Proving a series for the Watson Triple Integrals?

The sum

$$ \frac{4}{3} \sum_{n \ge 0} \frac{(4 n)!}{(n!)^4} z^4 = \frac{4}{3} \sum_{n \ge 0} \frac{ (\frac{1}{4})_n (\frac{3}{4})_n (\frac{1}{2})_n }{ (1)_n (1)_n} \frac{(256 z)^n}{n!} = \frac{4}{3} {}_3F_2\left( \left. \begin{array}{c} \frac{1}{4}, \frac{1}{2}, \frac{3}{4} \\ 1, 1 \end{array} \right| 256 z \right) $$

This particular hypergeometric function can be expressed in term of complete elliptic integral:

$$ {}_3F_2\left( \left. \begin{array}{c} \frac{1}{4}, \frac{1}{2}, \frac{3}{4} \\ 1, 1 \end{array} \right| w \right) = \frac{4 \sqrt{2} }{ \pi^2 } \frac{ K\left( \frac{1}{2}- { \frac{1}{\sqrt{2(1+ \sqrt{1-w} )} } } \right)^2}{ \sqrt{(1 + \sqrt{1 - w}) } } $$

For $z =\frac{1}{2 \cdot 18^2}$, $w =\frac{32}{81}$. Then, it results that

$$ I_1 = \frac{4 \sqrt{2}}{\pi^2} \left(K\left(\frac{1}{2} - \frac{3 \sqrt{2}}{8}\right)\right)^2 $$ Now elliptic integrals for certain moduli are known to related to gamma functions. These are known as singular values of elliptic integrals.

A beautiful well-explained solution may be found in the fantastic book, Inside Interesting Integrals by the author Paul J. Nahin, pages $206$-$212$.

The transformation presented in the solution turns the initial integral into a product of three integrals as you may see below, which looks fascinating,

$$\int_0^{\pi}\int_0^{\pi}\int_0^{\pi}\frac{\textrm{d}u \textrm{d}v \textrm{d}w}{1-\cos(u)\cos(v)\cos(w)}$$ $$=4\sqrt{2}\int_0^{\infty} \frac{\textrm{d}t}{1+t^4}\int_0^{\pi/2} \frac{\textrm{d}\theta}{\sqrt{\cos(\theta)}}\int_0^{\pi/2} \frac{\textrm{d}\psi}{\sqrt{\sin(\psi)}}.$$

Finding new solutions to the Watson triple integrals is always a very exciting mission.