<s>[INST] Closed form of $\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx$ 
Can a closed form solution for the following integral be found:
$$\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx\,?$$

I have tried all the standard tricks such as integration by parts, various substitutions, and parametric differentiation (Feynman's trick), but all to no avail.
An attempt is letting
$$f(t):=\int_0^\infty\,\arctan^2\left(\frac{2tx}{1+x^2}\right)\,\text{d}x\,.$$
Therefore,
$$f'(t)=\int_0^\infty\,\frac{8x^2(x^2+1)}{\big(x^4+2(2t^2+1)x^2+1\big)^2}\,\left(1+x^2-4tx\arctan\left(\frac{2tx}{1+x^2}\right)^{\vphantom{a^2}}\right)\,\text{d}x\,.$$
This doesn't seem to go anywhere.  Help!
 [/INST] $$\begin{align*}
I &= \int_0^\infty \arctan^2\left(\frac{2x}{x^2+1}\right) \, dx \\[1ex]
&= \int_0^1 \left(1+\frac1{x^2}\right) \arctan^2\left(\frac{2x}{x^2+1}\right) \, dx \tag1 \\[1ex]
&= 2 \int_0^1 \frac{\arctan^2(x)}{x^2\sqrt{1-x^2}} \, dx \tag{2} \\[1ex]
&= 4 \int_0^1 \frac{\arctan(x) \sqrt{1-x^2}}{x(1+x^2)} \, dx \tag{3} \\[1ex]
&= 4 \left(\underbrace{\int_0^1 \frac{\sqrt{1-x^2}}x \arctan(x) \, dx}_{J_1} - \underbrace{\int_0^1 \frac{x\sqrt{1-x^2}}{x^2+1} \arctan(x) \, dx}_{J_2}\right) \tag{4}
\end{align*}$$

$$\begin{align*}
J_1 &= \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \int_0^1 x^{2n} \sqrt{1-x^2} \, dx & \tag{5} \\[1ex]
&= \frac12 \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \int_0^1 x^{n-\frac12} (1-x)^{\frac12} \, dx \tag{6} \\[1ex]
&= \frac12 \Gamma\left(\frac32\right) \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)} \frac{\Gamma\left(n+\frac12\right)}{\Gamma(n+2)} \tag{7} \\[1ex]
&= \frac{\sqrt\pi}4 \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)(2n+1)} \frac{\Gamma\left(n+\frac12\right)}{\Gamma(n+1)} \tag{8} \\[1ex]
&= \frac{\sqrt\pi}2 \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \frac{(2n)!}{2^{2n}(n!)^2} - \frac{\sqrt\pi}4 \sum_{n=0}^\infty \frac{(-1)^n}{n+1} \frac{(2n)!}{2^{2n}(n!)^2} \tag{4/9} \\[1ex]
&= \frac\pi2 (\operatorname{arsinh}(1) + 1 - \sqrt2) \tag{10}
\end{align*}$$

$$\begin{align*}
J_2 &= \int_0^1 \int_0^1 \frac{x^2 \sqrt{1-x^2}}{(x^2+1)(1+y^2x^2)} \, dy \, dx \tag{11} \\[1ex]
&= \frac\pi2 \int_0^1 \frac{(\sqrt2-1)y^2+1-\sqrt{y^2+1}}{y^2(y^2-1)} \, dy \tag{12} \\[1ex]
&= \frac\pi{2\sqrt2} \left(\sqrt2 - 2 + \log(2)\right)
\end{align*}$$

Putting these together, we arrive at
$$I = \boxed{\pi\bigg(2\operatorname{arsinh}(1)-\sqrt2 \log(2)\bigg)}$$


*

*$(1)$ : substitute $x\mapsto\dfrac1x$ in the integral over $[1,\infty)$

*$(2)$ : substitute $x\mapsto\dfrac{1+\sqrt{1-x^2}}x$

*$(3)$ : integrate by parts

*$(4)$ : partial fractions

*$(5)$ : series expansion of $\arctan(x)$

*$(6)$ : substitute $x\mapsto\sqrt x$

*$(7)$ : beta function

*$(8)$ : $\Gamma(n+1)=n\,\Gamma(n)$

*$(9)$ : convert gammas to factorials in preparation for ...

*$(10)$ : series expansion of $\operatorname{arsinh}(x)$

*$(11)$ : integral definition of $\arctan(x)$

*$(12)$ : the last integral is elementary
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