How to evaluate $\sum _{n=1}^{\infty }\left(\frac{H_n^2+H_n^{\left(2\right)}}{n}\right)^2$ in a particular way. How to evaluate:
$$\sum _{n=1}^{\infty }\left(\frac{H_n^2+H_n^{\left(2\right)}}{n}\right)^2,$$
without splitting the expression into more sums.
Here $H_n^{\left(m\right)}=\sum _{k=1}^n\frac{1}{k^m}$ is the harmonic number of order $m$.
If one just wants to evaluate it if we split we have,
$$2\sum _{n=1}^{\infty }\frac{H_n^2H_n^{\left(2\right)}}{n^2}+\sum _{n=1}^{\infty }\frac{H_n^4}{n^2}+\sum _{n=1}^{\infty }\frac{\left(H_n^{\left(2\right)}\right)^2}{n^2},$$
Then making use of this results one only has to compute
$$\sum _{n=1}^{\infty }\frac{\left(H_n^{\left(2\right)}\right)^2}{n^2}$$
But I'd like to know if its possible to evaluate the series without splitting or expanding the terms.
 A: To find the desired series we must first consider the following integral.
$$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx$$
To evaluate it one can make use of the following trilogarithm identity.
$$\operatorname{Li}_3\left(\frac{x}{x-1}\right)=-\operatorname{Li}_3\left(x\right)-\operatorname{Li}_3\left(1-x\right)+\zeta \left(3\right)+\frac{1}{6}\ln ^3\left(1-x\right)$$
$$+\zeta \left(2\right)\ln \left(1-x\right)-\frac{1}{2}\ln \left(x\right)\ln ^2\left(1-x\right)$$
Using it on the previous integral yields:
$$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx=-\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(x\right)}{x}\:dx-\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(1-x\right)}{x}\:dx$$
$$+\zeta \left(3\right)\int _0^1\frac{\ln ^2\left(1-x\right)}{x}\:dx+\frac{1}{6}\int _0^1\frac{\ln ^5\left(1-x\right)}{x}\:dx+\zeta \left(2\right)\int _0^1\frac{\ln ^3\left(1-x\right)}{x}\:dx$$
$$-\frac{1}{2}\int _0^1\frac{\ln \left(x\right)\ln ^4\left(1-x\right)}{x}\:dx$$
$$=-\frac{81}{2}\zeta \left(6\right)+2\zeta ^2\left(3\right)+12\sum _{k=1}^{\infty }\frac{H_k}{k^5}-\sum _{k=1}^{\infty }\frac{H_k^2}{k^4}-\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{k^4}-2\sum _{k=1}^{\infty }\frac{H_k^{\left(3\right)}}{k^3}$$
The series remaining can be calculated quite easily and a nice thing to know is that to evaluate  them one does not have to cross paths with the series in the body of the OP.
Thus:
$$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx=-\frac{581}{24}\zeta \left(6\right)-4\zeta ^2\left(3\right)$$
Now what is left to do is to consider the following generating function.
$$\sum _{k=1}^{\infty }\frac{x^{k-1}}{k}\left(H_k^2+H_k^{\left(2\right)}\right)=-2\frac{\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}$$
Which can be found along other generating functions in the book (Almost) Impossible Integrals, Sums, and Series, page $\#285$.
Using it on the previously found integral means that,
$$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx=-\frac{1}{2}\sum _{k=1}^{\infty }\left(\frac{H_k^2+H_k^{\left(2\right)}}{k}\right)^2$$
Thus:
$$\sum _{k=1}^{\infty }\left(\frac{H_k^2+H_k^{\left(2\right)}}{k}\right)^2=\frac{581}{12}\zeta \left(6\right)+8\zeta ^2\left(3\right)$$
