Show that \begin{eqnarray*} I_2=\int_0^1 \frac{\ln(1-x) (\ln(1+x))^2}{x} dx = -\frac{\pi^4}{240}. \end{eqnarray*}
Motivation: I am actually interested in triple plums of the form \begin{eqnarray*} \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \frac{ (-1)^{\star}}{ijk(i+j+k)} \end{eqnarray*} where $\star$ is $0,i,i+j$ of $i+j+k$. (So with $\star=i+j$ we have the above integral.
My try to solve the problem:
It is easy to show (sub $y=1-x$, geometrically expand the denominator and integrate term by term) \begin{eqnarray*} I_0=\int_0^1 \frac{(\ln(1-x))^3 }{x} dx = -6\zeta(4)= -\frac{\pi^4}{15}. \end{eqnarray*}
Now look at Ali's answer here to see that: \begin{eqnarray*} I_1 =\int_0^1 \frac{(\ln(1-x))^2 \ln(1+x)}{x} dx = -\frac{5}{8}\zeta(4)+2 \left( \operatorname{Li_4}\left(\frac12\right)+\frac78\ln2\zeta(3)-\frac14(\ln 2)^2\zeta(2)+\frac{1}{24} (\ln 2)^4 \right). \end{eqnarray*}
Next use Wolfie to get: \begin{eqnarray*} I_3 =\int_0^1 \frac{ (\ln(1+x))^3}{x} dx = 6\zeta(4)-6 \left( \operatorname{Li_4}\left(\frac12\right)+\frac{7}{8}\ln2\zeta(3)-\frac14(\ln 2)^2\zeta(2)+\frac{1}{24} (\ln 2)^4 \right). \end{eqnarray*}
Finally consider the following linear combination of the above integrals \begin{eqnarray*} I_0+3I_1+3I_2+I_3= \int_0^1 \frac{(\ln(1-x^2))^3 }{x} dx = \frac{1}{2} I_0. \end{eqnarray*} Now just do the linear algebra & we are dumb.
So my question is: Is there an easier way that manages to avoid the fourth polylogarithm of a laugh?
