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.

\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 $$\left.\arctan\frac{y}{\sqrt{1-y^2}}\right|_0^x=\int_0^x\frac1{\sqrt{1-y^2}}\ dy$$
$$\arctan\frac{x}{\sqrt{1-x^2}}=\arcsin x$$
