I am trying to solve the following integral $$ \int_0^y \frac{x^{m-1}}{(1+x)^{m+k}}\, \exp\left(-\frac{m}{\gamma} x \right) \,dx. $$ I tried to look into different books such as Gradshteyn and Prudnikov but I did not find any interesting thing. I find some results when the higher bound of the integral is $\infty$, but I am actually calculating some Cumulative Function so the bounds have to be finite.
Any suggestion ?
Since your $y$ is a variable, you need an antiderivative
$$ J(m,k,x) = \int \dfrac{x^{m-1}}{(1+x)^{m+k}}\; \exp(-cx)\; dx$$
Assuming $m$ and $k$ are positive integers, use the change of variables $t = 1 + x$ and this becomes
$$\eqalign{ J(m,k,t-1) &= e^c \int \dfrac{(t-1)^{m-1}}{t^{m+k}} \exp(-ct)\; dt\cr &= e^c \sum_{j=0}^{m-1} {m-1 \choose j} (-1)^{j} \int t^{-1-j-k} \exp(-ct)\; dt\cr &= e^c \sum_{j=0}^{m-1} {m-1 \choose j} (-1)^{j+1} c^{j+k} \Gamma(-j-k,ct) + const}$$
where $\Gamma(\cdot,\cdot)$ is the incomplete Gamma function. You can also express this in terms of the Ei (exponential integral) function.