# How to find the indefinite integral of $\arctan((x^2+1)^{1/3})$?

I have to find the indefinite integral of $$\arctan((x^2+1)^{1/3})$$. I have tried to do different variable substitutions, then integration by parts, but in the end I always got a another integral that I could not calculate. Please give a hint how to calculate this integral.

• I'm not entirely certain but there doesn't seem to be a result for the indefinite integral in terms of standard mathematical functions. Commented Mar 7, 2021 at 19:08
• What do you mean with "have to find"? Many integrals cannot be evaluated in terms of elementary functions. This seems to be one of them.
– user
Commented Mar 7, 2021 at 19:30
• A CAS says it's a mess of Appell and Elliptic functions. Not likely there is a nice representation in elementary functions
– Sal
Commented Mar 7, 2021 at 19:45

If you do $$x=y^3-1$$ and $$\mathrm dx=3y^2\,\mathrm dy$$, then that indefinite integral becomes$$\int3y^2\arctan(y)\,\mathrm dy.$$You can deal with it by integration by parts$$\int3y^2\arctan(y)\,\mathrm dy=y^3\arctan(y)-\int\frac{y^3}{y^2+1}\,\mathrm dy.$$Can you take it from here?
• Should not it be $x^2=y^3-1$?
Change variables $$y^3 = 1+x^2$$ to get $$\int \arctan(\sqrt[3]{x^2+1}) \, dx = \frac{3}{2}\int\frac{y^2\arctan y}{\sqrt{y^3-1}}\;dy$$ Integrate by parts to get $${\frac {\arctan \left( y \right) \left( y-1 \right) \left( {y}^{2}+y +1 \right) }{\sqrt {{y}^{3}-1}}}-\int \!{\frac { \left( y-1 \right) \left( {y}^{2}+y+1 \right) }{ \left( {y}^{2}+1 \right) \sqrt {{y}^{3} -1}}}\,{\rm d}y \\= { {\arctan \left( y \right) \sqrt{ y^3-1 } }}-\int \!{\frac { \sqrt{y^3-1} }{ {y}^{2}+1 }}\,{\rm d}y$$ That is an elliptic integral. In Maple it is about a page long, in terms of $$F$$ and $$\Pi$$.
The best you can do is say that $$\int \arctan(\sqrt[3]{x^2+1}) \, dx = \int_{0}^{x} \arctan(\sqrt[3]{t^2+1}) \, dt+C \, .$$ This result follows directly from the fundamental theorem of calculus.