Integrating $\int (t+t^2)^\frac{1}{3}\,dt$

Question

Question: Find the integral below$$\int (t+t^2)^\frac {1}{3}\,\mathrm dt$$

I have tried to use the standard methods such as "by part" to solve this integral, only to end up in an algebraic mess.

Can someone suggest a method that would suffice in solving this and/or otherwise post their worked solution.

it might not be possible to find a known function whose derivative is your integrand . — user439545, Oct 08 '17 at 22:22
I'd try a change of variables with $x=t-\frac12$. To integrate $\left(x^2 -\frac14\right)^{1/3}$ I'd do a trig substitution; you might need to use the tangent of half-angle trick. — Mark Fischler, Oct 08 '17 at 22:22
It looks like it does not have an elementary primitive, according to WolframAlpha — Miguel, Oct 08 '17 at 22:23
@MarkFischler If I invent a name for this function $f(x)=\int_0^x (t+t^2)^{\frac{1}{3}} dt$, does it also count? (LOL LOL) — Miguel, Oct 08 '17 at 22:30
@Miguel Haha, no. The difference between hypergeometric functions and yours is simple: Is there any motivation to have hypergeometric functions for other purposes? Is there any motivation to have your "function" for any other purposes other than this problem? — Simply Beautiful Art, Oct 09 '17 at 00:18

score 0 · Accepted Answer · answered Oct 08 '17 at 22:24

There is no reason to think this integral is an elementary function.

Maple evaluates it in terms of a hypergeometric function $$ \int \!\sqrt [3]{{t}^{2}+t}\;{\rm d}t=\frac{3}{4}\,{t}^{4/3}\; {\mbox{$_2$F$_1$}\left(-\frac{1}{3},\frac{4}{3};\,\frac{7}{3};\,-t\right)} $$

Integrating $\int (t+t^2)^\frac{1}{3}\,dt$

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