Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^32^n}$

In the same spirit as Robert Israel's answer and continuing Raymond Manzoni's answer (both of them deserve the credit because of inspiring my answer) we have $$ \sum_{n=1}^\infty \frac{H_nx^n}{n^2}=\zeta(3)+\frac{1}{2}\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x). $$ Dividing equation above by $x$ and then integrating yields \begin{align} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}=&\zeta(3)\ln x+\frac12\color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}+\color{blue}{\int\frac{\ln(1-x)\operatorname{Li}_2(1-x)}x\ dx}\\&+\operatorname{Li}_4(x)-\color{green}{\int\frac{\operatorname{Li}_3(1-x)}x\ dx}.\tag1 \end{align} Using IBP to evaluate the green integral by setting $u=\operatorname{Li}_3(1-x)$ and $dv=\frac1x\ dx$, we obtain \begin{align} \color{green}{\int\frac{\operatorname{Li}_3(1-x)}x\ dx}&=\operatorname{Li}_3(1-x)\ln x+\int\frac{\ln x\operatorname{Li}_2(1-x)}{1-x}\ dx\qquad x\mapsto1-x\\ &=\operatorname{Li}_3(1-x)\ln x-\color{blue}{\int\frac{\ln (1-x)\operatorname{Li}_2(x)}{x}\ dx}.\tag2 \end{align} Using Euler's reflection formula for dilogarithm $$ \operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\frac{\pi^2}6-\ln x\ln(1-x), $$ then combining the blue integral in $(1)$ and $(2)$ yields $$ \frac{\pi^2}6\int\frac{\ln (1-x)}{x}\ dx-\color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}=-\frac{\pi^2}6\operatorname{Li}_2(x)-\color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}. $$ Setting $x\mapsto1-x$ and using the identity $H_{n+1}-H_n=\frac1{n+1}$, the red integral becomes \begin{align} \color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}&=-\int\frac{\ln (1-x)\ln^2 x}{1-x}\ dx\\ &=\int\sum_{n=1}^\infty H_n x^n\ln^2x\ dx\\ &=\sum_{n=1}^\infty H_n \int x^n\ln^2x\ dx\\ &=\sum_{n=1}^\infty H_n \frac{\partial^2}{\partial n^2}\left[\int x^n\ dx\right]\\ &=\sum_{n=1}^\infty H_n \frac{\partial^2}{\partial n^2}\left[\frac {x^{n+1}}{n+1}\right]\\ &=\sum_{n=1}^\infty H_n \left[\frac{x^{n+1}\ln^2x}{n+1}-2\frac{x^{n+1}\ln x}{(n+1)^2}+2\frac{x^{n+1}}{(n+1)^3}\right]\\ &=\ln^2x\sum_{n=1}^\infty\frac{H_n x^{n+1}}{n+1}-2\ln x\sum_{n=1}^\infty\frac{H_n x^{n+1}}{(n+1)^2}+2\sum_{n=1}^\infty\frac{H_n x^{n+1}}{(n+1)^3}\\ &=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n+1} x^{n+1}}{(n+1)^2}-\sum_{n=1}^\infty\frac{x^{n+1}}{(n+1)^3}\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n+1} x^{n+1}}{(n+1)^3}-\sum_{n=1}^\infty\frac{x^{n+1}}{(n+1)^4}\right]\\ &=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\sum_{n=1}^\infty\frac{x^{n}}{n^3}\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^3}-\sum_{n=1}^\infty\frac{x^{n}}{n^4}\right]\\ &=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^3}-\operatorname{Li}_4(x)\right]. \end{align} Putting all together, we have \begin{align} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+C.\tag3 \end{align} Setting $x=1$ to obtain the constant of integration, \begin{align} \sum_{n=1}^\infty \frac{H_n}{n^3}&=\operatorname{Li}_4(1)-\frac{\pi^2}{12}\operatorname{Li}_2(1)+C\\ \frac{\pi^4}{72}&=\frac{\pi^4}{90}-\frac{\pi^4}{72}+C\\ C&=\frac{\pi^4}{60}. \end{align} Thus \begin{align} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+\frac{\pi^4}{60}.\tag4 \end{align} Finally, setting $x=\frac12$, we obtain \begin{align} \sum_{n=1}^\infty \frac{H_n}{2^nn^3}=\color{purple}{\frac{\pi^4}{720}+\frac{\ln^42}{24}-\frac{\ln2}8\zeta(3)+\operatorname{Li}_4\left(\frac12\right)}, \end{align} which matches Cleo's answer.

