<s>[INST] A closed form for $\int_0^1 \frac{\left(\log (1+x)\right)^3}{x}dx$? I would like some help to find a closed form for the following integral:$$\int_0^1 \frac{\left(\log (1+x)\right)^3}{x}dx $$ I was told it could be calculated in a closed form. I've already proved that $$\int_0^1 \frac{\log (1+x)}{x}dx  = \frac{\pi^2}{12}$$ using power series expansion. 
Thank you.
 [/INST] First note that 
\begin{align}
\int \frac{\ln^{2}(1+x)}{x} \, dx = - 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x)  \, \ln^{2}(1+x)
\end{align}
for which
\begin{align}
I_{2} = \int_{0}^{1} \frac{\ln^{2}(1+x)}{x} \, dx = \frac{\zeta(3)}{4}.
\end{align}
Now,
\begin{align}
\int \frac{\ln^{3}(1+x)}{x} \, dx = \int \ln(1+x) \, \frac{\ln^{2}(1+x)}{x} \, dx 
\end{align}
can be integrated by parts. This leads to
\begin{align}
\int \frac{\ln^{3}(1+x)}{x} \, dx &= \ln(1+x) \left( - 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x)  \, \ln^{2}(1+x) \right) \\
& - \int \frac{- 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x)  \, \ln^{2}(1+x)}{1+x} \, dx \\
&= \ln(1+x) \left( - 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x)  \, \ln^{2}(1+x) \right) \\
& \hspace{10mm} + 6 Li_{4}(1+x) + Li_{2}(1+x) \, \ln^{2}(1+x) - 4 Li_{3}(1+x) \, \ln(1+x)
\end{align}
This leads to 
\begin{align}
I_{3} &= \int_{0}^{1} \frac{\ln^{3}(1+x)}{x} \, dx \\
&= 6 Li_{4}(2) + \ln^{2}(2) \, Li_{2}(2) - 4 \ln(2) \, Li_{3}(2) - 6 \zeta(4) - \frac{7}{4} \, \ln(2) \, \zeta(3) \\
&= \frac{\pi^{4}}{15} + \frac{\pi^{2}}{4} \, \ln^{2}(2) - \frac{21}{4} \, \zeta(3) \, \ln(2) - \frac{1}{4} \, \ln^{4}(2) - 6 Li_{4}\left(\frac{1}{2} \right)
\end{align}
</s>