In a course, my teacher told us that the following integral is convergent and used the comparison test to prove it; my question is how to find the antiderivative in closed form? It seems to exist; if, however, it doesn't exist, can someone prove it?

$$\int\sqrt{\dfrac1{1+x^3}}\mathrm dx$$

  • $\begingroup$ What do you mean by "exists"? To clarify, the antiderivative is a function, you can compute its values at various points (so long as you specify $C$), but not a so-called "elementary" function (the kind of functions you are probably used to). $\endgroup$ Apr 17, 2011 at 4:53
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    $\begingroup$ If anybody's interested, I'll write up the way to derive the solution in terms of elliptic integrals. Mathematica's results are damned messy... $\endgroup$ Apr 17, 2011 at 5:02
  • $\begingroup$ @J.M.- I will be more than interested to see this integral in action in an analytical method of solving. Thanks. $\endgroup$
    – night owl
    Apr 17, 2011 at 5:16
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    $\begingroup$ I'll give the details later... but here's the analytical solution: $$\frac1{\sqrt[4]{3}}F\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+x}-1\right)\mid\frac{2+\sqrt{3}}{4}\right)$$ $\endgroup$ Apr 17, 2011 at 5:36
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    $\begingroup$ @awllower: The final result is expressed in terms of "elliptic integrals", not "elliptic functions"; however, one can use elliptic functions to derive the elliptic integral result (which is what I did). $\endgroup$ Apr 17, 2011 at 16:19

3 Answers 3


The first thing to do is to note that


(one real and two complex conjugate roots). Using Jacobian elliptic functions requires having a quartic within the square root (the alternative of using Weierstrass elliptic functions is fine with square roots of cubics, but I'll leave that approach to someone else); the good thing is that by choosing a proper Möbius transformation, one can turn a cubic into a quartic (the algebraic geometers here might want to say a bit more than I have).

For the integral in question, the Möbius substitution needed is $x=\frac{-1+\sqrt{(-1)^2-(-1)+1}+(-1-\sqrt{(-1)^2-(-1)+1})v}{1+v}=\frac{2\sqrt{3}}{1+v}-(1+\sqrt{3})$; we then have

$$\int\frac{\mathrm dx}{\sqrt{x^3+1}}=-2\int\frac{\mathrm dv}{\sqrt{(1-v^2)(2\sqrt{3}-3+(2\sqrt{3}+3)v^2)}}$$

At this point, making use of the Jacobian elliptic function identity $\mathrm{sn}^2(u|m)+\mathrm{cn}^2(u|m)=1$ (nothing more than the usual Pythagorean identity in elliptic function garb), we could make either of the substitutions $v=\mathrm{sn}(u|m)$ or $v=\mathrm{cn}(u|m)$. The latter is a bit more convenient, since $\mathrm dv=-\mathrm{sn}(u|m)\mathrm{dn}(u|m)\mathrm du$, which can conveniently get rid of the negative sign in the integral. Thus, the integral turns into

$$2\int\frac{\mathrm{sn}(u|m)\mathrm{dn}(u|m)\mathrm du}{\sqrt{(1-\mathrm{cn}^2(u|m))(2\sqrt{3}-3+(2\sqrt{3}+3)\mathrm{cn}^2(u|m))}}$$

or (by using the Pythagorean identity)

$$2\int\frac{\mathrm{dn}(u|m)\mathrm du}{\sqrt{2\sqrt{3}-3+(2\sqrt{3}+3)\mathrm{cn}^2(u|m)}}$$

Here, one now chooses a proper value of $m$ such that the integrand reduces to a constant. Skipping the details, we let $m=\frac{2+\sqrt{3}}{4}$ such that

$$2\int\frac{\mathrm{dn}(u|m)\mathrm du}{\sqrt{2\sqrt{3}-3+(2\sqrt{3}+3)\mathrm{cn}^2(u|m)}}=\int\frac{\mathrm du}{\sqrt[4]{3}}$$

To undo the substitutions, we note that $u=F(\arccos(v)|m)$ and $v=\frac{2\sqrt{3}}{1+\sqrt{3}+x}-1$, giving the final result

$$\int\frac{\mathrm dx}{\sqrt{x^3+1}}=\frac1{\sqrt[4]{3}}F\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+x}-1\right)\mid\frac{2+\sqrt{3}}{4}\right)+C$$

This result can be verified by differentiating the right hand side (remember that $\frac{\mathrm d}{\mathrm d\phi}F(\phi|m)=\frac1{\sqrt{1-m\sin^2\phi}}$) and noting that it is the same as the integrand.

