Simplifying / approximating compound negative binomial - binomial distribution

Suppose we have that the distribution of a count $$x$$ is given by a binomial distribution: $$x\sim \mathcal{B}(n,\theta)$$. Further, suppose that the sample size $$n\sim NB(\lambda,\kappa)$$, where this represents a negative binomial distribution with mean $$\lambda$$ and variance $$\lambda^2/\kappa$$. I can write a Mathematica function that implements this parameterisation of the negative binomial as follows:

fCustomNB[λ_, κ_] := NegativeBinomialDistribution[κ, κ/(κ + λ)]

which has desired mean (variance also but omitting to keep question short):

FullSimplify@Mean[fCustomNB[λ, κ]]

λ

I can also use Mathematica to find the distribution of $$x$$ where I have marginalised $$n$$ out of the distribution (i.e. the distribution of $$x|\theta, \lambda, \kappa$$) as follows:

BinomialDistribution[n, θ],
n \[Distributed] fCustomNB[λ, κ]];

I can then evaluate the PDF of this using:

Assuming[{λ > 0, κ > 0, 0 < θ < 1, x >= 0}, PDF[aDist, x]]

which gives me this, somewhat heathen, function:

(1 - θ)^(-x + Ceiling[x]) θ^x κ^κ λ^ Ceiling[x] (κ + λ)^(-κ - Ceiling[x]) Binomial[Ceiling[x], x] Binomial[-1 + κ + Ceiling[x], -1 + κ] Hypergeometric2F1[ 1, κ + Ceiling[x], 1 - x + Ceiling[ x], (λ - θ λ)/(κ + λ)]

The part of this that is particularly difficult to translate into other languages (this distribution represents part of a larger model already coded up in Stan) is the Hypergeometric bit.

The bottom line is, can I simplify this expression in terms of more vanilla-like functions?

Edit: as finishing writing this question, I realised we can -- so I include my answer below in case it helps others!

Since $$x$$ is a count variable, just state it in the assumptions. Then everything is hunky dory (whatever hunky dory is):