Closed form for $\int_0^\infty\left(\int_0^1\frac1{\sqrt{1-y^2}\sqrt{1+x^2\,y^2}}\mathrm dy\right)^3\mathrm dx.$ I need to find a closed form for these nested definite integrals:
$$I=\int_0^\infty\left(\int_0^1\frac1{\sqrt{1-y^2}\sqrt{1+x^2\,y^2}}\mathrm dy\right)^3\mathrm dx.$$
The inner integral can be represented using the hypergeometric function $_2F_1$ or the complete elliptic integral of the 1st kind $K$ with an imaginary argument:
$$\int_0^1\frac1{\sqrt{1-y^2}\sqrt{1+x^2\,y^2}}\mathrm dy=\frac\pi2   {_2F_1}\left(\frac12,\frac12;1;-x^2\right)=K(x\,\sqrt{-1}).$$
My conjecture is the integral $I$ has a closed-form representation:$$I\stackrel{?}{=}\frac{3\,\Gamma (\frac14)^8}{1280\,\pi^2}=7.09022700484626946098980237...,$$
but I was neither able to find a proof of it, nor disprove the equality using numerical integration. Could you please help me with resolving this question?
 A: The first thing I would try is to replace $\int_{0}^{\infty}$ by $\frac12\int_{-\infty}^{\infty}$, then to consider this as a complex integral and pull the integration contour to $i\infty$. Since the only singularity of the hypergeometric function $_2F_1(\ldots,z)$ on the main sheet is the branch point $z=1$, in the end we would have to integrate the jump of $K^3(ix)$ on the cut $B=[i,i\infty)$. The jump of a function on the cut is often simpler than the function itself (like $2\pi$ vs $\ln z$), so I would hope for a simplification of the integral (as the branch point $x=i$ here seems to be logarithmic).
A: Using 
$$ K(ik) =
\frac{1}{\sqrt{1+k^2}}K\left(\sqrt{\frac{k^2}{k^2+1}}\right) $$
and a substitution $t^2 = \frac{k^2}{1+k^2}$, rewrite the integral as
$$ \int_0^\infty K(i k)^3\,dk = \int_0^1 K(t)^3\,dt. $$
There is a paper "Moments of elliptic integrals and critical L-values"
by Rogers, Wan and Zucker (http://arxiv.org/abs/1303.2259; also one of
the authors' earlier papers: http://arxiv.org/abs/1101.1132), and the
authors, by relating this integral to an L-series of a modular form (their theorems 1 and 2),
show that
$$ \int_0^1 K(k)^3\,dk = \frac{3}{5}K(1/\sqrt{2})^4 =
\frac{3\Gamma(\frac14)^8}{1280\pi^2}, $$
using $K(1/\sqrt{2}) = \frac14 \pi^{-1/2}\Gamma(\frac14)^2$. 
