Closed-form of $\int_0^1 \operatorname{Li}_3\left(1-x^2\right) dx$

Expanding my comment: the substitution $1-x^2\mapsto x$, followed by expanding the trilogarithm and keeping in mind Legrende's duplication formula $B(n+1,\frac12)=2^{2n+1}B(n+1,n+1)$, we arrive at $$I=\sum_{n=1}^{\infty} \frac{2^{2n}}{n^3(2n+1)\binom{2n}{n}}=\sum_{n=1}^{\infty}\frac{2^{2n}}{n^3\binom{2n}{n}}-\sum_{n=1}^{\infty}\frac{2^{2n+1}}{n^2(2n+1)\binom{2n}{n}}\\=4\int_0^1\frac{\arcsin^2x}{x}\,dx-4\int_0^1\arcsin^2x \,dx$$ where I used the fact that $\displaystyle \sum_{n=1}^{\infty} \frac{(2x)^{2n}}{n^2\binom{2n}{n}}=2\arcsin^2x$.

The second integral is easily evaluated by IBP twice: $$\int_0^1\arcsin^2x \,dx=\int_0^{\frac{\pi}{2}}x^2\cos x \,dx=\frac{\pi^2}{4}-2$$

The first integral may be evaluated by IBP and using that $\displaystyle -\ln\sin x=\ln2+\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n}$: $$\begin{align} \int_0^1\frac{\arcsin^2x}{x}\,dx\\&=\int_0^{\frac{\pi}{2}}x^2\cot x \,dx\\&=-2\int_0^{\frac{\pi}{2}}x\ln\sin x \,dx\\ &=2\int_0^{\frac{\pi}{2}}x\left(\ln2+\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n}\right)\,dx\\&=\frac{\pi^2}{4}\ln2+2\sum_{n=1}^{\infty}\frac1{n}\int_0^{\frac{\pi}{2}}x\cos(2nx)\,dx\\&=\frac{\pi^2}{4}\ln2+2\sum_{n=1}^{\infty}\frac1{n}\frac1{4n^2}((-1)^n-1)\\&=\frac{\pi^2}{4}\ln2-\frac{7}{8}\zeta(3). \end{align}$$

To prove L.G. result, one just needs to apply twice integration by parts, then prove through its favourite technique (for instance, differentiation under the integral sign and computation of a few derivatives of a Beta function) that: $$I_0=\int_{0}^{1}x^2 \log(x)\,\frac{dx}{1-x^2}=1-\frac{\pi^2}{8}, $$ $$I_{-}=\int_{0}^{1}x \log(x)\log(1-x)\,\frac{dx}{1-x^2}=\frac{\pi^2\log(4)-5\zeta(3)}{16}, $$ $$I_{+}=\int_{0}^{1}x \log(x)\log(1+x)\,\frac{dx}{1-x^2}=\frac{-\pi^2\log(4)+9\zeta(3)}{16}. $$ More details to come if wanted. Time to go to bed for me.

$$-\frac72\zeta\left(3\right)+\pi^2\left(\ln 2-1\right)+8$$