Integral of $\frac{x^9}{(1+x^3)^{1/3}}$, recurrence relation I need to find $$\int \frac{x^9}{(1+x^3)^{1/3}}$$
I have recurrence relation:
$$J_{m, p} = \int x^m (ax^n + b)^p dx$$
$$a(m+1+np)J_{m, p} = x^{m+1-n}(ax^n + b)^{p+1} - b(m+1-n) J_{m-n,p}$$
But when I tried to find the integral, I had problems with $$J_{0, p}$$,  because I got
$$0 \cdot J_{0,p} = x^{-2}(x^3+1)^{\frac{2}{3}} + 2J_{-3, p}$$ 

$$a = 1, n = 3, b = 1, p = -\frac{1}{3}$$
$$J_{m, p} = \int x^m (ax^n + b)^p dx$$
$m = 9$
$$9 J_{9, p} = x^{7}(x^3 + 1)^{\frac{2}{3}} - 7 J_{6,p}$$
$m = 6$
$$6 J_{6, p} = x^{4}(x^3 + 1)^{\frac{2}{3}} - 4 J_{3,p}$$
$m = 3$
$$3 J_{3, p} = x(x^3 + 1)^{\frac{2}{3}} - J_{0,p}$$
I think we have to stop the reccurrence at this stage and calculate the integral:
$$J_{0, p} = \int \frac{1}{(1+x^3)^{\frac{1}{3}}}dx =\dfrac{\ln\left(\left|\left(\frac{1}{x^3+1}-1\right)^\frac{2}{3}+\sqrt[3]{\frac{1}{-x^3-1}+1}+1\right|\right)-2\ln\left(\left|\sqrt[3]{\frac{1}{x^3+1}-1}+1\right|\right)}{6}-\dfrac{\arctan\left(\frac{2\sqrt[3]{\frac{1}{x^3+1}-1}-1}{\sqrt{3}}\right)}{\sqrt{3}}+C$$
