If a polynomial formula for the $a_n$ term is known, then the numerator of each term can be multipled by $x^{y-1}$ where $y$ is the denominator of that term.
$ \textstyle a_n=\frac{a}{a-n}+\frac{ n(n-1)a}{(a-n)(a-(n-1))}+\frac{n(n-1)(n-2)a}{(a-n)(a-(n-1))(a-(n-2))}+ ... +\frac{n!(a+a_0)(a-n)!}{(a-1)!} $
$\begin{eqnarray} a_n&=&\int_0^1 ( ax^{a-n-1}+nax^{(a-n)(a-(n-1))-1}\\ &&+n(n-1)ax^{(a-n)(a-(n-1))(a-(n-2))-1}+\dots\\ &&+n!(a+a_0)x^{\frac{(a-1)!}{(a-n)!}-1})dx\\ &=&\int_0^1 a_0 n!x^{\frac{(a-1)!}{(a-n)!}-1} + \sum_{k=0}^n k! {n\choose k} x^{k! {a-1\choose k}-1 } \text dx \end{eqnarray} $
This is not the simplest integral form, but at least it allows $a_n$ to be calculated when $n$ is not an integer.