Is there a elementary way to evaluate $$\int_0^1\frac{\ln(1-3x+x^2)\ln x}xdx=\frac85\zeta(3)+\frac{2}{5} \pi ^2 \ln\varphi-2 i \pi\ln^2\varphi$$ where $\varphi$ is the golden ratio?
(I posted it as an answer)
Factorizing $1-3x+x^2$ gives $1-3x+x^2=(x-\varphi^2)(x-\varphi^{-2})$, use the formula
$$\tiny\int\frac{\ln(x-a)\ln x}{x}dx=\operatorname{Li}_3\left(\frac{x}{a}\right)-\ln x \operatorname{Li}_2\left(\frac{x}{a}\right)-\frac{1}{2} \ln ^2x \ln \left(1-\frac{x}{a}\right)+\frac{1}{2} \ln ^2x \ln (x-a)+C$$
and use
$$\operatorname{Li}_3(\varphi^2)=\frac45 \zeta (3)-\frac 23 \ln ^3\varphi-2 i \pi \ln ^2\varphi+\frac{8}{15} \pi ^2 \ln\varphi$$
$$\operatorname{Li}_3(\varphi^{-2})=\frac45 \zeta (3)+\frac 23 \ln ^3\varphi-\frac{2}{15} \pi ^2 \ln\varphi$$
we can deduce the result.
Can it be done without using the value of $\operatorname{Li}_3$?
