Another way of doing integration What's your option for calculating this integral? No full solution is necessary, it's optional as usual.
Calculate 
$$\int_0^1 \frac{2 \zeta (3)\log ^3(1-x) \text{Li}_2(1-x) }{x}-\frac{2 \zeta (3) \log ^2(1-x) \text{Li}_3(1-x)}{x}+\frac{ \log (x) \log ^5(1-x)\text{Li}_2(1-x)}{x}+\frac{\log ^4(1-x)(\text{Li}_2(1-x){})^2 }{x}-\frac{ \log (x) \log ^4(1-x)\text{Li}_3(1-x)}{x}-\frac{2 \log ^3(1-x)\text{Li}_2(1-x) \text{Li}_3(1-x) }{x}+\frac{\log ^2(1-x)(\text{Li}_3(1-x){})^2 }{x} \textrm{d} x.$$
Did you meet it before? Some references, papers?
EDIT I: I'm going to remove the first 2 terms since I see the way to go there, and all reduces to 
$$\int_0^1 \frac{ \log (x) \log ^5(1-x)\text{Li}_2(1-x)}{x}+\frac{\log ^4(1-x)(\text{Li}_2(1-x){})^2 }{x}-\frac{ \log (x) \log ^4(1-x)\text{Li}_3(1-x)}{x}-\frac{2 \log ^3(1-x)\text{Li}_2(1-x) \text{Li}_3(1-x) }{x}+\frac{\log ^2(1-x)(\text{Li}_3(1-x){})^2 }{x} \textrm{d} x$$
which looks way simpler.
EDIT II: Thanks, I'm done with the question (I successfully finalized the calculations).
EDIT III: Supplementary question - Where is Cleo? I miss Cleo.
 A: Adding up the integrands so that all terms are over the common denominator $x$, we find that the resultant numerator factors in a very helpful way. As is, spotting a convenient way to group the terms isn't exactly easy, but if you're familiar with the Nielsen generalized polylogarithm you'll find that it tidies things up quite nicely.
In particular, note the following identity:
$$S_{1,2}{\left(1-x\right)}=-\operatorname{Li}_{3}{\left(x\right)}+\ln{\left(x\right)}\operatorname{Li}_{2}{\left(x\right)}+\frac12\ln^{2}{\left(x\right)}\ln{\left(1-x\right)}+\zeta{(3)}.$$
Them,
$$\begin{align}
\mathcal{J}
&=\int_{0}^{1}\frac{\ln{\left(x\right)}\ln^{5}{\left(1-x\right)}\operatorname{Li}_{2}{\left(1-x\right)}}{x}\,\mathrm{d}x\\
&~~~~~-\int_{0}^{1}\frac{\ln{\left(x\right)}\ln^{4}{\left(1-x\right)}\operatorname{Li}_{3}{\left(1-x\right)}}{x}\,\mathrm{d}x\\
&~~~~~+\int_{0}^{1}\frac{\ln^{4}{\left(1-x\right)}\left[\operatorname{Li}_{2}{\left(1-x\right)}\right]^{2}}{x}\,\mathrm{d}x\\
&~~~~~-2\int_{0}^{1}\frac{\ln^{3}{\left(1-x\right)}\operatorname{Li}_{2}{\left(1-x\right)}\operatorname{Li}_{3}{\left(1-x\right)}}{x}\,\mathrm{d}x\\
&~~~~~+\int_{0}^{1}\frac{\ln^{2}{\left(1-x\right)}\left[\operatorname{Li}_{3}{\left(1-x\right)}\right]^{2}}{x}\,\mathrm{d}x\\
&=\small{\int_{0}^{1}\frac{\ln^{2}{\left(1-x\right)}\left[\ln{\left(1-x\right)}\operatorname{Li}_{2}{\left(1-x\right)}-\operatorname{Li}_{3}{\left(1-x\right)}\right]\left[\ln{\left(1-x\right)}\operatorname{Li}_{2}{\left(1-x\right)}-\operatorname{Li}_{3}{\left(1-x\right)}+\ln{\left(x\right)}\ln^{2}{\left(1-x\right)}\right]}{x}\,\mathrm{d}x}\\
&=\small{\int_{0}^{1}\frac{\ln^{2}{\left(1-x\right)}\left[S_{1,2}{\left(x\right)}-\zeta{(3)}-\frac12\ln{\left(x\right)}\ln^{2}{\left(1-x\right)}\right]\left[S_{1,2}{\left(x\right)}-\zeta{(3)}+\frac12\ln{\left(x\right)}\ln^{2}{\left(1-x\right)}\right]}{x}\,\mathrm{d}x}\\
&=\int_{0}^{1}\frac{\ln^{2}{\left(1-x\right)}\left[\left(S_{1,2}{\left(x\right)}-\zeta{(3)}\right)^{2}-\frac14\ln^{2}{\left(x\right)}\ln^{4}{\left(1-x\right)}\right]}{x}\,\mathrm{d}x\\
&=2\int_{0}^{1}\frac{\ln^{2}{\left(1-x\right)}\left[S_{1,2}{\left(x\right)}-\zeta{(3)}\right]^{2}}{2x}\,\mathrm{d}x\\
&~~~~~-\frac{6!}{2}\int_{0}^{1}\frac{\ln^{2}{\left(x\right)}\ln^{6}{\left(1-x\right)}}{2\left(6!\right)x}\,\mathrm{d}x\\
&=2\int_{0}^{\zeta{(3)}}\left[S_{1,2}{\left(x\right)}-\zeta{(3)}\right]^{2}\,\mathrm{d}S_{1,2}{\left(x\right)}-\frac{6!}{2}\,S_{3,6}{(1)}\\
&=\frac23\,\zeta{(3)}-360\,S_{3,6}{(1)}.\\
\end{align}$$
Going this route is likely orders of magnitude quicker than trying to integrate each of initial terms individually.
