How to evaluate $ \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}$ I am having a difficult time evaulating
$$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}.$$
I have tried the following relation:
$$\frac{1}{2}\int_0^1 \frac{\mathrm{Li}_2(x)}{x(1-x)}\log^2{x}\ \mathrm{d}x.$$
Or, alternatively, we have
$$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3} = \sum_{n=1}^\infty \frac{1}{n^3}\sum_{m=1}^n \frac{1}{m^2};$$
from which I wrote
\begin{align}-\sum_{n=1}^\infty \frac{1}{n^3}\sum_{m=1}^n \int_0^1 x^{m-1}\log{x}\ \mathrm{d}x
&= -\sum_{n=1}^\infty \frac{1}{n^3}\int_0^1 \sum_{m=1}^n x^{m-1}\log{x}\ \mathrm{d}x,\\
&=-\sum_{n=1}^\infty \frac{1}{n^3}\int_0^1 \frac{x^n-1}{x-1}\log{x}\ \mathrm{d}x
\end{align}
followed by distributing the latter sum into the integral and integrating parts. Using this technique I obtained
$$-\zeta(2)\zeta(3)-\int_0^1\frac{\mathrm{Li}_2(x)\mathrm{Li}_2(1-x)}{x}\ \mathrm{d}x.$$
However, I do not know where to go. I know the value of the sum, but I am looking to evaulate the sum by the means of integrals. A hint/comment would be much appreciated. Thank you!
 A: Calculating your First integral:
$$\int_0^1 \frac{\mathrm{Li}_2(x)\ln^2(x)}{x(1-x)}dx=\int_0^1 \frac{\mathrm{Li}_2(x)\ln^2(x)}{x}dx+\int_0^1 \frac{\mathrm{Li}_2(x)\ln^2(x)}{1-x}dx=A+B$$
where
$$A=\sum_{n=1}^\infty\frac{1}{n^2}\int_0^1 x^{n-1}\ln^2(x)dx=2\sum_{n=1}^\infty\frac{1}{n^5}=2\zeta(5)$$
and
$$B\overset{IBP}{=}-\ln(1-x)\text{Li}_2(x)\ln^2(x)|_0^1+\int_0^1 \ln(1-x)\left[\frac{2\text{Li}_2(x)\ln(x)}{x}-\frac{\ln(1-x)\ln^2(x)}{x}\right]dx$$
$$=2\int_0^1\frac{\ln(x)\ln(1-x)\text{Li}_2(x)}{x}dx-\int_0^1\frac{\ln^2(1-x)\ln^2(x)}{x}dx$$
$$=B_1-B_2$$
$$B_1\overset{IBP}{=}-\text{Li}_2^2(x)\ln(x)|_0^1+\int_0^1\frac{\text{Li}_2^2(x)}{x}dx$$
$$=\sum_{n=1}^\infty\frac{1}{n^2}\int_0^1 x^{n-1}\text{Li}_2(x)dx=\sum_{n=1}^\infty\frac{1}{n^2}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)$$
$$=\zeta(2)\zeta(3)-\sum_{n=1}^\infty\frac{H_n}{n^4}=2\zeta(2)\zeta(3)-3\zeta(5)$$
$$B_2=2\sum_{n=1}^\infty \frac{H_{n-1}}{n}\int_0^1 x^{n-1}\ln^2(x)dx=4\sum_{n=1}^\infty \frac{H_{n-1}}{n^4}=4\sum_{n=1}^\infty \frac{H_n}{n^4}-4\zeta(5)$$
$$=8\zeta(5)-4\zeta(2)\zeta(3)$$
Combine all the pieces we get
$$\begin{equation}
\int_0^1 \frac{\mathrm{Li}_2(x)\ln^2(x)}{x(1-x)}dx=6\zeta(2)\zeta(3)-9\zeta(5)
\end{equation}$$

Calculating your second integral:
Make use of the dilog reflection identity $\text{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)-\text{Li}_2(x)$
$$\int_0^1\frac{\text{Li}_2(x)\text{Li}_2(1-x)}{x}dx$$
$$=\zeta(2)\int_0^1\frac{\text{Li}_2(x)}{x}dx-\underbrace{\int_0^1\frac{\ln x\ln(1-x)\text{Li}_2(x)}{x}dx}_{IBP}-\int_0^1\frac{\text{Li}_2^2(x)}{x}dx$$
$$=\zeta(2)\zeta(3)-\frac32\int_0^1\frac{\text{Li}_2^2(x)}{x}dx$$
$$=\zeta(2)\zeta(3)-\frac32\left(2\zeta(2)\zeta(3)-3\zeta(5)\right)$$
$$=\frac92\zeta(5)-2\zeta(2)\zeta(3)$$
A: The closed forms are\begin{equation}
\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)\label{H_n^(2)/n^3}
\end{equation}
\begin{equation}
\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\frac{11}{2}\zeta(5)-2\zeta(2)\zeta(3)\label{H_n^(3)/n^2}
\end{equation}
Proof: We are going to establish two relations and solve for the two sums by elimination.
The first relation: By Cauchy product we have
$$\text{Li}_2(x)\text{Li}_3(x)=6\sum_{n=1}^\infty \frac{H_n}{n^4}x^n+3\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}x^n+\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^2}-10\text{Li}_5(x)$$
and set $x=1$ and rearrange the terms we have
$$3\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}+\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^2}=\zeta(2)\zeta(3)+10\zeta(5)-6\sum_{n=1}^\infty \frac{H_n}{n^4}$$
$$=7\zeta(2)\zeta(3)-8\zeta(5)\tag1$$
where the last result follows from substituting $\sum_{n=1}^\infty \frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$
The second relation: By Abels' summation we have
$$\sum_{k=1}^\infty \frac{H_k^{(p)}}{k^q}+\sum_{k=1}^\infty \frac{H_k^{(q)}}{k^p}=\zeta(p)\zeta(q)+\zeta(p+q)$$
and set $p=2$ and $q=3$ we get
\begin{equation}
\sum_{k=1}^\infty \frac{H_k^{(3)}}{k^2}+\sum_{k=1}^\infty \frac{H_k^{(2)}}{k^3}=\zeta(2)\zeta(3)+\zeta(5)\tag2
\end{equation}
Solving $(1)$ and $(2)$ systematically, the closed forms follows.
