2.5.15 problem 15

Maple step by step solution
Maple trace
Maple dsolve solution
Mathematica DSolve solution

Internal problem ID [8402]
Book : Own collection of miscellaneous problems
Section : section 5.0
Problem number : 15
Date solved : Sunday, November 10, 2024 at 08:21:00 PM
CAS classification : [_quadrature]

Solve

\begin{align*} h^{2}+\frac {2 a h}{\sqrt {1+{h^{\prime }}^{2}}}&=b^{2} \end{align*}

Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} h^{\prime }&=-\frac {\sqrt {-h^{4}+4 a^{2} h^{2}+2 h^{2} b^{2}-b^{4}}}{\left (h+b \right ) \left (h-b \right )} \\ \tag{2} h^{\prime }&=\frac {\sqrt {-h^{4}+4 a^{2} h^{2}+2 h^{2} b^{2}-b^{4}}}{\left (h+b \right ) \left (h-b \right )} \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{h}-\frac {\left (b +\tau \right ) \left (\tau -b \right )}{\sqrt {4 a^{2} \tau ^{2}-b^{4}+2 b^{2} \tau ^{2}-\tau ^{4}}}d \tau = u +c_3 \]

Singular solutions are found by solving

\begin{align*} -\frac {\sqrt {4 a^{2} h^{2}-b^{4}+2 h^{2} b^{2}-h^{4}}}{\left (b +h \right ) \left (h -b \right )}&= 0 \end{align*}

for \(h\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} h = -a -\sqrt {a^{2}+b^{2}}\\ h = -a +\sqrt {a^{2}+b^{2}}\\ h = a -\sqrt {a^{2}+b^{2}}\\ h = a +\sqrt {a^{2}+b^{2}} \end{align*}

We now need to find the singular solutions, these are found by finding for what values \((-\frac {\sqrt {4 a^{2} h^{2}-b^{4}+2 h^{2} b^{2}-h^{4}}}{\left (b +h \right ) \left (h -b \right )})\) is zero. These give

\begin{align*} h&=-a -\sqrt {a^{2}+b^{2}} \\ h&=-a +\sqrt {a^{2}+b^{2}} \\ h&=a -\sqrt {a^{2}+b^{2}} \\ h&=a +\sqrt {a^{2}+b^{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.

The solution \(h = -a -\sqrt {a^{2}+b^{2}}\) satisfies the ode and initial conditions.

The solution \(h = -a +\sqrt {a^{2}+b^{2}}\) satisfies the ode and initial conditions.

The solution \(h = a -\sqrt {a^{2}+b^{2}}\) does not satisfiy the ode and initial conditions. Hence will not be used.

The solution \(h = a +\sqrt {a^{2}+b^{2}}\) does not satisfiy the ode and initial conditions. Hence will not be used.

Solving Eq. (2)

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{h}\frac {\left (b +\tau \right ) \left (\tau -b \right )}{\sqrt {4 a^{2} \tau ^{2}-b^{4}+2 b^{2} \tau ^{2}-\tau ^{4}}}d \tau = u +c_4 \]

Singular solutions are found by solving

\begin{align*} \frac {\sqrt {4 a^{2} h^{2}-b^{4}+2 h^{2} b^{2}-h^{4}}}{\left (b +h \right ) \left (h -b \right )}&= 0 \end{align*}

for \(h\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} h = -a -\sqrt {a^{2}+b^{2}}\\ h = -a +\sqrt {a^{2}+b^{2}}\\ h = a -\sqrt {a^{2}+b^{2}}\\ h = a +\sqrt {a^{2}+b^{2}} \end{align*}

We now need to find the singular solutions, these are found by finding for what values \((\frac {\sqrt {4 a^{2} h^{2}-b^{4}+2 h^{2} b^{2}-h^{4}}}{\left (b +h \right ) \left (h -b \right )})\) is zero. These give

\begin{align*} h&=-a -\sqrt {a^{2}+b^{2}} \\ h&=-a +\sqrt {a^{2}+b^{2}} \\ h&=a -\sqrt {a^{2}+b^{2}} \\ h&=a +\sqrt {a^{2}+b^{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.

The solution \(h = -a -\sqrt {a^{2}+b^{2}}\) satisfies the ode and initial conditions.

The solution \(h = -a +\sqrt {a^{2}+b^{2}}\) satisfies the ode and initial conditions.

