I am hoping to find a closed form for the following
$$\tag{1} \sum_{k\geq 1}\frac{H_k}{k^3} x^k $$
Using the generating function
$$\sum_{k\geq 1}H^{(n)}_k x^k = \frac{\operatorname{Li}_n(x)}{1-x}$$
I could find this by simple integration
So I am stuck at evaluating
$$\tag{2}\int^x_0 \frac{\operatorname{Li}_2(1-t)\log(1-t)}{t}\, dt$$
For $x=\pm 1$ the problem can be solved , but what about the general case ?
$$\begin{align} &\int^x_0 \frac{\operatorname{Li}_2(1-t)\log(1-t)}{t}\,dt=\\ &\frac{\operatorname{Li}_2^2(1-x)}2-2\operatorname{Li}_4\left(1-\frac1x\right)-2\operatorname{Li}_4(1-x)+2\operatorname{Li}_4(x)-\operatorname{Li}_2\left(1-\frac1x\right)\log^2\left(\frac1x-1\right)+\\ &\operatorname{Li}_2(x)\left(\log^2\left(\frac1x-1\right)+\log(1-x)\log(x)\right)+2\operatorname{Li}_3(x)\log\left(\frac1x\right)+\\&2\operatorname{Li}_3\left(1-\frac1x\right)\log\left(\frac1x-1\right)+2\operatorname{Li}_3(1-x)\left(\log\left(\frac1x-1\right)+\log(x)\right)-\\ &\frac14\log^4\left(\frac1x\right)+\frac13\log^3(1-x)\left(2\log(x)-\log\left(\frac1x\right)\right)-\\ &\log(1-x)\left(\log^3\left(\frac1x\right)+\frac13\pi^2\log\left(\frac1x\right)+\frac{\pi^2}6\log(x)\right)+\\&\log^2(1-x)\left(-\log^2\left(\frac1x\right)+\frac12\log^2(x)+\log(x)\log\left(\frac1x\right)-\frac{\pi^2}6\right)-\frac{11\pi^4}{360},\end{align}$$ that can be checked by taking derivatives from both sides.
See the corresponding indefinite integral at WolframAlpha.