Closed form for ${\large\int}_0^\infty\frac{\arctan(x)\,\operatorname{arccot}(x+1)}{x}dx$

The integrand has a closed-form antiderivative in terms of elementary functions and polylogarithms. It can be found using Mathematica after expressing inverse trig functions through logarithms of complex arguments, and can be manually checked for correctness using differentiation. After subtracting its limits at $\infty$ and $0$ and simplification, we can get this result: $$\begin{align}I&=\frac1{32}\Big[\!\operatorname{Li}_2\!\left(\tfrac15\right)\cdot\ln2+\operatorname{Li}_3\!\left(\tfrac15\right)+ \operatorname{Li}_3\!\left(\tfrac45\right)\!\Big]+\frac{\pi}{16}\,\left(3\operatorname{Ti}_2(2)-4\,G\right)\\&+\frac{\ln5}{192}\,\left(9\ln2\cdot\ln5-12\ln^22-2\ln^25\right)+\frac{\pi^2}{192}\,\left(\ln5-7\ln2\right)+\frac{17}{16}\zeta(3),\end{align}$$ where $\operatorname{Ti}_2(x)$ is the inverse tangent integral, $G=\operatorname{Ti}_2(1)$ is the Catalan constant.

Using Mathematica I have arrived at the result

$$I=\frac{1}{4} (-\text{Li}_3(-1-i)-\text{Li}_3(-1+i)+\text{Li}_3(1-i)+\text{Li}_3(1+i))+\frac{1}{4} \left(-\text{Li}_3\left(-\frac{1}{2}-\frac{i}{2}\right)-\text{Li}_3\left(-\frac{1}{2}+\frac{i}{2}\right)+\text{Li}_3\left(\frac{1}{2}-\frac{i}{2}\right)+\text{Li}_3\left(\frac{1}{2}+\frac{i }{2}\right)+\text{Li}_2\left(-\frac{1}{2}-\frac{i}{2}\right) \log \left(-\frac{1}{2}-\frac{i}{2}\right)+\text{Li}_2\left(-\frac{1}{2}+\frac{i}{2}\right) \log \left(-\frac{1}{2}+\frac{i}{2}\right)-\text{Li}_2\left(\frac{1}{2}-\frac{i}{2}\right) \log \left(\frac{1}{2}-\frac{i}{2}\right)-\text{Li}_2\left(\frac{1}{2}+\frac{i}{2}\right) \log \left(\frac{1}{2}+\frac{i}{2}\right)\right)$$

Which matches your numerical estimates.

$$I\approx1.3513049368715095284050230093075694014884142059538$$

I am unable to simplify it past the polylogs. Running full simplify we can get to

$$\frac{1}{384} \left(48 \pi C+3 \left(-64 \text{Li}_3\left(-\frac{1}{2}-\frac{i}{2}\right)-64 \text{Li}_3\left(-\frac{1}{2}+\frac{i}{2}\right)+32 (\text{Li}_3(1-i)+\text{Li}_3(1+i))+35 \zeta (3)\right)+\text{Li}_2\left(-\frac{1}{2}-\frac{i}{2}\right) (-48 \log (2)-72 i \pi )+24 \text{Li}_2\left(-\frac{1}{2}+\frac{i}{2}\right) (-\log (4)+3 i \pi )-4 \log ^3(2)+7 \pi ^2 \log (2)\right)$$ Where $C$ is Catalan's constant.

I think this is about as closed form as it is going to get.