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    $\begingroup$ Could you recommend a book containing the special functions and integration techniques you're using? These things have always been sort of a mystery to me. $\endgroup$
    – t.b.
    Apr 18, 2011 at 9:45
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    $\begingroup$ @Theo: at least for the elliptic integrals, I got my bag of tricks from Byrd/Friedman; I have to admit I'm still actively learning the tricks myself since one of my recent projects involves a great deal of elliptic integral manipulations. $\endgroup$ Apr 18, 2011 at 9:48
  • $\begingroup$ Also, I have been told that it might be more profitable to do manipulations with the Carlson symmetric elliptic integrals instead, but I have yet to study the requisite papers by Carlson. $\endgroup$ Apr 18, 2011 at 9:52
  • $\begingroup$ Thanks a lot. I'll check this book out, the big G doesn't let me have a peek, so I've ordered it in the library. Good luck with your project! (Ah, I didn't see your second comment before posting, thanks for that, too.) $\endgroup$
    – t.b.
    Apr 18, 2011 at 9:55
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    $\begingroup$ @jm324354, "If I could choose one area in mathematics to pursue it would probably be solving tough integrals." - sometimes the pursuit of a closed form is worthwhile, and sometimes it isn't. It depends. Nevertheless, in the case of the elliptic integrals, it's often worthwhile. $\endgroup$ May 1, 2015 at 13:46

For positive values of $x$, the integral is solved by $$ \int\sqrt{\dfrac1{1+x^3}}\mathrm \;dx = x \;_2F_1\left(\frac{1}{3},\frac{1}{2};\frac{4}{3},-x^3\right) $$ which contains a hypergeometric function. In general for positive $a$ $$ \int\sqrt{\dfrac1{a+x^3}}\mathrm \;dx = \frac{x}{\sqrt{a}} \;_2F_1\left(\frac{1}{3},\frac{1}{2};\frac{4}{3},-\frac{x^3}{a}\right) $$

To get this use a Mellin transform under the integral by introducing a virtual parameter (my pet adaptation of the Feynman trick): $$ \mathcal{M}_{a \to s}\left[\sqrt{\dfrac1{a+x^3}}\right] = \frac{\left(\frac{1}{x^3}\right)^{-s} \Gamma \left(\frac{1}{2}-s\right) \Gamma (s)}{\sqrt{\pi } \sqrt{x^3}} $$ $$ \int\mathcal{M}_{a \to s}\left[\sqrt{\dfrac1{a+x^3}}\right] \mathrm\; dx = \frac{x \left(\frac{1}{x^3}\right)^{-s} \Gamma \left(\frac{1}{2}-s\right) \Gamma (s)}{\sqrt{\pi } \left(3 s-\frac{1}{2}\right) \sqrt{x^3}} $$ Then take the inverse Mellin transform, and simplify $$ \mathcal{M}^{-1}_{s\to a}\left[ \frac{x \left(\frac{1}{x^3}\right)^{-s} \Gamma \left(\frac{1}{2}-s\right) \Gamma (s)}{\sqrt{\pi } \left(3 s-\frac{1}{2}\right) \sqrt{x^3}}\right] = \frac{x}{\sqrt{a}} \;_2F_1\left(\frac{1}{3},\frac{1}{2};\frac{4}{3},-\frac{x^3}{a}\right) $$

  • $\begingroup$ How did you find/prove these results? $\endgroup$ Oct 2, 2021 at 20:43
  • $\begingroup$ @Dr.WolfgangHintze I have updated the answer. Thanks for your input! $\endgroup$ Oct 7, 2021 at 10:53

Maybe it's off topic but I'd like to give the answer of Mathematica to the question "how to find the antiderivative in closed form, and how to prove it"

The antiderivative calculated by Mathematica is given by

ad = Integrate[1/Sqrt[1 + x^3 ], x]

Out[347]= (2 (-1)^(1/6) Sqrt[-(-1)^(1/6) ((-1)^(2/3) + x)] Sqrt[
 1 + (-1)^(1/3) x + (-1)^(2/3) x^2]
  EllipticF[ArcSin[Sqrt[-(-1)^(5/6) (1 + x)]/3^(1/4)], (-1)^(1/3)])/(3^(
 1/4) Sqrt[1 + x^3])

The result consists of an elliptic integral of the first kind and some square root factors. This looks a bit messy at first sight but it turns out that the square root factor cancel out.

Mathematica's (Full)Simplify does not simplify the expression, but my trick is to simplify the square of the expression in question.


Out[381]= (4 (-1)^(1/6)
  EllipticF[ArcSin[Sqrt[-(-1)^(5/6) (1 + x)]/3^(1/4)], (-1)^(1/3)]^2)/Sqrt[3]

and take the square root afterwards (by hand). Then the antiderivative is

ad1 = (2 (-1)^(1/12)/3^(1/4)) EllipticF[
   ArcSin[Sqrt[-(-1)^(5/6) (1 + x)]/3^(1/4)], (-1)^(1/3)]

Mathematica's help tells us "EllipticF is the inverse of JacobiAmplitude. If $\phi=\mathrm{am}(u|m)$ then $u=F(\phi|m)$". Hence ad1 can can be expressed differently (as J. M. has ingeniously elaborated here).

Ok, now the proof, that ad1 is in fact an antiderivative (up to an additive constant, of course), differentiating ad1 with respect to x gives

D[ad, x]; (* -> horrible expression *)

Simplifying this, using my "square sandwiching trick" again, gives


Out[392]= 1/(1 + x^3)

which completes the proof.


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