# Closed form for $\int\frac{\left((x + i) \beta\right)^\beta x^{\beta - 2}}{(x^2 + 1)^\beta} \exp\left(-\frac{\alpha}{A x}\right) \, \mathrm{d} x$

The following integral comes up in the solution of a differential equation when solved by Maple: $$\begin{equation} \int\frac{\left((x + i) \beta\right)^\beta x^{\beta - 2}}{(x^2 + 1)^\beta} \exp\left(-\frac{\alpha}{A x}\right) \, \mathrm{d} x \end{equation}$$ where $\alpha, \beta \in \mathbb{C}$, $A \in \mathbb{R}$ are parameters. Solving the same differential equation in Mathematica however gives a result without any integrals left to be solved. Neither Mathematica nor Maple seem to be able to find a closed form solution for the above.

The solutions given by the two programs are:

$\textbf{Maple:}$ \begin{align} \begin{split} F_{Map}(x) = & C_1[( x + i) \beta]^\beta + C_2[( x - i) \beta]^\beta + C_3 \left[[( x - i) \beta]^\beta\int\frac{[( x + i) \beta]^\beta x^{\beta - 2}}{( x^2 + 1)^\beta} \exp\left(-\frac{\alpha}{A x}\right) \, \mathrm{d} x \right. \\ & \left. - [( x + i) \beta]^\beta\int \frac{[( x - i) \beta]^\beta x^{\beta - 2}}{( x^2 + 1)^\beta} \exp\left(-\frac{\alpha}{A x}\right) \, \mathrm{d} x \right] \\ \end{split} \end{align} $\textbf{Mathematica:}$ \begin{align} \begin{split} F_{Mat}(x) = & \tilde{C}_1 ( x + i)^\beta + \tilde{C}_2 ( x - i) ^\beta + \tilde{C}_3\left[\alpha \left(\beta-1+x \left(1+\sqrt {-(\beta - 1 )^2}-\beta \right) \right. \right. \\ & \left. \left. +\sqrt {-(\beta - 1)^2} \right)+A (\beta - 2) \left(\beta-1+x \sqrt {-(\beta - 1)^2} \right) \right]^\beta \end{split} \end{align}

My questions about this are:

• Does this imply that the integral has a closed form?
• If not, is it still possible that both solutions are the same or at least both are correct?
• Too many parameters, and I won't bet a cent on the indefinite integral being an elementary function. Maybe the integral over $(0,+\infty)$ is manageable through the Laplace transform. – Jack D'Aurizio May 31 '16 at 18:19
• @JackD'Aurizio : Thanks for the input. Was expecting something like this, but still had some hope. Any thoughts on the two solutions looking so different? – TimD Jun 7 '16 at 21:11