Closed form expression for the harmonic sum $\sum\limits_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$ I'm wondering if one could derive a closed form expression for the series
$$\sum_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$$
$$\text{With } \text{ } \text{ } \text{ }H_n=\sum_{k=1}^{n}\frac{1}{k}\text{ } \text{ } \text{} \text{ } \text{ }\text{the } n^{th} \text{ harmonic number.}$$
Now, I know series involving harmonic numbers are well suited for a summation by part (or Abel's transformation) approach, but it doesn't lead anywere here, at least not in this state.
Any suggestions ?

Using the fact that $$\int_0^1x^{2n-1}\ln(1-x)\ dx=-\frac{H_{2n}}{2n}$$
multiply both sides by $\ \displaystyle-\frac{2}{n4^n}{2n \choose n}$ then take the sum, we get,
\begin{align}
S&=\sum_{n=1}^\infty\frac{H_{2n}}{n^24^n}{2n \choose n}=-2\int_0^1\frac{\ln(1-x)}{x}\left(\sum_{n=1}^\infty\frac{(x^2)^n}{n4^n}{2n \choose n}\right)\ dx
\end{align}
I managed here to prove: $$\quad\displaystyle\sum_{n=1}^\infty \frac{x^n}{n4^n}{2n \choose n}=-2 \tanh^{-1}{\sqrt{1-x}}-\ln x+2\ln2$$
which follows:
\begin{align}
S&=4\underbrace{\int_0^1\frac{\ln(1-x)\tanh^{-1}{\sqrt{1-x^2}}}{x}\ dx}_{\text{IBP}}+4\int_0^1\frac{\ln(1-x)\ln x}{x}\ dx-4\ln2\int_0^1\frac{\ln(1-x)}{x}\ dx\\
&=-4\int_0^1\frac{\operatorname{Li}_2(x)}{x\sqrt{1-x^2}}\ dx+4(\zeta(3))-4\ln2(-\zeta(2))\\
&=-4\left(\frac{3}{8}\pi^2 \ln(2) - \frac{7}{16} \zeta(3)\right)+4\zeta(3)+\frac{2}{3}\pi^2\ln(2)\\
&\boxed{=\frac{23}4\zeta(3)-\frac{5}{6}\pi^2\ln2}
\end{align}
Credit goes to ComplexYetTrivial for nicely evaluating $\ \displaystyle\int_0^1\frac{\operatorname{Li}_2(x)}{x\sqrt{1-x^2}}\ dx=\frac{3}{8}\pi^2 \ln(2) - \frac{7}{16} \zeta(3)$ 
