A twisted hypergeometric series $\sum_{n=1}^\infty\frac{H_n}{n}\left(\frac{(2n)!}{4^n(n!)^2}\right)^2$

Question

Question. I was given that $$S=\sum_{n=1}^\infty\frac{H_n}{n}\left(\frac{(2n)!}{4^n(n!)^2}\right)^2=\frac{32}\pi G\ln2+\frac{64}\pi\Im\operatorname{Li}_3\left(\frac{1+i}2\right)-2\ln^22-\frac53\pi^2$$ where $H_n$ harmonic numbers, $G$ Catalan and $\operatorname{Li}_n$ polylogarithm. How can it be proved?

My Approach. Using $\int_0^1x^{n-1}\ln(1-x)dx=-\frac{H_n}n$ one have $$S=\int_0^1-\ln(1-x)\sum_{n=1}^\infty\left(\frac{(2n)!}{4^n(n!)^2}\right)^2x^{n-1}dx =\int_0^1-\ln(1-x)\left(\frac2\pi\frac{\mathbf{K}(x)}x-\frac1x\right)dx$$ where $\mathbf{K}$ denotes elliptic integral of the first kind. The question boils down to finding $$\int_0^1\frac{\mathbf{K}(x)\ln(1-x)}xdx$$ For this integral, I tried to use the integral representation of the elliptic integral and got: $$\int_{(0,1)^3}\frac{dxdydz}{\sqrt{1-y^2}\sqrt{1-xy^2}(zx-1)}$$ This is the furthermost step I can get.

Answer 1

One have FL expansion $$\small K(x)=\sum _{n=0}^{\infty } \frac{2 P_n(2 x-1)}{2 n+1},\ \ \frac{\log (1-x)}{x}=\sum _{n=0}^{\infty } 2 (-1)^{n-1} (2 n+1) P_n(2 x-1) \left(\sum _{k=n+1}^{\infty } \frac{(-1)^{k-1}}{k^2}\right)$$ Thus by orthogonal relation of Legendre polynomial and series-integral conversion one have: $$\int_0^1 \frac{K(x) \log (1-x)}{x} \, dx=16 \int_0^1 \frac{\log (x) \tanh ^{-1}(x)}{x^2+1} \, dx$$ RHS has a polylogarithmic primitive (up to trilog) hence trivial.