A tricky integral to evaluate I came across this integral in some work. So, I would like to ask:

QUESTION. Can you evaluate this integral with proofs?
$$\int_0^1\frac{\log x\cdot\log(x+2)}{x+1}\,dx.$$


$$\int_0^1\frac{\ln x\cdot\ln(x+2)}{x+1}\,dx=$$
$$=\text{Li}_3\left(-\tfrac{1}{3}\right)-2 \,\text{Li}_3\left(\tfrac{1}{3}\right)+\tfrac{1}{2} \ln 3\left[ \text{Li}_2\left(\tfrac{1}{9}\right)-6\, \text{Li}_2\left(\tfrac{1}{3}\right)  -\tfrac{2}{3} \ln ^2 3\right]+\tfrac{13}{8} \zeta (3).$$
I checked that this combination of polylog's evaluates to $-0.651114$, equal to a numerical evaluation of the integral.
Update:
As Timothy Budd pointed out, that this combination of polylog's simplifies to $-\frac{13}{24}\zeta(3)$ is proven by Przemo at MSE.
The identities that enable this simplification are
$$\text{Li}_3\left(-\tfrac{1}{3}\right)-2 \,\text{Li}_3\left(\tfrac{1}{3}\right) = -\tfrac{1}{6} \ln^3 3 + \tfrac{1}{6}\pi^2 \ln 3 - \tfrac{13}{6} \zeta(3),$$
$$\text{Li}_2(\tfrac{1}{9})=2\,\text{Li}_2(-\tfrac{1}{3})+2\,\text{Li}_2(\tfrac{1}{3}),$$
$$2\text{Li}_2\left(-\tfrac{1}{3}\right)-4 \,\text{Li}_2\left(\tfrac{1}{3}\right) =
\ln^2 3 -\tfrac{1}{3}\pi^2 .$$
