Closed Form for  ∫10arctanh xtan(π2 x) dx~\int_0^1\frac{\text{arctanh }x}{\tan\left(\frac\pi2~x\right)}~dx

Here is an approach.

We give a preliminary result.

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A series of squares of logarithms

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Let us consider the poly-Hurwitz zeta function initially defined by the series
$$\begin{align} \displaystyle \zeta(s,t\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)^t}, \quad \Re a>-1, \, \Re b>-1, \, \Re (s+t)>1. \tag1 \end{align}$$
The function $\displaystyle \zeta(\cdot,\cdot \mid a,b)$ extends to a meromorphic function on $\mathbb{C}^2$ with only singularities on the set $\displaystyle \left\{(s,t) \in \mathbb{C}^2, \,\Re (s+t)=1\right\}$. It clearly generalizes the classic Hurwitz zeta function initially defined by the series
$$\begin{align} \displaystyle \zeta(s,a) := \sum_{n=0}^{+\infty} \frac{1}{(n+a)^s}, \quad \Re a>0, \, \Re s>1. \tag2 \end{align}$$

We have the following new result.

Theorem. Let $a, b$ be complex numbers such that $\Re a>-1$ and $\Re b>-1$.

Then
$$\begin{align} \sum_{n=1}^{+\infty}\log^2\!\left(\! \frac{n+a}{n+b}\!\right)= \zeta’’(0,a+1)+ \zeta’’(0,b+1)-2\zeta^{1,1}(0,0\mid a,b)\tag3 \end{align}$$

where $\log (z)$ denotes the principal value of the logarithm defined for all $z \neq 0$ by
$$\log (z) = \ln |z|+i \arg z, \quad -\pi<\arg z\leq \pi,$$
$\displaystyle \zeta(\cdot,a)$ and $\displaystyle \zeta(\cdot,\cdot \mid a,b)$ denoting the Hurwitz zeta function and the poly-Hurwitz zeta function respectively and where $$\zeta’’(0, a)=\partial_{s}^2\left.\zeta(s,a)\right|_{s=0},\qquad \zeta^{1,1}(0,0\mid a,b)=\partial_{st}^2\left.\zeta(s,t\mid a,b)\right|_{(s,t)=(0,0)}.$$

Proof.
On the one hand, one has
$$\begin{align} &\partial_a \left(\zeta''(0,a+1)+\zeta''(0,b+1)-2\zeta^{1,1}(0,0\mid a,b)\right)\\\\ &= \left.\partial_s^2 \left(\partial_a \zeta(s,a+1)\right)\right|_{s=0}-2\left.\partial_{st}^2 \left(\partial_a \zeta(s,t\mid a,b)\right)\right|_{(s,t)=(0,0)}\\\\ &= \left.\partial_s^2 \left(-s\zeta(s+1,a+1)\right)\right|_{s=0}-2\left.\partial_{st}^2 \left(-s\zeta(s+1,t\mid a,b)\right)\right|_{(s,t)=(0,0)}\\\\ &= -\left.\left(2\zeta'(s+1,a+1)+s\zeta''(s+1,a+1)\right)\right|_{s=0}+2\left.\partial_s \!\left(s\zeta^{0,1}(s+1,t\mid a,b)\right)\right|_{(s,t)=(0,0)}\\\\ &=2\gamma_1(a+1)-2\gamma_1(b,a), \end{align}$$ using Theorem $1$ here.

On the other hand, one has

$$\begin{align} \partial_a\! \left(\sum_{n=1}^{+\infty}\log^2\!\left(\! \frac{n+a}{n+b}\!\right)\right) \!= 2\sum_{n=1}^{+\infty} \frac{\log (n+a)-\log (n+b)}{n+a} =2\gamma_1(a+1)-2\gamma_1(b,a), \end{align}$$
using Theorem $2$ here.

Observing that
$$\zeta(s,t\mid 0,0)=\zeta(s+t), \quad \zeta(s,1)=\zeta(s),$$ where $\zeta(\cdot)$ is the Riemann zeta function, then
$$\zeta’’(0,1)-\zeta^{1,1}(0,0\mid 0,0)=0$$ and both sides of $(3)$ vanish at $a=b=0$.

Thus $(3)$ holds true. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$

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Lucian's integral

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We prove that Lucian's integral is related to the preceding family of logarithmic series.

