find closed form for following double integral containing radicals question:
To prove :
$\displaystyle\int_{0}^{1}\displaystyle \int_{0}^{1}\dfrac{dxdy}{\sqrt{1-x^2}{\sqrt{1-y^2}}{\sqrt{4x^2+y^2-x^2y^2}}}=\dfrac{3\left(\Gamma{\dfrac{1}{3}}\right)^6}{2^{\dfrac{17}{3}}\pi^2}$
i tried  attempting it by transforming into polar co-ordinates but it didn't 
helped
i couldn't get rid of radicals and also there is no scope for changing order of 
integration for it's evaluation.
please help . thank you  

This double integral is first transformed into a single integral of a complete elliptic integral:
\begin{align}
 I&=\displaystyle\int_{0}^{1}\displaystyle \int_{0}^{1}\dfrac{dxdy}{\sqrt{1-x^2}{\sqrt{1-y^2}}{\sqrt{4x^2+y^2-x^2y^2}}}\\
 &\stackrel{x=\cos t}{=}\int_0^1\frac{dy}{\sqrt{1-y^2}}\int_{0}^{\pi/2}\frac{dt}{\sqrt{y^2+\left( 1-\sin^2t \right)\left( 4-y^2 \right)}}\\
 &=\frac{1}{2}\int_0^1\frac{dy}{\sqrt{1-y^2}}\int_{0}^{\pi/2}\frac{dt}{\sqrt{1-\left( 1-y^2/4 \right)\sin^2t}}
 \end{align}
We introduce the complete elliptic integral using its integral representation (DLMF):
\begin{align}
 I&=\frac{1}{2}\int_0^1\frac{\mathbf{K}\left( \sqrt{1-y^2/4} \right)}{\sqrt{1-y^2}}dy\\
 &\stackrel{y=\cos \phi}{=}\frac{1}{2}\int_0^{\pi/2}\mathbf{K}\left( \sqrt{1-\frac{1}{4}\cos^2\phi} \right)\,d\phi
\end{align} 
(the argument of the elliptic function corresponds to its modulus). Similar integrals are calculated in these articles here and here  where the following result is given:
\begin{equation}
 \int_0^{\pi/2}\mathbf{K}\left( \sqrt{1-\sin^2\alpha\cos^2\phi} \right)\,d\phi=
 \mathbf{K}\left(\sin \frac{\alpha}{2}  \right)\mathbf{K}\left(\cos \frac{\alpha}{2}  \right)
\end{equation} 
Here we choose $\alpha=\pi/6$ and then
\begin{equation}
 I=\frac{1}{2} \mathbf{K}\left(\sin \frac{\pi}{12}  \right)\mathbf{K}\left(\cos \frac{\pi}{12}  \right)
\end{equation} 
But $\sin \frac{\pi}{12}=\frac{\sqrt{2}}{4}\left( \sqrt{3}-1 \right)=\lambda^*(3)$, where $\lambda^*(r)$ gives a singular value of the elliptic modulus $k_r$ for which $\mathbf{K}\left( \sqrt{1-k_r^2} \right)=\sqrt{r}\mathbf{K}(k_r)$ see here, here and here. Then
\begin{align}
 \mathbf{K}\left(\sin \frac{\pi}{12}  \right)&=\frac{3^{1/4}\Gamma^3\left( \frac{1}{3} \right)}{2^{7/3}\pi}\\
 \mathbf{K}\left(\cos \frac{\pi}{12}  \right)&=\frac{3^{3/4}\Gamma^3\left( \frac{1}{3} \right)}{2^{7/3}\pi}
\end{align}
and thus 
\begin{equation}
 I=\frac{3\Gamma^6\left( \frac{1}{3} \right)}{2^{17/3}\pi^2}
\end{equation} 
as expected. This result is related to some Random Walk Integrals.
