This integral is not so bad if we use Feynman's trick.

$$I(a)=\int_{0}^{\frac 12} \frac{1}{x} \, \ln(1+ax) \, \log\left(\frac{1}{x} -1\right)\,dx$$ $$I'(a)=\int_{0}^{\frac 12} \frac 1 {1+ax}\, \log\left(\frac{1}{x} -1\right)\,dx$$ $$I'(a)=\frac{12 \text{Li}_2\left(\frac{a+1}{a+2}\right)+6 \log ^2(a+2)-\pi ^2}{12 a}$$ $$I(a)=\int_0^1 \frac{12 \text{Li}_2\left(\frac{a+1}{a+2}\right)+6 \log ^2(a+2)-\pi ^2}{12 a}\,da$$ The corresponding antiderivative is not bad (in terms of logarithms and polylogarithms) and $$\color{blue}{I(1)=\frac{\log ^3(3)}{3}-\frac{13}{8} \zeta (3)+\left(2 \text{Li}_2\left(\frac{1}{3}\right)-\text{Li}_2\left(-\frac{1} {3}\right)\right) \log (3)+}$$ $$\color{blue}{\left(2 \text{Li}_3\left(\frac{1}{3}\right)-\text{Li}_3\left(-\frac{1} {3}\right)\right)}$$

This integral is not so bad if we use Feynman's trick.

$$I(a)=\int_{0}^{\frac 12} \frac{1}{x} \, \ln(1+ax) \, \log\left(\frac{1}{x} -1\right)\,dx$$ $$I'(a)=\int_{0}^{\frac 12} \frac 1 {1+ax}\, \log\left(\frac{1}{x} -1\right)\,dx$$ $$I'(a)=\frac{12 \text{Li}_2\left(\frac{a+1}{a+2}\right)+6 \log ^2(a+2)-\pi ^2}{12 a}$$ $$I(a)=\int_0^1 \frac{12 \text{Li}_2\left(\frac{a+1}{a+2}\right)+6 \log ^2(a+2)-\pi ^2}{12 a}\,da$$ The corresponding antiderivative is not bad (in terms of logarithms and polylogarithms) and $$\color{blue}{I(1)=\frac{\log ^3(3)}{3}-\frac{13}{8} \zeta (3)+\left(2 \text{Li}_2\left(\frac{1}{3}\right)-\text{Li}_2\left(-\frac{1} {3}\right)\right) \log (3)+}$$ $$\color{blue}{\left(2 \text{Li}_3\left(\frac{1}{3}\right)-\text{Li}_3\left(-\frac{1} {3}\right)\right)}$$

Edit

I do not know how to reduce the result to $\frac{13 }{24}\zeta (3)$ which is exact for at least $10,000$ significant figures.