Taylor Series

$\begin{aligned} \frac {x^4}{3-x^2} & = \frac {x^4}{(\sqrt 3-x)(\sqrt 3+x)} \\ & = \frac {x^3}2 \left(\frac 1{\sqrt 3-x}-\frac 1{\sqrt 3+x}\right) \\ & = \frac {x^3}{2\sqrt 3} \left(\frac 1{1-\frac x{\sqrt 3}}-\frac 1{1+\frac x{\sqrt 3}}\right) \\ & = \frac {x^3}{2\sqrt 3} \left[ \left(1+\frac x{\sqrt 3} + \frac {x^2}3 + \frac {x^3}{3\sqrt 3} + \frac {x^4}9 + \cdots \right) - \left(1-\frac x{\sqrt 3} + \frac {x^2}3 - \frac {x^3}{3\sqrt 3} + \frac {x^4}9 - \cdots \right) \right] & \small \color{#3D99F6} \text{for }|x| < \sqrt 3 \\ & = \frac {x^3}{2\sqrt 3} \left[ \frac {2x}{\sqrt 3} + \frac {2x^3}{3\sqrt 3} + \frac {2x^5}{9\sqrt 3} + \cdots \right] \\ & = x^3 \left[ \frac x3 + \frac {x^3}{3^2} + \frac {x^5}{3^3} + \cdots \right] \\ & = \sum_{n=1}^\infty \frac {x^{2n+2}}{3^n} \end{aligned}$

$\implies a+b = 2+3 = \boxed{5}$

Dividing the terms in both the numerator and the denominator of the given rational expression by 3 reveals it to be a closed form of an infinite geometric series.