I found this double integral in this group
$$\displaystyle\int_{0}^{1}\int_{0}^{1}\frac{1}{\sqrt{x(1-x)}\sqrt{y(1-y)}\sqrt{1-xy}}dxdy$$
After integrating with respects to $x$ (or $y$, the order is not important because of the symmetry), I get this integral
$$\displaystyle2\int_{0}^{1}\frac{K(y)}{\sqrt{y(1-y)}}dy$$
Where $K(y)$ is the complete elliptic integral of the first kind.
WolframAlpha says its closed form is $$\frac{256\pi^3}{\Gamma^4{\left(\frac{-1}{4}\right)}}$$ or it is equavalent $$\frac{\pi^3}{\Gamma^4{\left(\frac{3}{4}\right)}}$$
I also try to use series expansion of $K(y)$ but the sum is more complicated. Any another approach is welcome, and thank you for reading.
Best Answer
It is not bad to use the series $$K(y)=\frac \pi 2\sum_{n=0}^\infty2^{-4 n}\frac{ ((2 n)!)^2}{(n!)^4}\,y^n$$ $$\int \frac {y^k}{\sqrt{y(1-y)}}\,dy=-2 \sqrt{1-y} \,\,\, _2F_1\left(\frac{1}{2},\frac{1}{2}-k;\frac{3}{2};1-y\right)$$ $$\int_0^1 \frac {y^k}{\sqrt{y(1-y)}}\,dy=\sqrt{\pi } \,\frac{\Gamma \left(k+\frac{1}{2}\right)}{\Gamma (k+1)}$$ which make the integral to be $$\sqrt{\pi } \sum_{n=0}^\infty\frac{\Gamma \left(n+\frac{1}{2}\right)^3}{\Gamma (n+1)^3}$$
Now, considering $$\sqrt{\pi } \sum_{n=0}^\infty\frac{\Gamma \left(n+\frac{1}{2}\right)^3}{\Gamma (n+1)^3}x^n=4 K\left(\frac{1}{2} \left(1-\sqrt{1-x}\right)\right)^2$$ Using $x=1$ $$4 K\left(\frac{1}{2}\right)^2=\frac{\pi \Gamma \left(\frac{1}{4}\right)^2}{2 \Gamma \left(\frac{3}{4}\right)^2}$$ and several othe forms.