What is $\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx$? I encountered the integral $$\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx = -0.9393323982...$$ while researching the evaluation of harmonic sums. Mathematica 11 is not able to evaluate this integral, and is not able to evaluate the underlying indefinite integral.
I considered using the Maclaurin series for the expression $\sin ^{-1}(x)^2$ to evaluate this integral. Using this Maclaurin series, it is easily seen that the problem of evaluating the above integral is equivalent to the problem of evaluating the series $$\sum _{n=1}^{\infty } \frac{2^{2 n-1} H_{n-\frac{1}{2}}}{(1-2 n) n^2 \binom{2 n}{n}}=-0.9393323982...$$ which Mathematica 11 is not able to evaluate directly.
I also considered using the Maclaurin series for the expression $\log \left(1-x^2\right)$ to evaluate the above integral. Using this Macluarin series, it is easily seen that this integral is equal to $$\frac{1}{4} \left(16 \pi  C-\pi ^3-\pi ^2 \log (4)\right)+\sum _{n=1}^{\infty } \frac{\, _3F_2\left(\frac{1}{2},\frac{1}{2},1;\frac{3}{2},n+1;1\right)}{n^2-2 n^3}=-0.939332...,$$ but Mathematica is unable to evaluate the above infinite series.
What is a closed-form evaluation of $\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx$? What techniques can be applied to evaluate this definite integral? 

As an addendum to the previous answer, it can be shown (through the FL-expansion of $\frac{\arcsin\sqrt{x}}{\sqrt{x}}$) that
$$\phantom{}_4 F_3\left(\tfrac{1}{2},\tfrac{1}{2},1,1;\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};1\right)=4\text{ Im } \text{Li}_3\left(\frac{1+i}{2}\right)-\frac{\pi^3}{32}-\frac{\pi}{8}\log^2(2).$$
