We have to show that $\mathcal{L}\{t^2e^{-3t}\cos{at}\}=-\frac{s^3-9s^2+(3a^2+9)s+5a^2+33}{(s^2+2s+a^2+9)^3}$.
Steps : $\mathcal{L}\{t^2\cos(at)\}=(-1)^2\frac{d^2}{ds^2}\left(\frac{s}{s^2+a^2}\right)=\frac{d}{ds}\left(\frac{(s^2+a^2)\cdot 1- s\cdot 2s}{(s^2+a^2)^2}\right)=\frac{d}{ds}\frac{a^2-s^2}{(s^2+a^2)^2}=\frac{(s^2+a^2)^2(-2s)-(a^2-s^2)\cdot 2(s^2+a^2)\cdot 2s}{(s^2+a^2)^4}$ How can I find the remaining part?
Let be $f(t)=e^{-3t}\cos(at)$ so that $$ \mathcal{L}\{f(t)\}=F(s)=\frac{s+3}{(s+3)^2+a^2} $$ and then $$\mathcal{L}\{t^2e^{-3t}\cos(at)\}=\mathcal{L}\{t^2f(t)\}=(-1)^2F''(s)=\frac{2 (s+3)\left[ (s+3)^2-3 a^2)\right]}{\big[a^2+(s+3)^2\big]^3}$$