Context:
This question is related to Calculate $\sum_{n = 0}^\infty \frac{C_n^2}{16^n}$ and Is there a closed form for a give infinite sum?. We have also: $$\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^3}{64^n(n+1)^2}=\frac{\Gamma(\frac{1}{4})^4}{2\pi^3}-\frac{96\pi}{\Gamma(\frac{1}{4})^4},\tag{1}$$ $$\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^3}{64^n(n+1)^3}=8-\frac{384\pi}{\Gamma{(\frac{1}{4})}^4}\tag{2}.$$ Both can be obtained in an elementary way (avoiding hypergeometric functions). I have derived: $$I=-\frac{32}{\pi^2}\int_{0}^{\pi/2}\int_{0}^{\pi/2}\sqrt{1-\sin^2{(k)}\sin^2{(x)}}dxdk\\=4\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^3}{64^n(n+1)}-\frac{2\Gamma(\frac{1}{4})^4}{\pi^3}.\tag{3}$$ I tried several different ways to compute: $$S=\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^3}{64^n(n+1)}\tag{4}.$$ It seems that is easier than $(1)$ and $(2)$, but it still eludes me.
Updated 1: Thanks to Mariusz Iwaniuk and KStarGamer we have (see https://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/03/08/05/01/01/06/0002/): $$I=\int_{0}^{\pi/2}E(\sin({k}))dk=\frac{\Gamma(\frac{3}{4})^4}{2\pi}+\frac{\pi^3}{8\Gamma(\frac{3}{4})^4} ,$$ where $E(k)$ is the complete elliptic integral of the second kind. This integral implies: $${_3F_2(-1/2,1/2,1/2;1,1;1)}=\frac{\pi}{2\Gamma(3/4)^4}+\frac{2\Gamma(3/4)^4}{\pi^3},$$ and Wolfram and Mathematica are unable to deal with it.
Updated 2: The natural question is to ask about the closed form of:
$$S'=\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^3}{64^n(n+1)^ 4}={_5F_4(1/2,1/2,1/2,1,1;2,2,2,2;1)}\tag{5},$$ If we find the closed form of $S'$ also we will get the closed form for: $$I'=\int_{0}^{\pi/2}k\cot(k)K(\sin(k))dk\\=\frac{3\pi^2\log{2}}{4}+\frac{5\Gamma(3/4)^4}{\pi}+\frac{\pi^3}{4\Gamma(3/4)^4}-\pi G+\frac{\pi^2S'}{64}-\frac{3\pi^2}{4} \tag{6},$$ where $K(k)$ is the complete elliptic integral of the first kind and G is Catalan's constant.
Updated 3: It can be done using $(6)$ also: $$ I'=\frac{3\pi^2\log{2}}{4}-\frac{\pi^2 S_{2}}{64}-\pi G \tag{7}, $$ where: $$S_{2}={_5F_4(1,1,3/2,3/2,3/2;2,2,2,2;1)}.$$ As @SetnessRamesory has suggested probably $S'$ and $S_{2}$ don't have easy closed forms. But at least we know from $(6)$ and $(7)$ this nice relation: $${_5F_4(1/2,1/2,1/2,1,1;2,2,2,2;1)}+{_5F_4(1,1,3/2,3/2,3/2;2,2,2,2;1)}=-\frac{320\Gamma{(3/4)}^4}{\pi^3}-\frac{16\pi}{\Gamma{(3/4)}^4}+48.\tag{8}$$
This question is going to be updated for a while. Thanks for your feedback!
