Internal problem ID [10394]
Internal file name [OUTPUT/9342_Monday_June_06_2022_02_14_05_PM_28407073/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 1, section 1.2. Riccati Equation. 1.2.2. Equations Containing Power
Functions
Problem number: 65.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_rational, _Riccati]
\[ \boxed {x^{3} y^{\prime }-a \,x^{3} y^{2}-\left (b \,x^{2}+c \right ) y=s x} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {a \,x^{3} y^{2}+b \,x^{2} y +y c +s x}{x^{3}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,y^{2}+\frac {y b}{x}+\frac {y c}{x^{3}}+\frac {s}{x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {s}{x^{2}}\), \(f_1(x)=\frac {b \,x^{2}+c}{x^{3}}\) and \(f_2(x)=a\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{a u} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=0\\ f_1 f_2 &=\frac {\left (b \,x^{2}+c \right ) a}{x^{3}}\\ f_2^2 f_0 &=\frac {a^{2} s}{x^{2}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a u^{\prime \prime }\left (x \right )-\frac {\left (b \,x^{2}+c \right ) a u^{\prime }\left (x \right )}{x^{3}}+\frac {a^{2} s u \left (x \right )}{x^{2}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \frac {4 \,{\mathrm e}^{-\frac {c}{2 x^{2}}} \left (\frac {\left (\left (-1+b \right ) x^{2}+c \right ) \left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right ) c_{1} \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )}{2}+x^{2} c_{1} \left (\frac {1}{2}+\frac {\left (-1+b \right ) \sqrt {-4 a s +b^{2}+2 b +1}}{2}+a s -\frac {b^{2}}{2}\right ) \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}-\frac {3}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )-4 \left (\frac {\left (\left (-b +1\right ) x^{2}-c \right ) \operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )}{2}+\operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}-\frac {3}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right ) x^{2}\right ) c_{2} \right ) x^{-\frac {3}{2}+\frac {b}{2}-\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}}}{\left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right ) \left (\sqrt {-4 a s +b^{2}+2 b +1}+b +1\right )} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {8 \,{\mathrm e}^{-\frac {c}{2 x^{2}}} x^{-\frac {1}{2}+\frac {b}{2}-\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}} \left (c_{1} \left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right ) \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )+4 \operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right ) c_{2} \right ) s^{2} a^{2}}{\left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right )^{2} \left (\sqrt {-4 a s +b^{2}+2 b +1}+b +1\right )^{2}} \] Using the above in (1) gives the solution \[ y = -\frac {2 x^{-\frac {1}{2}+\frac {b}{2}-\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}} \left (c_{1} \left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right ) \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )+4 \operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right ) c_{2} \right ) s^{2} a \,x^{\frac {3}{2}-\frac {b}{2}+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}}}{\left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right ) \left (\sqrt {-4 a s +b^{2}+2 b +1}+b +1\right ) \left (\frac {\left (\left (-1+b \right ) x^{2}+c \right ) \left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right ) c_{1} \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )}{2}+x^{2} c_{1} \left (\frac {1}{2}+\frac {\left (-1+b \right ) \sqrt {-4 a s +b^{2}+2 b +1}}{2}+a s -\frac {b^{2}}{2}\right ) \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}-\frac {3}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )-4 \left (\frac {\left (\left (-b +1\right ) x^{2}-c \right ) \operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )}{2}+\operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}-\frac {3}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right ) x^{2}\right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {8 \left (\frac {c_{3} \left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right ) \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )}{4}+\operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )\right ) x \,s^{2} a}{\left (\frac {\left (\left (-1+b \right ) x^{2}+c \right ) \left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right ) c_{3} \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )}{2}+x^{2} c_{3} \left (\frac {1}{2}+\frac {\left (-1+b \right ) \sqrt {-4 a s +b^{2}+2 b +1}}{2}+a s -\frac {b^{2}}{2}\right ) \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}-\frac {3}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )+2 \operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right ) \left (\left (-1+b \right ) x^{2}+c \right )-4 \operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}-\frac {3}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right ) x^{2}\right ) \left (\sqrt {-4 a s +b^{2}+2 b +1}+b +1\right ) \left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {8 \left (\frac {c_{3} \left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right ) \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )}{4}+\operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )\right ) x \,s^{2} a}{\left (\frac {\left (\left (-1+b \right ) x^{2}+c \right ) \left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right ) c_{3} \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )}{2}+x^{2} c_{3} \left (\frac {1}{2}+\frac {\left (-1+b \right ) \sqrt {-4 a s +b^{2}+2 b +1}}{2}+a s -\frac {b^{2}}{2}\right ) \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}-\frac {3}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )+2 \operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right ) \left (\left (-1+b \right ) x^{2}+c \right )-4 \operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}-\frac {3}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right ) x^{2}\right ) \left (\sqrt {-4 a s +b^{2}+2 b +1}+b +1\right ) \left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {8 \left (\frac {c_{3} \left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right ) \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )}{4}+\operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )\right ) x \,s^{2} a}{\left (\frac {\left (\left (-1+b \right ) x^{2}+c \right ) \left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right ) c_{3} \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )}{2}+x^{2} c_{3} \left (\frac {1}{2}+\frac {\left (-1+b \right ) \sqrt {-4 a s +b^{2}+2 b +1}}{2}+a s -\frac {b^{2}}{2}\right ) \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}-\frac {3}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )+2 \operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right ) \left (\left (-1+b \right ) x^{2}+c \right )-4 \operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}-\frac {3}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right ) x^{2}\right ) \left (\sqrt {-4 a s +b^{2}+2 b +1}+b +1\right ) \left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{3} y^{\prime }-a \,x^{3} y^{2}-\left (b \,x^{2}+c \right ) y=s x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a \,x^{3} y^{2}+\left (b \,x^{2}+c \right ) y+s x}{x^{3}} \end {array} \]
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 Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying Riccati trying Riccati sub-methods: trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (b*x^2+c)*(diff(y(x), x))/x^3-a*s*y(x)/x^2, y(x)` *** Sublevel 2 Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Kummer successful <- special function solution successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 435
