 Zero-based geometric sum of exponential variables

If independent summands have exponential distribution of mean $\mu$ and the independent count has a zero-based geometric distribution of parameter $p$, their random sum of independent claims has an atom $p$ at zero and an exponential positive component of rate $p/\mu$. Its Laplace transform of a nonnegative random variable is $p/[1-(1-p)/(1+\mu s)]=p+(1-p)p/(p+\mu s)$. Thus $\mathbb{P}(S>x)=(1-p)e^{-px/\mu }$ for $x>0$. Unlike the geometric sum of exponential variables with a positive count, this law is not purely exponential.