$\int_0^\infty\dfrac{e^{-at}}{\sqrt{A+Bt+Ct^2}}dt$
$=\int_0^\infty\dfrac{e^{-at}}{\sqrt{C\left(t^2+\dfrac{Bt}{C}+\dfrac{A}{C}\right)}}dt$
$=\int_0^\infty\dfrac{e^{-at}}{\sqrt{C\biggl(t^2+\dfrac{Bt}{C}+\dfrac{B^2}{4C^2}+\dfrac{A}{C}-\dfrac{B^2}{4C^2}\biggr)}}dt$
$=\int_0^\infty\dfrac{e^{-at}}{\sqrt{C\biggl(t+\dfrac{B}{2C}\biggr)^2-\dfrac{B^2-4AC}{4C}}}dt$
$=e^\frac{aB}{2C}\int_\frac{B}{2C}^\infty\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$
Case $1$: $C>0$ and $B^2-4AC>0$
Then $e^\frac{aB}{2C}\int_\frac{B}{2C}^\infty\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$
$=e^\frac{aB}{2C}\int_\frac{B}{2C}^\sqrt\frac{B^2-4AC}{4C^2}\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt+e^\frac{aB}{2C}\int_\sqrt\frac{B^2-4AC}{4C^2}^\infty\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$
For $\int_\sqrt\frac{B^2-4AC}{4C^2}^\infty\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$ ,
$\int_\sqrt\frac{B^2-4AC}{4C^2}^\infty\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$
$=\int_0^\infty\dfrac{e^{-a\sqrt{\frac{B^2-4AC}{4C^2}}\cosh t}}{\sqrt{C\left(\sqrt{\dfrac{B^2-4AC}{4C^2}}\cosh t\right)^2-\dfrac{B^2-4AC}{4C}}}d\left(\sqrt{\dfrac{B^2-4AC}{4C^2}}\cosh t\right)$
$=\int_0^\infty\dfrac{e^{-a\sqrt{\frac{B^2-4AC}{4C^2}}\cosh t}\sqrt{\dfrac{B^2-4AC}{4C^2}}\sinh t}{\sqrt{\dfrac{B^2-4AC}{4C}\cosh^2t-\dfrac{B^2-4AC}{4C}}}dt$
$=\int_0^\infty e^{-a\sqrt{\frac{B^2-4AC}{4C^2}}\cosh t}~dt$
$=K_0\left(a\sqrt{\dfrac{B^2-4AC}{4C^2}}\right)$
But for $\int_\frac{B}{2C}^\sqrt\frac{B^2-4AC}{4C^2}\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$ , it seems that this is already the simplest approach:
$\int_\frac{B}{2C}^\sqrt\frac{B^2-4AC}{4C^2}\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt=\int_\frac{B}{2C}^\sqrt\frac{B^2-4AC}{4C^2}\sum\limits_{n=0}^\infty\dfrac{(-1)^na^nt^n}{n!\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$
Then handle $\int\dfrac{t^n}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$ , where $n$ is any non-negative integer.
Case $2$: $C>0$ and $B^2-4AC<0$
Then $e^\frac{aB}{2C}\int_\frac{B}{2C}^\infty\dfrac{e^{-at}}{\sqrt{Ct^2-\dfrac{B^2-4AC}{4C}}}dt$
$=e^\frac{aB}{2C}\int_\frac{B}{2C}^0\dfrac{e^{-at}}{\sqrt{Ct^2+\dfrac{4AC-B^2}{4C}}}dt+e^\frac{aB}{2C}\int_0^\infty\dfrac{e^{-at}}{\sqrt{Ct^2+\dfrac{4AC-B^2}{4C}}}dt$
For $\int_0^\infty\dfrac{e^{-at}}{\sqrt{Ct^2+\dfrac{4AC-B^2}{4C}}}dt$ ,
$\int_0^\infty\dfrac{e^{-at}}{\sqrt{Ct^2+\dfrac{4AC-B^2}{4C}}}dt$
$=\int_0^\infty\dfrac{e^{-a\sqrt{\frac{4AC-B^2}{4C^2}}\sinh t}}{\sqrt{C\left(\sqrt{\dfrac{4AC-B^2}{4C^2}}\sinh t\right)^2+\dfrac{4AC-B^2}{4C}}}d\left(\sqrt{\dfrac{4AC-B^2}{4C^2}}\sinh t\right)$
$=\int_0^\infty\dfrac{e^{-a\sqrt{\frac{4AC-B^2}{4C^2}}\sinh t}\sqrt{\dfrac{4AC-B^2}{4C^2}}\cosh t}{\sqrt{\dfrac{4AC-B^2}{4C}\sinh^2t+\dfrac{4AC-B^2}{4C}}}dt$
$=\int_0^\infty e^{-a\sqrt{\frac{4AC-B^2}{4C^2}}\sinh t}~dt$
$=\dfrac{\pi}{2}\mathbf{K}_0\left(a\sqrt{\dfrac{4AC-B^2}{4C^2}}\right)$
But for $\int_\frac{B}{2C}^0\dfrac{e^{-at}}{\sqrt{Ct^2+\dfrac{4AC-B^2}{4C}}}dt$ , it seems that this is already the simplest approach:
$\int_\frac{B}{2C}^0\dfrac{e^{-at}}{\sqrt{Ct^2+\dfrac{4AC-B^2}{4C}}}dt=\int_\frac{B}{2C}^0\sum\limits_{n=0}^\infty\dfrac{(-1)^na^nt^n}{n!\sqrt{Ct^2+\dfrac{4AC-B^2}{4C}}}dt$
Then handle $\int\dfrac{t^n}{\sqrt{Ct^2+\dfrac{4AC-B^2}{4C}}}dt$ , where $n$ is any non-negative integer.
Best Answer
The more general formula is $\;\displaystyle \sum_{n=1}^\infty \frac 1{n^2 + a^2} = \frac{a \pi \coth(a \pi) - 1}{2 a^2}$.
This may be obtained using the Fourier series for $\cosh(z x)$ as in this answer for $\cos(z x)$
Substitute $z=i a\,$ in the answer to convert
$$\cot(\pi z)=\frac 1{\pi}\left[\frac1{z}-\sum_{k=1}^{\infty}\frac{2z}{k^2-z^2}\right]$$
to (since $\;\cot(\pi ia)=-i \coth(\pi a)$) $$\coth(\pi a)=\frac 1{\pi}\left[\frac1{a}+\sum_{k=1}^{\infty}\frac{2a}{k^2+a^2}\right]$$