Integral $\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}\mathrm dx$ Is it possible to evaluate this integral in a closed form?
$$I=\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}\mathrm dx$$
It also can be represented as
$$I=\int_0^{\pi/4}\frac{\phi^2}{\cos \phi\,\sqrt{\cos 2\phi}}\mathrm d\phi$$
 A: Okay, finally I was able to prove it.
Step 0. Observations. In view of the following identity
$$ \int_{0}^{\frac{\pi}{2}} \arctan (r \sin\theta) \, d\theta = 2 \chi_{2} \left( \frac{\sqrt{1+r^{2}} - 1}{r} \right), $$
Vladimir's result suggests that there may exists a general formula connecting
$$ I(r, s) = \int_{0}^{\frac{\pi}{2}} \arctan (r \sin\theta) \arctan (s \sin\theta) \, d\theta $$
and the Legendre chi function $\chi_{2}$. Indeed, inspired by Vladimir's result, I conjectured that
$$ I(r, s) = \pi \chi_{2} \left( \frac{\sqrt{1+r^{2}} - 1}{r} \cdot \frac{\sqrt{1+s^{2}} - 1}{s} \right). \tag{1} $$
I succeeded in proving this identity, so I post a solution here.
Step 1. Proof of the identity $\text{(1)}$. It is easy to check that the following identity holds:
$$ \arctan(ab) = \int_{1/b}^{\infty} \frac{a \, dx}{a^{2} + x^{2}}. $$
So it follows that
\begin{align*}
I(r, s)
&= \int_{1/r}^{\infty} \int_{1/s}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{\sin^{2}\theta}{(x^{2} + \sin^{2}\theta)(y^{2} + \sin^{2}\theta)} \, d\theta dy dx \\
&= \int_{1/r}^{\infty} \int_{1/s}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{1}{x^{2} - y^{2}} \left( \frac{x^{2}}{x^{2} + \sin^{2}\theta} - \frac{y^{2}}{y^{2} + \sin^{2}\theta} \right) \, d\theta dy dx \\
&= \frac{\pi}{2} \int_{1/r}^{\infty} \int_{1/s}^{\infty} \frac{1}{x^{2} - y^{2}} \left( \frac{x}{\sqrt{x^{2} + 1}} - \frac{y}{\sqrt{y^{2} + 1}} \right) \, dy dx.
\end{align*}
For the convenience of notation, we put
$$ \alpha = \frac{\sqrt{r^{2} + 1} - 1}{r} \quad \text{and} \quad \beta = \frac{\sqrt{s^{2} + 1} - 1}{s}. $$
Then it is easy to check that $\mathrm{arsinh}(1/r) = - \log \alpha$ and likewise for $s$ and $\beta$. Thus with the substitution $x \mapsto \sinh x$ and $y \mapsto \sinh y$, we have
\begin{align*}
I(r, s)
&= \frac{\pi}{2} \int_{-\log\alpha}^{\infty} \int_{-\log\beta}^{\infty} \frac{\sinh x \cosh y - \sinh y \cosh x}{\sinh^{2}x - \sinh^{2}y} \, dy dx.
\end{align*}
Applying the substitution $e^{-x} \mapsto x$ and $e^{-y} \mapsto y$, it follows that
\begin{align*}
I(r, s)
&= \pi \int_{0}^{\alpha} \int_{0}^{\beta} \frac{dydx}{1 - x^{2}y^{2}} \\
&= \pi \sum_{n=0}^{\infty} \left( \int_{0}^{\alpha} x^{2n} \, dx \right) \left( \int_{0}^{\beta} y^{2n} \, dx \right)
 = \pi \sum_{0}^{\infty} \frac{(\alpha \beta)^{2n+1}}{(2n+1)^{2}} \\
&= \pi \chi_{2}(\alpha \beta)
\end{align*}
as desired, proving the identity $\text{(1)}$.

EDIT. I found a much simpler and intuitive proof of $\text{(1)}$. We first observe that $\text{(1)}$ is equivalent to the following identity
$$ \int_{0}^{\frac{\pi}{2}} \arctan\left( \frac{2r\sin\theta}{1-r^{2}} \right) \arctan\left( \frac{2s\sin\theta}{1-s^{2}} \right) \, d\theta = \pi \chi_{2}(rs). $$
Now we first observe that from the addition formula for the hyperbolic tangent, we obtain the following formula
$$ 
\operatorname{artanh}x - \operatorname{artanh} y = \operatorname{artanh} \left( \frac{x - y}{1 - xy} \right) $$
which holds for sufficiently small $x, y$. Thus
\begin{align*}
\arctan\left( \frac{2r\sin\theta}{1-r^{2}} \right)
&= \frac{1}{i} \operatorname{artanh}\left( \frac{2ir\sin\theta}{1-r^{2}} \right) 
 = \frac{\operatorname{artanh}(re^{i\theta}) - \operatorname{artanh}(re^{-i\theta})}{i} \\
&= 2 \Im \operatorname{artanh}(re^{i\theta})
 = 2 \sum_{n=0}^{\infty} \frac{\sin(2n+1)\theta}{2n+1} r^{2n+1}.
\end{align*}
We readily check this holds for any $|r| < 1$. Therefore
\begin{align*}
&\int_{0}^{\frac{\pi}{2}} \arctan\left( \frac{2r\sin\theta}{1-r^{2}} \right) \arctan\left( \frac{2s\sin\theta}{1-s^{2}} \right) \, d\theta \\
&\quad = 4 \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{r^{2m+1}s^{2n+1}}{(2m+1)(2n+1)} \int_{0}^{\frac{\pi}{2}} \sin(2m+1)\theta \sin(2n+1)\theta \, d\theta\\
&\quad = 2 \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{r^{2m+1}s^{2n+1}}{(2m+1)(2n+1)} \int_{0}^{\frac{\pi}{2}} \{ \cos(2m-2n)\theta - \cos(2m+2n+2)\theta \} \, d\theta\\
&\quad = \pi \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{r^{2m+1}s^{2n+1}}{(2m+1)(2n+1)} \delta_{m,n} \\
&\quad = \pi \chi_{2}(rs).
\end{align*}
A: This is a beautiful integral with a beautiful result:
$$\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}dx=\frac\pi2\Big(\operatorname{Li}_2\left(3-\sqrt8\right)-\operatorname{Li}_2\left(\sqrt8-3\right)\Big).$$
One could expect it to have an elegant solution too, giving an insight into why the result is of this form. 
Alas, I was not able to find such a solution. Mine was a long, semi-manual with a lot of assistance from Mathematica, lookups into Gradshteyn-Ryzhik Table of Integrals and huge intermediate expressions not fitting on a single screen. I would rather not post it here now. Instead, I hope somebody comes up with an elegant one.

Update: An alternative and perhaps simpler form of the result is:
$$\frac{7\pi^3}{48}-\frac\pi8\,\ln^22-\frac\pi2\,\ln(1+\sqrt2)\cdot\ln2-\pi  \operatorname{Li}_2\left(\tfrac1{\sqrt{2}}\right)$$
