Integral of $\sqrt{x^3 + 8}$?

For any real number $x$:

When $|x|\leq2$ ,

$$\begin{array}\int\sqrt{x^3+8}\,dx &=\int2\sqrt2\sqrt{\dfrac{x^3}{8}+1}\,dx\\ &=\int\sum\limits_{n=0}^\infty\dfrac{2\sqrt2(-1)^n(2n)!x^{3n}}{8^n4^n(n!)^2(1-2n)}\,dx\\ &=\int\sum\limits_{n=0}^\infty\dfrac{2\sqrt2(-1)^n(2n)!x^{3n}}{32^n(n!)^2(1-2n)}\,dx\\ &=\sum\limits_{n=0}^\infty\dfrac{2\sqrt2(-1)^n(2n)!x^{3n+1}}{32^n(n!)^2(1-2n)(3n+1)}+C\end{array}$$

When $|x|\geq2$ ,

$$\begin{array}\int\sqrt{x^3+8}\,dx &=\int x^\frac{3}{2}\sqrt{1+\dfrac{8}{x^3}}\,dx\\ &=\int x^\frac{3}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!8^n}{4^n(n!)^2(1-2n)x^{3n}}\,dx\\ &=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^nx^{\frac{3}{2}-3n}}{(n!)^2(1-2n)}dx\\ &=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^nx^{\frac{5}{2}-3n}}{(n!)^2(1-2n)\left(\dfrac{5}{2}-3n\right)}+C\\ &=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^{n+1}}{(n!)^2(2n-1)(6n-5)x^{3n-\frac{5}{2}}}+C\end{array}$$