Evaluating $\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}$ 
Question: How can we evaluate $$\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n},$$where $H_n=\frac11+\frac12+\cdots+\frac1n$?

Quick Results
This series converges because $$\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}=O\left(\frac{\ln^2n}{n^{3/2}}\right).$$
My Attempt
Recall the integral representation of harmonic number $$H_n=\int_0^1\frac{1-x^n}{1-x}d x$$
we have $$
S=\sum_{n=1}^\infty\frac1n\frac{\binom{2n}n}{4^n}\iint_{[0,1]^2}\frac{(1-x^n)(1-y^n)}{(1-x)(1-y)}d xd y\\
=\tiny\iint_{[0,1]^2}\frac{x y \log (4)-2 x y \log \left(\sqrt{1-x}+1\right)-2 x y \log \left(\sqrt{1-y}+1\right)+2 x y \log \left(\frac{1}{2} \left(\sqrt{1-x y}+1\right)\right)}{\left(\sqrt{1-x y}-1\right) \left(\sqrt{1-x y}+1\right)}dxdy\\
$$
This integral is too hard for me and Mathematica to compute. Numerical integration returns $12.6178$, it agrees with the numerical summation of the original series. I tried to integrate with respect to $x$, but failed.

From this paper, Eq $(13)$ page 4, we have
$$\sum_{n=1}^\infty H_n\binom{2n}n x^n=\frac{2}{\sqrt{1-4x}}\ln\left(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}\right).$$
Replace $x$ by $\frac{x}{4}$, multiply both sides by $-\frac{\ln(1-x)}{x}$ then integrate using $-\int_0^1 x^{n-1} \ln(1-x)dx=\frac{H_n}{n}$, we have
$$\sum_{n=1}^\infty \frac{H_n^2}{n4^n}\binom{2n}n=-\int_0^1\frac{\ln(1-x)}{x\sqrt{1-x}}\ln\left(\frac{1+\sqrt{1-x}}{2\sqrt{1-x}}\right)dx$$
$$\overset{\sqrt{1-x}=y}{=}-8\int_0^1\frac{\ln y}{1-y^2}\ln\left(\frac{1+y}{2y}\right)dy$$
$$\overset{y=\frac{1-x}{1+x}}{=}4\int_0^1\frac{\ln^2(1-x)}{x}dx-4\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}dx$$
$$=4[2\zeta(3)]-4\left[-\frac58\zeta(3)\right]=\frac{21}{2}\zeta(3).$$
