TITLE: How to find the value of $I_1=\int_0^\infty\frac{\sqrt{x}\arctan{x}\log^2({1+x^2})}{1+x^2}dx$
QUESTION [11 upvotes]: How to find the value of  $$I_1=\int_0^\infty\frac{\sqrt{x}\arctan{x}\log^2({1+x^2})}{1+x^2}dx$$ 
If we put $$I_2=\int_0^\infty\frac{\arctan^2({x})\log({1+x^2})}{\sqrt{x}(1+x^2)}dx$$
After long calculations,I have found the following relationship.
 $$I_1-2I_2=\frac{\pi\sqrt{2}}{4}\left(\frac{5}{4}{\pi^3}-4{\pi}G+4{\pi}\ln^{2}2+\frac{5}{2}{\pi^2}\ln2+24G\ln2+12\ln^{3}2+\frac{27}{2}{\zeta(3)}\right)$$
 Where $G$ is Catalan's number.

REPLY [7 votes]: Sketch of a particular method for integral 1.
Consider the series
\begin{align}
\tan^{-1}(x) = \sum_{n=0}^{\infty} \frac{(1)_{n} (1)_{n}}{n! \, (3/2)_{n}} \, \frac{x^{2n+1}}{(1+x^{2})^{n+1}}.
\end{align}
Now consider the integral
\begin{align}
I(\mu, \nu) = \int_{0}^{\infty} \frac{x^{\nu} \, tan^{-1}(x) }{ (1+x^{2})^{\mu +1}} \, dx.
\end{align}
From this it is seen that:
\begin{align}
I(\mu, \nu) &= \sum_{n=0}^{\infty} \frac{ (1)_{n} (1)_{n}}{n! (3/2)_{n}} \, \int_{0}^{\infty} \frac{x^{2n+\nu+1} \, dx}{(1+x^{2})^{n+\mu+2}} \\
&= \frac{1}{2} \, \sum_{n=0}^{\infty} \frac{ (1)_{n} (1)_{n}}{n! (3/2)_{n}} \, B(n+ \nu/2 +1, \mu -\nu/2 +1) \\
&= \frac{1}{2} \, B\left( \frac{\nu}{2} +1, \mu - \frac{\nu}{2} + 1\right) \, {}_{3}F_{1}\left( 1, 1, \frac{\nu}{2}+1; \frac{3}{2}, \mu+2; 1\right).
\end{align}
For the case of $\nu = 1/2$ this reduces to
\begin{align}
\int_{0}^{\infty} \frac{\sqrt{x} \, tan^{-1}(x) }{ (1+x^{2})^{\mu +1}} \, dx
= \frac{\Gamma\left(\frac{5}{4}\right) \Gamma\left(\mu + \frac{3}{4}\right)}{2 \Gamma(\mu + 2)} {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{4}, \mu+2; 1\right)
\end{align}
Taking the derivative of both sides with respect to $\mu$ leads to
\begin{align}
\int_{0}^{\infty} \frac{\sqrt{x} \, tan^{-1}(x) \, \ln(1+x^{2})}{ (1+x^{2})^{\mu +1}} \, dx 
&= \frac{\Gamma\left(\frac{5}{4}\right) \Gamma\left(\mu + \frac{3}{2}\right)}{2 \, \Gamma(\mu + 2)} \, \left[ \psi\left( \mu + \frac{3}{2}\right) - \psi(\mu+2) \right] \, {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2}, \mu + 2; 1 \right) \\
& \hspace{10mm} + \frac{\Gamma\left(\frac{5}{4}\right) \Gamma\left( \mu + \frac{3}{2}\right)}{2 \, \Gamma(\mu + 2) } \, \partial_{\mu} \left[ {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2} , \mu + 2; 1 \right) \right]
\end{align}
and evaluating at $\mu =0$ leads to
\begin{align}
\int_{0}^{\infty} \frac{\sqrt{x} \, tan^{-1}(x) \, \ln(1+x^{2})}{ (1+x^{2})^{\mu +1}} \, dx 
&= \frac{\pi}{4\sqrt{2}} \left( 1 + 3\ln(2) - \frac{\pi}{2} \right) \, {}_{3}F_{2}\left( 1 ,1, \frac{5}{4}; \frac{3}{2}, 2; 1 \right) \\
& \hspace{10mm} - \frac{\pi}{4 \sqrt{2}} \partial_{\mu} \left[ {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2}, \mu+2; 1\right) \right]_{\mu=0}
\end{align}
Repeating this process leads to
\begin{align}
\int_{0}^{\infty} \frac{\sqrt{x} \, tan^{-1}(x) \, \ln^{2}(1+x^{2})}{ (1+x^{2})^{\mu +1}} \, dx 
&= \frac{\Gamma\left(\frac{5}{4}\right) \Gamma\left(\mu + \frac{3}{2}\right)}{2 \, \Gamma(\mu + 2)} \, \left[ \psi^{'}\left( \mu + \frac{3}{2}\right) - \psi^{'}(\mu+2) + \left( \psi\left(\mu + \frac{3}{2}\right) - \psi(\mu+2) \right)^{2} \right] \\
& \hspace{10mm} \cdot \, {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2}, \mu + 2; 1 \right) \\
& \hspace{2mm} + \frac{\Gamma\left(\frac{5}{4}\right) \Gamma\left( \mu + \frac{3}{2}\right)}{ \Gamma(\mu + 2) } \, \left[ \psi\left(\mu + \frac{3}{2}\right) - \psi(\mu+2) \right] \, \partial_{\mu} \left[ {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2} , \mu + 2; 1 \right) \right] \\
& \hspace{10mm} + \frac{\Gamma\left(\frac{5}{4}\right) \Gamma\left( \mu + \frac{3}{2}\right)}{ \Gamma(\mu + 2) } \, \partial_{\mu}^{2} \left[ {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2} , \mu + 2; 1 \right) \right]
\end{align}
When $\mu = 0$ this reduces to
\begin{align}
\int_{0}^{\infty} \frac{\sqrt{x} \, tan^{-1}(x) \, \ln^{2}(1+x^{2})}{1+x^{2}} \, dx 
&= \frac{\pi}{4 \sqrt{2}} \, \left[ - 8 G + 2 - \pi + (6 - 3 \pi) \ln(2) + 9 \ln^{2}(2) + \frac{13 \pi^{2}}{12}  \right] \\
& \hspace{10mm} \cdot \, {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2}, 2; 1 \right) \\
& + \frac{\pi}{ 2\sqrt{2} } \, \left[ \frac{\pi}{2} - 1 - 3 \ln(2) \right] \, \partial_{\mu} \left[ {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2} , \mu + 2; 1 \right) \right]_{\mu = 0} \\
& \hspace{10mm} + \frac{\pi}{ 4 \sqrt{2} } \, \partial_{\mu}^{2} \left[ {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2} , \mu + 2; 1 \right) \right]_{\mu = 0}
\end{align}