If you just do simple algebra, and use (x+a)(x-a)=x^2-a^2 on the original poster's "more horrifying expression," a lot of terms cancel and you are left with:
E=2kqa^2\frac{3x^2-a^2}{x^2(x^2-a^2)^2}
This is still exact, no approximations. As x\rightarrow\infty:
E=2kqa^2\frac{3}{x^4}
which is...