How to evaluate $$\int (x^{6}+x^{9})^{\frac{1}{6}}dx?$$
I am trying to evaluate this integral by taking out a common factor of $x^{6}$. Now the integral is $$\int x(1+x^{3})^{\frac{1}{6}}dx.$$
Now I am thinking of writing $x=\frac{x^{2}}{x}$. So, the above integral is becoming $$\int \frac{x^{2}(1+x^{3})^{\frac{1}{6}}}{x}dx$$. Now I substituted $x^{3}=u$. So, $x^{2}dx=\frac{1}{3}du$.
Therefore, the integral will be $$\frac{1}{3}\int\frac{(1+u)^{\frac{1}{6}}}{u^{\frac{1}{3}}}du.$$
Now, I am thinking of substituting $u=tan^{3}\theta$. So, $du=3(tan^{2}\theta)(sec^{2}\theta)d\theta$. So, the above integral will be $$\int (1+tan^{3}\theta)^{\frac{1}{6}}(tan\theta)(sec^{2}\theta)d\theta.$$ But after this step I can't proceed further. Please help me out.
You cannot avoid hypergeometric function or the binomial expansion.
Making the problem more general $$I=\int \left(x^n+x^{3 n/2}\right)^{\frac{1}{n}}\,dx=\frac{2 }{5}x^{5/2} \, _2F_1\left(-\frac{5}{n},-\frac{1}{n};\frac{n-5}{n};-x^{-n/2} \right)$$ which is $$\frac{2\,x^{5/2}}{\Gamma \left(-\frac{1}{n}\right) }\sum_{k=0}^\infty (-1)^{k+1}\,\frac{ \Gamma \left(k-\frac{1}{n}\right)}{(k n-5)\, \Gamma (k+1)}\,\,x^{-\frac{k n}{2}}$$