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?

$\bf{Modified:}$
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.
