Series expansion with irrational power

After you define your function execute this line:

Series[f, {x, \[Infinity], 1}] // Normal // PowerExpand

Following your $x \gg a$ we can rewrite it as

% /. (-a + x) -> x // FullSimplify

which gives:

Now you have to be very careful dealing with derivative. It is not enough to keep 0th and 1st terms (as you did) - you will loose information. It is very interesting that to get correct result you need to keep as much as 4 terms:

Series[D[f, x], {x, \[Infinity], 3}] // Normal // PowerExpand // FullSimplify

Carefully examining the above expression with respect to $x \gg a$ we rewrite it as

% /. a -> 0 // FullSimplify

which gives:

And now it is correct because derivative of the function series is equal to the series of the derivative:

You can do a series expansion about $a=0$.

Series[f, {a, 0, 1}] // Normal // PowerExpand // Collect[#, x, Simplify] &

$$ -\frac{a}{2}+x+b x^{\frac{1}{12} \left(7-\sqrt{73}\right)}-\frac{\left(-91+\sqrt{73}\right) a b x^{-1+\frac{1}{12} \left(7-\sqrt{73}\right)}}{48 \left(6+\sqrt{73}\right)} $$

Series[D[f, x], {a, 0, 1}] // Normal // PowerExpand // Collect[#, x, Simplify] &

$$ 1-\frac{\left(191+43 \sqrt{73}\right) a b x^{-2+\frac{1}{12} \left(7-\sqrt{73}\right)}}{288 \left(6+\sqrt{73}\right)}-\frac{1}{12} \left(-7+\sqrt{73}\right) b x^{-1+\frac{1}{12} \left(7-\sqrt{73}\right)} $$

Then you can manually discard the $O(\frac{a}{x})$ correction to the $x^{-p}$ term.

I did a little bit more exploration, and found that if you series expand to higher order in $a$ then the extra terms are all $O((\frac{a}{x})^n)$ corrections to the $x^{-p}$ term.