Integral $\int_0^1 \frac{\ln(1+x+x^2)\ln(1-x+x^2)}{x}dx$

As shown in the question we have: $$\sf I=-\frac14\int_0^1\frac{\ln^2(1+x+x^2)}{x}dx-\frac12\int_0^1\frac{\ln^2(1-x+x^2)}{x}dx$$ For the first integral we can write: $$\sf (a-b)^2=a^2-b^2-2b(a-b);\ a=\ln(1-x^3),b=\ln(1-x)$$ $$\sf \Rightarrow \int_0^1\frac{\ln^2(1+x+x^2)}{x}dx=\int_0^1 \frac{\left(\ln(1-x^3)-\ln(1-x)\right)^2}{x}dx$$ $$\sf =\color{blue}{\int_0^1 \frac{\ln^2(1-x^3)}{x}dx}-\int_0^1 \frac{\ln^2(1-x)}{x}dx-2\int_0^1 \frac{\ln(1-x)\ln(1+x+x^2)}{x}dx$$ $$\sf \overset{\color{blue}{x^3\to x}}=-\frac23\int_0^1 \frac{\ln^2(1-x)}{x}dx-2\int_0^1 \frac{\ln(1-x)\ln(1+x+x^2)}{x}dx=-\frac43\zeta(3)-2J$$ Note also that: $$\sf \int_0^1 \frac{\ln^2(1-x)}{x}dx=\int_0^1 \frac{\ln^2 x}{1-x}dx=\sum_{n=1}^\infty \int_0^1 x^{n-1}\ln^2 x\, dx=2\sum_{n=1}^\infty \frac{1}{n^3}=2\zeta(3)$$ The latter integral can be found here: $$\sf J=\int_0^1 \frac{\ln(1-x)\ln(1-x+x^2)}{x}dx=-\frac{\pi}{9\sqrt 3}\psi_1\left(\frac13\right)+\frac{2\pi^3}{27\sqrt 3}-\frac13\zeta(3) $$ $$\Rightarrow \boxed{\sf \int_0^1\frac{\ln^2(1+x+x^2)}{x}dx=\frac{2\pi}{9\sqrt3}\psi_1\left(\frac13\right)-\frac{4\pi^3}{27\sqrt3}-\frac23\zeta(3)}$$ The second integral can be found here: $$\boxed{\sf\int_0^1 \frac{\ln^2(1-x+x^2)}{x}dx=-\frac{4\pi}{9\sqrt{3}}\psi_1\left(\frac{1}{3}\right)+\frac{8\pi^3}{27\sqrt{3}}+\frac{22}{9}\zeta(3)}$$ Combining the boxed results yields: $$\boxed{\sf \int_0^1 \frac{\ln(1+x+x^2)\ln(1-x+x^2)}{x}dx=\frac{\pi}{6\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{\pi^3}{9\sqrt{3}}-\frac{19}{18}\zeta(3)}$$