Computing $\sum_{n=1}^\infty\frac{2^{2n}H_{n+1}}{(n+1)^2{2n\choose n}}$

An advanced sum proposed by Cornel Valean:

$$S=\sum_{n=1}^\infty\frac{2^{2n}H_{n+1}}{(n+1)^2{2n\choose n}}$$ $$=4\text{Li}_4\left(\frac12\right)-\frac12\zeta(4)+\frac72\zeta(3)-4\ln^22\zeta(2)+6\ln2\zeta(2)+\frac16\ln^42-1$$

I managed to find the integral representation of $\ \displaystyle\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^2{2n\choose n}}\ $ but not $S$:

Since

$$\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^\infty\frac{(2x)^{2n-1}}{n{2n\choose n}}$$

we can write

$$\frac{2\sqrt{x}\arcsin \sqrt{x}}{\sqrt{1-x}}=\sum_{n=1}^\infty\frac{2^{2n}x^{n}}{n{2n\choose n}}$$

now multiply both sides by $-\frac{\ln(1-x)}{x}$ then $\int_0^1$ and use that $-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$ we have

$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^2{2n\choose n}}=-2\int_0^1 \frac{\arcsin \sqrt{x}\ln(1-x)}{\sqrt{x}\sqrt{1-x}}dx\tag1$$

But I could not get the integral representation of $S$. Any idea?

In case you find the integral, I prefer solutions that do not use contour integration or you can leave it to me to give it a try. Thank you.

In case the reader is curious about computing the integral in $(1)$, set $x=\sin^2\theta$ then use the Fourier series of $\ln(\cos \theta)$.

Following @Felix's idea above:

$$S=\sum_{n=1}^\infty\frac{2^{2n}H_{n+1}}{(n+1)^2{2n\choose n}}=\sum_{n=2}^\infty\frac{2^{2n-2}H_n}{n^2{2n-2\choose n-1}}$$

Note that

$$\frac{{2n+2\choose n+1}}{{2n\choose n}}=\frac{\frac{\Gamma(2n+3)}{\Gamma^2(n+2)}}{\frac{\Gamma(2n+1)}{\Gamma^2(n+1)}}=\frac{\frac{(2n+2)(2n+1)\Gamma(2n+1)}{((n+1)\Gamma(n+1))^2}}{\frac{\Gamma(2n+1)}{\Gamma^2(n+1)}}=\frac{(2n+2)(2n+1)}{(n+1)^2}=\frac{2(2n+1)}{n+1}$$

replace $n$ by $n-1$ we get

$$\frac{1}{{2n-2\choose n-1}}=\frac{2(2n-1)}{n{2n\choose n}}$$

Therefore

$$S=\sum_{n=2}^\infty\frac{2^{2n-1}(2n-1)H_n}{n^3{2n\choose n}}=\sum _{n=1}^{\infty } \frac{2^{2n} H_n}{n^2 {2n\choose n}}-\frac12 \sum _{n=1}^{\infty } \frac{2^{2n} H_n}{n^3 {2n\choose n}}-1\tag1$$

In the question body we have

$$\sum _{n=1}^{\infty } \frac{2^{2n} H_n}{n^2 {2n\choose n}}=-2\int_0^1 \frac{\arcsin \sqrt{x}\ln(1-x)}{\sqrt{x}\sqrt{1-x}}dx\overset{\sqrt{x}=\sin\theta}{=}-8\int_0^{\pi/2} \theta \ln(\cos\theta)d\theta$$

$$=-8\int_0^{\pi/2}\theta\left(-\ln(2)-\sum_{n=1}^\infty\frac{(-1)^n\cos(2n\theta)}{n}\right)d\theta=6\ln(2)\zeta(2)+\frac72\zeta(3)\tag2$$

and here we already showed

$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}=-8\text{Li}_4\left(\frac12\right)+\zeta(4)+8\ln^2(2)\zeta(2)-\frac{1}{3}\ln^4(2)\tag3$$

