How to prove that $\int_{0}^{1}\ln{(x/(1-x))}\ln{(1+x-x^2)}\frac{dx}{x}=-\frac{2}{5}\zeta{(3)}$ 
$$\int_{0}^{1}\ln{\big(\frac{x}{1-x}\big)}\ln{(1+x-x^2)}\frac{dx}{x}=-\frac{2}{5}\zeta{(3)}$$

Put  $$\frac{x}{1-x}=y$$
 $$I=\int_{0}^{\infty}\ln{y}\ln{(1+3y+y^2)}\frac{dy}{y(y+1)}=\frac{8}{5}\zeta{(3)}$$
 Simple integral at first sight, however I cannot prove that. I would appreciate your help.

I'll prove the second integral in the post.
Let $I(a)=\int^{\infty}_{0}\frac{\log y\log(1+ay)}{y(y+1)}dy$, we have $I'(a)=\int^{\infty}_{0}\frac{\log y\,dy}{(1+y)(1+ay)}$, and $I=I(\frac{3+\sqrt{5}}{2})+I(\frac{3-\sqrt{5}}{2})$.
Now $I(0)=0$, and $I'(a)=\frac{(\log a)^2}{2(1-a)}$(Proof: Substitute $y\rightarrow\frac{1}{ay}$), so for any $0<a<1$ we have 
$\begin{align*}
I(a)&=\int^a_0\frac{(\log b)^2}{2(1-b)}db\\
&=\frac12(\log a)^2\log(1-a)+\log a Li_2(a)-Li_3(a)
\end{align*}$
and 
$\begin{align*}
I(1/a)&=\int^{1/a}_0\frac{(\log b)^2}{2(1-b)}db\\
&=\int^{a}_{+\infty}\frac{(\log b)^2}{2b(1-b)}db\\
&=-\frac12(\log a)^2\log(1-a)+\frac16(\log a)^3-\log a Li_2(a)+Li_3(a)
\end{align*}$
therefore 
$\begin{align*}
I(a)+I(1/a)&=-(\log a)^2\log(1-a)+\frac16(\log a)^3-2\log a Li_2(a)+2Li_3(a).
\end{align*}$
and we put $a=\frac{3-\sqrt{5}}{2}$ to get
$\begin{align*}
I&=-(\log \frac{3-\sqrt{5}}{2})^2\log(\frac{1+\sqrt{5}}{2})+\frac16(\log\frac{3-\sqrt{5}}{2})^3-2\log \frac{3-\sqrt{5}}{2} Li_2(\frac{3-\sqrt{5}}{2})+2Li_3(\frac{3-\sqrt{5}}{2})\\
&=\frac{8}{3}(\log\frac{1+\sqrt{5}}{2})^3+4\log \frac{1+\sqrt{5}}{2} Li_2(\frac{3-\sqrt{5}}{2})+2Li_3(\frac{3-\sqrt{5}}{2}).
\end{align*}$
The claim follows from the specific values $Li_2(\frac{3-\sqrt{5}}{2})=\frac{\pi^2}{15}-(\log\frac{1+\sqrt{5}}{2})^2$ and $Li_3(\frac{3-\sqrt{5}}{2})=-\frac{2\pi^2}{15}\log\frac{1+\sqrt{5}}{2}+\frac23(\log\frac{1+\sqrt{5}}{2})^3+\frac{4}{5}\zeta(3)$.
