Closed form for $\int z^n\ln{(z)}\ln{(1-z)}\,\mathrm{d}z$? 
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?
 A: For the second antiderivative, it can be simplified to $$\int z^n\operatorname{Li}_2{\left(z\right)}\,\mathrm{d}z=\frac{z^{n+1} ((n+1) \text{Li}_2(z)+\log (1-z))+B_z(n+2,0)}{(n+1)^2}$$ However, for the third one, I did not arrive to anything significantly different from your own result.
I am pretty sure that there is a way to make all of that simpler since, defining $$I_n=\int z^n\ln{\left(z\right)}\ln{\left(1-z\right)}\,\mathrm{d}z$$ for any positive integer value of $n$, a "quite simple" expression is obtained (in which $\zeta(2)$ never appears).
So, basically, I wonder if the use Euler's reflection identity for dilogarithms states does not make the problem more complex in terms of resulting functions.
A: 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. :)
