<s>[INST] What is the quickest method of finding the inverse laplace transform of $ 7e^{-6}/(s^2+6)^4 $ Solving for a differential equation, I found this which I need to obtain the inverse laplace of. What method should I use to solve this as quick as possible? Partial fractions and convolution seem to both take quite a long time. 
 [/INST] You could start with a few derivatives:
$$\frac d{ds}\frac1{s^2+6}=-\frac{2s}{(s^2+6)^2}$$
$$\frac{d^2}{ds^2}\frac1{s^2+6}=\frac6{(s^2+6)^2}-\frac{48}{(s^2+6)^3}$$
$$\frac d{ds}\frac s{s^2+6}=-\frac1{s^2+6}+\frac{12}{(s^2+6)^2}$$
$$\frac{d^2}{ds^2}\frac s{s^2+6}=\frac{2s}{(s^2+6)^2}-\frac{48s}{(s^2+6)^3}$$
$$\begin{align}\frac{d^3}{ds^3}\frac s{s^2+6}&=-\frac{1728}{(s^2+6)^4}+\frac{288}{(s^2+6)^3}-\frac6{(s^2+6)^2}\\
&=-\frac{1728}{(s^2+6)^4}-6\frac{d^2}{ds^2}\frac1{s^2+6}+\frac{30}{(s^2+6)^2}\\
&=-\frac{1728}{(s^2+6)^4}-6\frac{d^2}{ds^2}\frac1{s^2+6}+\frac52\frac d{ds}\frac s{s^2+6}+\frac52\frac1{s^2+6}\end{align}$$
Or
$$\frac{7e^{-6}}{(s^2+6)^4}=-\frac{7e^{-6}}{1728}\left\{\frac{d^3}{ds^3}\frac s{s^2+6}+6\frac{d^2}{ds^2}\frac1{s^2+6}-\frac52\frac d{ds}\frac s{s^2+6}-\frac52\frac1{s^2+6}\right\}$$
Now we can find
$$f(t)=-\frac{7e^{-6}}{1728}\left\{-t^3\cos(\sqrt6t)+\sqrt6t^2\sin(\sqrt6t)+\frac52\cos(\sqrt6t)-\frac{5\sqrt6}{12}\sin(\sqrt6t)\right\}$$
Just look at the method; don't trust the arithmetic.  
EDIT: As one might have suspected, there were arithmetic mistakes. After correction, my results now agree with Wolfram|Alpha.
</s>