dsolve(x^3*diff(y(x),x)=a*x^3*y(x)^2+(b*x^2+c)*y(x)+s*x,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\left (\frac {\left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right ) \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )}{4}+c_{1} \operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )\right ) \left (\sqrt {-4 a s +b^{2}+2 b +1}+b +1\right ) x \left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right )}{2 a \left (\frac {\left (\left (b -1\right ) x^{2}+c \right ) \left (1-\sqrt {-4 a s +b^{2}+2 b +1}+b \right ) \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )}{2}+\left (\frac {1}{2}+\frac {\left (b -1\right ) \sqrt {-4 a s +b^{2}+2 b +1}}{2}+a s -\frac {b^{2}}{2}\right ) x^{2} \operatorname {KummerM}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}-\frac {3}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )-4 c_{1} \left (\frac {\left (\left (-b +1\right ) x^{2}-c \right ) \operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}+\frac {1}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right )}{2}+\operatorname {KummerU}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{4}+\frac {b}{4}-\frac {3}{4}, 1+\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{2}, \frac {c}{2 x^{2}}\right ) x^{2}\right )\right )} \]
✓ Solution by Mathematica
Time used: 3.199 (sec). Leaf size: 907
DSolve[x^3*y'[x]==a*x^3*y[x]^2+(b*x^2+c)*y[x]+s*x,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {-\left (\left (\sqrt {-4 a s+b^2+2 b+1}-b-1\right ) c^{\frac {1}{2} \sqrt {-4 a s+b^2+2 b+1}} \left (\frac {1}{x}\right )^{\sqrt {-4 a s+b^2+2 b+1}} \operatorname {Hypergeometric1F1}\left (\frac {1}{4} \left (-b+\sqrt {b^2+2 b-4 a s+1}-1\right ),\frac {1}{2} \left (\sqrt {b^2+2 b-4 a s+1}+2\right ),-\frac {c}{2 x^2}\right )\right )+\frac {\left (\sqrt {-4 a s+b^2+2 b+1}-b-1\right ) c^{\frac {1}{2} \sqrt {-4 a s+b^2+2 b+1}+1} \left (\frac {1}{x}\right )^{\sqrt {-4 a s+b^2+2 b+1}+2} \operatorname {Hypergeometric1F1}\left (\frac {1}{4} \left (-b+\sqrt {b^2+2 b-4 a s+1}+3\right ),\frac {1}{2} \left (\sqrt {b^2+2 b-4 a s+1}+4\right ),-\frac {c}{2 x^2}\right )}{\sqrt {-4 a s+b^2+2 b+1}+2}+\frac {c_1 2^{\frac {1}{2} \sqrt {-4 a s+b^2+2 b+1}} \left (\sqrt {-4 a s+b^2+2 b+1}+b+1\right ) \left (x^2 \left (\sqrt {-4 a s+b^2+2 b+1}-2\right ) \operatorname {Hypergeometric1F1}\left (\frac {1}{4} \left (-b-\sqrt {b^2+2 b-4 a s+1}-1\right ),1-\frac {1}{2} \sqrt {b^2+2 b-4 a s+1},-\frac {c}{2 x^2}\right )+c \operatorname {Hypergeometric1F1}\left (\frac {1}{4} \left (-b-\sqrt {b^2+2 b-4 a s+1}+3\right ),2-\frac {1}{2} \sqrt {b^2+2 b-4 a s+1},-\frac {c}{2 x^2}\right )\right )}{x^2 \left (\sqrt {-4 a s+b^2+2 b+1}-2\right )}}{2 a x \left (c^{\frac {1}{2} \sqrt {-4 a s+b^2+2 b+1}} \left (\frac {1}{x}\right )^{\sqrt {-4 a s+b^2+2 b+1}} \operatorname {Hypergeometric1F1}\left (\frac {1}{4} \left (-b+\sqrt {b^2+2 b-4 a s+1}-1\right ),\frac {1}{2} \left (\sqrt {b^2+2 b-4 a s+1}+2\right ),-\frac {c}{2 x^2}\right )+c_1 2^{\frac {1}{2} \sqrt {-4 a s+b^2+2 b+1}} \operatorname {Hypergeometric1F1}\left (\frac {1}{4} \left (-b-\sqrt {b^2+2 b-4 a s+1}-1\right ),1-\frac {1}{2} \sqrt {b^2+2 b-4 a s+1},-\frac {c}{2 x^2}\right )\right )} \\ y(x)\to \frac {\frac {c \left (b \left (\sqrt {-4 a s+b^2+2 b+1}+4\right )+3 \sqrt {-4 a s+b^2+2 b+1}-4 a s+b^2+3\right ) \operatorname {Hypergeometric1F1}\left (\frac {1}{4} \left (-b-\sqrt {b^2+2 b-4 a s+1}+3\right ),2-\frac {1}{2} \sqrt {b^2+2 b-4 a s+1},-\frac {c}{2 x^2}\right )}{\operatorname {Hypergeometric1F1}\left (\frac {1}{4} \left (-b-\sqrt {b^2+2 b-4 a s+1}-1\right ),1-\frac {1}{2} \sqrt {b^2+2 b-4 a s+1},-\frac {c}{2 x^2}\right )}+x^2 \left (-4 a s+b^2+2 b-3\right ) \left (\sqrt {-4 a s+b^2+2 b+1}+b+1\right )}{2 a x^3 \left (4 a s-b^2-2 b+3\right )} \\ \end{align*}