Is it possible to get the equation below into closed form? I have tried using integration tables but I haven't found anything that matches. Are there any other methods to achieve a closed form expression of a bessel function with two exponentials? If so could someone please advise on how I would go about this?
Kind regards.


$$\int^\infty\_0 \frac{1}{2}\frac{x}{a}^\frac{b-2}{4}\exp\left(-\frac{a+x}{2}\right)\left(1-\exp\left(-\frac{x}{2}\right)\right)^{n} I\_{\frac{b}{2}-1}\left(\sqrt{xa}\right) \mathrm{d}x$$


To clarify a, b and n are all integer values $\ \ge 1 $.



Is it possible to get an equation with two exponentials and a bessel function in closed form?
Maybe you can accept the following result as a closed form. Transforming the variable of integration in your integral
$$
I\_{n}(a,b):= \frac{1}{2 a}\ e^{-a/2}\int\_{0}^{\infty} dx\ x^{(b-2)/4} e^{-x/2}(1-e^{-x/2})^{n} I\_{\frac{b}{2}-1}(\sqrt{a x})
$$
to $y:=\sqrt{a x}$, we get 
$$
I\_{n}(a,b)=e^{-a/2}\ a^{-\frac{b}{4}-\frac{3}{2}} \int\_{0}^{\infty}dy\ y^{b/2}\ e^{-y^{2}/(2 a)} I\_{\frac{b}{2}-1}(y) \sum\_{m=0}^{n} {n \choose m} (-1)^m e^{-m\frac{y^{2}}{2a}}
$$
where the $n$th power is expanded using the binomial formula. After interchanging summation and integral use DLMF formula <http://dlmf.nist.gov/10.43.E23> to get
$$
I\_{n}(a,b)=e^{-a/2}\ a^{\frac{b}{4}-\frac{3}{2}}\sum\_{m=0}^{n}{n \choose m} (-1)^{m} (m+1)^{-b/2} \exp\left(\frac{a}{2(m+1))}\right).
$$


