$$\newcommand{\arctanh}{~\mathrm{arctanh}~}\newcommand{\sech}{~\mathrm{sech}~}$$
$$I=\int_{-1}^1\frac{\arctan x}{\arctanh x}\,\mathrm{d}x$$
Mathematica gives an approximate result of $I=1.581949621806183890451628…$, but no exact form. I predict it's a function of $e$ and $\pi$, and perhaps even the Golden Ratio $\phi$ (It certainly wouldn't be the first time)
The motivation behind this question is pure curiosity. I thought the shape looked nice 🙂
1st edit: Substitutions of $x=\tan u$ and $x=\tanh u$ respectively yield
$$I= 2\int_{-\pi/4}^{\pi/4}\dfrac{u\sec^2u}{\ln|\frac{1+\tan u}{1-\tan u}|}\,\mathrm{d}u$$
$$I= \int_{-\infty}^{\infty}\dfrac{\arctan(\tanh u)}{u}\sech^2u\,\mathrm{d}u$$
2nd edit: I've considered another approach starting with parameterizing the desired integral by
$$I_a=\int_{-1}^1\frac{\arctan ax}{\arctanh x}\,\mathrm{d}x$$
so that
$$\frac{\partial I_a}{\partial a}=\int_{-1}^1\frac{x}{(1+(ax)^2)\arctanh x}\,\mathrm{d}x$$
Integrating by parts with
$$\begin{matrix}u=\dfrac{1}{\arctanh x}&&\mathrm{d}v=\dfrac{x}{1+(ax)^2}\,\mathrm{d}x\\[1ex]
\mathrm{d}u=\dfrac{\mathrm{d}x}{(x^2-1)\arctanh^2x}&&v=\dfrac{1}{2a^2}\log(1+(ax)^2)\end{matrix}$$
yields the following integral:
$$\frac{\partial I_a}{\partial a}=\frac{1}{2a^2}\int_{-1}^1\frac{\log(1+(ax)^2)}{(1-x^2)\arctanh^2x}\,\mathrm{d}x$$
which can be modified by a substitution of $y=\arctanh x$ to obtain
$$\frac{\partial I_a}{\partial a}=\frac{1}{2a^2}\int_{-\infty}^\infty \frac{\log(1+(a\tanh y)^2)}{y^2}\,\mathrm{d}y$$
I have an idea of approaching the remaining integral using the series expansion of $\log(1+x)$; namely, the integral would become
$$\frac{\partial I_a}{\partial a}=\frac{1}{2a^2}\int_{-\infty}^\infty \frac{\mathrm{d}y}{y^2}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}(a\tanh y)^{2k}=-\frac{1}{2a^2}\sum_{k=1}^\infty \frac{a^{2k}(-1)^k}{k}\underbrace{\int_{-\infty}^\infty \frac{\tanh^{2k}y}{y^2}\,\mathrm{d}y}_{J_k}$$
According to this question, we have a closed from $J_k$ in the case of $k=1$ and potentially all $k>1$ in terms of the Riemann zeta function, but I have yet to do any more investigation.
Another method that occurred to me was to consider a keyhole contour to tackle $\dfrac{\partial I_a}{\partial a}$ but I'm afraid I'm not familiar enough with complex analysis to make that jump just yet.
Best Answer
I tried 2 ways to find a closed form, though unsuccessful up to this point.
1st trial. Let $I$ denote the integral, and write
$$ I = 8 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{\infty} \frac{e^{-x}(1 - e^{-(2n+1)x})}{x (1 + e^{-x})^{2}} \, dx. $$
In order to evaluate the integral inside the summation, we introduce new functions $I(s)$ and $J_n(s)$ by
$$ I(s) = 8 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{\infty} \frac{x^{s-1} e^{-x}(1 - e^{-(2n+1)x})}{(1 + e^{-x})^{2}} \, dx =: 8 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} J_n(s) $$
so that $I = I(0)$. Then it is easy to calculate that for $\Re(s) > 1$, $J_n(s)$ is written as
$$ J_n(s) = \Gamma(s) \left( \eta(s-1) + \sum_{k=2n+1}^{\infty} \frac{(-1)^{k-1}}{k^{s-1}} - (2n+1) \sum_{k=2n+1}^{\infty} \frac{(-1)^{k-1}}{k^{s}} \right), $$
where $\eta$ is the Dirichlet eta function. Plugging this back to $I(s)$ and manipulating a little bit, we obtain
$$ I(s) = 8\Gamma(s) \left( \frac{\pi}{4} \eta(s-1) - 4^{-s}\left( \zeta(s, \tfrac{1}{4}) - \zeta(s, \tfrac{1}{2}) \right) + \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \sum_{k=2n+1}^{\infty} \frac{(-1)^{k-1}}{k^{s-1}} \right). $$
This is valid for $\Re(s) > 1$. But if we can somehow manage to find an analytic continuation of the last summation part, then we may find the value of $I = I(0)$.
2nd trial. I began with the following representation
\begin{align*} I &= -2 \int_{0}^{\infty} \frac{1-e^{-t}}{1+e^{-t}} \left( \frac{1}{\cosh t} - \frac{2}{t} ( \arctan(1) - \arctan (e^{-t})) \right) \, \frac{dt}{t} \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)(2n+2)} \int_{-\infty}^{\infty} \frac{\tanh^{2(n+1)} x}{x^{2}} \, dx. \end{align*}
With some residue calculation, we can find that
\begin{align*} \int_{-\infty}^{\infty} \frac{\tanh^{2n} x}{x^{2}} \, dx &= \frac{2}{i\pi} \, \underset{z=0}{\mathrm{Res}} \left[ \psi_{1}\left(\tfrac{1}{2} + \tfrac{1}{i\pi} z\right) \coth^{2n} z \right] \\ &= 2^{2n+3} \sum_{m=1}^{n} (-1)^{m-1}m (1-2^{-2m-1}) A_{n-m}^{(2n)} \, \frac{\zeta(2m+1)}{\pi^{2m}}, \end{align*}
where $A_m^{(n)}$ is defined by the following combinatoric sum
$$ A_m^{(n)} = \sum_{\substack{ k_1 + \cdots + k_n = m \\ k_1, \cdots, k_n \geq 0 }} \frac{B_{2k_1} \cdots B_{2k_n}}{(2k_1)! \cdots (2k_n)!} = 2^{-2m} [z^{2m}](z \coth z)^{n} \in \Bbb{Q}, $$
where $B_k$ are Bernoulli numbers. Still the final output is egregiously complicated, so I stopped here.
3rd trial. The following yet another representation may be helpful, I guess.
$$ I = \int_{0}^{1/2} \frac{1 - \cot(\pi u/2)}{2} \left\{ \psi_1\left(\tfrac{1+u}{2}\right) - \psi_1\left(\tfrac{1-u}{2}\right) \right\} \, du. $$