A closed form for $\int_0^1{_2F_1}\left(-\frac{1}{4},\frac{5}{4};\,1;\,\frac{x}{2}\right)^2dx$ 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
$$S\approx0.8263551866500213413164525287...$$
 A: 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.
A: Your integral has an elementary closed form, that was correctly stated by Cleo in her answer without proof:
$$S=\int_0^1\left({_2F_1}\!\left(-\tfrac14,\tfrac54;\,1;\,\tfrac{x}2\right)\right)^2dx=\frac{8\sqrt2+4\ln\left(\sqrt2-1\right)}{3\pi}.\tag1$$

Proof:  Using DLMF 14.3.6 we can express the hypergeometric function in the integrand as the Legendre function of the $1^{st}$ kind (also known as the Ferrers function of the $1^{st}$ kind) with fractional index:
$${_2F_1}\!\left(-\tfrac14,\tfrac54;\,1;\,\tfrac{x}2\right)=P_{\small1/4}(1-x).\tag2$$
Now the integral can be written as
$$S=\int_0^1\left(P_{\small1/4}(1-x)\right)^2dx=\int_0^1\left(P_{\small1/4}(x)\right)^2dx.\tag3$$
To evaluate it, we use formula 7.113 on page 769 in Gradshteyn & Ryzhyk:
$$\int_0^1P_\nu(x)\,P_\sigma(x)\,dx=\\\frac{\frac{\Gamma\left(\frac12+\frac\nu2\right)\,\Gamma\left(1+\frac\sigma2\right)}{\Gamma\left(\frac12+\frac\sigma2\right)\,\Gamma\left(1+\frac\nu2\right)}\sin\!\left(\frac{\pi\sigma}2\right)\cos\!\left(\frac{\pi\nu}2\right)-\frac{\Gamma\left(\frac12+\frac\sigma2\right)\,\Gamma\left(1+\frac\nu2\right)}{\Gamma\left(\frac12+\frac\nu2\right)\,\Gamma\left(1+\frac\sigma2\right)}\sin\!\left(\frac{\pi\nu}2\right)\cos\!\left(\frac{\pi\sigma}2\right)}{\frac\pi2(\sigma-\nu)(\sigma+\nu+1)}.\tag4$$
Note that in our case $\nu=\sigma=\frac14$, so we cannot use this formula directly because of the term $(\sigma-\nu)$ in the denominator. Instead, we let $\nu=\frac14$ and find the limit for $\sigma\to\frac14$:
$$S=\lim\limits_{\sigma\to{\small1/4}}\int_0^1P_{\small1/4}(x)\,P_\sigma(x)\,dx=\\\lim\limits_{\sigma\to{\small1/4}}\frac{\frac{\Gamma\left(\frac58\right)\,\Gamma\left(1+\frac\sigma2\right)}{\Gamma\left(\frac12+\frac\sigma2\right)\,\Gamma\left(\frac98\right)}\sin\!\left(\frac{\pi\sigma}2\right)\cos\!\left(\frac\pi8\right)-\frac{\Gamma\left(\frac12+\frac\sigma2\right)\,\Gamma\left(\frac98\right)}{\Gamma\left(\frac58\right)\,\Gamma\left(1+\frac\sigma2\right)}\sin\!\left(\frac\pi8\right)\cos\!\left(\frac{\pi\sigma}2\right)}{\frac\pi2(\sigma-\frac14)(\sigma+\frac54)}.\tag5$$
To evaluate the limit, we use l'Hôpital's rule. This gives quite a big expression that I will not copy here. It contains values of the gamma and digamma functions at rational points, that could be simplified to elementary terms using the Gauss digamma theorem and identities given in the MathWorld and in the famous Vidūnas paper, yielding the desired result $(1)$.

Indeed, we can have a more general result:
$$\int_0^1\left({_2F_1}\!\left(-\nu,\nu+1;\,1;\,\tfrac x2\right)\right)^2dx=\int_0^1\left(P_\nu\left(x\right)\right)^2dx=\\\frac{1+\!\left[\psi_0\!\left(1+\frac\nu2\right)-\psi_0\!\left(\frac12+\frac\nu2\right)\right]\frac{\sin(\pi\nu)}\pi}{1+2\nu}.\tag6$$
A: Yes, it is possible in some cases, for example,
$$S=\frac{8\sqrt2+4\ln\left(\sqrt2-1\right)}{3\pi}$$
