Integral of a rational function I need a closed form for the following integral 
$$\int^\infty_0\frac{t+b}{(t^2+a)(k^2+(t+b)^2)}dt$$
where $a,b \in \mathbb{C}$ and $k \in \mathbb{N}$. That $t+b$ makes the integral complicated for me.
Any simple approaches for that ? 
I tried wolfram |Alpha but it fails. A final solution using CAS is acceptable for me.
I guess a closed form can be obtained using Incomplete Beta function, But I need a simpler one ?

Use partial fractions to do the calculation. Let
$$ \frac{t+b}{(t^2+a)(k^2+(t+b)^2)}=\frac{At+B}{t^2+a}+\frac{Ct+D}{(t+b)^2+k^2}$$
and then one has
$$ (At+B)((t+b)^2+k^2)+(Ct+D)(t^2+a)=t+b.$$
It is not hard to obtain
\begin{eqnarray}
A&=&\frac{k^2-a-b^2}{a^2+2 a b^2-2 ak^2+b^4+2 b^2 k^2+k^4},\\
B&=&\frac{a b+b^3+b k^2}{a^2+2 a b^2-2 a
   k^2+b^4+2 b^2 k^2+k^4},\\
C&=&-A,\\
D&=&\frac{a b+b^3-3 b k^2}{a^2+2 a b^2-2 a k^2+b^4+2 b^2
   k^2+k^4}.
\end{eqnarray}
(Suppose $a,b,k>0$.) It is not hard to get
\begin{eqnarray}
&&\int^\infty_0\frac{t+b}{(t^2+a)(k^2+(t+b)^2)}dt\\
&=&\int^\infty_0\left(\frac{At+B}{t^2+a}+\frac{-At+D}{(t+b)^2+k^2}\right)\\
&=&\frac A2\ln\frac{t^2+a}{k^2+(t+b)^2}+\frac{B}{\sqrt a}\arctan\frac{t}{\sqrt a}+\frac{-bC+D}{k}\arctan\frac{b+t}{k}\bigg|_0^\infty\\
&=&(\frac{B}{\sqrt a}+\frac{-bC+D}{k})\frac{\pi}{2}-\frac{A}{2}\ln\frac{a}{k^2+b^2}-\frac{-bC+D}{k}\arctan\frac{b}{k}.
\end{eqnarray}
