Evaluation of a dilogarithmic integral 
Problem. Prove that the following dilogarithmic integral has the indicated value:
  $$\int_{0}^{1}\mathrm{d}x \frac{\ln^2{(x)}\operatorname{Li}_2{(x)}}{1-x}\stackrel{?}{=}-11\zeta{(5)}+6\zeta{(3)}\zeta{(2)}.$$


My attempt:
I began by using the polylogarithmic expansion in terms of generalized harmonic numbers,
$$\frac{\operatorname{Li}_r{(x)}}{1-x}=\sum_{n=1}^{\infty}H_{n,r}\,x^n;~~r=2.$$
Then I switched the order of summation and integration and used the substitution $u=-\ln{x}$ to evaluate the integral:
$$\begin{align}
\int_{0}^{1}\mathrm{d}x \frac{\ln^2{(x)}\operatorname{Li}_2{(x)}}{1-x}
&=\int_{0}^{1}\mathrm{d}x\ln^2{(x)}\sum_{n=1}^{\infty}H_{n,2}x^n\\
&=\sum_{n=1}^{\infty}H_{n,2}\int_{0}^{1}\mathrm{d}x\,x^n\ln^2{(x)}\\
&=\sum_{n=1}^{\infty}H_{n,2}\int_{0}^{\infty}\mathrm{d}u\,u^2e^{-(n+1)u}\\
&=\sum_{n=1}^{\infty}H_{n,2}\frac{2}{(n+1)^3}\\
&=2\sum_{n=1}^{\infty}\frac{H_{n,2}}{(n+1)^3}.
\end{align}$$
So I've reduced the integral to an Euler sum, but unfortunately I've never quite got the knack for evaluating Euler sums. How to proceed from here?

From $$\sum_{n=1}^\infty H_n^{(2)}x^n=\frac{\operatorname{Li}_2(x)}{1-x}$$
it follows that
$$I=\int_0^1\frac{\ln^2(x)\operatorname{Li}_2(x)}{1-x}dx=\sum_{n=1}^\infty H_n^{(2)}\int_0^1 x^n \ln^2(x)dx\\=2\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)^3}=2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}-2\zeta(5)\tag1$$
By Cauchy product we have
$$\operatorname{Li}_2(x)\operatorname{Li}_3(x)=\sum_{n=1}^\infty\left(\frac{6H_n}{n^4}+\frac{3H_n^{(2)}}{n^3}+\frac{H_n^{(3)}}{n^2}-\frac{10}{n^5}\right)x^n$$ 
set $x=1$ to get
$$\zeta(2)\zeta(3)=6\sum_{n=1}^\infty\frac{H_n}{n^4}+3\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}+\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}-10\zeta(5)\tag{2}$$
Now lets use the well-known identity 
$$\sum_{n=1}^\infty\frac{H_n^{(p)}}{n^q}+\sum_{n=1}^\infty\frac{H_n^{(q)}}{n^p}=\zeta(p)\zeta(q)+\zeta(p+q)$$
set $p=2$ and $q=3$ 
$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\zeta(2)\zeta(3)+\zeta(5)-\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}$$
Plugging this result in $(2)$ yields
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)$$
and fianlly, plugging this result in $(1)$, the closed form of $I$ follows. Note that the value $\sum_{n=1}^\infty 
\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$ was used in our calculations which can be found using Euler identity.
