Assuming $0 \leq x\leq 1$ a CAS gives for the antiderivative
$$\text{Li}_3\left(1+\frac{1}{2
x}\right)-\text{Li}_3\left(2+\frac{1}{x}\right)+\text{Li}_3(-2 x-1)+\text{Li}_3(-2
x)+\text{Li}_3(-x)+\left(\text{Li}_2\left(2+\frac{1}{x}\right)-\text{Li}_2\left(1+
\frac{1}{2 x}\right)\right) \log \left(\frac{1}{x}+2\right)-\text{Li}_2(-2 x-1)
\log (2 x+1)-(\text{Li}_2(-2 x)+\text{Li}_2(-x)) \log (x)+\frac{1}{2} \log (2)
\left(\log ^2(x)-\pi ^2\right)-\log (x) \log (2 (x+1)) \log (2 x+1)$$ I must confess that using the bounds I have a quite ugly expression
$$\text{Li}_3(-3)+\text{Li}_3(-2)+\text{Li}_3\left(\frac{3}{2}\right)-\text{Li}_3(3)-
\text{Li}_2(-3) \log (3)+2 \text{Li}_2(3) \log (3)-\frac{\log ^3(3)}{2}+(\log (2)+i
\pi ) \log ^2(3)+\frac{1}{6} \left(-\log ^3(2)+3 i \pi \log ^2(2)+2\pi ^2 \log
(2)\right)+\frac{1}{2} \left(\log \left(\frac{3}{2}\right)+i \pi \right)^2 \log
(3)$$ which numerically is the rhs.
If we use partial fraction decomposition,
$$\int \frac{\log (x) \log (2 x+1)}{x (x+1)}\,dx=\int \frac{\log (x) \log (2 x+1)}{x }\,dx-\int \frac{\log (x) \log (2 x+1)}{ (x+1)}\,dx$$
we have for the simplest one
$$int \frac{\log (x) \log (2 x+1)}{x }\,dx=\text{Li}_3(-2 x)-\text{Li}_2(-2 x) \log (x)$$ but, as you wrote, the second is not pleasant
$$\int \frac{\log (x) \log (2 x+1)}{ (x+1)}\,dx=-\text{Li}_3\left(1+\frac{1}{2
x}\right)+\text{Li}_3\left(2+\frac{1}{x}\right)-\text{Li}_3(-2
x-1)-\text{Li}_3(-x)+\left(\text{Li}_2\left(1+\frac{1}{2
x}\right)-\text{Li}_2\left(2+\frac{1}{x}\right)\right) \log
\left(\frac{1}{x}+2\right)+\text{Li}_2(-2 x-1) \log (2 x+1)+\text{Li}_2(-x) \log
(x)+\frac{1}{2} \log (2) \left(\pi ^2-\log ^2(x)\right)+\log (x) \log (2 (x+1))
\log (2 x+1)$$ and still the same problem when using the bounds.