Is $\int_0^{\infty} e^{-t} (1+Ct)^{p} t^{q} dt$ convergent? Fix $C>0$ and $q, p>0.$

Can we find the upper  bound of  the following integral:
  $$\int_0^{\infty} e^{-t} (1+Ct)^{p} t^{q} dt?$$
  If so, how that upper bound will depend on $C$?

 A: As others have mentioned it is clearly convergent and also has a closed form expression. However, if you are looking for a simpler upper bound, then you can do something like this. Note that if $p \in (0,1)$ then $(1 + Ct)^p \leq 1 + (Ct)^p$ and if $p \geq 1$ then $(1 + Ct)^p \leq p(1 + (Ct)^p)$. Thus, we can combine both of them to write $(1 + Ct)^p \leq \max \{ 1, p\}(1 + (Ct)^p)$. Using this, we can write,
\begin{align}
\int_0^{\infty}e^{-t} (1 + Ct)^p t^q & \leq \int_0^{\infty}e^{-t} \max \{ 1, p\}(1 + (Ct)^p) t^q \\
& \leq  \max \{ 1, p\} \left( \int_0^{\infty}e^{-t}t^q +  C^p \int_0^{\infty}e^{-t} t^{p+q}  \right) \\
& \leq  \max \{ 1, p\} \left( \Gamma(q+1) +  C^p \Gamma(p + q + 1)  \right) \\
\end{align}
This also gives you the required dependence on $C$ which is $O(C^p)$ as one would intuitively expect.
A: Just for the fun of it.
A CAS produces the "nice" expression
$$I=\pi  \csc (\pi  (p+q)) \left(\frac{q \, \Gamma (q) \,
   _1\tilde{F}_1\left(q+1;p+q+2;\frac{1}{c}\right)}{c^{q+1}\Gamma (-p)}-c^p \,
   _1\tilde{F}_1\left(-p;-p-q;\frac{1}{c}\right)\right)$$  where appears the regularized confluent hypergeometric function under the condtions $(\Re(c)\geq 0\lor c\notin \mathbb{R})\land \Re(q)>-1$
Enjoy it !
