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?
 A: $$\newcommand{\b}[1]{\left(#1\right)}
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This belongs to a large category of functions that are easily solvable with integration by parts like:

$$\int\frac{x\arctan(x)}{\sqrt{1+x^2}}\d x=\sqrt{1+x^2}\arctan(x)-\ln|x+\sqrt{x^2+1}|+c$$


So let our functions be:
$$f(x)=\f{x^3}{(3-x^2)^2\s{1-x^2}};g(x)=\arctan(x)\\\int f(x)g(x)\d x=g(x)\int f(x)\d x-\int g'(x)\u{\int f(x)\d x}_{h(x)}\d x$$
And(Beware I used Reduction Formula for evaluating last integral!):
$$\int f(x)\d x=\f12\int\f{t\d t}{(3-t)^2\s{1-t}}\quad(t=x^2)\\
=\int\f{u^2-1}{(u^2+2)^2}\d u=\int\f1{(u^2+2)}\d u-3\int\f{1}{(u^2+2)^2}\d u\quad(u^2=1-t)\\
=\frac1{\sqrt2}\arctan\frac u{\sqrt2}-3\b{\frac u{4(u^2+2)}+\f1{4\s2}\arctan\f u{\s2}}+{\cal C}\\
=\f1{4\s2}\arctan\s{\f{1-x^2}2}-\f34\f{\s{1-x^2}}{3-x^2}+{\cal C}=\t{(let) }h(x)$$
And thus:
$$\int g'(x)h(x)\d x=\f1{4\s2}\int \f1{1+x^2}\arctan\s{\f{1-x^2}2}\d x-\f34\int \f{\s{1-x^2}}{(3-x^2)(1+x^2)}\d x$$
So:
$$I=\int_0^1f(x)g(x)\d x=\b{\arctan (x)\b{\f1{4\s2}\arctan\s{\f{1-x^2}2}-\f34\f{\s{1-x^2}}{3-x^2}}}_0^1\\-\f1{4\s2}\int_0^1 \f1{1+x^2}\arctan\s{\f{1-x^2}2}\d x+\f34\int_0^1 \f{\s{1-x^2}}{(3-x^2)(1+x^2)}\d x\\
I=\f34\u{\int_0^1 \f{\s{1-x^2}}{(3-x^2)(1+x^2)}\d x}_{I_1}-\f1{4\s2}\int_0^1 \f1{1+x^2}\arctan\s{\f{1-x^2}2}\d x$$
Now:
$$I_1=\int_0^\infty\f{\d y}{4y^4+8y^2+3}\quad\b{x=\f{y^2}{y^2+1}\t{ or }y=\f{x}{\s{1-x^2}}=\tan\arcsin x \\\t{ which can also be obtained after two sunstitutions viz., }x=\sin z\t{, then }y=\tan z}\\
\f12\b{\f1{2y^2+1}-\f1{2y^2+3}}_0^{\infty}\\
=\f{3-\s3}{24}\pi\s2=\f{3-\s3}{24}\pi\s2$$
Still working on the last integral!!
BTW the final form is:
$$I=\frac{\pi\sqrt{2}}{192}(18-6\sqrt{3})-\f1{4\s2}\int_0^1 \f1{1+x^2}\arctan\s{\f{1-x^2}2}\d x$$
So it remains to prove:
$$\int_0^1 \f1{1+x^2}\arctan\s{\f{1-x^2}2}\d x=\pi^2/24$$
A: 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.
A: $\bf{Modified:}$
Based on an idea from the solution of  @user111187:
Consider a general integral of the form
$$I_b =\int_0^1 \frac{R(x)}{\sqrt{1-x^2}} \arctan(b x)dx$$
with derivative 
$$J_b \colon =\frac{dI_b}{db} = \int_0^1 \frac{x\, R(x)}{(1+b^2 x^2)\sqrt{1-x^2}} dx$$
and $I_0=0$. 
The integral $J_b$ is doable if $R$ is a rational function. We get 
$$I =I_1 = \int_{0}^1 J_b\, db$$
This last integral is doable in some cases.
In our case $R(x) = \frac{x^3}{(3-x^2)^2}$ so we get
$$J_b = \int_0^1 \frac{x^4}{(1+b^2 x^2)(3-x^2)^2  \sqrt{1-x^2}} dx$$
We find first the indefinite integral
$$\int \frac{x^4}{(1+b^2 x^2)(3-x^2)^2  \sqrt{1-x^2}} dx $$
One can check that it equals
$$
F_b(x):=\frac{3 x \sqrt{1-x^2}}{4 \left(1+3 b^2\right) \left(-3+x^2\right)}+\frac{3 \sqrt{\frac{3}{2}} \left(-1+b^2\right) \text{ArcTan}\left[\frac{\sqrt{\frac{2}{3}}
x}{\sqrt{1-x^2}}\right]}{4 \left(1+3 b^2\right)^2}+\frac{\text{ArcTan}\left[\frac{\sqrt{1+b^2} x}{\sqrt{1-x^2}}\right]}{\sqrt{1+b^2} \left(1+3 b^2\right)^2}$$
Indeed, by direct calculations 
$$F_b'(x) = \frac{x^4}{(1+b^2 x^2)(3-x^2)^2  \sqrt{1-x^2}}$$
We get 
$$J_b = F_b(1_{-})- F_b(0) = \frac{\left(8-3 \sqrt{6}(1+b^2)^{\frac{3}{2}}\right) \pi }{16 \sqrt{1+b^2} \left(1+3 b^2\right)^2}$$
and therefore 
$$I = I_1 = \int_0^1 \frac{\left(8-3 \sqrt{6}(1+b^2)^{\frac{3}{2}}\right) \pi }{16 \sqrt{1+b^2} \left(1+3 b^2\right)^2} db = \frac{\pi \sqrt{2}}{192}\cdot (\,18 - 6 \sqrt{3} - \pi) $$
This approach works for all integrals of form
$$\int_0^1 \frac{R(x)}{\sqrt{1-x^2}} \arctan(b x)dx=\int_0^1 \frac{x \, Q(x^2)}{\sqrt{1-x^2}} \arctan(x) dx$$
where $R(x) = xQ(x^2)$ is an odd rational function.
