Evaluate the following integral \begin{equation} J = \int \frac{dx}{\sqrt{x^3+2x+3}} \end{equation}
I do not find suitable substitution to compute the above indefinite integral. Since $x^3+2x+3=(x+1)(x^2-x+3)$, substituting $z=\sqrt{x+1}$, we have $$J= 2\int \frac{dz}{\sqrt{z^4-3z^2+5}}.$$ I think this is not a good substitution. Any help is appreciated.
Since the integral is just a cubic raised to the power $-\frac12$ and the cubic has no repeated roots, it is not elementary but may be expressed in terms of elliptic integrals.
Suppose we want to find the integral from the pole at $-1$ to some number $y$ greater than that – the indefinite integral from $0$ follows from this easily. The cubic's factorisation as given in the question is $x^3+2x+3=(x+1)(x^2-x+3)=(x+1)((x-1/2)^2+11/4)$, so Byrd and Friedman 239.00 gives for this $$\int_{-1}^y\frac1{\sqrt{x^3+2x+3}}\,dx=5^{-1/4}F\left(\cos^{-1}\frac{\sqrt5-1-y}{\sqrt5+1+y},m=\frac{10+3\sqrt5}{20}\right)$$