Since $5t^{2}\sin\!\left(at\right)+9t^{2}e^{mt} = t^2\left(5\sin\!\left(at\right)+9e^{mt}\right)$, to find the Laplace transform of $5t^{2}\sin\!\left(at\right)+9t^{2}e^{mt}$ you take the second derivative of .
Therefore $L\lbrace 5t^{2}\sin\!\left(at\right)+9t^{2}e^{mt}\rbrace (s) =$ .