Integral $\int_0^1 \frac{2x-1}{1+x-x^2}\left(4\ln x\ln(1+x)-\ln^2(1+x)\right)dx$ The following problem was posted earlier this year by Cornel Ioan Valean:

Prove that
  $$I=\int_0^1 \frac{2x-1}{1+x-x^2}\left(4\ln x\ln(1+x)-\ln^2(1+x)\right)dx=\frac{127}{20}\zeta(3)-\frac{8\pi^2}{5}\ln(\varphi)$$

My idea was to consider the following integral:
$$\mathcal J(a)=\int_0^1 \frac{2x-1}{1+x-x^2}\ln(a+x)\ln(1+x)dx$$
So $I=4\mathcal J(0)-\mathcal J(1)$. In order to evaluate $\mathcal J(a)$ I tried to apply Feynman's trick:
$$\mathcal J'(a)=\int_0^1 \frac{\ln(1+x)}{a+x}\frac{dx}{\varphi-x}-\int_0^1 \frac{\ln(1+x)}{a+x}\frac{dx}{\frac{1}{\varphi}+x},\quad \varphi=\frac{1+\sqrt 5}{2}$$
$$\small =\frac{1}{a+\varphi}\int_0^1 \frac{\ln(1+x)}{a+x}dx+\frac{1}{a+\varphi}\int_0^1 \frac{\ln(1+x)}{\varphi-x}dx+\frac{1}{a-\frac{1}{\varphi}}\int_0^1 \frac{\ln(1+x)}{a+x}dx-\frac{1}{a-\frac{1}{\varphi}}\int_0^1 \frac{\ln(1+x)}{\frac{1}{\varphi}+x}dx$$
But I gave up on this idea after I realised that:
$$\int_0^1 \frac{\ln(1+x)}{t+x}dx=\ln 2\ln \left(\frac{t+1}{t-1}\right)+\operatorname{Li}_2\left(\frac{2}{1-t}\right)-\operatorname{Li}_2\left(\frac{1}{1-t}\right)$$
Where $\operatorname{Li}_2(x)$ is the Dilogarithm.
Other methods weren't promising either, such as the substitution $x=\frac{1-t}{1+t}$, to combine the integral with it's sister one that has the denominator $1-x+x^2$, or to integrate by parts which gives:
$$\small 2I=2\int_0^1 \frac{\ln(1+x-x^2)\ln x}{1+x}dx+2\int_0^1 \frac{\ln(1+x-x^2)\ln(1+x)}{x}dx-\int_0^1 \frac{\ln(1+x-x^2)\ln (1+x)}{1+x}dx$$
I believe that the factor of $4$ plays a big role into obtain nicely this result and one shouldn't split the integral into two parts, but I had no success and I would appreciate some help.
 A: We start by writing
\begin{align}
I &= \int \limits_0^1 \frac{1-2x}{1+x -x^2} \log(1+x) [\log(1+x)-4\log(x)] \, \mathrm{d} x \\
&= \int \limits_0^1 \frac{1-2x}{1+x-x^2} \left[\log^2\left(\frac{x^2}{1+x}\right) - 4 \log^2(x)\right] \mathrm{d} x \equiv J -4K \, .
\end{align}
Then
\begin{align}
J &= \int \limits_0^1 \frac{1-2x}{1+x-x^2} \log^2\left(\frac{x^2}{1+x}\right) \, \mathrm{d} x = \int \limits_0^1 \left[\frac{1}{1+x} - \frac{x (2+x)}{(1+x)(1+x-x^2)}\right] \log^2\left(\frac{x^2}{1+x}\right) \, \mathrm{d} x \\
&= \int \limits_0^1 \frac{\log^2\left[\frac{\left(\frac{x}{1+x}\right)^2}{1-\frac{x}{1+x}}\right]}{1 - \frac{x}{1+x}} \, \mathrm{d} \frac{x}{1+x} - \int \limits_0^1 \frac{\log^2\left(\frac{x^2}{1+x}\right)}{1 - \frac{x^2}{1+x}} \, \mathrm{d} \frac{x^2}{1+x} = \int \limits_0^{1/2} \frac{\log^2\left(\frac{t^2}{1-t}\right) - \log^2(t)}{1-t} \, \mathrm{d} t \\
&= \int \limits_0^{1/2} \frac{3 \log^2(t) + \log^2(1-t) - 4 \log(t) \log(1-t)}{1-t} \, \mathrm{d} t \\
&= \left[6 \operatorname{Li}_3(t) - 6 \log(t) \operatorname{Li}_2(t)-3\log(1-t)\log^2(t) \vphantom{\frac{1}{3}}\right. \\
&\phantom{= \left[\vphantom{\frac{1}{3}}\right.}\left.-\frac{1}{3}\log^3(1-t) + 4 \operatorname{Li}_3(1-t) - 4 \log(1-t) \operatorname{Li}_2(1-t) \right]_{t=0}^{t=1/2}\\
&= \frac{19}{4} \zeta(3)
\end{align}
follows from the known polylogarithm values $\operatorname{Li}_2(1),\operatorname{Li}_3(1),\operatorname{Li}_2(1/2),\operatorname{Li}_3(1/2)$ and we find 
\begin{align}
K &= \int \limits_0^1 \frac{1-2x}{1+x-x^2} \log^2(x) \, \mathrm{d} x = \int \limits_0^1 \left[\frac{\log^2(x)}{\varphi^{-1} + x} - \frac{\log^2(x)}{\varphi - x}\right] \mathrm{d} x \\
&= 2 \left[- \operatorname{Li}_3(-\varphi) -\operatorname{Li}_3 (\varphi^{-1})\right] = 2\left[- \operatorname{Li}_3(-\varphi) + \operatorname{Li}_3 (-\varphi^{-1}) - \frac{1}{4} \operatorname{Li}_3 (\varphi^{-2})\right] \\
&= 2 \left[\frac{1}{6} \log^3(\varphi) + \frac{\pi^2}{6} \log(\varphi) - \frac{1}{4} \left(\frac{4}{5} \zeta(3) +\frac{2}{3} \log^3(\varphi) - \frac{2\pi^2}{15} \log(\varphi)\right)\right] \\
&= \frac{2}{5} \left[\pi^2 \log(\varphi) - \zeta(3)\right] \, .
\end{align}
using trilogarithm functional equations and the known value of $\operatorname{Li}_3(\varphi^{-2})$ (from here).
Therefore,
$$ I = J - 4 K = \frac{19}{4} \zeta(3) - \frac{8}{5} \left[\pi^2 \log(\varphi) - \zeta(3)\right] = \frac{127}{20} \zeta(3) - \frac{8}{5} \pi^2 \log(\varphi) \, .$$
