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.
 A: The change of variables $x=\frac{t-t^{-1}}{2}$ reduces the integral to
$$I=\int_1^{1+\sqrt2}\frac{(t^2-1)^4\arctan\frac{t-t^{-1}}{2}}{16t^5}dt$$
Writing $\arctan$ in the logarithmic form we get an integral of the form $\int R_1(t)\ln R_2(t)\,dt$ with rational $R_{1,2}(t)$ which has antiderivative in terms of dilogarithms. It will be a little simpler if we first integrate by parts to kill $\arctan$ by differentiation. This gives
$$I=\frac{\pi\left(3\operatorname{arcsinh} 1-\sqrt2\right)}{32}-
\int_1^{1+\sqrt2}\left(\frac{t^6-9t^4+9t^2-1}{32t^4}+\frac34\frac{\ln t}{1+t^2}\right)dt.$$
The first part of the integral is elementary, so we end up with
$$I=\frac{\pi\left(3\operatorname{arcsinh} 1-\sqrt2\right)}{32}-\frac{13-11\sqrt2}{24}-\frac34\int_1^{1+\sqrt2}\frac{\ln t\, dt}{1+t^2}.$$
Antiderivative of the last part can be expressed in terms of dilogarithms:
$$\int\frac{\ln t\, dt}{1+t^2}=\arctan t\ln t-\frac{\operatorname{Li}_2\left(it\right)-\operatorname{Li}_2\left(-it\right)}{2i},$$
which leads to 
\begin{align}I=\frac34\left[\Im\operatorname{Li}_2\Bigl(i(1+\sqrt2)\Bigr)-\mathbf C\right]-\frac{\pi\left(6\operatorname{arcsinh} 1+\sqrt2\right)}{32}-\frac{13-11\sqrt2}{24},
\end{align}
where $\mathbf C$ denotes Catalan's constant. Actually the particular dilogarithm value appearing in this expression can be expressed in terms of trigamma function, which finally yields
$$I=\frac{3\left[\psi_1\left(\frac18\right)+\psi_1\left(\frac38\right)\right]}{128\sqrt2}-\frac{3\mathbf C}{4}-\frac{3\pi^2}{32\sqrt2}+\frac{\pi\left(3\operatorname{arcsinh} 1-\sqrt2\right)}{32}-\frac{13-11\sqrt2}{24}.$$
A: 
How to evaluate $\displaystyle\int_{0}^{1}\frac{x^{4}\arctan x}{\sqrt{1+x^{2}}}~dx$ ?

Start with the obvious trigonometric substitution $x=\tan t,$ inspired by the presence of 
both $\arctan x$ and $1+x^2,$ since $1+\tan^2t$ simplifies to $\sec^2t=\dfrac1{\cos^2t},~$ as does $\tan't.$ 
Then try an integration by parts with regard to t $($on one hand$)$ and the trigonometric 
part $($on the other hand$)$. This implies evaluating the anti-derivative of $$\dfrac{\tan^4t}{\cos t}~=~\dfrac{\sin^4t}{\cos^5t}~=~\dfrac{\sin^4t\cdot\cos t}{\cos^6t}~=~\dfrac{\sin^4t\cdot\sin't}{(1-\sin^2t)^3},$$ which is trivial, following a simple trigonometric substitution and some partial fraction 
decomposition. The non-trivial part consists in integrating $\ln(1\pm\sin t),$ which, for all I 
care, can be done by expanding the integrand into its Mercator series, and reversing the 
order of summation and integration. Alternately, transform $\sin t$ into $\cos2u$ using the 
fact that $\sin t=\cos\bigg(\dfrac\pi2-t\bigg),$ and then employ the well-known trigonometric formulas 
for $1\pm\cos2u,$ along with the properties of the natural logarithm. Now use the fact that 
$\displaystyle\int_0^\tfrac\pi4\ln\cot u~du~=~$ Catalan's constant, and we are left with expressing $$\int_0^\tfrac\pi8\ln\cot u~du~=~\int_0^\tfrac\pi8\ln\cos u~du-\int_0^\tfrac\pi8\ln\sin u~du$$ in terms of dilogarithms of argument $\pm~i~\tan\dfrac\pi8~=~\pm~i~(\sqrt2-1)$. The final expression 
can be shown to be equivalent to the others presented in this thread by appealing to the 
various polylogarithmic properties.
A: 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.
A: Using Adaptive Gaussian Quadrature, we obtain the integral to be $0.104946775042126$, which is accurate unto $14$ digits.
