Closed form solution? Can a closed form solution be found for:
$$\intop_{0}^{K}e^{-\frac{\rho^{2}}{2c}}{}I_{0}\left(\sqrt{\frac{\left(a^{2}+b^{2}\right)\rho^{2}}{2c^{2}}}\right)d\rho,$$
where $a,b,c,K\in\mathbb{R}^+$ and $I_0$ is the modified Bessel function of the first kind?
 A: Hint: let us use the power series expansion of the modified Bessel function as suggested by what'sup. A bit of notation to start:
$$ \frac{\sqrt{(a^2+b^2)\rho^2}}{2c^2}:=\alpha\rho,$$
with $\rho\in[0,K]$ and $K>0$.
We need to compute the integrals
$$(k!)^2\mathcal I_k:=(-1)^k\frac{\alpha^{2k}}{4^k}\int_0^K \rho^{2k}e^{-\frac{\rho^2}{2c^2}}d\rho $$
for all $k\geq 0$. For $k=0$ we have
$$\mathcal I_0:=\int_0^K e^{-\frac{\rho^2}{2c^2}}d\rho=\frac{\sqrt{\pi}}{2\sqrt{2}|c|}
\operatorname{erf}\left(\frac{K}{\sqrt{2}|c|}\right).$$
and for $k=1$
$$\mathcal I_1:=-\frac{\alpha^2}{4}\int_0^K \rho^2e^{-\frac{\rho^2}{2c^2}}d\rho=
-\frac{\alpha^2}{4}\int_0^K (-c^2\rho)\left(-\frac{1}{c^2}\rho\rho e^{-\frac{\rho^2}{2c^2}}\right)d\rho= \\-\frac{\alpha^2}{4}\left(-c^2\rho e^{-\frac{\rho^2}{2c^2}}|^K_0+c^2
\int_0^K e^{-\frac{\rho^2}{2c^2}}d\rho\right)=-\frac{\alpha^2}{4}\left(-c^2Ke^{-\frac{K^2}{2c^2}}+c^2\mathcal I_0\right).
$$
For $k=2$ we have
$$2!\mathcal I_2:=\frac{\alpha^4}{4^2}\int_0^K \rho^4e^{-\frac{\rho^2}{2c^2}}d\rho=
\frac{\alpha^4}{4^2}\int_0^K (-c^2\rho^3)\left(-\frac{1}{c^2}\rho\rho e^{-\frac{\rho^2}{2c^2}}\right)d\rho= \\
\frac{\alpha^4}{4^2}\left(-c^2\rho^3 e^{-\frac{\rho^2}{2c^2}}|^K_0+3c^2
\int_0^K \rho^2e^{-\frac{\rho^2}{2c^2}}d\rho\right)
=\frac{\alpha^4}{4^2}\left(-c^2K^3e^{-\frac{K^2}{2c^2}}\right)-3c^2\frac{\alpha^2}{4}\mathcal I_1.
$$
Can you go on and find a recursive formula for the $(k!)^2\mathcal I_k$?
EDIT
The recursive formula for $(k!)^2\mathcal I_k$ is
$$(k!)^2\mathcal I_k=\sum_{n=1}^k(-1)^{k-1}\frac{(2k-1)!!}{(2n-1)!!}\frac{1}{4^k}K^{2n-1}e^{-\frac{K^2}{2}} +(-1)^k(2k-1)!!\frac{\mathcal I_0}{4^k},$$
in the easier case $\alpha=c=1.$
A: Looking at only the case $K=\infty$:
By first checking the special case in GEdgar's comment, by comparing to the formula of the (-1)st moment of the Rice distribution (totally not allowed, I know...) we see that 
$\int_0^\infty \exp(-x^2)I_0(x)dx = \frac{1}{2}e^{1/8}\sqrt{\pi}I_0(\frac{1}{8})$,
which is verified with Wolfram Alpha.
In the general case, through some algebraic manipulations, using the same comparison we can posit that the answer to your integral, for $K = \infty$, MIGHT be (denoting $\gamma = \sqrt{a^2+b^2}$):
$ \dots = \sqrt{\dfrac{c}{2}}\exp\left(\dfrac{\gamma^2}{4c}\right)\sqrt{\pi}L_{-\frac{1}{2}}\left(\dfrac{\gamma^2}{4c}\right)$,
where $L$ is a Laguerre polynomial.
Again, this has no reason to be correct (but it might be...)
