<s>[INST] Integral of $\sqrt{x^3 + 8}$? I have issues solving the following integral:
$$\int\sqrt{x^3+8}~dx$$
I tried substitution and integration by parts, but with no use. 
I'm guessing I have to use some trigonometric substitution. 
Can anybody help solve this integral?
 [/INST] 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 $|x|\geq2$ ,
$$\begin{array}\int\sqrt{x^3+8}\,dx
&=\int x^\frac{3}{2}\sqrt{1+\dfrac{8}{x^3}}\,dx\\
&=\int x^\frac{3}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!8^n}{4^n(n!)^2(1-2n)x^{3n}}\,dx\\
&=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^nx^{\frac{3}{2}-3n}}{(n!)^2(1-2n)}dx\\
&=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^nx^{\frac{5}{2}-3n}}{(n!)^2(1-2n)\left(\dfrac{5}{2}-3n\right)}+C\\
&=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^{n+1}}{(n!)^2(2n-1)(6n-5)x^{3n-\frac{5}{2}}}+C\end{array}$$
</s>