Proposition 1. We have
$$\begin{align} \int_0^1 \frac{\text{arctanh}\: x}{\tan \left( \frac{\pi}2x\right)}\:{\rm d}x=\frac\pi4-\frac1{2\pi}\sum_{n=1}^{+\infty}\log^2\!\left(\! \frac{2n-1 }{ 2n+1}\!\right).\tag{4} \end{align}$$

Proof. Let us proceed on Jack D'Aurizio's route which starts by using the standard expansion
$$\frac1{\tan \left( \frac{\pi}2x\right)}=\frac{2}{\pi x}-\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{\zeta(2n+2)}{2^{2n}}x^{2n+1},\quad 0<x<1,\tag{5}$$ then integrating termwise using
$$\begin{align} &\int_{0}^{1}x^{2n+1}\:\text{arctanh} \:x \:{\rm d}x\\ &=\frac1{2(n+1)(2n+1)}+\frac{\ln2}{2(n+1)}+\frac1{4(n+1)}\left(\gamma+\psi \left(n+\frac12 \right) \right)\tag{6} \end{align}$$
to get
$$\begin{align} \int_0^1 \frac{\text{arctanh}\: x}{\tan \left( \frac{\pi}2x\right)}\:{\rm d}x &=\frac{\pi }{4}+\frac{2}{\pi }(1-\ln 2)\ln\left(\frac{\pi }{2}\right)\\\\&-\frac{1}{\pi }\sum_{n=0}^{\infty}\frac{\zeta(2n+2)}{(2n+1)2^{2n}}-\frac{1}{\pi }\sum_{n=0}^{\infty}\frac{\zeta(2n+2)\left(\psi\left(n+\frac12\right)+\gamma\right)}{(n+1)2^{2n+2}}. \tag7 \end{align}$$
We are left with two non trivial series to evaluate.

We prove that each series may be evaluated using the poly-Stieltjes constants.

One may write
$$\require{cancel} \begin{align} \sum_{n=0}^{\infty}\frac{\zeta(2n+2)}{(2n+1)2^{2n}}&=\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\frac1{k^{2n+2}}\frac1{(2n+1)2^{2n}}\\ &=4\sum_{k=1}^{\infty}\sum_{n=0}^{\infty}\frac1{(2n+1)}\frac1{(2k)^{2n+2}}\\ &=\sum_{k=1}^{\infty}\frac1k\left(\log \left(1 + \frac1{2k}\right)-\log \left(1 - \frac1{2k}\right)\right)\\ &=\sum_{k=1}^{\infty}\frac1k\left(\log \left(k + \frac12\right)-\log \left(k - \frac12\right)\right)\\ &=\gamma_1\Big({\small\frac12,0}\Big)-\gamma_1\Big({\small-\frac12,0}\Big) \tag{8} \end{align}$$ using Theorem $2$ here.

To evaluate the last series on the right hand side of $(7)$, one may check with some algebra that, for any complex number $z$ satisfying $|z|<1$, the following identity holds true:
$$\begin{align} &\sum_{n=0}^{\infty}\frac{\psi\left(n+\frac12\right)+\gamma}{n+1}z^{2n+2}\\ &=2z\log\left(\frac{1-z}{1+z} \right)-2\left(1- \ln 2 \right)\log (1-z^2)+\frac12\log^2\left(\frac{1-z}{1+z} \right). \tag9 \end{align}$$
Then
$$\require{cancel} \begin{align} &\sum_{n=0}^{\infty}\frac{\zeta(2n+2)\left(\psi\left(n+\frac12\right)+\gamma\right)}{(n+1)2^{2n+2}} \\ &=\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\frac1{k^{2n+2}}\frac{\psi\left(n+\frac12\right)+\gamma}{(n+1)2^{2n+2}}\\ &=\sum_{k=1}^{\infty}\sum_{n=0}^{\infty}\frac{\psi\left(n+\frac12\right)+\gamma}{n+1}\frac1{(2k)^{2n+2}}\\ &=\sum_{k=1}^{\infty}\frac1k \log \left(\frac{1-\frac1{2k}}{1+\frac1{2k}}\right)-2\left(1- \ln 2 \right)\sum_{k=1}^{\infty}\log \left(1 - \frac1{4k^2}\right)+\frac12\sum_{k=1}^{\infty}\log^2 \left(\frac{1-\frac1{2k}}{1+\frac1{2k}}\right)\\ &=\sum_{k=1}^{\infty}\frac1k \log \left(\frac{k-\frac12}{k+\frac12}\right)+2\left(1- \ln 2 \right)\ln\left(\frac{\pi }{2}\right)+\frac12\sum_{k=1}^{\infty}\log^2\!\left(\! \frac{2k-1 }{ 2k+1}\!\right)\\ &=\gamma_1\Big({\small-\frac12,0}\Big)-\gamma_1\Big({\small\frac12,0}\Big)+2\left(1- \ln 2 \right)\ln\left(\frac{\pi }{2}\right)+\frac12\sum_{k=1}^{\infty}\log^2\!\left(\! \frac{2k-1 }{ 2k+1}\!\right).\tag{10} \end{align}$$
Inserting $(10)$ and $(8)$ into $(7)$ gives the announced result $(4)$. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$

