Laplace Transforms with Non-repeated Irreducible quadratic factors I'm solving the following differential equation with a Laplace Transform:
$$x″+ 9x = \cos(t) + \delta(t-\pi)$$
The initial conditions are that x(0) and x'(0) are equal to 0.
After after the transform, we obtain:
$$F(S) = \frac{S}{(S^2+9)[(S^2+1)+9]} + \frac{e^{-\pi\cdot t}}{S^2+9}$$
The problem is that there are two unique irreducible quadratic factors in the denominator of that first term: therefore I cannot seem to find a way to simplify this to find equivalent components which would make the inverse transform easier. How do I find the inverse of that first term?
$$\frac{S}{(S^2+9)[(S^2+1)+9]}$$
The answer includes simple cos(t) and cos(3t) terms with coefficients of $\frac{1}{8}$ and $\frac{-1}{8}$ respectively. What is the process of going from this complex laplace transform to something so simple?
 A: Alternative Method
$$
F(s)=\underbrace{\frac{s}{(s^2+1)(s^2+9)}}_{\mathbf{(I)}}+\underbrace{\frac{e^{-\pi s}}{s^2+9}}_{\mathbf{(II)}} 
$$
Alternative (and fast) method for $\mathbf{(I)}$:

We can also decompose $\mathbf{(I)}$ using partial fractions by the following way:
$$
\frac{s}{(s^2+1)(s^2+9)}\equiv\frac{Es+F}{s^2+1}+\frac{Gs+H}{s^2+9}
$$
Finding the constants of partial fractions decomposition
$$
\frac{s}{(s^2+1)(s^2+9)}\equiv\frac{(Es+F)\cdot(s^2+9)+(Gs+H)\cdot(s^2+1)}{(s^2+1)(s^2+9)}
$$
Now, working only with the numerators:
$$s\equiv (E+G)\cdot s^3+(F+H)\cdot s^2+(9E+G)\cdot s+(9F+H)$$
Thus, we can find $E$, $F$, $G$ and $H$ solving the following system of equations:
$$\mbox{$
\begin{cases}
\begin{alignat}{1}
E+G&=0 
\\F+H&=0 
\\9E+G&=1
\\9F+H&=0 
\end{alignat}
\end{cases} 
$}
\implies
\mbox{$
\begin{cases}
\begin{alignat}{1}
E&=1/8 
\\F&=0 
\\G&=-1/8
\\H&=0 
\end{alignat}
\end{cases}
$}
$$
Thus, we get:
$$
\frac{s}{(s^2+1)(s^2+9)}\equiv \frac{s}{8\cdot(s^2+1)}-\frac{s}{8\cdot(s^2+9)}
$$

Now we can apply the Inverse Laplace Transform:
Remember that the Laplace transform of cosine function is:
$$
\cos(\omega t) \iff \frac{s}{s^2+\omega^2}
$$
So, we have:
$$
\begin{alignat}{1}
\mathscr{L}^{-1}\left[\frac{s}{8\cdot(s^2+1)}-\frac{s}{8\cdot(s^2+9)}\right]&=\frac 1 8 \,\mathscr{L}^{-1}\left[\frac{s}{s^2+1^2}\right]-\frac 1 8 \,\mathscr{L}^{-1}\left[\frac{s}{s^2+3^2}\right]
\\&=\frac{1}{8}\left[cos(t)-cos(3t)\right]
\end{alignat}
$$

Same answer here
$$
\bbox[5px,border:1.1px solid black] {
x(t)=\frac{1}{8}\left[cos(t)-cos(3t)\right]-\frac 1 3\sin(3t)\cdot\operatorname{H}(t-\pi)
}
$$