References :

$[1]\ $ Harmonic number

$[2]\ $ Polylogarithm

$$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}=\frac{\pi^4}{720}+\frac{\ln^42}{24}-\frac{\ln2}8\zeta(3)+\operatorname{Li}_4\left(\frac12\right).$$

Note: Please note the top voted answer by @Tunk-Fey is regrettably not correct. Contrary to his claim his final expression (4) when evaluated at $x=\frac{1}{2}$ does not match @Cleo's answer but differs by $\frac{\pi^4}{120}$ from the correct identity: \begin{align*} \sum_{n=1}^\infty \frac{H_n}{n^32^n}&=-\frac{1}{8}\ln 2\zeta(3)+\frac{1}{24}\ln^4(2)+\frac{\pi^4}{720}+ \operatorname{Li}_4\left(\frac{1}{2}\right)\\ &\stackrel{.}{=}0.55824 \end{align*} A rather detailed analysis of the deviation from the correct result is provided in this answer.

Nevertheless it was a pleasure to review his answer which contains nice and instructive aspects. Here I provide a solution in a similar spirit which hopefully overcomes the problems of his answer.

Raymond Manzoni's has nicely demonstrated that for $|x|<1$ \begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^2}&=\zeta(3)+\frac{1}{2}\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)\\ &\qquad+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x) \end{align*}

This result is our starting point.

\begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}&=\int\sum_{n=1}^\infty \frac{H_nx^{n-1}}{n^2}dx\\ &=\zeta(3)\ln(x)+\frac{1}{2}\int\frac{1}{x}\ln x\ln^2(1-x)dx+\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\\ &\qquad+\int\frac{1}{x}\operatorname{Li}_3(x)dx-\int\frac{1}{x}\operatorname{Li}_3(1-x)dx+C\tag{1}\\ \end{align*}

At first we consider $\int\frac{1}{x}\operatorname{Li}_3(1-x)dx$. Integration by parts with $u=\frac{1}{x}$ and $dv=\operatorname{Li}_3(1-x)dx$ gives

\begin{align*} \int\frac{1}{x}\operatorname{Li}_3(1-x)dx&=\ln x\operatorname{Li}_3(1-x)+\int\frac{\ln x}{1-x}\operatorname{Li}_2(1-x)dx\\ &=\ln x\operatorname{Li}_3(1-x)+\frac{1}{2}\operatorname{Li}_2^2(1-x)+C \end{align*} Once again integration by parts on the RHS with $u=\frac{\ln x}{1-x}$ and $dv=\operatorname{Li}_2(1-x)dx$ gives \begin{align*} \int\frac{\ln x}{1-x}\operatorname{Li}_2(1-x)dx&=\operatorname{Li}_2^2(1-x) -\int\frac{\ln x}{1-x}\operatorname{Li}_2(1-x)dx\\ \Longrightarrow\int\frac{\ln x}{1-x}\operatorname{Li}_2(1-x)dx&=\frac{1}{2}\operatorname{Li}_2^2(1-x)+C \end{align*}

It follows \begin{align*} \int\frac{1}{x}\operatorname{Li}_3(1-x)dx&=\operatorname{Li}_3(1-x)\ln x+\frac{1}{2}\operatorname{Li}_2^2(1-x)+C \end{align*}

and we obtain substituting this result in (1) and noting that \begin{align*} \int\frac{1}{x}\operatorname{Li}_3(x)dx=\operatorname{Li}_4(x)+C \end{align*}

\begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}&=\zeta(3)\ln x+\frac{1}{2}\int\frac{1}{x}\ln x\ln^2(1-x)dx+\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\\ &\qquad+\operatorname{Li}_4(x)-\left(\operatorname{Li}_3(1-x)\ln x+\frac{1}{2}\operatorname{Li}_2^2(1-x)\right)+C\tag{2}\\ \end{align*}

