Evaluating this integral in Mathematica
$$i_2(z)=\int_0^z \frac{\log ^2(x) \log (1-x)}{1-x} \, dx\tag{1}$$
returns a mixture of polylogs up to order 4 and several log-terms.
In the region of interest $0 \le z \le 1$ we can simplify the expression using the simple functional substitutions
$$\log \left(\frac{1}{z}-1\right)\to \log (1-z)- \log (z)\tag{2a}$$ $$\log \left(\frac{1}{z}\right)\to - \log (z)\tag{2b}$$
and the more sophistcated ones ([1], [2])
$$\operatorname{Li}_2\left(\frac{z-1}{z}\right)\to - \operatorname{Li}_2(1-z)-\frac{1}{2} \log ^2(z)\tag{3a}$$ $$\operatorname{Li}_3\left(\frac{z-1}{z}\right)\to \left(-\operatorname{Li}_3(1-z)-\operatorname{Li}_3(z)\\ +\zeta (2) \log (z)+\frac{\log ^3(z)}{6}-\frac{1}{2} \log (1-z) \log ^2(z)+\zeta (3)\right)\tag{3b}$$
The expression for $i_2(z)$ now has, in the region of interest, been tranformed to a beautiful form defined as containing only explicitly real terms with simple linear arguments (except for one term) :
$$i_2(z) = -\text{Li}_4(1-z)-\text{Li}_4\left(\frac{z-1}{z}\right)+\text{Li}_4(z)+\frac{1}{2} \text{Li}_2(1-z) \log ^2(z)+\frac{1}{2} \text{Li}_2(z) \log ^2(z)-\text{Li}_2(1-z) \log (1-z) \log (z)+\text{Li}_3(1-z) \log (z)-\text{Li}_3(z) \log (1-z)-\zeta (3) \log (z)+\zeta (3) \log (1-z)-\frac{1}{24} \log ^4(z)\\+\frac{2}{3} \log (1-z) \log ^3(z)-\frac{1}{6} \pi ^2 \log ^2(z)-\frac{1}{2} \log ^2(1-z) \log ^2(z)\\ +\frac{1}{6} \pi ^2 \log (1-z) \log (z)-\frac{\pi ^4}{120}\tag{4}$$
The exceptional "ugly" term left is
$$u_4(z) =\operatorname{Li}_4\left(\frac{z-1}{z}\right)\tag{5}$$
We have seen that for lower order there are functional equations which "beautify" the expression.
Unfortunately, I have neither found a reference in the literature nor have I found the functional transformation myself
Question: can you transform $u_4$ to a beautiful form defined above?
My attempts
1) Starting with
$$u_4(z)=\int_0^{1-\frac{1}{z}} \frac{\operatorname{Li}_3(x)}{x} \, dx$$
Substituting $x\to 1-\frac{1}{t}$ gives
$$u_4(z)=\int_1^z \frac{\operatorname{Li}_3\left(\frac{t-1}{t}\right)}{(t-1) t} \, dt$$
The integral on the r.h.s. is readily calculated by Mathematica, but the result which consists of many terms, unfortunately contains exactly $u_4(z)$, i.e. it cancels out. Maybe it is just a trivial tautology. Here I got stuck.
2) Expanding $u_2(z)$ about $z=0$ and trying to guess the result. I wasn't good enough at guessing to find anything resonable up to now.
References