Let $n\ge0$ and $m\ge0$ be integers. We intend to find the following integral: \begin{equation} {\mathfrak I}^{(m,n)}(z) := \int\limits_0^z [\log(1-\xi)]^m \cdot [\log(\xi)]^n d\xi \end{equation}
Now, by using the Taylor expansion of $[\log(1-\xi)]^m$ given in the answer to Compute an integral containing a product of powers of logarithms. combined with the following identity: \begin{equation} \int\limits_0^z \xi^m [\log(\xi)]^n d\xi = \frac{z^m}{m+1} \cdot \sum\limits_{l=0}^n [\log(z)^{n-l} \cdot \binom{n}{l} l! \cdot \frac{(-1)^l}{(m+1)^l} \end{equation} we have found the following results: \begin{eqnarray} &&{\mathfrak I}^{(0,n)}(z) = +1\sum\limits_{l=0}^n \left( z \right) \cdot[\log(z)]^{n-l} (-1)^l \binom{n}{l} l!\\ &&{\mathfrak I}^{(1,n)}(z) = -1\sum\limits_{l=0}^n \left( z Li_1(z) + (l+1) z - \sum\limits_{p=1}^{l+1} Li_p(z) \right) \cdot[\log(z)]^{n-l} (-1)^l \binom{n}{l} l!\\ &&{\mathfrak I}^{(2,n)}(z) = +2\sum\limits_{l=0}^n \left( \frac{(z-1)}{2} [\log(1-z)]^2 + (z-1) Li_1(z) (l+1) + z \binom{l+2}{2} - \right.\\ &&\left.\sum\limits_{p=2}^{l+1} \left\{ S_{p-1,2}(z) + (l+2-p) Li_p(z) \right\}\right) \cdot[\log(z)]^{n-l} (-1)^l \binom{n}{l} l!\\ \end{eqnarray} where \begin{equation} S_{n,p}(z) := \frac{(-1)^{n-1+p}}{(n-1)! p!} \int\limits_0^1 \frac{[\log(\xi)]^{n-1} [\log(1-z \xi)]^p}{\xi} d\xi \end{equation} is the Nielsen generalized poly-logarithm.
Now of course the question is how do we write down the result for generic $m\ge 2$ ?
