How to calculate inverse Laplace transform of $\frac{7s^2 + 3s +5}{(s^2-4s+29)(s^2+25)}$?

$$\mathcal{L}^{-1} \left\{ \frac{7s^2 + 3s +5}{(s^2-4s+29)(s^2+25)} \right\} = \text{?}$$

Anyone know how to calculate the inverse Laplace of this behemoth? The only approaches to calculating inverse Laplace transforms I know of are:

-factoring the denominator completely

-usig partial fractions to split up fraction

-adjusting form using algebraic manipulations (e.g. completing the square)

And after doing all that I'd hope that the form of the fraction matches that of one of the functions that are in my table. But doing these things doesn't seem to be effective for this fraction (doesn't put it into a known form), so I'm lost. I'm guessing there is some obscure manipulation I can do, but I can't find it

Edit: partial fractions gives:

$$\frac{167s+305}{104(s^2 - 4s +29)} - \frac{167s+245}{104(s^2+25)}$$Not sure where to go from there...

• After partial fractions and completing the squares, you should be able to get it. Are you aware of the Laplace Transforms of $e^{at}\sin bt$ and $e^{at}\cos bt$? Jun 20 '20 at 17:41
• I know of those transformations yes, but I'm not sure where to go from the partial fraction stage. I've edited my post to show the very messy result I get from partial fractions. I could use the functions you suggested for the second term, but not for the first term, even after completing the square? Jun 20 '20 at 18:20
• The first term that you have written can be written as $$\dfrac{167(s-2)}{104((s-2)^2 + 5^2)} + \dfrac{639}{104((s-2)^2 + 5^2)}.$$ Can you see how you can handle each term above separately? Jun 20 '20 at 18:25

We need to go a little further with the partial fractions. Note that$$\frac{As+B}{s^2+5^2}=\frac{A/2-iB/10}{s-5i}+\frac{A/2+iB/10}{s+5i}$$has inverse Laplace transform$$(A/2-iB/10)e^{5it}+(A/2+iB/10)e^{-5it}=A\cos5t+\frac{B}{5}\sin5t.$$Since $$s^2-4s+29=(s-2)^2+5^2$$, the other partial fraction admits a similar analysis.