Another polylog integral In the interest of housekeeping, I recently took a look at what what polylogarithm integrals are still in the unanswered questions list. Some of those questions have probably languished there because the solutions methods are presumably too tedious and too similar to previously answered questions make carrying out a solution worth while.
A few of the integrals with products of four or more logarithms in the numerator gave more trouble than I anticipated. After playing around with substitutions/integration-by-parts/et-cetera, it seems each the unsolved integrals I looked at can be boiled down to the following integral:

For $\left|z\right|\le1$, define $\mathcal{I}{\left(z\right)}$ via the integral representation
  $$\mathcal{I}{\left(z\right)}:=-\frac16\int_{0}^{z}\frac{\ln^{3}{\left(1-x\right)}\ln{\left(1+x\right)}}{x}\,\mathrm{d}x.\tag{1}$$

Question: Can integral $(1)$ be evaluated in terms of polylogarithms?

Notes: My best idea for a place to begin was to somehow reduce the integral to one with a single fourth-power logarithm in the numerator so as to make subsequent substitutions less of a hassle. I succeeded in reducing $\mathcal{I}{\left(z\right)}$ to the following integral:

$$J_{1}{\left(z\right)}=\int_{0}^{z}\frac{\ln^{4}{\left(\frac{1-y}{\left(1+y\right)^2}\right)}}{y}\,\mathrm{d}y.\tag{2}$$

Everything I've tried after that doesn't appear to be going anywhere though. Can somebody perhaps help me out?
Thanks.
 A: Each logarithm in the integrand can be written as an integral of a rational function. If you expand everything out, $\mathcal{I}(z)$ is a five-fold nested integral of rational functions.
It is a theorem of Kummer that three-fold nested integrals of rational functions can be expressed in terms of the logs, dilogs, and trilogs. For four or more nested integral, in general one will need the multiple polylogs
$$
\mathrm{Li}_{s_1,\ldots,s_k}(z_1,\ldots,z_k):=\sum_{n_1>\ldots>n_k\geq 1}\frac{z_1^{n_1}\ldots z_k^{n_k}}{n_1^{s_1}\ldots n_k^{s_k}},
$$
where $s_1,\ldots,s_k$ are positive integers. For $k=1$ this is the ordinary polylog. An $N$-fold nested integral of rational functions can be expressed in terms of polylogs with $s_1+\ldots+s_k\leq N$.
Erik Panzer has written a nice Maple package "HyperInt" than can perform this kind of computation (you can find the package on his webpage). For your integral, HyperInt produces
$$
\mathcal{I}(z)=\mathrm{Li}_{1,1,1,1,1}\left({\frac {-1+z}{z}},1,1,{\frac {1+z}{-1+z}},{\frac {z}{1+z}}\right)+\mathrm{Li}_{1,1,1,1,1}\left({\frac {-1+z}{z}},1,{\frac {1+z}{-1+z}},{\frac {-1+z}{1+z}},{\frac {z}{-1+z}}\right)\\
+\mathrm{Li}_{1,1,1,1,1}\left({\frac {-1+z}{z}},{\frac {1+z}{-1+z}},{\frac {-1+z}{1+z}},1,{\frac {z}{-1+z}}\right)+\mathrm{Li}_{1,1,1,1,1}\left({\frac {1+z}{z}},{\frac {-1+z}{1+z}},1,1,{\frac {z}{-1+z}}\right).
$$
This still leaves the question: can $\mathcal{I}(z)$ be expressed in terms of ordinary polylogs? I do not know.
