Is this distribution known? I have something like 
$F(t)=\int_0^{t} e^{-kx} x^\alpha (1+x)^\beta dx$
Is this a form of some known distribution (more specifically density)?
EDIT:
I was asked where I encountered the problem.
Its is as follows...
$\theta|\lambda \sim\mathcal{N}_p(0,\frac{1-\lambda}{\lambda}I)$ and $\lambda$ follows density $f(x)\sim(1-b)x^{-b}$ for $(b<1)$.
I calculated the density of $\theta$ (unconditional).
(While fiddling about 5.4 of this paper)
 A: I will interpret your question as if the positive function given by
$$ f(x)= e^{-kx} x^\alpha (1+x)^\beta $$ (removing the superfluous integral) is proportional to some known (named) distribution. 
First, I will rewrite (and rename, reparametrize, so note that $\alpha$ below has new meaning) this function to show that it is a multiplicative modification of a gamma density:
$$
   f(x) \propto \frac{k^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-kx} \cdot (1+x)^\beta \quad \text{for $x>0$}
$$ which has a finite integral for $k>0, \alpha>0, \beta\in\mathbb{R}$. For $\beta=0$ it is a gamma density. 
Attempting to find a closed form for the normalization constant (with the help of maple) result in an expression to large to be very useful, including Laguerre polynomials or Kummer functions. Part of the nonusefulness of the resulting expression is because it includes, for some parameter values, gamma functions evaluated at negative integer arguments. 
But numerical integration works without problems. I don't know any standard name for this distribution. Where did you encounter it, why are you interested?
EDIT

After some more work with maple (doing first some by hand), I get this expression for the $\int_0^\infty \cdot \; dx$:
$$
{\frac {{\beta\choose -\alpha}
{\mbox{$_1$F$_1$}(\alpha;\,\alpha+\beta+1;\,k)}\pi\,\sin \left( \pi\,
\beta \right) {k}^{\alpha}}{\sin \left( \pi\,\alpha+\pi\,\beta
 \right) \sin \left( \pi\,\alpha \right) \Gamma \left( \alpha \right) 
}}
$$
which is much simpler than what I got before, but still needs more work to simplify for some difficult integer parameter cases. (for instance, for $\beta=0$ plugging directly into that expression gives 0)
