Rewrite rational function $f(x)$ as a series if the quadratic expression in the denominator has no roots A function of the type 
$$f(x)=\frac{ex+f}{ax^2+bx+c}$$
with $b^2-4 a c \geq 0$ can be written as a series using partial fraction decomposition and geometric series.
But if one has the same function with $b^2-4 a c < 0$? Partial fraction decomposition does not help and geometric series neither, so what is the strategy to rewrite $f(x)$ as a series in that case?

Example
$$f(x)=\frac{x+1}{x^2+x+3}$$
 A: Partial fraction decomposition still works, just that the roots are now a conjugate pair of complex numbers, $$ax^2+bx+c=a(x-z)(x-\bar z)=c(1-\frac{x}z)(1-\frac{x}{\bar z}).$$
In the end you get a formula
$$
\frac{A}{1-\frac{x}z}+\frac{\bar A}{1-\frac{x}{\bar z}}=2\Re\left(\sum_{k=0}^\infty\frac{A}{z^k}x^k\right)
$$
where for each term there is the conjugate complex term so that in the end the result is again real.
A: More than likely, too simplistic !
You wrote the function $$F(x)=\frac{ex+f}{ax^2+bx+c}$$ Since you look for an expansion around $x=0$, let us rewrite it as $$F(x)=\frac{f+ex}{c+bx+ax^2}$$ and now use the long division to get $$F(x)=\frac{f}{c}+\frac{ c e-b f}{c^2}x+\frac{ b^2f-a c f-b c
   e}{c^3}x^2+O\left(x^3\right)$$ I did not write the next terms because the expression becomes too long.
Using your example, we should get $$\frac{x+1}{x^2+x+3}=\frac{1+x}{3+x+x^2}=\frac{1}{3}+\frac{2 x}{9}-\frac{5 x^2}{27}-\frac{x^3}{81}+\frac{16
   x^4}{243}+O\left(x^5\right)$$ I am sure that you notice the powers of $3$ as denominators of the coefficients.
Let us try an easy value $x=\frac 1{10}$. The exact value is $\frac{110}{311}\approx 0.3536977492$ while the expansion gives $\frac{429743}{1215000}\approx 0.3536979424$.