Finally, plug $(2)$ and $(3)$ in $(1)$ we obtain

$$S=4\text{Li}_4\left(\frac12\right)-\frac12\zeta(4)+\frac72\zeta(3)-4\ln^2(2)\zeta(2)+6\ln(2)\zeta(2)+\frac16\ln^4(2)-1$$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}{2^{2n}H_{n + 1} \over \pars{n + 1}^{2}{2n \choose n}}} = \sum_{n = 2}^{\infty}H_{n}\,{4^{n - 1} \over n^{2}{2n - 2 \choose n - 1}} = -1 + \sum_{n = 1}^{\infty} H_{n}\,{4^{n - 1} \over n^{2}}\,{\Gamma\pars{n}\Gamma\pars{n} \over \Gamma\pars{2n - 1}} \\[5mm] = &\ -1 + \sum_{n = 1}^{\infty} H_{n}\, 4^{n - 1}\pars{{2 \over n} - {1 \over n^{2}}}\,{\Gamma\pars{n}\Gamma\pars{n} \over\Gamma\pars{2n}} \\[5mm] = &\ -1 + 2\sum_{n = 1}^{\infty}H_{n}\, 4^{n - 1} \pars{\int_{0}^{1}x^{n - 1}\,\dd x} \int_{0}^{1}y^{n - 1}\pars{1 - y}^{n - 1}\,\dd y \\[2mm] &\ -\sum_{n = 1}^{\infty}H_{n}\, 4^{n - 1} \bracks{-\int_{0}^{1}\ln\pars{x}x^{n - 1}\,\dd x} \int_{0}^{1}y^{n - 1}\pars{1 - y}^{n - 1}\,\dd y \\[5mm] = &\ -1 + 2\int_{0}^{1}\int_{0}^{1} \sum_{n = 1}^{\infty}H_{n}\pars{4xy \over 1 - y}^{n - 1} \,\dd x\,\dd y \\[2mm] &\ + \int_{0}^{1}\ln\pars{y}\int_{0}^{1} \sum_{n = 1}^{\infty}H_{n}\, \pars{4xy \over 1 - y}^{n - 1} \,\dd x\,\dd y \\[5mm] = &\ -1 + 2\int_{0}^{1}\int_{0}^{4y/\pars{1 - y}} \sum_{n = 1}^{\infty}H_{n}x^{n - 1}\, {1 - y \over 4y}\,\dd x\,\dd y \\[2mm] &\ + \int_{0}^{1}\ln\pars{y}\int_{0}^{4y} \sum_{n = 1}^{\infty}H_{n}\, x^{n - 1}\,{y - 1 \over 4y} \,\dd x\,\dd y \\[5mm] = &\ -1 + {1 \over 2}\int_{0}^{1}{1 - y \over y}\int_{0}^{4y/\pars{1 - y}} \bracks{-\,{\ln\pars{1 - x} \over 1 - x}} \,{\dd x \over x}\,\dd y \\[2mm] &\ + {1 \over 4}\int_{0}^{1}{\pars{1 - y}\ln\pars{y} \over y}\int_{0}^{4y/\pars{1 - y}} \bracks{-\,{\ln\pars{1 - x} \over 1 - x}} \,{\dd x \over x}\,\dd y \\[5mm] = &\ -1 - {1 \over 2}\int_{0}^{1}{\ln\pars{1 - x} \over x\pars{1 - x}} \int_{0}^{x/\pars{x + 4}}{1 - y \over y}\,\dd y\,\dd x \\[2mm] &\ - {1 \over 4}\int_{0}^{1}{\ln\pars{1 - x} \over x\pars{1 - x}} \int_{0}^{x/\pars{x + 4}}{\pars{1 - y}\ln\pars{y} \over y} \,\dd y\,\dd x \\[5mm] = &\ -1 - {1 \over 4}\int_{0}^{1}{\ln\pars{1 - x} \over x\pars{1 - x}} \int_{0}^{x/\pars{x + 4}} {\pars{1 - y}\bracks{2 + \ln\pars{y}} \over y}\,\dd y\,\dd x \end{align} The $\ds{y}$-integration becomes: $$ -2\ln\pars{x \over 4 + x} - {1 \over 2}\ln^{2}\pars{x \over 4 + x} - {4 \over 4 + x} - {x \over 4 + x}\ln\pars{4 + x \over x} $$ It seems to be a nasty job !!!. I hope somebody else can take it from here.