# Evaluating an indefinite integral $\int \frac{dx}{\sqrt{x^3+2x+3}}$

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.

• Are you sure that it's possible in elementary functions? Feb 14, 2019 at 21:01
• @MichaelRozenberg It's not. Feb 14, 2019 at 21:02
• I do not have any idea. Feb 14, 2019 at 21:02
• @user149418 This integral does not have an indefinite solution as an elementary function. Where (or how) did you come across this problem ? Feb 14, 2019 at 21:03
• @Rebellos You are welcome to show us your proof of your statement. Feb 14, 2019 at 21:04

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 $$\int_{-1}^y\frac1{\sqrt{x^3+2x+3}}\,dx=5^{-1/4}F\left(\cos^{-1}\frac{\sqrt5-1-y}{\sqrt5+1+y},m=\frac{10+3\sqrt5}{20}\right)$$