A generalized Ahmed's integral Let $\vec{A}:=(A_1,A_2,A_3)$ be a vector where all its components are positive real numbers.
In the context of this question An integral involving error functions and a Gaussian we came across  a following integral.
\begin{equation}
I(\vec{A}) := A_1\int\limits_0^{A_3} \frac{ \arctan\left(\frac{A_2}{\sqrt{1+A_1^2 +\xi^2}}\right)}{(1+\xi^2)\sqrt{1+A_1^2+\xi^2}} d\xi
\end{equation}
By performing the following transformations, firstly by substituting $\xi = \sqrt{1+A_1^2} \tan(\theta)$and then by substituting $t=\tan(\theta/2)$ we brought the quantity being sought for  to the following form:
\begin{eqnarray}
&&I(\vec{A}) = \arctan(\frac{A_1 A_3}{\sqrt{1+A_1^2+A_3^2}}) \arctan(\frac{A_2}{\sqrt{1+A_1^2+A_3^2}})+\\
&&4 A_2\sqrt{1+A_1^2}\int\limits_0^{B} \arctan(\frac{t}{\sqrt{1+A_1^2}-A_1}) \cdot \frac{ t}{A_2^2 (1-t^2)^2 + (1+A_1^2) (1+t^2)^2} dt-\\
&&4 A_2\sqrt{1+A_1^2}\int\limits_0^{B} \arctan(\frac{t}{\sqrt{1+A_1^2}+A_1}) \cdot \frac{ t}{A_2^2 (1-t^2)^2 + (1+A_1^2) (1+t^2)^2} dt
\end{eqnarray}
where $B:= (-\sqrt{1+A_1^2} + \sqrt{1+A_1^2+A_3^2})/A_3$.
Now the question is how do we complete the calculation? Is the result expressed through elementary functions only and if not what kind of special functions enter the result?
 A: Define $\phi:= \arccos(A_2/\sqrt{1+A_1^2+A_2^2})$ and $\alpha := \sqrt{1+A_1^2}-A_1$ and $\beta:=\sqrt{1+A_1^2}+A_1$ and
\begin{eqnarray}
&&{\mathcal F}^{(a,b)}(t):=\int \arctan(\frac{t}{a}) \frac{1}{t-b} dt = \log(t-b) \arctan(\frac{t}{a})\\
&&-\frac{1}{2 \imath} \left( \log(t-b) \left[ \log(\frac{t-\imath a}{b-\imath a}) - \log(\frac{t+\imath a}{b+\imath a})\right] + Li_2(\frac{b-t}{b-\imath a}) -  Li_2(\frac{b-t}{b+\imath a})\right)
\end{eqnarray}
By using the following partial fraction decomposition :
\begin{eqnarray}
\frac{t}{A_2^2(1-t^2)^2 + (1+A_1^2) (1+t^2)^2} = 
\frac{I}{8 A_2 \sqrt{1+A_1^2}} \left( -\frac{1}{t-e^{\imath\phi}} + \frac{1}{t-e^{-\imath\phi}}+\frac{1}{t+e^{-\imath\phi}}-\frac{1}{t+e^{\imath\phi}}\right)
\end{eqnarray}
and by integrating each fraction over $t$ using the antiderivative above we arrive at the following result:
\begin{eqnarray}
&&I(\vec{A})= \arctan\left( \frac{A_1 A_3}{\sqrt{1+A_1^2+A_3^2}}\right) + \arctan\left( \frac{A_2}{\sqrt{1+A_1^2+A_3^2}}\right)+\\
&&\left.\frac{\imath}{2} \left(
{\mathcal F}^{\alpha,+\exp(-\imath \phi))}(t) +
{\mathcal F}^{\alpha,-\exp(-\imath \phi))}(t) -
{\mathcal F}^{\alpha,-\exp(+\imath \phi))}(t) -
{\mathcal F}^{\alpha,+\exp(+\imath \phi))}(t) 
\right)\right|_0^B -\\
&&\left.\frac{\imath}{2} \left(
{\mathcal F}^{\beta,+\exp(-\imath \phi))}(t) +
{\mathcal F}^{\beta,-\exp(-\imath \phi))}(t) -
{\mathcal F}^{\beta,-\exp(+\imath \phi))}(t) -
{\mathcal F}^{\beta,+\exp(+\imath \phi))}(t) 
\right)\right|_0^B
\end{eqnarray}
