Compute $\int \frac{x+3}{\sqrt{x^3+1}} dx$

Question

Compute $\int \frac{x+3}{\sqrt{x^3+1}} dx$

Was thinking about the taylor expansion of $1/\sqrt{x^3+1}$ but is also rather hairy. Any other tricks?

Answer 1

By Byrd & Friedman 239.07 and 239.00 $$I=\int_{-1}^y\frac{x+3}{\sqrt{x^3+1}}\,dx=g\left((-1-\sqrt3)F(\varphi,m)+2\sqrt3\int_0^{F(\varphi,m)}\frac1{1+\operatorname{cn}u}\,du+3F(\varphi,m)\right)$$ where the parameter $m$ is $\frac12+\frac{\sqrt3}4$, $\cos\varphi=\operatorname{cn}u=\frac{\sqrt3-1-y}{\sqrt3+1+y}$ and $g=3^{-1/4}$. B&F 341.53 solves the second integral, yielding after some simplification $$I=g((2+\sqrt3)F(\varphi,m)-2\sqrt3E(\varphi,m))+\frac{2\sqrt{y^3+1}}{\sqrt3+1+y}$$

Answer 2

Using series is not bad since $$\frac{1}{\sqrt{x^3+1}}=\frac 1{\sqrt{\pi } }\sum_{n=0}^\infty (-1)^n\frac{ \left(n-\frac{1}{2}\right)! }{ n!} x^{3n}$$ So $$\frac{x+3}{\sqrt{x^3+1}}=\frac 1{\sqrt{\pi } }\sum_{n=0}^\infty (-1)^n\frac{ \left(n-\frac{1}{2}\right)! }{ n!} x^{3n+1}+\frac 3{\sqrt{\pi } }\sum_{n=0}^\infty (-1)^n\frac{ \left(n-\frac{1}{2}\right)! }{ n!} x^{3n}$$ Just integrate termwise and this gives you the expansion of some nasty elliptic integral.