The next step is to calculate $\int\frac{1}{x}\ln x\ln^2(1-x)dx$. We use Euler's reflection formula \begin{align*} \operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\frac{\pi^2}{6}-\ln x\ln(1-x) \end{align*} to split the integral into parts which can either be directly calculated or which can be transformed to the remaining integral. We obtain using the reflection formula

\begin{align*} \int&\frac{1}{x}\ln x\ln^2(1-x)dx\\ &=\int\frac{\ln(1-x)}{x}\left(\frac{\pi^2}{6}-\operatorname{Li}_2(x)-\operatorname{Li}_2(1-x)\right)\\ &=-\frac{\pi^2}{6}\operatorname{Li}_2(x)-\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(x)dx -\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\\ &=-\frac{\pi^2}{6}\operatorname{Li}_2(x)+\frac{1}{2}\operatorname{Li}_2^2(x)dx -\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx \end{align*}

Putting this result into (2) we get

\begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}&=\zeta(3)\ln x +\frac{1}{2}\left(-\frac{\pi^2}{6}\operatorname{Li}_2(x)+\frac{1}{2}\operatorname{Li}_2^2(x) -\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\right)\\ &\qquad+\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\\ &\qquad+\operatorname{Li}_4(x)-\left(\operatorname{Li}_3(1-x)\ln x+\frac{1}{2}\operatorname{Li}_2^2(1-x)\right)+C\\ &=\zeta(3)\ln x-\frac{\pi^2}{12}\operatorname{Li}_2(x)+\frac{1}{4}\operatorname{Li}_2^2(x) -\frac{1}{2}\operatorname{Li}_2^2(1-x)\\ &\qquad-\operatorname{Li}_3(1-x)\ln x+\operatorname{Li}_4(x)\\ &\qquad+\frac{1}{2}\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx+C\tag{3}\\ \end{align*}

The most complex and cumbersome part is the remaining integral in (3). With the help of Wolfram Alpha a rather lengthy result is provided. After some simplifications we obtain \begin{align*} \int&\frac{\ln(1-x)}{x}\operatorname{Li}_2{(1-x)}dx\\ &=-\frac{1}{2}\ln^2(1-x)\ln^2x+\ln(1-x)\ln^3x-\frac{1}{4}\ln^4x\\ &\qquad-\operatorname{Li}_2(1-x)\left(\ln^2(1-x)-\ln(1-x)\ln x\right)+\operatorname{Li}_2(x)\ln^2 x\\ &\qquad-\operatorname{Li}_2\left(1-\frac{1}{x}\right)\left(\ln^2(1-x)-2\ln(1-x)\ln x+\ln^2 x\right)+\frac{1}{2}\operatorname{Li}_2^2(1-x)\\ &\qquad+2\left(\operatorname{Li}_3\left(1-\frac{1}{x}\right)\left(\ln(1-x)-\ln x\right)+\operatorname{Li}_3(1-x)\ln(1-x) -\operatorname{Li}_3(x)\ln x\right)\\ &\qquad-2\left(\operatorname{Li}_4(1-x)+\operatorname{Li}_4\left(1-\frac{1}{x}\right)-\operatorname{Li}_4(x)\right)+C\\ \end{align*}

Finally substituting this expression into (3) and doing some more simplifications we obtain

\begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}&=\zeta(3)\ln x-\frac{\pi^2}{12}\operatorname{Li}_2(x)+\frac{1}{4}\operatorname{Li}_2^2(x) -\frac{1}{2}\operatorname{Li}_2^2(1-x)\\ &\quad-\operatorname{Li}_3(1-x)\ln x+\operatorname{Li}_4(x)\\ &\quad+\frac{1}{2}\left(-\frac{1}{2}\ln^2(1-x)\ln^2x+\ln(1-x)\ln^3x-\frac{1}{4}\ln^4x\right.\\ &\quad\quad-\operatorname{Li}_2(1-x)\left(\ln^2(1-x)-\ln(1-x)\ln x\right)+\operatorname{Li}_2(x)\ln^2 x\\ &\quad\quad-\operatorname{Li}_2\left(1-\frac{1}{x}\right)\left(\ln^2(1-x)-2\ln(1-x)\ln x+\ln^2 x\right)\\ &\quad\quad+\frac{1}{2}\operatorname{Li}_2^2(1-x)\\ &\quad\quad+2\left(\operatorname{Li}_3\left(1-\frac{1}{x}\right)\left(\ln(1-x)-\ln x\right)\right.\\ &\quad\quad\quad+\left.\operatorname{Li}_3(1-x)\ln(1-x)-\operatorname{Li}_3(x)\ln x\right)\\ &\quad\quad\left.-2\left(\operatorname{Li}_4(1-x)+\operatorname{Li}_4\left(1-\frac{1}{x}\right)-\operatorname{Li}_4(x)\right)\right)+C\\ &=\zeta(3)\ln x-\frac{1}{4}\ln^2(1-x)\ln^2x+\frac{1}{2}\ln(1-x)\ln^3x-\frac{1}{8}\ln^4x\\ &\quad-\frac{1}{2}\operatorname{Li}_2(1-x)\left(\ln^2(1-x)-\ln(1-x)\ln x\right)+\frac{1}{2}\operatorname{Li}_2(x)\left(\ln^2 x-\frac{\pi^2}{6}\right)\\ &\quad-\frac{1}{2}\operatorname{Li}_2\left(1-\frac{1}{x}\right)\left(\ln^2(1-x)-2\ln(1-x)\ln x+\ln^2 x\right)\\ &\quad+\frac{1}{4}\operatorname{Li}^2_2(x)-\frac{1}{4}\operatorname{Li}^2_2(1-x)-\operatorname{Li}_3(x)\ln x\\ &\quad+\operatorname{Li}_3\left(1-\frac{1}{x}\right)\left(\ln(1-x)-\ln x\right)+\operatorname{Li}_3(1-x)\left(\ln(1-x)-\ln(x)\right)\\ &\quad-\operatorname{Li}_4(1-x)-\operatorname{Li}_4\left(1-\frac{1}{x}\right)+2\operatorname{Li}_4(x)+C\tag{4} \end{align*}

From (4) we can now determine the integration constant $C$. In order to do so we calculate $C$ by taking the limit as $x\rightarrow 1$. Most of the terms vanish and noting that according to this answer \begin{align*} \sum_{n=1}^\infty \frac{H_n}{n^3}=\frac{\pi^4}{72} \end{align*} we obtain respecting that $\operatorname{Li}_2(1)=\frac{\pi^2}{6}$ and $\operatorname{Li}_4(1)=\frac{\pi^4}{90}$

\begin{align*} \frac{\pi^4}{72}&=\frac{1}{2}\operatorname{Li}_2(1)\left(-\frac{\pi^2}{6}\right)+\frac{1}{4}\operatorname{Li}^2_2(1)+2\operatorname{Li}_4(1)+C\\ &=-\frac{\pi^4}{72}+\frac{\pi^4}{144}+\frac{2\pi^4}{90}+C\\ \text{it follows}\qquad C&=-\frac{\pi^4}{720} \end{align*}

Setting $x=\frac{1}{2}$ in (4) we finally obtain with $C=-\frac{\pi^4}{720}$ and noting that \begin{align*} \operatorname{Li}_2\left(\frac{1}{2}\right)&=\frac{\pi^{2}}{12}-\frac{1}{2}\ln^2(2)\\ \operatorname{Li}_3\left(\frac{1}{2}\right)&=\frac{7}{8}\zeta(3)+\frac{1}{6}\ln^3(2)-\frac{\pi^{2}}{12}\ln 2\\ \operatorname{Li}_4(-1)&=-\frac{7\pi^4}{720} \end{align*}

\begin{align*} \sum_{n=1}^\infty \frac{H_n}{n^32^n}&=-\zeta(3)\ln(2)+\frac{1}{8}\ln^4(2) +\frac{1}{2}\operatorname{Li}_2\left(\frac{1}{2}\right)\left(\ln^2(2)-\frac{\pi^2}{6}\right)\\ &\qquad+\operatorname{Li}_3\left(\frac{1}{2}\right)\ln 2-\operatorname{Li}_4(-1)+\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{\pi^4}{720}\\ &=-\frac{1}{8}\ln 2\zeta(3)+\frac{1}{24}\ln^4(2)+\frac{\pi^4}{720}+ \operatorname{Li}_4\left(\frac{1}{2}\right)\\ &\stackrel{.}{=}0.55824 \end{align*} and the claim follows.

Note: Two aspects remain open. The important one is a derivation of \begin{align*} \int&\frac{\ln(1-x)}{x}\operatorname{Li}_2{(1-x)}dx \end{align*} without support from WA. It would also be nice to find some further simplifications of the final expression (4).