The solution \(h = a -\sqrt {a^{2}+b^{2}}\) does not satisfiy the ode and initial conditions. Hence will not be used.

The solution \(h = a +\sqrt {a^{2}+b^{2}}\) does not satisfiy the ode and initial conditions. Hence will not be used.

Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & h \left (u \right )^{2}+\frac {2 a h \left (u \right )}{\sqrt {1+\left (\frac {d}{d u}h \left (u \right )\right )^{2}}}=b^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d u}h \left (u \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d u}h \left (u \right )=\frac {\sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}{\left (h \left (u \right )+b \right ) \left (h \left (u \right )-b \right )}, \frac {d}{d u}h \left (u \right )=-\frac {\sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}{\left (h \left (u \right )+b \right ) \left (h \left (u \right )-b \right )}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d u}h \left (u \right )=\frac {\sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}{\left (h \left (u \right )+b \right ) \left (h \left (u \right )-b \right )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\left (\frac {d}{d u}h \left (u \right )\right ) \left (h \left (u \right )+b \right ) \left (h \left (u \right )-b \right )}{\sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} u \\ {} & {} & \int \frac {\left (\frac {d}{d u}h \left (u \right )\right ) \left (h \left (u \right )+b \right ) \left (h \left (u \right )-b \right )}{\sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}d u =\int 1d u +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {2 b^{4} \sqrt {1+\frac {h \left (u \right )^{2} \left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right )}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h \left (u \right )^{2}}{b^{4}}}\, \left (\mathit {EllipticF}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )-\mathit {EllipticE}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )\right )}{\sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}\, \sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}\, \left (4 a^{2}+2 b^{2}+4 a \sqrt {a^{2}+b^{2}}\right )}-\frac {b^{2} \sqrt {1+\frac {h \left (u \right )^{2} \left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right )}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h \left (u \right )^{2}}{b^{4}}}\, \mathit {EllipticF}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )}{\sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}\, \sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}=u +\textit {\_C1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d u}h \left (u \right )=-\frac {\sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}{\left (h \left (u \right )+b \right ) \left (h \left (u \right )-b \right )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\left (\frac {d}{d u}h \left (u \right )\right ) \left (h \left (u \right )+b \right ) \left (h \left (u \right )-b \right )}{\sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}=-1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} u \\ {} & {} & \int \frac {\left (\frac {d}{d u}h \left (u \right )\right ) \left (h \left (u \right )+b \right ) \left (h \left (u \right )-b \right )}{\sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}d u =\int \left (-1\right )d u +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {2 b^{4} \sqrt {1+\frac {h \left (u \right )^{2} \left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right )}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h \left (u \right )^{2}}{b^{4}}}\, \left (\mathit {EllipticF}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )-\mathit {EllipticE}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )\right )}{\sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}\, \sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}\, \left (4 a^{2}+2 b^{2}+4 a \sqrt {a^{2}+b^{2}}\right )}-\frac {b^{2} \sqrt {1+\frac {h \left (u \right )^{2} \left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right )}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h \left (u \right )^{2}}{b^{4}}}\, \mathit {EllipticF}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )}{\sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}\, \sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}=-u +\textit {\_C1} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\frac {2 b^{4} \sqrt {1+\frac {h \left (u \right )^{2} \left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right )}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h \left (u \right )^{2}}{b^{4}}}\, \left (\mathit {EllipticF}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )-\mathit {EllipticE}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )\right )}{\sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}\, \sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}\, \left (4 a^{2}+2 b^{2}+4 a \sqrt {a^{2}+b^{2}}\right )}-\frac {b^{2} \sqrt {1+\frac {h \left (u \right )^{2} \left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right )}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h \left (u \right )^{2}}{b^{4}}}\, \mathit {EllipticF}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )}{\sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}\, \sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}=-u +\mathit {C1} , \frac {2 b^{4} \sqrt {1+\frac {h \left (u \right )^{2} \left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right )}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h \left (u \right )^{2}}{b^{4}}}\, \left (\mathit {EllipticF}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )-\mathit {EllipticE}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )\right )}{\sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}\, \sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}\, \left (4 a^{2}+2 b^{2}+4 a \sqrt {a^{2}+b^{2}}\right )}-\frac {b^{2} \sqrt {1+\frac {h \left (u \right )^{2} \left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right )}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h \left (u \right )^{2}}{b^{4}}}\, \mathit {EllipticF}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )}{\sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}\, \sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}=u +\mathit {C1} \right \} \end {array} \]

