Evaluate $\int_0^1 \log (1-x)\ _3F_2\left(1,1,1;\frac{3}{2},\frac{3}{2};x\right) \, dx$

Question

I encountered a hypergeometric integral while investigating harmonic sums $$\int_0^1 \log (1-x)\ _3F_2\left(1,1,1;\frac{3}{2},\frac{3}{2};x\right) \, dx$$ Based on my experience I suspect a nice closed-form exists but have found none. Any kind of help will be appreciated.

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Update: To complete the solution following @Jack D'Aurizio's derivation,

$1$. Let $uz\to u$ in expression $f(z)=\int_{0}^{1}\frac{\arcsin\sqrt{uz}}{\sqrt{uz(1-uz)}}\log(1-u)\,du$

$2$. Apply Fubini to $\int_0^1 f(4x(1-x))dx$, then it become $\int_0^1 du \int_{\frac{1-\sqrt{1-u}}2}^{\frac{1+\sqrt{1-u}}2}dx\cdots$

$3$. Integrate w.r.t $x$ by brute force, then let $u\to \frac{4t^2}{1+2t^2+t^4}$

$4$. These integrals are evaluated using method of arXiv $2007.03957$. Whence

$$-\frac{1}4 \sum _{n=1}^{\infty } \left(\frac{4^n}{\binom{2 n}{n}}\right)^2\frac{ H_n}{n^3}=\int_0^1 \log (1-x)\ _3F_2\left(1,1,1;\frac{3}{2},\frac{3}{2};x\right) \, dx=-8 C^2+8 \pi C \log (2)-32 \pi \Im(\text{Li}_3(1+i))-16 \text{Li}_4\left(\frac{1}{2}\right)+\frac{413 \pi ^4}{360}-\frac{2}{3} \log ^4(2)+\frac{8}{3} \pi ^2 \log ^2(2)$$

Answer 1

We have $$\begin{eqnarray*} I=\int_{0}^{1}\log(1-x)\;\phantom{}_3 F_2\left(1,1,1;\tfrac{3}{2},\tfrac{3}{2};x\right)\,dx &=& \sum_{n\geq 0}\int_{0}^{1}\left(\frac{4^n}{\binom{2n}{n}}\right)^2\frac{x^n \log(1-x)}{(2n+1)^2}\,dx\\&=&-\sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{H_{n+1}}{n+1}\end{eqnarray*}$$ and $\frac{4^n}{(2n+1)\binom{2n}{n}}= 4^n B(n+1,n+1)$, so the integral is given by $$ I = -\int_{0}^{1} \sum_{n\geq 0}\frac{4^n H_{n+1}}{(n+1)(2n+1)\binom{2n}{n}}(4x(1-x))^n\,dx $$ and the problem boils down to computing $$ f(z)=\sum_{n\geq 0}\frac{4^n H_{n+1}}{(n+1)(2n+1)\binom{2n}{n}}z^n=- \int_{0}^{1}\frac{\arcsin\sqrt{uz}}{\sqrt{uz(1-uz)}}\log(1-u)\,du$$ leading to $$\begin{eqnarray*} I &=& \iint_{(0,1)^2}\frac{\arcsin\sqrt{4xu(1-x)}}{\sqrt{4ux(1-x)(1-4ux(1-x))}}\log(1-u)\,du\,dx\\&=&2\iint_{(0,1)^2}\frac{\arcsin(ux)\log(1-u^2)}{\sqrt{1-(ux)^2}\sqrt{1-x^2}}\,du\,dx \end{eqnarray*}$$ This is a polylogarithmic integral which is expected to break down into Euler sums with weight four.
I guess that a reasonable approach for finishing the computation is to enforce the substitutions $x=\sin\vartheta, u=\sin\varphi $ and then exploit well-known Fourier series to convert the double integral into a combination of Euler sums.