Integral of polylogarithms and logs in closed form: $\int_0^1 \frac{du}{u}\text{Li}_2(u)^2(\log u)^2$ Is it possible to evaluate this integral in closed form?
$$ \int_0^1 \frac{du}{u}\text{Li}_2(u)^2\log u \stackrel{?}{=} -\frac{\zeta(6)}{3}.$$
I found the possible closed form using an integer relation algorithm.
I found several other possible forms for similar integrals, including
$$ \int_0^1 \frac{du}{u}\text{Li}_2(u)^2(\log u)^2 \stackrel{?}{=} -20\zeta(7)+12\zeta(2)\zeta(5).$$
There doesn't seem to be an equivalent form when the integrand contains $(\log u)^3$, at least not just in terms of $\zeta$.
Does anybody know a trick for evaluating these integrals?
Update. The derivation of the closed form for the second integral follows easily along the ideas O.L. used in the answer for the first integral.
Introduce the functions
$$ I(a,b,c) = \int_0^1 \frac{du}{u}(\log u)^c \text{Li}_a(u)\text{Li}_b(u) $$
and
$$ S(a,b,c) = \sum_{n,m\geq1} \frac{1}{n^am^b(n+m)^c}. $$
Using integration by parts, the expansion of polylogarithms from their power series definition and also that
$$ \int_0^1 (\log u)^s u^{t-1}\,du = \frac{(-1)^s s!}{t^{s+1}},$$
check that
$$ I(2,2,2) = -\frac23 I(1,2,3) = 4S(1,2,4). $$
Now use binomial theorem and the fact that $S(a,b,c)=S(b,a,c)$ to write
$$ 6S(1,2,4) + 2S(3,0,4) = 3S(1,2,4) + 3S(2,1,4)+S(0,3,4)+S(3,0,4) = S(3,3,1). $$
Now, using Mathematica, 
$$ S(3,3,1) = \sum_{n,m\geq1}\frac{1}{n^3m^3(n+m)} = \sum_{m\geq1}\frac{H_m}{m^6} - \frac{\zeta(2)}{m^5} + \frac{\zeta(3)}{m^4}, $$
and
$$ \sum_{m\geq1}\frac{H_m}{m^6} = -\zeta(4)\zeta(3)-\zeta(2)\zeta(5)+4\zeta(7), $$
so
$$ S(3,3,1) = 4\zeta(7)-2\zeta(2)\zeta(5). $$
Also,
$$ S(0,3,4) = \zeta(3)\zeta(4) - \sum_{m\geq1} \frac{H_{n,4}}{m^3} = -17\zeta(7)+10\zeta(2)\zeta(5)+\zeta(3)\zeta(4), $$
from which it follows that
$$ I(2,2,2) = \frac23\left(S(3,3,1)-2S(0,3,4)\right) = -20\zeta(7)+12\zeta(2)\zeta(5). $$

I've decided to publish my work so far - I do not promise a solution, but I've made some progress that others may find interesting and/or helpful.
$$\text{Let } I_{n,k}=\int_{0}^{1}\frac{\text{Li}_{k}(u)}{u}\log(u)^{n}du$$
Integrating by parts gives $$I_{n,k}=\left[\text{Li}_{k+1}(u)\log(u)^{n}\right]_{u=0}^{u=1}-\int_{0}^{1}\frac{\text{Li}_{k+1}(u)}{u}n\log(u)^{n-1}du$$
$$\text{Hence, }I_{n,k}=-nI_{n-1,k+1} \implies I_{n,k}=(-1)^{r}\frac{n!}{(n-r)!}I_{n-r,k+r}$$
Taking $r=n$ gives $I_{n,k}=(-1)^{n}n!I_{0,n+k}$.
$$\text{But obviously } I_{0,n+k}=\int_{0}^{1}\frac{\text{Li}_{n+k}(u)}{u}du=\text{Li}_{n+k+1}(1)-\text{Li}_{n+k+1}(0)=\zeta(n+k+1)$$
$$\text{Now consider }J_{n,k,l}=\int_{0}^{1}\frac{\text{Li}_{k}(u)}{u}\text{Li}_{l}(u)\log(u)^{n}du$$
Integrating by parts again,
$$J_{n,k,l}=\left[\text{Li}_{k+1}(u)\text{Li}_{l}(u)\log(u)^{n}\right]_{0}^{1}-\int_{0}^{1}\frac{\text{Li}_{l-1}(u)}{u}\text{Li}_{k+1}(u)\log(u)^{n}-\int_{0}^{1}\frac{n\log(u)^{n-1}}{u}\text{Li}_{k+1}(u)\text{Li}_{l}(u) du$$
So $J_{n,k,l}=-J_{n,k+1,l-1}-nJ_{n-1,k+1,l}$; continuing in the spirit of the first part suggests that we ought to try to increase the first and second indices, while decreasing the third. If we can succeed in this, we have found a closed form.
