Closed-form of $\int_0^1\left(\frac{\arctan x}{x}\right)^n\,dx$ Inspired by this question, is there a closed-form of
$$\int_0^1\left(\frac{\arctan x}{x}\right)^n\,dx\,?$$
Here $n \in \mathbb{N_+}$. In the answers to the question above we could find proofs of cases $n=2,3$.
I state here some specific cases.
$$\begin{align}
\int_0^1\frac{\arctan x}{x}\,dx & = G, \\
\int_0^1\left(\frac{\arctan x}{x}\right)^2\,dx & =G-\frac{\pi^2}{16}+\frac{\pi}{4}\ln2,\\
\int_0^1\left(\frac{\arctan x}{x}\right)^3\,dx & = \frac{3G}{2}-\frac{\pi^3}{64}-\frac{3\pi^2}{32}+\frac{3\pi}{8}\ln2.\\
\end{align}$$
Furtheremore I've evaluated $n=4,5$ cases.
$$\int_{0}^{1}\left(\frac{\arctan(x)}{x}\right)^4dx$$
equals to
$$2G-\frac{3\pi^4}{256}-\frac{\pi^3}{48}-\frac{\pi^2}{8}-\frac{\pi^2G}{8}+\frac{‌​3\pi}{64}\zeta(3)-\frac{\pi^3}{96} \ln2+\frac{\pi}{2} \ln2+\frac{1}{768}\psi_3\left(\frac{1}{4}\right),$$
and
$$\int_{0}^{1}\left(\frac{\arctan(x)}{x}\right)^5dx$$
equals to
$$\frac{5G}{2}-\frac{25\pi^4}{512}-\frac{5\pi^3}{192}-\frac{5\pi^2}{32}-\frac{5\pi^2 G}{8}+\frac{15\pi}{64}\zeta(3)-\frac{5\pi^3}{96}\ln 2+\frac{5\pi}{8}\ln 2 + \frac{5}{768}\psi_3\left(\frac{1}{4}\right).$$
Here $G$ is Catalan's constant, $\zeta$ is the Riemann zeta function, $\psi_3$ is the polygamma function of order $3$, and $\pi$ is also a famous constant.
Note that the problem is related to Dirichlet beta function, since
$$\begin{align}
\beta(2) & = G \\
\beta(3) & = \frac{\pi^3}{32} \\
\beta(4) & = \frac{1}{768}\left(\psi_3\left(\frac{1}{4}\right)-8\pi^4\right).
\end{align}$$

Using substitution and integration by parts,
\begin{align}&\int_0^1\left(\frac{\tan^{-1}x}{x}\right)^n\,dx=\int_0^{\pi/4}\frac{x^n}{\tan^n x}\sec^2 x\,dx\\&=\left[\frac{x^n}{\tan^n x}\tan x\right]_0^{\pi/4}-\int_0^{\pi/4}\frac{nx^{n-1}}{\tan^{n-1}x}-\frac{nx^n}{\tan^n x}\sec^2x\,dx\end{align}
So we have $$\int_0^1\left(\frac{\tan^{-1}x}{x}\right)^n\,dx=\frac{n}{n-1}\int_0^{\pi/4}\frac{x^{n-1}}{\tan^{n-1}x}\,dx-\frac{1}{n-1}\left(\frac{\pi}{4}\right)^n$$
Now define $\displaystyle I(m,n)=\int_0^{\pi/4}x^m\cot^nx\,dx$ for $m\ge n$. Using integration by parts, we get a recurrence formula.
\begin{align}I(m,n)&=\int_0^{\pi/4}x^{m}\cot^{n}x\,dx\\&=\left[\frac{1}{m+1}x^{m+1}\cot^n x\right]_0^{\pi/4}+\frac{n}{m+1}\int_0^{\pi/4}x^{m+1}\cot^{n-1}x(\cot^2x+1)\,dx\end{align}
So we have $\displaystyle I(m,n)=-I(m,n-2)+\frac{m}{n-1}I(m-1,n-1)-\frac{1}{n-1}\left(\frac\pi4\right)^m$. We now need to determine $I(m,0)$ and $I(m,1)$ to find $I(n-1,n-1)$. It is easily seen $$I(m,0)=\frac{1}{m+1}\left(\frac\pi4\right)^{m+1}$$ $I(m,1)$ is a bit of a problem. This is where things start to get messy.(All zetas, catalans, and polygammas appear here)
\begin{align}I(m,1)&=\int_0^{\pi/4}x^m\cot x\,dx\\&=\left[x^m\log\sin x\right]_0^{\pi/4}-m\int_0^{\pi/4}x^{m-1}\log\sin x\,dx\\&=\frac{\log 2}{2}\left(\frac{\pi}{4}\right)^m+\sum_{n=1}^\infty\frac mn\int_0^{\pi/4}x^{m-1}\cos 2nx\,dx\end{align} Now using integration by parts,$$\int_0^{\pi/4}x^{m-1}\cos 2nx\,dx=\frac{1}{2n}\left(\frac\pi4\right)^{m-1}\sin\frac{n\pi}2+\frac{m-1}{(2n)^2}\left(\frac\pi4\right)^{m-2}\cos\frac{n\pi}2-\frac{(m-1)(m-2)}{(2n)^2}\int_0^{\pi/4}x^{m-3}\cos 2nx\,dx$$
This yields
\begin{align}\int_0^{\pi/4}x^{m-1}\cos 2nx\,dx=&\sin\frac{n\pi}{2}\sum_{k=1}^{[\frac{m+1}2]}\frac{(-1)^{k-1}}{(2n)^{2k-1}}\left(\frac\pi4\right)^{m-2k+1}\frac{(m-1)!}{(m-2k+1)!}\\&+\cos\frac{n\pi}{2}\sum_{k=1}^{[\frac m2]}\frac{(-1)^{k-1}}{(2n)^{2k}}\left(\frac\pi4\right)^{m-2k}\frac{(m-1)!}{(m-2k)!}+\left(\frac{(m-1)!}{(2n)^m}(-1)^{m/2}\right)_{2|m}\end{align}
where the last term $\displaystyle\left(\frac{(m-1)!}{(2n)^m}(-1)^{m/2}\right)$ is present iff $m$ is even. Now plug this up on the above summation and we get the $I(m,1)$.
\begin{align}I(m,1)=&\sum_{k=1}^{[\frac{m+1}2]}\frac{(-1)^{k-1}\pi^{m+1-2k}}{2^{2m+1+4k}}\binom{m}{2k-1}\left(\psi_{2k-1}\left(\frac14\right)-\psi_{2k-1}\left(\frac34\right)\right)\\&+\sum_{k=1}^{[\frac m2]}\frac{(-1)^{k-1}\pi^{m-2k}(2^{2k}-1)}{2^{2m+2k+1}}\frac{m!}{(m-2k)!}\zeta(2k+1)+\left(\frac{(-1)^{m/2}m!}{2^m}\zeta(m+1)\right)_{2|m}\end{align}
