Integral $\int_0^{1/\phi}x\log(x)\log(1+x)\log(1-x)\,dx$ How can we evaluate this definite integral
$$I=\int_0^{1/\phi}x\log(x)\log(1+x)\log(1-x)\,dx,$$
where $\displaystyle\phi=\frac{1+\sqrt5}2$ is the golden ratio?
 A: How can we evaluate ?
Integrating by parts, we  get
\begin{align}
\mathcal{I}(x)&=\int x \ln x\ln(1+x)\ln(1-x)\,dx=\\
&=-\frac12\int  \ln x\ln(1+x)\ln(1-x)\,d(1-x^2)=\\
&=-\frac12 (1-x^2)\ln x\ln(1+x)\ln(1-x)\\
&\quad+\frac12\int\frac{\ln (1-x)\ln(1+x)}{x}dx\\&\quad-\frac12\underbrace{\int x\ln (1-x)\ln(1+x)\,dx}_{\text{elementary}}\\
&\quad +\underbrace{\frac12\int(1-x)\ln x\ln(1-x)dx-\frac12\int(1+x)\ln x\ln(1+x)dx}_{-\frac18\operatorname{Li}_2(x^2)+\text{elementary}}.
\end{align}
The second term antiderivative is the most complicated as naively it involves trilogarithms. However it can be simplified using the same trick that I used answering your other question: 
\begin{align*}
\int\frac{\ln (1-x)\ln(1+x)}{x}dx=\frac12\int\frac{\ln^2(1-x^2)}{x}dx-
\frac12\int\frac{\ln^2\frac{1+x}{1-x}}{x}dx
\end{align*}
Calculating these two antiderivatives separately, Mathematica simplifies them to a much more compact (one-line) expression, still involving trilogarithms. 
It may well happen that due to special integration bounds the tri- (and?) dilogarithms may be lifted using various polylogarithm identities (see equation (2) here and equations (12)-(19) here for a few examples). However I find such calculations rather boring, so for that you should wait for an answer of You-Know-Who.
A: Just an idea (for now)
$$
\lim_{a,b,c\to0}\partial_{abc}\int_0^{1/\phi}x^{a+1}(1+x)^b(1-x)^cdx
$$
changing variables to $x\phi = y$
$$
\frac{1}{\phi^{a+2}}\int_0^{1}y^{a+1}(1+\frac{1}{\phi}y)^b(1-\frac{1}{\phi}y)^cdy
$$
using the Appel Hyper geometric we find
$$
\int_0^1x^{\alpha-1}(1-a_1x)^{-\beta}(1-b_1x)^{-\beta'}dx = \frac{\Gamma(\alpha)\Gamma(1)}{\Gamma(\alpha+1)}F_1\left(\alpha,\beta,\beta';\gamma;a_1,b_1\right)
$$
where $\gamma-\alpha = 1$.
comparing terms we find
$$
a+1 = \alpha-1\to \alpha = a+2\\
\beta = -b\\
\beta' = -c\\
a_1 = \frac{1}{\phi}\\
b_1 = -\frac{1}{\phi}
$$
thus the integral is
$$
\lim_{a,b,c\to0}\partial_{abc}\frac{1}{\phi^{a+2}}\frac{\Gamma(a+2)\Gamma(1)}{\Gamma(a+3)}F_1\left(a+1,-b,-c;a+3;\frac{1}{\phi},-\frac{1}{\phi}\right)=\lim_{a,b,c\to0}\partial_{abc}\frac{1}{\phi^{a+2}}\frac{1}{a+2}F_1\left(a+1,-b,-c;a+3;\frac{1}{\phi},-\frac{1}{\phi}\right)
$$
