Fast partial-fraction decomposition I'm studying Laplace transformations for my differential equations class and typically there's a partial fraction decomposition involved, which can be very long and demanding for calculations by hand, if done the standard way. 
I am aware of some of the tricks used to speed-up this procedure (like the usage of limits at infinity, or multiplying with denominators and taking particular values of $s$), however I am not able to apply them at this example:
$$\frac {s} {(s^2+2s+5)(s^2+4)} = \frac {\alpha s + \beta} {s^2+2s+5} + \frac {\gamma s + \delta} {s^2+4}$$
For example, if I attempt to extract a relation for $\gamma$ and $\delta$, by multiplying with $s^2+4$ and taking $s=2i$, I get and equation that involves complex numbers; that makes me feel I have not gained much in terms of number of operations.
Are there any better tricks for this example?
 A: Well, the following is certainly a trick. Whether it's better or not remains to be decided....
First, I want to make the factors in the denominator symmetric with respect to the origin (right now they're of the form $s^2+4$ and $(s+1)^2+4$, hence symmetric with respect to $1/2$). So setting $s=t-1/2$, we have the equivalent problem of doing partial fractions on
$$
\frac{16 t-8}{(4 t^2-4t+17)(4 t^2+4t+17)}.
$$
Let's start by noting (using the symmetry) that
$$
\frac{1}{4 t^2-4t+17} - \frac{1}{4 t^2+4t+17} = \frac{8t}{(4 t^2-4t+17)(4 t^2+4t+17)}.
$$
How can we divide everything in sight by $t$? Notice that the inverse of $t$ modulo $4t^2-4t+17$ is $(-4t+4)/17$, while the inverse of $t$ modulo $4t^2+4t+17$ is $(-4t-4)/17$. So we get
\begin{align*}
\frac{8}{(4 t^2-4t+17)(4 t^2+4t+17)} &= \frac{1/t}{4 t^2-4t+17} - \frac{1/t}{4 t^2+4t+17} \\
&\equiv \frac{-4t+4}{17(4 t^2-4t+17)} + \frac{4t+4}{17(4 t^2+4t+17)} \pmod1.
\end{align*}
But this (mod 1)-equivalence must actually be an equality(!): by the theory of partial fractions, the left-hand side (whose numberator has larger degree than the denominator) has a representation in the form of the right-hand side.
Between the last two identities, we see that
$$
\frac{16 t-8}{(4 t^2-4t+17)(4 t^2+4t+17)} = \frac{42 t+30}{17 (4 t^2-4 t+17)}-\frac{4t+38}{17 (4 t^2+4 t+17)};
$$
now plugging $t=s+1/2$ back in yields
$$
\frac s{(s^2+2s+5)(s^2+4)} = \frac{-s-10}{17 (s^2+2 s+5)}+\frac{s+8}{17 (s^2+4)}.
$$
A: Let's apply the Heaviside cover-up method, nonlinear version, as described by Bill Dubuque.
$$\rm \frac{x}{(x^2\!+2x+5)(x^2\!+4)} \ =\ \frac{ax+b}{x^2\!+2x+5}\, +\ \frac{cx+d}{x^2\!+4}$$
Clearing denominators yields
$$\rm x\, =\, (x^2\!+4)(ax+b)\, +\, (x^2\!+2x+5)(cx+d) $$
Evaluating this mod $\rm\ x^2\! +\! 4\,\: $ i.e.$\:$ iteratively applying the rewrite rule $\rm\ x^2 \to -4\,\ $ yields
$$\rm x\, =\, (c\!+\!2d)\ x\,+\, d\!-\!8c\ \ \Rightarrow\ \ c = 1/17,\,\ d= 8/17$$
Evaluating it mod $\rm\ x^2\! +\! 2x\!+\!5\:,\: $ i.e.$\:$ iteratively applying  rewrite rule $\rm\ x^2 \to -2x\!-\!5\: $ yields
$$\rm x\, =\, (3a\!-\!2b)\ x \,+\, 10a\!-\!b\ \ \Rightarrow\ \ a = -1/17,\,\ b = -10/17$$
That seems easy enough to me.
A: I know this is a late answer but I thought it would be useful. First, you can use cover up method by plugging in $x=2i$. You end up getting $$\frac{(2i+8)}{17}= C(2i)+D\implies  \frac{2i}{17}= C(2i) \implies  C=\frac1{17}\text{ and }D=\frac8{17}.$$ Easy as that!
To get $A$ and $B$, use cover up with $x= -1+2i$ which is one of the complex roots of $x^2+2x+5$. So $$\frac{-1+2i}{(-1+2i)^2+4} = A(-1+2i)+B \implies  \frac{-1+2i}{1-4i}= A(-1+2i)+B.$$
Finally multiply by conjugate and we get 
$$\frac{(-1+2i)(1+4i)}{17}= A(-1+2i)+B\implies\frac{-9-2i}{17}= B-A+2iA \implies  -\frac2{17}=2iA \implies A=-\frac1{17}$$
and so $B-A= -\frac9{17}\implies  B= -\frac1{17}-\frac9{17}=-\frac{10}{17}$. Hence $A=-\frac1{17}$ and $B=-\frac{10}{17}$.
