Integral $\int^{\infty}_0 \exp\left[-\left(4x+\frac{9}{x}\right)\right] \sqrt{x}\,dx$ 
How do I evaluate $$\displaystyle\int^{\infty}_0 \exp\left[-\left(4x+\dfrac{9}{x}\right)\right] \sqrt{x}\;dx?$$ 

To my knowledge the following integral should be related to the Gamma function.
I have tried using the substitution $t^2 = x$, and I got 
$$
2e^{12}\displaystyle \int^{\infty}_0 \exp\left[-\left(2t + \dfrac{3}{t}\right)^2\right] t^2 \; dt
$$ 
after substitution. But it seems like I can do nothing about this integral anymore. Can anyone kindly give me a hint, or guide me to the answer?

(I started an answer involving only "plain computations", but was not quick enough, maybe it is time now to complete and submit, rather then removed the typed formulas and quit the post. I am posting an alternative solution in the hope it looks simpler from some point of view, although there is a lot to be typed.)

We have to compute the integral:
$$
\begin{aligned}
J&=
\int_0^{\infty} \exp\left(-\left(4x+\frac{9}{x}\right)\right) \; \sqrt{x}\;dx
\\
&\qquad\text{Substitution, so formally: $t=2\sqrt x$, $t^2=4x$, $x=t^2/4$, $dx=\frac 12t\; dt$}
\\
&=
\int_0^{\infty} 
\exp\left(-\left(t^2+\frac{36}{t^2}\right)\right) \; \frac 12 t\cdot
\frac 12t\; dt
\\
&=
\frac 14
e^{-12}
\underbrace{
\int_0^{\infty} 
\exp\left(-\left(t-\frac{6}t\right)^2\right)
t^2\; dt}_{\text{Notation: }K}
\\[3mm]
&\qquad\text{ and we want to show the above is equal to...}
\\
&\overset{(?)}=
\frac 14
e^{-12}\cdot\frac14\cdot 13\sqrt \pi\ .
\\[3mm]
&\qquad\text{ So we consider the integral...}
\\
K&=
\int_0^{\infty} 
\exp\left(-\left(t-\frac{6}t\right)^2\right)
t^2\; dt
\\
&\qquad\text{ Substitution $\displaystyle
s =t-\frac 6t
$, so formally $t^2-st-6=0$,}
\\
&\qquad\text{ we use $t=\frac 12(s+\sqrt{s^2+24})$,
formally 
$\displaystyle dt=\frac12
\left(1+\frac s{\sqrt{s^2+24}}\right)\; ds
$...}
\\
&=
\int_{-\infty}^{\infty} 
e^{-s^2}\cdot
\frac 14
(s^2+\color{blue}{2s}\sqrt{s^2+24}+(s^2+24))
\;
\frac 12
\left(1+\frac {\color{red}{s}}{\sqrt{s^2+24}}\right)\; ds
\\
&\qquad\text{ now expand the parentheses, and ignore the odd part...}
\\
&=
\frac14\cdot\frac 12
\int_{\Bbb R}
e^{-s^2}\;\Big(\ 
s^2\ +\ 
(s^2+24)\ + \ 
\color{blue}{2s}\cdot\color{red}{s}
\ \Big)
\;ds
\\
&=
\frac 14\cdot\frac 12
\cdot26\sqrt\pi\ .
\\[3mm]
&\qquad\text{ Putting all together:}
\\
J&=
\frac 14
e^{-12}
\cdot K
\\
&=
\frac 14
e^{-12}
\cdot 
\frac 14\cdot\frac 12
\cdot26\sqrt\pi
\\
&=
\color{magenta}{
\frac {13}{16}\cdot
e^{-12}
\cdot\sqrt\pi}\ .
\end{aligned}
$$

Numerical validation, sage code:
sage: J = integral( exp(-4*x-9/x) * sqrt(x), x, 0, oo )
sage: J.n()
8.848395438034755e-06
sage: ( 13. / 16. * exp(-12) * sqrt(pi) ).n()
8.84839543773073e-6

sage: var('s');
sage: integral( exp(-s^2) * (s^2+ (s^2+24) + 2*s*s), s, -oo, +oo )
26*sqrt(pi)