Maple trace
`Methods for first order ODEs: 
-> Solving 1st order ODE of high degree, 1st attempt 
trying 1st order WeierstrassP solution for high degree ODE 
trying 1st order WeierstrassPPrime solution for high degree ODE 
trying 1st order JacobiSN solution for high degree ODE 
trying 1st order ODE linearizable_by_differentiation 
trying differential order: 1; missing variables 
<- differential order: 1; missing  x  successful`
 
Maple dsolve solution

Solving time : 3.867 (sec)
Leaf size : 103

dsolve(h(u)^2+2*a*h(u)/(1+diff(h(u),u)^2)^(1/2) = b^2, 
       h(u),singsol=all)
 
\begin{align*} u -\left (\int _{}^{h}\frac {\textit {\_a}^{2}-b^{2}}{\sqrt {-\textit {\_a}^{4}+\left (4 a^{2}+2 b^{2}\right ) \textit {\_a}^{2}-b^{4}}}d \textit {\_a} \right )-c_{1} &= 0 \\ u +\int _{}^{h}\frac {\textit {\_a}^{2}-b^{2}}{\sqrt {-\textit {\_a}^{4}+\left (4 a^{2}+2 b^{2}\right ) \textit {\_a}^{2}-b^{4}}}d \textit {\_a} -c_{1} &= 0 \\ \end{align*}
Mathematica DSolve solution

Solving time : 24.227 (sec)
Leaf size : 913

DSolve[{h[u]^2 + 2*a*h[u]/Sqrt[1 + (D[ h[u],u])^2] == b^2,{}}, 
       h[u],u,IncludeSingularSolutions->True]
 
\begin{align*} h(u)\to \text {InverseFunction}\left [-\frac {i \sqrt {\left (b^2-\text {$\#$1}^2\right )^2} \sqrt {1-\frac {\text {$\#$1}^2}{-2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2}} \sqrt {1-\frac {\text {$\#$1}^2}{2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2}} \left (\left (2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2\right ) E\left (i \text {arcsinh}\left (\sqrt {\frac {1}{-2 a^2-b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}} \text {$\#$1}\right )|\frac {2 a^2+b^2-2 \sqrt {a^2 \left (a^2+b^2\right )}}{2 a^2+b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}\right )-2 \left (\sqrt {a^2 \left (a^2+b^2\right )}+a^2\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {1}{-2 a^2-b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}} \text {$\#$1}\right ),\frac {2 a^2+b^2-2 \sqrt {a^2 \left (a^2+b^2\right )}}{2 a^2+b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}\right )\right )}{\left (b^2-\text {$\#$1}^2\right ) \sqrt {\frac {1}{2 \sqrt {a^2 \left (a^2+b^2\right )}-2 a^2-b^2}} \sqrt {-\text {$\#$1}^4+4 \text {$\#$1}^2 a^2+2 \text {$\#$1}^2 b^2-b^4}}\&\right ][-u+c_1] \\ h(u)\to \text {InverseFunction}\left [-\frac {i \sqrt {\left (b^2-\text {$\#$1}^2\right )^2} \sqrt {1-\frac {\text {$\#$1}^2}{-2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2}} \sqrt {1-\frac {\text {$\#$1}^2}{2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2}} \left (\left (2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2\right ) E\left (i \text {arcsinh}\left (\sqrt {\frac {1}{-2 a^2-b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}} \text {$\#$1}\right )|\frac {2 a^2+b^2-2 \sqrt {a^2 \left (a^2+b^2\right )}}{2 a^2+b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}\right )-2 \left (\sqrt {a^2 \left (a^2+b^2\right )}+a^2\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {1}{-2 a^2-b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}} \text {$\#$1}\right ),\frac {2 a^2+b^2-2 \sqrt {a^2 \left (a^2+b^2\right )}}{2 a^2+b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}\right )\right )}{\left (b^2-\text {$\#$1}^2\right ) \sqrt {\frac {1}{2 \sqrt {a^2 \left (a^2+b^2\right )}-2 a^2-b^2}} \sqrt {-\text {$\#$1}^4+4 \text {$\#$1}^2 a^2+2 \text {$\#$1}^2 b^2-b^4}}\&\right ][u+c_1] \\ h(u)\to -\sqrt {a^2+b^2}-a \\ h(u)\to \sqrt {a^2+b^2}-a \\ \end{align*}