How to integrate $ x = \int \frac{ky^2}{\sqrt{1-k^2y^4}} dy $? The problem is to make the following integral stationary:
$$ \int_{x_1}^{x_2} \frac{\sqrt{1+y'^2}}{y^2}dx $$
to simplify the Euler equation, I tried to change the independent variable:
$$ \int_{y_1}^{y_2} \frac{\sqrt{1+x'^2}}{y^2}dy, \: \: \: \: \: \: \: \: y=y\left( x \right),\: y'=\frac{dy}{dx} $$
with the correspondent Euler equation:
$$ \frac{d}{dy}\frac{\partial F}{\partial x'}-\frac{\partial F}{\partial x}=0 $$
thus
$$
\begin{aligned}
\frac{\partial F}{\partial x}=0 \Rightarrow \frac{\partial F}{\partial x'} &= k\\
                                               \frac{x'}{y^2 \sqrt{1+x^2}} &= k\\
                                                        \frac{dx}{dy} = x' &= \frac{ky^2}{\sqrt{1-k^2y^4}}
\end{aligned}
$$
and I get:
$$ x = \int \frac{ky^2}{\sqrt{1-k^2y^4}} dy $$
Now, can I change $ ky^2 $ to a new single arbitrary variabel to simplify the integrand? Or are there a more effective method?
 A: Do it, substitute $u=ky^2,\frac{du}{dy}=2\sqrt{ku}$ (assuming $k>0$):
$$x=\frac1{2\sqrt k}\int\sqrt{\frac u{1-u^2}}\,du$$
This is an elliptic integral. Say we're integrating from $0$ to $u$, then Byrd and Friedman 235.06 gives
$$x=\sqrt{\frac2k}\left(E(\varphi,m)-F(\varphi,m)/2-\sqrt{\frac{u(1-u)}{2(1+u)}}\right)+C$$
where $\sin^2\varphi=\frac{2u}{1+u}$ and $m=1/2$. You will have to use numerical methods to get $y$ in terms of $x$.
A: As said in comments and answer, using $\frac{u}{\sqrt{k}}$ you will end with
$$\int \frac{ky^2}{\sqrt{1-k^2y^4}}\, dy=\frac{E\left(\left.\sin ^{-1}(u)\right|-1\right)-F\left(\left.\sin
   ^{-1}(u)\right|-1\right)}{\sqrt{k}}$$
If you need to solve for $u$ the equation
$$f(u)=E\left(\left.\sin ^{-1}(u)\right|-1\right)-F\left(\left.\sin
   ^{-1}(u)\right|-1\right)=a$$ you could use the series expansion
$$f(u)=\sum_{n=0}^\infty b_n \,u^{4n+3}$$ where the $b_n$'s form the sequence
$$\left\{\frac{1}{3},\frac{1}{14},\frac{3}{88},\frac{1}{48},\frac{35}{2432},\frac{63}{5
   888},\frac{77}{9216},\frac{429}{63488},\frac{1287}{229376},\frac{935}{196608},\cdots\right\}$$ Using the above coefficients, the fit is quite good : the absolute error is

*

*$<0.1$% if $u \lt 0.921$

*$<0.01$% if $u \lt 0.874$

*$<0.001$% if $u \lt 0.829$
Using series reversion
$$u=t \left(1-\sum_{n=1}^\infty c_n \,t^{4n} \right)\qquad \text{where} \qquad t=\sqrt[3]{3a}$$ the first $c_n$'s forming the sequence
$$\left\{\frac{1}{14},\frac{15}{4312},\frac{61}{181104},\frac{85325}{2119641216},\frac{
   1214019}{227508157184}\right\}$$
For a test example, using $a=0.5$ the above would give as estimate
$u=0.99010375$ while the "exact" solution is
$u=0.99010360$
