The main ingredient here is the integral representation
$$\operatorname{Li}_n(z)=\frac{(-1)^{n-1}}{(n-2)!}\int_0^1
\frac{\ln\left(1-zx\right)\ln^{n-2}x\,dx}{x},\tag{$\spadesuit$}$$
valid for $|z|<1,n\in\mathbb{N}_{\ge 2}$.
The derivation goes as follows:
Rewrite the initial integral as
\begin{align*}
\mathcal{I}&=\int_0^1\ln(x+2)\underbrace{\left[\ln x\ln^2(1+x)-\ln^2x\ln(1+x)\right]}_{=\frac13\left(\ln x-\ln(x+1)\right)^3-\frac13\ln^3 x+\frac13\ln^3 (x+1)} dx=\\
&=\frac13\biggl[\underbrace{\int_0^1\ln(x+2)\ln^3(x+1)dx}_{\mathcal{I}_1}-\underbrace{\int_0^1\ln(x+2)\ln^3x\,dx}_{\mathcal{I}_2}-\underbrace{\int_0^1\ln(x+2)\ln^3\frac{x+1}{x}dx}_{\mathcal{I}_3}\biggr].
\end{align*}
The integrals $\mathcal{I}_{1,2}$ have antiderivatives that can be expressed in terms of polylogarithms (say, with Mathematica), therefore we concentrate on $\mathcal{I}_3$. After the change of variables $t=\frac{2x}{x+1}$, we obtain
\begin{align*}\mathcal{I_3}&=-2\int_0^1\frac{\ln\frac{4-t}{2-t}\ln^3\frac t2\,dt}{(2-t)^2}=\\&=-2\int_0^1\frac{\ln\frac{4-t}{2-t}\ln^3 t\,dt}{(2-t)^2}+
6\ln 2 \int_0^1\frac{\ln\frac{4-t}{2-t}\ln^2 t\,dt}{(2-t)^2} \tag{$\clubsuit$}\\&\quad -6\ln^22\int_0^1\frac{\ln\frac{4-t}{2-t}\ln t\,dt}{(2-t)^2}+
2\ln^32\int_0^1\frac{\ln\frac{4-t}{2-t}dt}{(2-t)^2}.
\end{align*}
Now let me explain how these integrals can be computed. Consider, for instance, the first term in ($\clubsuit$):
\begin{align*}
2\int_0^1\frac{\ln\frac{4-t}{2-t}\ln^3 t\,dt}{(2-t)^2}&=\int_0^1\ln\frac{4-t}{2-t}\ln^3 t\,d\left(\frac{t}{2-t}\right)=-
\int_0^1\frac{t}{2-t}d\left(\ln\frac{4-t}{2-t}\ln^3 t\right)=\\
&=-
\int_0^1\frac{t}{2-t}\left[\color{red}{-\frac{\ln^3 t}{4-t}+\frac{\ln^3 t}{2-t}}+\frac{3}{t}\ln\frac{4-t}{2-t}\ln^2 t\right]dt
\end{align*}
The terms shown in red lead to integrals computable with the help of ($\spadesuit$) (e.g. differentiate it with respect to $z$ and see what happens). The remaining nontrivial piece is thus
$$\int_0^1\frac{\ln\frac{4-t}{2-t}\ln^2 t}{2-t}dt=
\int_0^1\frac{\ln(4-t)\ln^2 t}{2-t}dt-\int_0^1\frac{\ln(2-t)\ln^2 t}{2-t}dt$$
The second part again has again a polylogarithmic antiderivative computable with Mathematica, so it remains to compute
$$\mathcal{I}_4=\int_0^1\frac{\ln(4-t)\ln^2 t}{2-t}dt.$$
Note that the same procedure applied to the other three terms in ($\clubsuit$) leads to easily computable integrals (as instead of $\ln^2t$ in the analog of $\mathcal{I}_4$ we have $\ln t$ or $1$).
Thus it remains to compute $\mathcal{I}_4$. And this is the only place where a certain miracle takes place, which indicates that there should be an easier way to do the initial integral. Making the change of variables $s=2-t$, we get
\begin{align*}
\mathcal{I}_4&=\int_1^2\frac{\ln(2+s)\ln^2(2-s)\,ds}{s}=\\
&=\frac16\int_1^2\frac{\left[\ln(2+s)+\ln(2-s)\right]^3+\left[\ln(2+s)-\ln(2-s)\right]^3-2\ln^3(2+s)}{s}ds=\\
&=\frac16\int_1^2\frac{\ln^3(4-s^2)}{s}ds+\frac16\int_1^2\frac{\ln^3\frac{2+s}{2-s}}{s}ds-\frac13\int_1^2\frac{\ln^3(s+2)}{s}ds.
\end{align*}
Each of these three pieces again has polylogarithmic antiderivatives that can be computed by Mathematica. This becomes obvious after change of variables $u=s^2$ in the first integral (the miracle is here: due to special parameter values we don't have an additional linear term under logarithm which would spoil the things) and $u=\frac{2+s}{2-s}$ in the second.
So the conclusion is that indeed, the integral $\mathcal{I}$ can be expressed in terms of polylogarithms (up to $\operatorname{Li}_4$), but I was too lazy to type the answer. Fortunately, for that we have Cleo.
Added: As suggested by Vladimir Reshetnikov, the integration bounds $(0,1)$ are not really important: the above approach yields an explicit antiderivative which I posted at https://gist.github.com/anonymous/4c35e5617cf846e8f517