<s>[INST] Evaluate $\int_0^1 \arctan^3 x\,dx$ I don't want to use the Fourier series. My work \begin{align}J&=\int_0^1 \arctan^3 x\,dx\\
 &=[x\arctan^3 x]_0^1 -3\int_0^1
 \frac{x\arctan^2 x}{1+x^2}\,dx\\
 &=\frac{\pi^3 }{64}-3\int_0^1 \frac{x\arctan^2
 x}{1+x^2}\,dx\\
 &= \frac{\pi^3
 }{64}-\frac{3}{2}\left[\ln(1+x^2)\arctan^2
 x\right]_0^1 +3\int_0^1 \frac{\ln(1+x^2)\arctan
 x}{1+x^2}\,dx\\
 &=\frac{\pi^3 }{64}-\frac{3\pi^2\ln
2}{32}+3\int_0^1 \frac{\ln(1+x^2)\arctan
 x}{1+x^2}\,dx\\
\end{align}
How to continue?
 [/INST] Not an answer, but Wolfram Alpha finds the amazing $$\int (\arctan(x))^3\mathrm{d}x$$ $$=\frac{3}{2}\operatorname{Li}_3(-e^{2i\arctan(x)})-3i\arctan(x)\operatorname{Li}_2(-e^{2i\arctan(x)})+(\arctan(x))^2(x\arctan(x)-i\arctan(x)+3\ln(1+e^{2i\arctan(x)}))+c$$
And the even more beautiful
$$\int_0^1 (\arctan(x))^3\mathrm{d}x=\frac{1}{64}(\pi^2(\pi+\ln(64))+63\zeta(3)-48\pi C)$$
With Catalan's constant $C$ and the Riemann zeta function $\zeta$.
But, continuing from your last line,
$$\int_0^1 \frac{\ln(1+x^2)\arctan(x)}{1+x^2}\mathrm{d}x$$
We can make the substitution $x=\tan\theta, \mathrm{d}x=\sec^2(\theta)\mathrm{d}\theta$ to get
$$\int_0^{\arctan(1)} \theta\ln(\sec^2(\theta))\mathrm{d}\theta$$
Although knowledge of the polylogarithm will still be needed.
</s>