I would like to calculate the definite integral $$I_1 = \int_{0}^{\infty} \exp(-\sqrt{x^2+bx+c})dx$$ where $b$ and $c$ are reals. I feel that the solution, if there is one, has something to do with Bessel functions. I've first tried the following substitution $$t = \sqrt{x^2+bx+c} $$ which gave me $$I_1 = \int_{0}^{\infty} \exp(-t) \frac{2t}{\sqrt{b^2-4(c-t^2)}}dt$$ but I don't know where to go after that step. It doesn't look like any of the famous integrals of exponential functions and I haven't been able to integrate it by parts.
EDIT : Accelerator pointed out to me that the integration bounds shouldn't be $[0,\infty[$ (there isn't always a solution for $t=0$ if $b$ and $c$ are reals), which is absolutely right. I had overlooked this point and will try to modify my question in line with this remark.
I also tried the substitution $t = x+b/2$ to shift the parabola and obtain $$I_1 = \int_{b/2}^{\infty} \exp(-\sqrt{t^2+z})dt,$$ with $z=c-b^2/4$, which is the same integral as in this question but with a non-zero lower bound.
I also know that my problem can be formulated with another integral, $$I_2 = \int_{\theta_0}^{\pi/2} \exp\left(-\frac{k}{cos(\theta)}\right )d\theta,$$ where $k$ is a real number, but I don't know if this can be helpful as we again have a non-zero lower bound.
Ideally I would like to obtain the indefinite integral but the integral between zero and infinity would already be really useful.
One of my past PhD students faced almost he sam problem for the case whare $x^2+bx+c$ is always positive and $\left(c-\frac{b^2}{4}\right)$ being "small".
Completing the square and changing notations, the integrand write $$e^{-\sqrt{(x+\alpha )^2+\beta}}$$
What she did was to expand it as a series around $\beta=0$ $$e^{-\sqrt{(x+\alpha )^2+\beta}}=\sum_{n=0}^\infty A_n\,\beta^n$$ with $$A_0=e^{-(x+\alpha)} \qquad \qquad A_1=-\frac{e^{-(x+\alpha)}}{2 (x+\alpha )}$$ $$A_n=-\frac{2 (n-1) (2 n-3) A_{n-1}-A_{n-2} } {4 n(n-1)(x+\alpha )^2 }$$
In other words, $$e^{-\sqrt{(x+\alpha )^2+\beta}}=e^{-(x+\alpha )} \Big[\cdots\Big]$$ where $$\Big[\cdots\Big]=1-\frac{1}{2 (x+\alpha )}\beta+\left(\frac{1}{8 (x+\alpha )^2}+\frac{1}{8 (x+\alpha )^3}\right)\beta ^2 -$$ $$ \left(\frac{1}{48 (x+\alpha )^3}+\frac{1}{16 (x+\alpha )^4}+\frac{1}{16 (x+\alpha )^5}\right)\beta ^3 +O\left(\beta ^4\right)$$ leading to simple integrals since $$I_n=\int_0^\infty \frac{e^{-(x+\alpha)}} {(x+\alpha)^{n}}=\Gamma (1-n,\alpha )$$
For a quick test $(\alpha=2,\beta=3)$, using the very truncated series given above leads to $0.0798$ to be compared to the "exact" $0.0832$. Using twice more terms leads to $0.0836$