# Integrating $\int (t^2+7)^{\frac{1}{3}}$dt

I have to integrate $\int (t^2+7)^{\frac{1}{3}}$dt

I tried some trigonometric substitution and then rationalisation but didn't get anything

Please provide a hint on how to proceed with this question

• I don't think you can expect an elementary anti-derivative. Are you sure you're supposed to find one (by hand)? Jan 24 '18 at 15:29
• Are you sure about the power $\frac 13$ ? This makes the problem very difficult (no closed form solution). I bet for $\frac 12$. Jan 24 '18 at 15:30
• No it is 1/3 for sure Jan 24 '18 at 15:30
• Do you want to bet ? I am sure that, once more, there is a typo in a textbook. Jan 24 '18 at 15:31
• Any answer is appreciated@StackTD Jan 24 '18 at 15:32

Setting $$t=\sqrt{7}\cosh x\implies dt =\sqrt{7}\sinh xdx$$

$$\int (t^2+7)^{\frac{1}{3}}dt = 7^{2/3}\int (\cosh^2x+1)^{\frac{1}{3}} \sinh x\, dx = 7^{2/3}\int \sinh^{5/3} x\, dx$$

this integral cannot be written as elementary function. It belongs to the class of hypergeometric functions

• I tried that , but power is 1/3 , this is not helping for 1/3 power Jan 24 '18 at 15:35
• I bet this is the hint for exponent $\frac12$. Jan 24 '18 at 15:35
• This is very fine for $\frac 12$ but $\frac 13$ leads to hypergeometric function, I guess. Jan 24 '18 at 15:36
• @ClaudeLeibovici ya I just check it does not an obvious integral Jan 24 '18 at 15:37
• @Isham it will make it worse Jan 24 '18 at 15:42

$$\begin{equation} \int \left(t^2 + 7 \right)^{\frac{1}{3}}\:dt = \int_0^t \left(w^2 + 7 \right)^{\frac{1}{3}}\:dw = \int_0^t \frac{1}{ \left(w^2 + 7 \right)^{\frac{1}{3}}}\:dw \end{equation}$$
We can employ the solution that I've spoken to here: $$\begin{equation} \int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt = \frac{1}{n}a^{\frac{k + 1}{n} - m} \left[B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right) - B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}, \frac{1}{1 + ax^n} \right)\right] \end{equation}$$
Here $$a = 7$$, $$m = -\frac{1}{3}$$, $$k = 0$$, and $$n = 2$$. Thus,
\begin{align} \int \left(t^2 + 7 \right)^{\frac{1}{3}}\:dt &= \frac{1}{2}\cdot 7^{\frac{0 + 1}{2} --\frac{1}{3}}\left[B\left(-\frac{1}{3} - \frac{0 + 1}{2}, \frac{0 + 1}{2}\right) - B\left(-\frac{1}{3}- \frac{0 + 1}{2}, \frac{0 + 1}{2}, \frac{1}{1 + 7t^2} \right)\right] \\ &= \frac{7^{\frac{5}{6}}}{2}\left[B\left(-\frac{5}{6}, \frac{1}{2}\right) - B\left(-\frac{5}{6}, \frac{1}{2}, \frac{1}{1 + 7t^2} \right)\right] \end{align}