Expectation of $\left( 1+x \right)^p$ for positive constant $p>0$

Question

I want to calculate the expectation of $\mathbb{E}\left[\left( 1+x \right)^p\right]$ where $x\sim \Gamma\left(k,\theta\right)$ is a gamma distributed random variable.

I know that one can use the general binomial expansion $\left(1+x\right)^p = \sum_{n=1}^{\infty} \binom{p}{n}x^n$ where $\binom{p}{n} = \frac{p(p-1)\cdots (p-n+1)}{n!}$ and then using the well known moment result for Gamma distribution $\mathbb{E}[x^v] = \theta^v\frac{\Gamma(k+v)}{\Gamma(k)}$ where $\Gamma(s) = \int_{0}^{\infty}t^{s-1}e^{-t}dt$ is the gamma function. However, I want to know is there a way to get a closed form solution or an infinite series is the best we can do?

Thank you!

Answer 1

Mathematica:

Mean@Assuming[p > 1,
  TransformedDistribution[(1 + x)^p, 
   x \[Distributed] GammaDistribution[k, \[Theta]]]]

$$\theta ^{-k} U\left(k,k+p+1,\frac{1}{\theta }\right)$$

where $U$ is the confluent hypergeometric function.