Generalized form of this Harmonic Number series $\sum_{n=1}^{\infty} \frac{{H_n}x^{n+1}}{(n+1)^3}$

Question

i've tried to Evaluate $$\int_{0}^{\frac{\pi}{6}}x\ln^2(2\sin(x))dx$$ without using Contour integral first i changed $2\sin(x)$ into polar form ,and i got $$\int_{0}^{\frac{\pi}{6}}x\ln^2(2\sin(x))dx = \Re(\int_{0}^{\frac{\pi}{6}}x\ln^2(1-e^{2ix})dx)+\int_{0}^{\frac{\pi}{6}} x(x-\frac{\pi}{2})^2 dx $$ then to find $$\Re(\int_{0}^{\frac{\pi}{6}}x\ln^2(1-e^{2ix})dx)$$ So i used Harmonic Number identities $$\sum_{n=1}^{\infty} \frac{H_{n}x^{n+1}}{(n+1)} = \frac{\ln^2(1-x)}{2}$$ then substituted $x$ with $e^{2ix}$ and multiply x on both sides and take an integral $$\int_{0}^{\frac{\pi}{6}}\sum_{n=1}^{\infty} \frac{e^{2i(n+1)x}xH_n}{n+1} = \frac{1}{2}\int_{0}^{\frac{\pi}{6}}x\ln^2(1-e^{2ix})dx $$ and considered only real part $$ \Re(\frac{1}{2}\int_{0}^{\frac{\pi}{6}}x\ln^2(1-e^{2ix})dx) = \Re(\sum_{n=1}^{\infty} \frac{H_{n}}{n+1}\bigg[\frac{e^{\frac{i\pi(n+1)}{3}}(2i\pi(n+1))-6}{-24(n+1)^2} \bigg])$$ So i got $$\Re(\sum_{n=1}^{\infty} \frac{H_{n}}{n+1}\bigg[\frac{e^{\frac{i\pi(n+1)}{3}}(2i\pi(n+1))-6}{-24(n+1)^2} \bigg]) = \frac{1}{4}\bigg[\sum_{n=1}^{\infty}\frac{H_{n}\cos(\frac{\pi(n+1)}{3})}{(n+1)^3}-\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)^3}\bigg]+\sum_{n=1}^{\infty}\frac{H_{n}\sin(\frac{\pi(n+1)}{3})}{12(n+1)^2}$$ last but not least,So i change $\cos(\frac{\pi(n+1)}{3})$ into a polar form again then i've no idea How to evaluate it properly with out any huge expansion So i think i might need to use $$\sum_{n=1}^{\infty} \frac{{H_n}x^{n+1}}{(n+1)^3}$$ to complete the whole progress ,but i'm stuck at this point Could someone give me a hand please.