<s>[INST] How to prove $\int_0^1\frac{x^3\arctan x}{(3-x^2)^2}\frac{\mathrm dx}{\sqrt{1-x^2}}=\frac{\pi\sqrt{2}}{192}\left(18-\pi-6\sqrt{3}\,\right)$? How to prove the following result?

$$\int_0^1\frac{x^3\arctan x}{(3-x^2)^2}\frac{\mathrm dx}{\sqrt{1-x^2}}=\frac{\pi\sqrt{2}}{192}\left(18-\pi-6\sqrt{3}\,\right)$$

For my part no idea?
 [/INST] Although integration by parts will work, a nice general way to attack the integral is by introducing
$$
J(a,b) := \int_0^1 \frac{x \arctan(ax)}{1-b^2 x^2} \frac{dx}{\sqrt{1-x^2}},
$$
so $J(0,b) = 0$ and
$$
\begin{align}
\partial_a J &= \int_0^1 \frac{x^2}{(1-b^2 x^2)(1+a^2x^2)} \frac{dx}{\sqrt{1-x^2}}
= \int_0^{\pi/2} \frac{du\sec^2 u}{(\sec^2 u -b^2)(\sec^2 u +a^2)} 
\\&= \int_0^{\infty} \frac{dt}{(t^2 + 1-b^2)(t^2 + 1+a^2)} = \frac{\pi/2}{\sqrt{1-b^2}\sqrt{1+a^2}} \frac{1}{\sqrt{1-b^2}+\sqrt{1+a^2}},
\end{align}
$$
where $ x = \cos u, t = \tan u$, and the last equality follows easily using partial fractions.
Hence
$$
\begin{align}
J(1,b) &= \frac{\pi/2}{\sqrt{1-b^2}} \int_0^1 \frac{da}{\sqrt{1+a^2}} \frac{1}{\sqrt{1-b^2}+\sqrt{1+a^2}}
\\&= \left.\frac{\pi/2}{\sqrt{1-b^2}} \frac{1}{b} \left[\arctan\left(\frac{b}{\sqrt{1-b^2}}\frac{\sqrt{1+a^2}}{a} \right)- \arctan \frac b a\right] \right\lvert_{a=0}^{a=1}
\\&= \frac{\pi/2}{\sqrt{1-b^2}} \frac{1}{b} \left[\arctan\sqrt\frac{2b^2}{1-b^2}- \arctan b\right].
\end{align}
$$
This implies that
$$
\begin{align}
&2b^{-3} \int_0^1 \frac{x^3 \arctan(x)}{(b^{-2} - x^2)^2} \frac{dx}{\sqrt{1-x^2}} = \partial_b J(1,b)
\\&= \frac{\pi}{2b^2(1-b^2)^{3/2}} \left\{b\frac{ \sqrt 2\sqrt{1- b^2}+2}{1+b^2}+\left(1-2b^2\right) \left[\arctan b-\arctan \sqrt{\frac{2b^2}{1-b^2}}\right]-b\right\}.
\end{align}
$$
Plugging in $b = \dfrac{1}{\sqrt 3}$ simplifies this a lot, giving
$$
\int_0^1 \frac{x^3 \arctan(x)}{(3 - x^2)^2} \frac{dx}{\sqrt{1-x^2}} =\frac{\pi \sqrt 2}{192}\left[18-\pi - 6\sqrt 3 \right],
$$
as desired.
</s>