TITLE: Closed form for $\int z^n\ln{(z)}\ln{(1-z)}\,\mathrm{d}z$?
QUESTION [8 upvotes]: Problem. Find an anti-derivative for the following indefinite integral, where $n$ is a non-negative integer:
  $$\int z^n\ln{\left(z\right)}\ln{\left(1-z\right)}\,\mathrm{d}z=~???$$


My attempt:
Euler's reflection identity for dilogarithms states,
$$\ln{\left(z\right)}\ln{\left(1-z\right)}=\zeta{(2)}-\operatorname{Li}_2{\left(z\right)}-\operatorname{Li}_2{\left(1-z\right)}.\tag{1}$$
Using identity $(1)$, our integral may be split up into three integrals, the first of which is elementary:
$$\begin{align}
\int z^n\ln{\left(z\right)}\ln{\left(1-z\right)}\,\mathrm{d}z
&=\int z^n\left[\zeta{(2)}-\operatorname{Li}_2{\left(z\right)}-\operatorname{Li}_2{\left(1-z\right)}\right]\,\mathrm{d}z\\
&=\zeta{(2)}\int z^n\,\mathrm{d}z-\int z^n\operatorname{Li}_2{\left(z\right)}\,\mathrm{d}z-\int z^n\operatorname{Li}_2{\left(1-z\right)}\,\mathrm{d}z\\
&=\zeta{(2)}\frac{z^{n+1}}{n+1}-\int z^n\operatorname{Li}_2{\left(z\right)}\,\mathrm{d}z-\int z^n\operatorname{Li}_2{\left(1-z\right)}\,\mathrm{d}z.\tag{2}\\
\end{align}$$
Wolfram Alpha gives anti-derivatives for the two dilogarithmic integrals in terms of hypergeometric functions:
$$\small{\int z^n\operatorname{Li}_2{\left(z\right)}\,\mathrm{d}z=\frac{z^{n+1}}{(n+1)^2}\ln{(1-z)}+\frac{z^{n+1}}{n+1}\operatorname{Li}_2{\left(z\right)}+\frac{z^{n+2}}{(n+1)^2(n+2)}{_2F_1}{\left(1,n+2;n+3;z\right)}+\color{grey}{constant}};$$
$$\small{\int z^n\operatorname{Li}_2{\left(1-z\right)}\,\mathrm{d}z=-\frac{z^{n+1}}{(n+1)^3}+\frac{z^{n+1}\ln{(z)}}{(n+1)^2}+\frac{z^{n+1}\operatorname{Li}_2{\left(1-z\right)}}{n+1}-\frac{z^{n+1}\ln{(z)}}{(n+1)^2}{_2F_1}{\left(1,n+2;n+3;z\right)}+\frac{z^{n+1}}{(n+1)^3}{_3F_2}{\left(1,n+1,n+1;n+2,n+2;z\right)}+\color{grey}{constant}}.$$
Now, if we permit closed forms in terms of hypergeometric functions, then we might consider the problem solved. However, in my experience hypergeometric functions of integer parameters often have representations in terms of simpler functions, and I wonder if that might not be the case here. Does the anti-derivative sought have a more mundane representation that avoids hypergeometric functions?

REPLY [10 votes]: Integration by parts seems enough.
$$\int z^n\ln z\ln(1-z)\,\mathrm{d}z=\frac{z^{n+1}}{n+1}\ln z\ln(1-z)-\frac{1}{n+1}\int z^{n+1}\left(\frac{\ln(1-z)}{z}-\frac{\ln z}{1-z}\right)\,\mathrm{d}z$$
Now
\begin{align}
\int z^n\ln(1-z)\,\mathrm{d}z&=\frac{z^{n+1}}{n+1}\ln(1-z)-\int \frac{z^{n+1}}{(n+1)(z-1)}\,\mathrm{d}z\\
&=\frac{z^{n+1}}{n+1}\ln(1-z)-\frac{\ln(1-z)}{n+1}-\frac{1}{n+1}\int\frac{z^{n+1}-1}{z-1}\,\mathrm{d}z\\
&=\frac{z^{n+1}}{n+1}\ln(1-z)-\frac{\ln(1-z)}{n+1}-\frac{1}{n+1}\sum_{k=1}^{n+1}\frac{z^k}{k}
\end{align}
and
\begin{align}
\int \frac{z^{n+1}}{1-z}\ln z\,\mathrm{d}z &= \int\frac{\ln z}{1-z}-\frac{1-z^{n+1}}{1-z}\ln z\,\mathrm{d}z\\
&=\operatorname{Li}_2(1-z)-\sum_{k=1}^{n+1}\left(\frac{z^k}{k}\ln z-\frac{z^k}{k^2}\right)
\end{align}
Putting these together, we get
$$\int z^n\ln z\ln(1-z)\,\mathrm{d}z=\frac{z^{n+1}}{n+1}\ln z\ln(1-z)-\frac{z^{n+1}}{n+1}\ln(1-z)+\frac{\ln(1-z)}{(n+1)^2}+\frac{\operatorname{Li}_2(1-z)}{n+1}-\frac{1}{n+1}\sum_{k=1}^{n+1}\left(\frac{z^k}{k}\ln z-\frac{k+n+1}{k^2(n+1)}z^k\right)$$
I hope there aren't any typos. :)
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