Compute (or approximate) $\int_{0}^{+\infty} \exp(-bx)\sqrt{x(x+a)} dx$ The following integral arises in a physics problem I am trying to solve, ($a,b$ are positive real numbers):
$\int_{0}^{+\infty} \exp(-bx)\sqrt{x(x+a)} dx.$
I tried solving it using Wolfram Alpha to no avail.
Is this a computable integral? It would be great to have an explicit solution.
If it is not, could you please suggest a method to obtain a reasonable approximation for it?
 A: By http://dlmf.nist.gov/10.32.E8, we have
\begin{align*}
\int_0^{ + \infty } {e^{ - bx} \sqrt {x(x + a)} dx} & \mathop  = \limits^{x = at} a^2 \int_0^{ + \infty } {e^{ - abt} \sqrt {t(t + 1)} dt} 
\\ & \! \mathop  = \limits^{t = \frac{{w - 1}}{2}} 
\frac{{a^2 }}{4}e^{\frac{{ab}}{2}} \int_1^{ + \infty } {e^{ - \frac{{ab}}{2}w} \sqrt {w^2  - 1} d} w =\frac{a}{{2b}}e^{\frac{{ab}}{2}} K_1 \!\left( {\frac{{ab}}{2}} \right).
\end{align*} This result is valid whenever $\Re a>0$ and $\Re b>0$.
A: Let’a use an $e^y$ Maclaurin Series integrating each term to interchange sum and integral:
$$\int_0^\infty\frac{ \sqrt{x(x+a)}}{e^{bx} }dx=\int_0^\infty \sum_{k=0}^\infty \frac{(-1)^k b^k x^k\sqrt{x(x+a)}}{k!}dx=\sum_{k=0}^\infty \frac{(-1)^kb^k}{k!}\int_0^\infty x^{k+\frac12}(x+a)^\frac12 dx $$
Next use Wolfram Functions 7.23.7.2.1 or some machine help with the Gauss hypergeometric function which probably can be simplified. The substitution of $0$ is just $0$ in the summand with $a\ne 0$. The other restrictions seem to be permissible:
$$\sum_{k=0}^\infty \frac{(-1)^kb^k}{k!}\int_0^\infty x^{k+\frac12}(x+a)^\frac12 dx =\frac23 \lim_{x\to\infty}\sum_{k=0}^\infty \frac{(-1)^k b^k (x+a)^\frac32 \left(-\frac1a\right)^{-k-\frac12}\,_2F_1\left(\frac32,-k-\frac12,\frac52,\frac{x}{a}+1 \right)}{k!}= -\frac{2\sqrt ai}3 \lim_{x\to\infty}\sum_{k=0}^\infty \frac{ (ab)^k (x+a)^\frac32\,_2F_1\left(\frac32,-k-\frac12,\frac52,\frac{x}{a}+1 \right)}{k!} $$
Using @Raymond’s method, one can find that:
$$\int_0^\infty \frac{\sqrt{x(x+a)}}{e^{-bx}}dx=\frac{ae^\frac{a b}{2}\text K_1\left(\frac{ab}{2}\right)}{2b}$$
It comes from Wolfram functions 3.4.7.1.1:
$$\text K_1(z)=\frac{z}{2}\int_1^\infty \frac{\sqrt{x^2-1}}{e^{z x}}dx$$
Where appears the Modified Bessel function of the Second Kind.
Let me work on this more. Please correct me and give me feedback!