We deduce the following closed form.

Proposition 2. We have

$$\begin{align} \int_0^1 \frac{\text{arctanh} x}{\tan \left( \frac{\pi}2x\right)}\:{\rm d}x=\frac\pi4+\frac2\pi\ln^2 2+\frac1\pi\ln 2\ln \pi+\frac1\pi\zeta^{1,1}\Big(0,0 \:\Bigr\rvert {\small-\frac12,\frac12}\Big). \tag{11} \end{align}$$

Proof. One may observe that
$$\begin{align} \zeta\left(s,\frac12 \right) & = \left(2^s-1 \right)\zeta(s) \tag{12}\\ \zeta\left(s,\frac32 \right) & = \left(2^s-1 \right)\zeta(s)-2^s, \tag{13} \end{align}$$ and recalling that $\zeta'(0)=-\frac12 \ln (2 \pi)$, one may obtain
$$\begin{align} \zeta''\left(0,\frac12 \right) & = -\frac32 \ln^2 2 - \ln 2 \ln \pi \tag{14}\\ \zeta''\left(0,\frac32 \right) & = -\frac52 \ln^2 2 - \ln 2 \ln \pi. \tag{15} \end{align}$$
From $(4)$ and $(3)$, we have
$$\require{cancel} \begin{align} \int_0^1 \frac{\text{arctanh}\: x}{\tan \left( \frac{\pi}2x\right)}\:{\rm d}x &=\frac\pi4-\frac1{2\pi}\sum_{n=1}^{+\infty}\log^2\!\left(\! \frac{2n-1 }{ 2n+1}\!\right)\\\\ &=\frac\pi4-\frac1{2\pi}\sum_{n=1}^{+\infty}\log^2\!\left(\! \frac{n-\frac12 }{ n+\frac12}\!\right)\\\\ &=\frac\pi4-\frac1{2\pi}\left( \zeta''\left(0,-\frac12+1 \right)+ \zeta''\left(0,\frac12+1 \right)-2\zeta^{1,1}\Big(0,0 \:\Bigr\rvert {\small-\frac12,\frac12}\Big)\right)\\\\ &=\frac\pi4-\frac1{2\pi}\left( \zeta''\left(0,\frac12 \right)+ \zeta''\left(0,\frac32 \right)-2\zeta^{1,1}\Big(0,0 \:\Bigr\rvert {\small-\frac12,\frac12}\Big)\right), \end{align}$$ by appealing to $(14)$ and $(15)$, we get $(11)$. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$

By combining Proposition $2$ and pisco125's derivation we obtain the following new closed forms.

Proposition 3. We have

$$\begin{align} \int_{0}^{\infty} {\ln (1+x^2)\over {e^{2\pi x}+1}}\:{\rm d}x =&\:\frac\pi4+\frac1{2\pi}\ln^2 2+\frac1\pi\zeta^{1,1}\Big(0,0 \:\Bigr\rvert {\small-\frac12,\frac12}\Big) \tag{16} \\\\ \sum_{k=1}^{\infty} {(-1)^k\over k} \text{Ci} (2k\pi)=&\:\frac{\pi^2}4+\frac12\ln^2 2+\:\zeta^{1,1}\Big(0,0 \:\Bigr\rvert {\small-\frac12,\frac12}\Big) \tag{17} \end{align}$$

where $\text{Ci} (\cdot)$ is the cosine integral and where $\displaystyle \zeta(\cdot,\cdot\mid a,b)$ is the poly-Hurwitz zeta function.