Integral of $\frac{x^9}{(1+x^3)^{1/3}}$, recurrence relation

Question

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 $$0 \cdot J_{0,p} = x^{-2}(x^3+1)^{\frac{2}{3}} + 2J_{-3, p}$$

Answer 1

$$a = 1, n = 3, b = 1, p = -\frac{1}{3}$$

$$J_{m, p} = \int x^m (ax^n + b)^p dx$$ $m = 9$ $$9 J_{9, p} = x^{7}(x^3 + 1)^{\frac{2}{3}} - 7 J_{6,p}$$

$m = 6$ $$6 J_{6, p} = x^{4}(x^3 + 1)^{\frac{2}{3}} - 4 J_{3,p}$$

$m = 3$ $$3 J_{3, p} = x(x^3 + 1)^{\frac{2}{3}} - J_{0,p}$$

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