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

Evaluate the following integral $$$$J = \int \frac{dx}{\sqrt{x^3+2x+3}}$$$$

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? – Michael Rozenberg Feb 14 '19 at 21:01
• @MichaelRozenberg It's not. – Rebellos Feb 14 '19 at 21:02
• I do not have any idea. – user149418 Feb 14 '19 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 ? – Rebellos Feb 14 '19 at 21:03
• @Rebellos You are welcome to show us your proof of your statement. – Michael Rozenberg Feb 14 '19 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 $$\int_{-1}^y\frac1{\sqrt{x^3+2x+3}}\,dx=5^{-1/4}F\left(\cos^{-1}\frac{\sqrt5-1-y}{\sqrt5+1+y},m=\frac{10+3\sqrt5}{20}\right)$$