Anti-derivative of a function that involves poly-logarithms. Let $n\ge 1$ be an integer and let $0 < z < a$ be real numbers. 
Let $Li_n(x):= \sum\limits_{l=1}^\infty z^l/l^n$ by the polylogarithm of order $n$. 
The question is to find the following anti-derivative:
\begin{equation}
{\mathfrak J}^{(n)}_a(z):=\int\frac{Li_n(a-z)}{z} dz=?
\end{equation}
By using integration by parts we have found the result for $n<= 3$. We have:
\begin{eqnarray}
{\mathfrak J}^{(1)}_a(z)&=&-\text{Li}_2\left(\frac{-a+z+1}{1-a}\right)-\log \left(\frac{z}{a-1}\right) \log (-a+z+1)\\
{\mathfrak J}^{(2)}_a(z)&=&\frac{1}{6} \pi ^2 \log (z)-\frac{1}{2} \log (a) \log ^2\left(\frac{z}{a}\right)+\text{Li}_2\left(\frac{z}{a}\right) \log \left(\frac{z}{a}\right)-\text{Li}_3\left(\frac{z}{a}\right)+\\
&&\left(\text{Li}_2(-a+z+1)+\text{Li}_2\left(\frac{-a+z+1}{z}\right)-\text{Li}_2\left(\frac{a (-a+z+1)}{z}\right)\right) \log \left(\frac{a (-a+z+1)}{z}\right)+\\
&&-\text{Li}_3(-a+z+1)-\text{Li}_3\left(\frac{-a+z+1}{z}\right)+\text{Li}_3\left(\frac{a (-a+z+1)}{z}\right)\\
{\mathfrak J}^{(3)}_a(z)&=&\text{Li}_2(a-z) \left(-\text{Li}_2\left(\frac{z}{a}\right)+\log \left(\frac{a-z}{a}\right) \log \left(\frac{a}{z}\right)+\frac{\pi ^2}{6}\right)+\\
&&\text{Li}_3(a-z) \log
   \left(\frac{z}{a}\right)+\text{Li}_3\left(1-\frac{z}{a}\right) \log (-a+z+1)-\\
&& {\mathfrak J}^{(3)}_{\frac{1}{a}}(\frac{1-a+z}{a})
\end{eqnarray}
What is the result for higher values of $n$?

Here we provide an answer for $n=3$. The idea is to expand the function to be sought for in a series about the value $a=1$. Clearly we have:
\begin{equation}
-{\mathfrak J}^{(n)}_a(a-x)=\sum\limits_{m=0}^\infty \int \frac{Li_3(x)}{(1-x)^{m+1}}dx \cdot (1-a)^m
\end{equation}
Now, using integration by parts we can derive the anti-derivative in the sum. There are no essential problems in this process it is just that it is a tedious task and, as usual, requires a lot of double-checking to avoid errors. We just state the result:
\begin{eqnarray}
\int\frac{Li_3(x)}{(1-x)^{m+1}}dx=Li_1(x) Li_3(x)-\frac{1}{2} Li_2(x)^2
\end{eqnarray}
for $m=0$ and
\begin{eqnarray}
&&m \int\frac{Li_3(x)}{(1-x)^{m+1}}dx=\\
&&
(((1-x)^{-m}-1) \text{Li}_3(x)-2 \text{Li}_3(1-x)+
\log(1-x)(-\text{Li}_2(x) -\log(1-x) \log (x)+\frac{1}{3} \pi ^2 ))+\\
&&H_{m-1} \cdot(\frac{1}{2} \log(1-x)^2+\text{Li}_2(x))-\\
&&\sum\limits_{j=2}^m \frac{1}{j-1} \cdot \frac{Li_2(x)}{(1-x)^{j-1}}+
\sum\limits_{j=2}^{m-1} \frac{ \left(H_{m-1}-H_{j-1}\right) }{(j-1)(1-x)^{j-1}}\cdot \left(-\frac{1}{j-1}-\log (1-x)\right)
\end{eqnarray}
for $m\ge 1$.
Now the only thing we need to do is to insert the above expressions into the sum and do the summations. Again, even a glimpse at those equations suffices to see that it is possible to carry out the summations. We state the result:
\begin{eqnarray}
&&\int\limits_0^x \frac{Li_3(t)}{a-t}dt=\\
&&-\frac{\text{Li}_2(x){}^2}{2}-\text{Li}_3(x) \log (1-x)+\\
&&-\log (a) \left(-2 \text{Li}_3(1-x)-\text{Li}_3(x)+
\log(1-x)(-\text{Li}_2(x) -\log (x) \log(1-x)+\frac{1}{3} \pi ^2 )\right)+\\
&&\text{Li}_3(x)
   \left(-\log \left(1-\frac{1-a}{1-x}\right)\right)+\\
&&\frac{1}{2} \log ^2(a) \left(\text{Li}_2(x)+\frac{1}{2} \log ^2(1-x)\right)-\\
&&\text{Li}_2(x) \left(\text{Li}_2\left(1-\frac{a}{x}\right)+\log \left(\frac{a-x}{1-x}\right)
   \log \left(\frac{a}{x}\right)-\text{Li}_2\left(\frac{x-1}{x}\right)\right)+\\
&&2 Li_3(1) Li_1(1-a)+\\
&&\sum\limits_{j=2}^{\infty}\sum\limits_{m=j+1}^\infty \frac{ \left(H_{m-1}-H_{j-1}\right) }{(j-1)}\cdot \left(\frac{1}{j-1}-\frac{1}{(j-1)(1-x)^{j-1}}-\frac{\log (1-x)}{(1-x)^{j-1}}\right)\cdot \frac{(1-a)^m}{m}
\end{eqnarray}
We will calculate  the last double sum  and simplify the whole result later on.
