How to evaluate $\int_{0}^{1}\frac{x^{4}\arctan x}{\sqrt{1+x^{2}}}\mathrm{d}x$ How to evaluate$$\int_{0}^{1}\frac{x^{4}\arctan x}{\sqrt{1+x^{2}}}\mathrm{d}x$$
I tried to define:$$I\left ( \alpha  \right )=\int_{0}^{1}\frac{x^{4}\arctan \left ( \alpha x \right )}{\sqrt{1+x^{2}}}\mathrm{d}x$$
and
$$I'\left ( \alpha  \right )=\int_{0}^{1}\frac{x^{5}}{\sqrt{1+x^{2}}\left ( 1+\alpha ^{2}x^{2} \right )}\mathrm{d}x$$
but I have no idea how to do next.

Starting from $$I'\left ( a \right )=\int_{0}^{1}\frac{x^{5}}{\sqrt{1+x^{2}}\left ( 1+a ^{2}x^{2} \right )}\mathrm{d}x$$ change variable $x=\sqrt t$ and get  $$I'\left (a  \right )=\int_{0}^{1}\frac{t^2}{2 \sqrt{t+1} \left(a^2 t+1\right)}\,dt$$ Now $$\frac{t^2}{2 \sqrt{t+1} \left(a^2 t+1\right)}=\frac{\sqrt{t+1}}{2 a^2 \left(a^2-1\right) \left(a^2 t+1\right)}+\frac{\sqrt{t+1}}{2
   a^2}-\frac{1}{2 \left(a^2-1\right) \sqrt{t+1}}$$ For each piece, the antiderivative can be established (even if the first one is quite tricky but doable). So, $I'(a)$ can be computed but the result is a real monster.
At this point, I give up and, as Leg answered, go to numerical integration.
Edit
For the given initial integral, a CAS found (enjoy !) $$\int_{0}^{1}\frac{x^{4}\arctan x}{\sqrt{1+x^{2}}}\mathrm{d}x=\frac{1}{96} \left(-72 C+4 \left(9 i \text{Li}_2\left(-(-1)^{3/4}\right)-9 i
   \text{Li}_2\left((-1)^{3/4}\right)-13+11 \sqrt{2}\right)-3 \pi  \left(\sqrt{2}+6
   \tanh ^{-1}\left((-1)^{3/4}\right)\right)\right)$$
Edit
This was done after Start wearing purple's elegant answer
If we look at the antiderivative, without change of variable, integrating by parts gives $$I=\int\frac{x^{4}\arctan x}{\sqrt{1+x^{2}}}\, dx=\frac{1}{8} \tan ^{-1}(x) \left(x \sqrt{x^2+1} \left(2 x^2-3\right)+3 \sinh
   ^{-1}(x)\right)- J$$ using $$J=\int \frac{x \sqrt{x^2+1} \left(2 x^2-3\right)+3 \sinh ^{-1}(x)}{8 \left(x^2+1\right)}\,dx$$ Looking at the integrand, it can rewrite  $$\frac{x \sqrt{x^2+1} \left(2 x^2-3\right)+3 \sinh ^{-1}(x)}{8 \left(x^2+1\right)}=\frac{1}{4} x\sqrt{x^2+1} -\frac{5 x}{8 \sqrt{x^2+1}}+\frac{3 \sinh ^{-1}(x)}{8
   \left(x^2+1\right)}$$ The first and second antiderivatives are quite simple $$\int\frac{1}{4} x\sqrt{x^2+1}\,dx=\frac{1}{12} \left(x^2+1\right)^{3/2}$$ $$\int \frac{5 x}{8 \sqrt{x^2+1}}\,dx=\frac{5 \sqrt{x^2+1}}{8}$$ and we are left with the problem of $$\int\frac{3 \sinh ^{-1}(x)}{8
   \left(x^2+1\right)}\,dx$$ which creates the monster.
