I was wondering whether the below indefinite integral, $$\int\frac{1}{\sqrt{x^3+2x+1}}\,dx$$

is analytically solvable? If not, how could we prove that it cannot be written in terms of elementary functions?

I came across this question while discussing integrals with a friend, and whether or not these seemingly simple expressions can be integrated reasonably.

Naturally, some similar integrals are solvable, especially when the numerator lends itself well to substitution, but this does not seem to be the case here.

I am not very experienced with proofs of the non-elementary nature of integrals.

Any help is much appreciated.

  • 2
    $\begingroup$ Have you tried plugging it into wolfram alpha? That would give a good indication (not perfect, but good). $\endgroup$
    – Deepak
    May 18, 2021 at 0:58
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – Will Jagy
    May 18, 2021 at 0:59
  • 1
    $\begingroup$ Hello : it is possible to facilitate reopening of the question if you address the following queries (a) where did you encounter this integral? In a book/PDF/ conversation with friend/social media? (b) Have you solved similar-looking integrals before? Perhaps some with square roots in the denominator? How did you do those? (c) Have you seen proofs that anti-derivatives are NOT elementary? Like Liouville's theorem on elementary antiderivatives? Answer these questions in the question and call me. We can reopen the question, which will give it increased attention and better quality answers. $\endgroup$ May 18, 2021 at 13:36

1 Answer 1


When you face an integral $$I=\int \frac {dx}{\sqrt{x^3+\alpha x^2+\beta x+\gamma}}$$ you will, most of the time obtain very nasty expressions (such as in this case).

However, if you isolate the roots $(a,b,c)$ of the cubic polynomial and write $$I=\int \frac {dx}{\sqrt{(x-a)(x-b)(x-c)}}=-\frac{2 }{\sqrt{b- a}}F\left(\sin ^{-1}\left(\frac{\sqrt{b-a}}{\sqrt{x-a}}\right)|\frac{a-c}{a-b}\right)$$ where appears the elliptic integral of the first kind.

The same with quartics.


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