Sum $\sum\limits_{n=1}^\infty\frac{H_n^2}{n^22^n}$ Where $ H_n$ is the harmonic number, $\ \displaystyle H_n=1+\frac12+\frac13+...+\frac1n$.
I am going to present my solution as I need it as a reference.
Other approaches are appreciated.
here is the closed form $$\sum_{n=1}^\infty\frac{H_n^2}{n^22^n}=-\frac1{24}\ln^42+\frac14\ln^22\zeta(2)-\frac74\ln2\zeta(3)+\frac{37}{16}\zeta(4)-\operatorname{Li}_4\left(\frac12\right)$$
 A: The second derivative of beta function gives $\ \displaystyle \int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2}{n}+\frac{H_n^{(2)}}{n}$
divide both sides by $\ n2^n$ and take the sum, we get
\begin{align}
\sum_{n=1}^\infty\frac{H_n^2}{n^22^n}+\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}&=\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty\frac{(x/2)^n}{n} dx=-\int_0^1\frac{\ln^2(1-x)\ln(1-x/2)}{x} dx\\
&=-\int_0^1\frac{\ln^2(1-x)\left[\ln(2-x)-\ln2\right]}{x}\ dx, \quad 1-x=y\\
&=\ln2\int_0^1\frac{\ln^2x}{1-x} dx-\int_0^1\frac{\ln^2x\ln(1+x)}{1-x}\ dx\\
&=2\ln2\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1\frac{x^n\ln^2x}{1-x}\ dx\\
&=2\ln2\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^n}{n}\left(2\zeta(3)-2H_n^{(3)}\right)\\
&=2\ln2\zeta(3)-2\ln2\zeta(3)-2\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n}\\
&=-2\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n}
\end{align}
then $$\sum_{n=1}^\infty\frac{H_n^2}{n^22^n}=-\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}-2\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n}$$
the first sum can be found here, as for the second one, can be calculated as follows:
using the generating function $\displaystyle\sum_{n=1}^\infty z^nH_n^{(3)}=\frac{\operatorname{Li}_3(z)}{1-z}$, divide both sides by $z$ and integrate from $z=0$ to $x$,
then $\quad\displaystyle\sum_{n=1}^\infty\frac{x^nH_n^{(3)}}{n}=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}_2^2(x)\ $ and by taking $x=-1$, we get
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n}=\frac34\ln2\zeta(3)-\frac{19}{16}\zeta(4)$$
plugging the closed forms of these two sums, we get
$$\sum_{n=1}^\infty\frac{H_n^2}{n^22^n}=-\frac1{24}\ln^42+\frac14\ln^22\zeta(2)-\frac74\ln2\zeta(3)+\frac{37}{16}\zeta(4)-\operatorname{Li}_4\left(\frac12\right)$$
A: The problem reduces to the main integral in OP's post (see his answer). 
Here's an alternative approach. Consider the following integrals: $$I=\int_0^1 \frac{\ln^2 x\ln(1+x)}{1-x}dx$$
$$J=\int_0^1 \frac{\ln^2 x\ln(1-x)}{1-x}dx=-\frac12\zeta(4)$$

$$X=I+J=\int_0^1 \frac{\ln^2 x\ln(1-x^2)}{1-x}dx$$
$$Y=\int_0^1 \frac{\ln^2 x\ln(1-x^2)}{1+x}dx$$
$$X+Y=2\int_0^1 \frac{\ln^2 x\ln(1-x^2)}{1-x^2}dx$$
$$X-Y=2\int_0^1 \frac{x\ln^2 x\ln(1-x^2)}{1-x^2}dx\overset{x^2\to x}=\frac14J$$

$$\Rightarrow I = \underbrace{\frac12\left((X+Y)+(X-Y)\right)}_{=X=I+J}-J=\frac12 (X+Y)-\frac78 J$$
$$=\int_0^1 \frac{\ln^2 x\ln(1-x^2)}{1-x^2}dx+\frac7{16}\zeta(4)$$
$$=\frac72\ln 2 \zeta(3)-\frac{45}{16}\zeta(4)+\frac7{16}\zeta(4)=\boxed{\frac72\ln 2 \zeta(3)-\frac{19}{8}\zeta(4)}$$

The last integral can be computed either using Beta function or with the power series
$$\frac{\ln(1-x^2)}{1-x^2}=-\sum_{n=1}^\infty H_n x^{2n}$$
