Closed-form of $\int_0^1 \operatorname{Li}_3\left(1-x^2\right) dx$ By using dilogarithm functional equations we can show that
$$
\int_0^1 \operatorname{Li}_2\left(1-x^2\right)\,dx = \frac{\pi^2}{2}-4,
$$
where $\operatorname{Li}_2$ is the dilogarithm function.
Could we evaluate in closed-form the following integral?

$$
I = \int_0^1 \operatorname{Li}_3\left(1-x^2\right)\,dx,
$$

where $\operatorname{Li}_3$ is the trilogarithm function.
A related integral with known closed-form is
$$\int_0^1 \operatorname{Li}_3\left(\frac{1}{x^2}\right)\,dx = \zeta(3)+\frac{\pi^2}{3}-8\ln2 - 4\pi\,i,$$
where $\zeta$ is the Riemann zeta function.
 A: 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.
A: $$-\frac72\zeta\left(3\right)+\pi^2\left(\ln 2-1\right)+8$$
A: \begin{align}
I&=\int_0^1\operatorname{Li}_3(1-x^2)\ dx\overset{IBP}{=}2\int_0^1\frac{x^2\operatorname{Li}_2(1-x^2)}{1-x^2}\ dx\\
&=2\int_0^1\left(\frac1{1-x^2}-1\right)\operatorname{Li}_2(1-x^2)\ dx\\
&=2\int_0^1\frac{\operatorname{Li}_2(1-x^2)}{1-x^2}\ dx-2\int_0^1\operatorname{Li}_2(1-x^2)\ dx\tag{1}
\end{align}
By the OP, the second integral is $\boxed{\frac{\pi^2}{2}-4}$.
To calculate the first integral, we are going to use the generalized integral expression of the polylogrithmic function which can be found in the book (Almost) Impossible Integrals, Sums and series page 4.
$$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$ and by setting $n=1$ and replacing $x$ with $1-x^2$ we get
$$\frac{\operatorname{Li}_{2}(1-x^2)}{1-x^2}=-\int_0^1\frac{\ln(u)}{1-ux+ux^2}\ du$$
Now we can write 
$$\int_0^1\frac{\operatorname{Li}_2(1-x^2)}{1-x^2}\ dx=-\int_0^1\ln u\left(\int_0^1\frac{dx}{1-ux+ux^2}\right)\ du$$
$$=-\int_0^1\ln u\left(\frac{\arctan\sqrt{\frac{u}{1-u}}}{\sqrt{u-u^2}}\right)\ du, \quad \color{red}{\arctan\sqrt{\frac{u}{1-u}}=\arcsin\sqrt{u}=x}$$
$$=-4\int_0^{\pi/2}x\ln(\sin x)\ dx=-4\left(\frac7{16}\zeta(3)-\frac{\pi^2}{8}\ln2\right)=\boxed{\frac{\pi^2}{2}\ln2-\frac74\zeta(3)}$$
where the last result follows from the Fourier series of $\ln(\sin x)=-\ln2-\sum_{n=1}^\infty \frac{(-1)^n \cos(2nx)}{n}$.
Plugging the boxed results of the two integrals in $(1)$, we get

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


Note: Since $$\arctan x=-\frac{i}{2}\ln\left(\frac{1+ix}{1-ix}\right)$$
Then 
\begin{align}
\arctan\frac{x}{\sqrt{1-x^2}}&=-\frac{i}{2}\ln\left(\frac{1+\frac{ix}{\sqrt{1-x^2}}}{1-\frac{ix}{\sqrt{1-x^2}}}\right)\\
&=-\frac{i}{2}\ln\left(\frac{\sqrt{1-x^2}+ix}{\sqrt{1-x^2}-ix}*\color{red}{\frac{\sqrt{1-x^2}+ix}{\sqrt{1-x^2}+ix}}\right)\\
&=-\frac{i}{2}\ln\left(\frac{(\sqrt{1-x^2}+ix)^2}{1}\right)\\
&=-i\ln\left(\sqrt{1-x^2}+ix\right)\\
&=\arcsin x
\end{align}
and if we replace $x$ with $\sqrt{x}$, we get
$$\arctan\sqrt{\frac{x}{1-x}}=\arcsin\sqrt{x}$$

Here is a different proof:
Since $$\frac{d}{dy}\arctan\frac{y}{\sqrt{1-y^2}}=\frac1{\sqrt{1-y^2}}$$
Then $$\left.\arctan\frac{y}{\sqrt{1-y^2}}\right|_0^x=\int_0^x\frac1{\sqrt{1-y^2}}\ dy$$
$$\arctan\frac{x}{\sqrt{1-x^2}}=\arcsin x$$
A: 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 $\displaystyle -\ln\sin x=\ln2+\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n}$:
$$\begin{align} \int_0^1\frac{\arcsin^2x}{x}\,dx\\&=\int_0^{\frac{\pi}{2}}x^2\cot x \,dx\\&=-2\int_0^{\frac{\pi}{2}}x\ln\sin x \,dx\\
&=2\int_0^{\frac{\pi}{2}}x\left(\ln2+\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n}\right)\,dx\\&=\frac{\pi^2}{4}\ln2+2\sum_{n=1}^{\infty}\frac1{n}\int_0^{\frac{\pi}{2}}x\cos(2nx)\,dx\\&=\frac{\pi^2}{4}\ln2+2\sum_{n=1}^{\infty}\frac1{n}\frac1{4n^2}((-1)^n-1)\\&=\frac{\pi^2}{4}\ln2-\frac{7}{8}\zeta(3).
\end{align}$$
