Taylor expansion of charge density in Jackson's book

I am learning from Jackson (3r edition), where I found one concept very confusing, that is Taylor expansion of charge density. (This is given in section "1.7 Poisson and Laplace equations" p.n:35)

I will write some equations first.

$$ {Phi}_a(x) = frac{1}{4{pi}{epsilon_0}}int frac{{rho}(x’)}{sqrt{(x – x’)^2 + a^2}}d^3x’ $$

Now we want to find out potential such that a tends to zero.
$$ nabla^2{Phi}_a(x) = frac{1}{4{pi}{epsilon_0}}int{rho}(x’) nabla^2frac{1}{(r^2 + a^2)^frac{1}{2}}d^3x’ $$

$$ nabla^2{Phi}_a(x) = -frac{1}{4{pi}{epsilon_0}}int{rho}(x’) frac{3a^2}{(r^2 + a^2)^frac{5}{2}}d^3x’ $$

I understood this until above step, but now I didn’t got the next step

$$ nabla^2{Phi}_a(x) = -frac{1}{{epsilon_0}}int_0^R frac{3a^2}{(r^2 + a^2)^frac{5}{2}} left[{rho}(x) + frac{r^2}{6}nabla^2{rho} + ……. right]r^2 dr + O(a^2) $$

Jackson says that we gonna expand ${rho}(x’)$ around x’ = x., but the expansion of ${rho}(x’)$ should also contain the first order derivative of ${rho}(x’)$ like $nabla{rho}$, also the taylor expansion of the second term should contain 2 at the denominator but it’s 6, and how the last term of $O(a^2)$

So what I am thinking is that Taylor expansion should be $left[{rho}(x) + rnabla{rho} + frac{r^2}{2}nabla^2{rho}right]$

I know I am wrong but I don’t know what is the answer. Any help is appreciated.