The function to be expanded is: $$f(x) = \frac{x}{x-a\sqrt{1-e^{bx^2}}}$$ I would like to expand it up to second order terms, something like $c_0+c_1 x+c_2 x^2$. $x$ is not a relatively small quantity. $a$ and $b$ are real numbers (negative values).
Related information: this is a variant of getting formulae (14) (15) from (13) in Laskin's paper doi: 10.1117/12.2063388, and the author uploaded preprint here: https://www.researchgate.net/publication/266385491_Designing_refractive_beam_shapers_via_aberration_theory
I tried to get the Maclaurin term but the derivatives are cumbersome (mathDF.com may help with getting derivatives) and the square root part is annoying (didn't get through it by L'Hôpital's rule).
To avoid sign problems, I think that it is easier to consider $$f(x) = \frac{x}{x+a\sqrt{1-e^{-bx^2}}}$$ where $a$ and $b$ are positive and above all that $x$ is positive.
Using steps $$1-e^{-bx^2}=b x^2-\frac{b^2 }{2}x^4+\frac{b^3 }{6}x^6+O\left(x^{8}\right)$$ Using binomial expansion $$\sqrt{1-e^{-bx^2}}=b^{1/2} x-\frac{ b^{3/2}}{4} x^3+\frac{5b^{5/2}}{96} x^5+O\left(x^6\right)$$ $$x+a\sqrt{1-e^{-bx^2}}= \left(1+a \sqrt{b}\right)x-\frac{a b^{3/2}}{4} x^3 +\frac{5 a b^{5/2}}{96} x^5+O\left(x^6\right)$$ Long division leads to $$f(x)=\frac{1}{1+a \sqrt{b}}+\frac{a b^{3/2} }{4 \left(a \sqrt{b}+1\right)^2}x^2+\frac{a b^{5/2} \left(a \sqrt{b}-5\right)}{96 \left(a \sqrt{b}+1\right)^3}x^4+O\left(x^5\right)$$