A closed form for $\int_0^1{_2F_1}\left(-\frac{1}{4},\frac{5}{4};\,1;\,\frac{x}{2}\right)^2dx$

Question

Is it possible to evaluate in a closed form integrals containing a squared hypergeometric function, like in this example? $$\begin{align}S&=\int_0^1{_2F_1}\left(-\frac{1}{4},\frac{5}{4};\,1;\,\frac{x}{2}\right)^2dx\\\vphantom{=}\\&=\frac{1}{4\pi}\int_0^1\left(\sum_{n=0}^\infty\frac{4n+1}{8^n}\cdot\frac{\Gamma\left(2n-\frac{1}{2}\right)}{\Gamma(n+1)^2}\cdot x^n\right)^2dx\end{align}$$ It is approximately $$S\approx0.8263551866500213413164525287...$$

Answer 1

Yes, it is possible in some cases, for example, $$S=\frac{8\sqrt2+4\ln\left(\sqrt2-1\right)}{3\pi}$$

Answer 2

Incomplete solution:

In fact, we have $${}_2F_1(-\frac14,\frac54;1;\frac{x}{2})=\frac{8\sqrt{2}}{\pi\sqrt{2+\sqrt{2x}}}((2+\sqrt{2x})E(\frac{2\sqrt{x}}{\sqrt{2}+\sqrt{x}})-K(\frac{2\sqrt{x}}{\sqrt{2}+\sqrt{x}})).$$

Here $E(m)=\int^1_0\sqrt{\frac{1-mt^2}{1-t^2}}~dt$ and $K(m)=\int^1_0\frac{dt}{\sqrt{(1-t^2)(1-mt^2)}}$ are the complete elliptic integrals

Therefore, by substituting $y=\frac{2\sqrt{x}}{\sqrt{2}+\sqrt{x}}$ (so that $x=\frac{y^2}{(2-y)^2}$), we have $$S=\frac{16}{\pi^2}\int^{2\sqrt{2}-2}_{0}\frac{y}{(2-y)^4}\left(4E(y)-(2-y)K(y)\right)^2~dy.$$

$$%E'(y)=\frac{E(y)-K(y)}{2y}\\K'(y)=\frac{E(y)-(1-y)K(y)}{2y(1-y)}\\\int E(y)dy=\frac{2(1+y)E(y)-2(1-y)K(y)}{3}\\\int K(y)dy=2E(y)-(1-y)K(y)$$

This may be doable with some clever integration by parts.