{\textit {\_Z}}\frac {\left (\textit {\_a} a -b \right )^{2}}{\textit {\_a}^{3} a^{2}-2 \textit {\_a}^{2} a b -\textit {\_a}^{2} \alpha ^{2}+2 \textit {\_a} \alpha \beta +\textit {\_a} \,b^{2}-\beta ^{2}}d \textit {\_a} +\ln \left (a \beta x -\alpha b x +a \gamma -\alpha c \right )+c_{1} \right )+b \gamma -\beta c}{a \beta -b \alpha } \]

1.291.6 Mathematica DSolve solution

Solving time : 25.145 (sec)
Leaf size : 1653

DSolve[{(a*y[x]+b*x+c)^2*D[y[x],x]+(\[Alpha]*y[x]+\[Beta]*x+\[Gamma])^2==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [\beta (b \alpha -a \beta ) \text {RootSum}\left [-\gamma ^3 b^3-\alpha ^3 y(x)^3 b^3-\gamma \text {$\#$1}^2 b^3-3 \alpha ^2 \gamma y(x)^2 b^3+2 \alpha ^2 \text {$\#$1} y(x)^2 b^3+2 \gamma ^2 \text {$\#$1} b^3-3 \alpha \gamma ^2 y(x) b^3-\alpha \text {$\#$1}^2 y(x) b^3+4 \alpha \gamma \text {$\#$1} y(x) b^3+3 a \alpha ^2 \beta y(x)^3 b^2+3 c \beta \gamma ^2 b^2+c \beta \text {$\#$1}^2 b^2+3 c \alpha ^2 \beta y(x)^2 b^2+6 a \alpha \beta \gamma y(x)^2 b^2-4 a \alpha \beta \text {$\#$1} y(x)^2 b^2-4 c \beta \gamma \text {$\#$1} b^2+3 a \beta \gamma ^2 y(x) b^2+a \beta \text {$\#$1}^2 y(x) b^2+6 c \alpha \beta \gamma y(x) b^2-4 c \alpha \beta \text {$\#$1} y(x) b^2-4 a \beta \gamma \text {$\#$1} y(x) b^2-\alpha \beta \text {$\#$1}^3 b-3 a^2 \alpha \beta ^2 y(x)^3 b+\alpha \beta \gamma \text {$\#$1}^2 b-6 a c \alpha \beta ^2 y(x)^2 b-3 a^2 \beta ^2 \gamma y(x)^2 b+2 a^2 \beta ^2 \text {$\#$1} y(x)^2 b-3 c^2 \beta ^2 \gamma b+2 c^2 \beta ^2 \text {$\#$1} b-3 c^2 \alpha \beta ^2 y(x) b+\alpha ^2 \beta \text {$\#$1}^2 y(x) b-6 a c \beta ^2 \gamma y(x) b+4 a c \beta ^2 \text {$\#$1} y(x) b+c^3 \beta ^3+a \beta ^2 \text {$\#$1}^3+a^3 \beta ^3 y(x)^3-c \alpha \beta ^2 \text {$\#$1}^2+3 a^2 c \beta ^3 y(x)^2+3 a c^2 \beta ^3 y(x)-a \alpha \beta ^2 \text {$\#$1}^2 y(x)\&,\frac {\log (x \beta +\gamma -\text {$\#$1}+\alpha y(x)) \text {$\#$1}^2}{-2 \gamma ^2 b^3-2 \alpha ^2 y(x)^2 b^3+2 \gamma \text {$\#$1} b^3-4 \alpha \gamma y(x) b^3+2 \alpha \text {$\#$1} y(x) b^3+4 a \alpha \beta y(x)^2 b^2+4 c \beta \gamma b^2-2 c \beta \text {$\#$1} b^2+4 c \alpha \beta y(x) b^2+4 a \beta \gamma y(x) b^2-2 a \beta \text {$\#$1} y(x) b^2-2 c^2 \beta ^2 b+3 \alpha \beta \text {$\#$1}^2 b-2 a^2 \beta ^2 y(x)^2 b-2 \alpha \beta \gamma \text {$\#$1} b-4 a c \beta ^2 y(x) b-2 \alpha ^2 \beta \text {$\#$1} y(x) b-3 a \beta ^2 \text {$\#$1}^2+2 c \alpha \beta ^2 \text {$\#$1}+2 a \alpha \beta ^2 \text {$\#$1} y(x)}\&\right ]+\int _1^{y(x)}\left (\frac {-\beta K[1]^2 a^3+b \alpha K[1]^2 a^2-2 c \beta K[1] a^2-2 b x \beta K[1] a^2-c^2 \beta a-b^2 x^2 \beta a-2 b c x \beta a+2 b c \alpha K[1] a+2 b^2 x \alpha K[1] a+b c^2 \alpha +b^3 x^2 \alpha +2 b^2 c x \alpha }{x^2 \gamma b^3+x^2 \alpha K[1] b^3+2 a x \alpha K[1]^2 b^2-c x^2 \beta b^2+2 c x \gamma b^2+2 c x \alpha K[1] b^2-a x^2 \beta K[1] b^2+2 a x \gamma K[1] b^2+a^2 \alpha K[1]^3 b+x^3 \alpha \beta ^2 b+x \alpha \gamma ^2 b+x \alpha ^3 K[1]^2 b+2 a c \alpha K[1]^2 b-2 a^2 x \beta K[1]^2 b+a^2 \gamma K[1]^2 b-2 c^2 x \beta b+c^2 \gamma b+2 x^2 \alpha \beta \gamma b+c^2 \alpha K[1] b+2 x^2 \alpha ^2 \beta K[1] b-4 a c x \beta K[1] b+2 x \alpha ^2 \gamma K[1] b+2 a c \gamma K[1] b-a x^3 \beta ^3-a \gamma ^3-a^3 \beta K[1]^3+c x^2 \alpha \beta ^2+c \alpha \gamma ^2-3 a x \beta \gamma ^2+c \alpha ^3 K[1]^2-a x \alpha ^2 \beta K[1]^2-3 a^2 c \beta K[1]^2-a \alpha ^2 \gamma K[1]^2-c^3 \beta -3 a x^2 \beta ^2 \gamma +2 c x \alpha \beta \gamma -2 a x^2 \alpha \beta ^2 K[1]-2 a \alpha \gamma ^2 K[1]-3 a c^2 \beta K[1]+2 c x \alpha ^2 \beta K[1]+2 c \alpha ^2 \gamma K[1]-4 a x \alpha \beta \gamma K[1]}-\frac {(a \beta -b \alpha ) \left (c^2+2 b x c+2 a K[1] c+b^2 x^2+a^2 K[1]^2+2 a b x K[1]\right )}{-x^2 \gamma b^3-x^2 \alpha K[1] b^3-2 a x \alpha K[1]^2 b^2+c x^2 \beta b^2-2 c x \gamma b^2-2 c x \alpha K[1] b^2+a x^2 \beta K[1] b^2-2 a x \gamma K[1] b^2-a^2 \alpha K[1]^3 b-x^3 \alpha \beta ^2 b-x \alpha \gamma ^2 b-x \alpha ^3 K[1]^2 b-2 a c \alpha K[1]^2 b+2 a^2 x \beta K[1]^2 b-a^2 \gamma K[1]^2 b+2 c^2 x \beta b-c^2 \gamma b-2 x^2 \alpha \beta \gamma b-c^2 \alpha K[1] b-2 x^2 \alpha ^2 \beta K[1] b+4 a c x \beta K[1] b-2 x \alpha ^2 \gamma K[1] b-2 a c \gamma K[1] b+a x^3 \beta ^3+a \gamma ^3+a^3 \beta K[1]^3-c x^2 \alpha \beta ^2-c \alpha \gamma ^2+3 a x \beta \gamma ^2-c \alpha ^3 K[1]^2+a x \alpha ^2 \beta K[1]^2+3 a^2 c \beta K[1]^2+a \alpha ^2 \gamma K[1]^2+c^3 \beta +3 a x^2 \beta ^2 \gamma -2 c x \alpha \beta \gamma +2 a x^2 \alpha \beta ^2 K[1]+2 a \alpha \gamma ^2 K[1]+3 a c^2 \beta K[1]-2 c x \alpha ^2 \beta K[1]-2 c \alpha ^2 \gamma K[1]+4 a x \alpha \beta \gamma K[1]}\right )dK[1]=c_1,y(x)\right ] \]