\(\int \frac {(1+x^3)^{2/3} (-2+x^3+x^6)}{x^9} \, dx\) [1512]

Optimal result

Integrand size = 21, antiderivative size = 105 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (-2+x^3+x^6\right )}{x^9} \, dx=\frac {\left (1+x^3\right )^{2/3} \left (5-2 x^3-17 x^6\right )}{20 x^8}+\frac {\arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{1+x^3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (-x+\sqrt [3]{1+x^3}\right )+\frac {1}{6} \log \left (x^2+x \sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right ) \]

[Out]

1/20*(x^3+1)^(2/3)*(-17*x^6-2*x^3+5)/x^8+1/3*arctan(3^(1/2)*x/(x+2*(x^3+1)^(1/3)))*3^(1/2)-1/3*ln(-x+(x^3+1)^(
1/3))+1/6*ln(x^2+x*(x^3+1)^(1/3)+(x^3+1)^(2/3))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1502, 277, 270, 283, 245} \[ \int \frac {\left (1+x^3\right )^{2/3} \left (-2+x^3+x^6\right )}{x^9} \, dx=\frac {\arctan \left (\frac {\frac {2 x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (\sqrt [3]{x^3+1}-x\right )+\frac {\left (x^3+1\right )^{5/3}}{4 x^8}-\frac {7 \left (x^3+1\right )^{5/3}}{20 x^5}-\frac {\left (x^3+1\right )^{2/3}}{2 x^2} \]

[In]

Int[((1 + x^3)^(2/3)*(-2 + x^3 + x^6))/x^9,x]

[Out]

-1/2*(1 + x^3)^(2/3)/x^2 + (1 + x^3)^(5/3)/(4*x^8) - (7*(1 + x^3)^(5/3))/(20*x^5) + ArcTan[(1 + (2*x)/(1 + x^3
)^(1/3))/Sqrt[3]]/Sqrt[3] - Log[-x + (1 + x^3)^(1/3)]/2

Rule 245

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 1502

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Sy
mbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^n)^q*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e,
f, m, q}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 \left (1+x^3\right )^{2/3}}{x^9}+\frac {\left (1+x^3\right )^{2/3}}{x^6}+\frac {\left (1+x^3\right )^{2/3}}{x^3}\right ) \, dx \\ & = -\left (2 \int \frac {\left (1+x^3\right )^{2/3}}{x^9} \, dx\right )+\int \frac {\left (1+x^3\right )^{2/3}}{x^6} \, dx+\int \frac {\left (1+x^3\right )^{2/3}}{x^3} \, dx \\ & = -\frac {\left (1+x^3\right )^{2/3}}{2 x^2}+\frac {\left (1+x^3\right )^{5/3}}{4 x^8}-\frac {\left (1+x^3\right )^{5/3}}{5 x^5}+\frac {3}{4} \int \frac {\left (1+x^3\right )^{2/3}}{x^6} \, dx+\int \frac {1}{\sqrt [3]{1+x^3}} \, dx \\ & = -\frac {\left (1+x^3\right )^{2/3}}{2 x^2}+\frac {\left (1+x^3\right )^{5/3}}{4 x^8}-\frac {7 \left (1+x^3\right )^{5/3}}{20 x^5}+\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (-x+\sqrt [3]{1+x^3}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (-2+x^3+x^6\right )}{x^9} \, dx=\frac {\left (1+x^3\right )^{2/3} \left (5-2 x^3-17 x^6\right )}{20 x^8}+\frac {\arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{1+x^3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (-x+\sqrt [3]{1+x^3}\right )+\frac {1}{6} \log \left (x^2+x \sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right ) \]

[In]

Integrate[((1 + x^3)^(2/3)*(-2 + x^3 + x^6))/x^9,x]

[Out]

((1 + x^3)^(2/3)*(5 - 2*x^3 - 17*x^6))/(20*x^8) + ArcTan[(Sqrt[3]*x)/(x + 2*(1 + x^3)^(1/3))]/Sqrt[3] - Log[-x
+ (1 + x^3)^(1/3)]/3 + Log[x^2 + x*(1 + x^3)^(1/3) + (1 + x^3)^(2/3)]/6

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 1.87 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.42

method	result	size
risch	\(-\frac {17 x^{9}+19 x^{6}-3 x^{3}-5}{20 x^{8} \left (x^{3}+1\right )^{\frac {1}{3}}}+x \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{3}\right )\)	\(44\)
meijerg	\(-\frac {\operatorname {hypergeom}\left (\left [-\frac {2}{3}, -\frac {2}{3}\right ], \left [\frac {1}{3}\right ], -x^{3}\right )}{2 x^{2}}-\frac {\left (x^{3}+1\right )^{\frac {5}{3}}}{5 x^{5}}+\frac {\left (-\frac {3}{5} x^{6}+\frac {2}{5} x^{3}+1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}}{4 x^{8}}\)	\(54\)
pseudoelliptic	\(\frac {10 \ln \left (\frac {x^{2}+x \left (x^{3}+1\right )^{\frac {1}{3}}+\left (x^{3}+1\right )^{\frac {2}{3}}}{x^{2}}\right ) x^{8}-20 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x +2 \left (x^{3}+1\right )^{\frac {1}{3}}\right )}{3 x}\right ) x^{8}-20 \ln \left (\frac {-x +\left (x^{3}+1\right )^{\frac {1}{3}}}{x}\right ) x^{8}+\left (-51 x^{6}-6 x^{3}+15\right ) \left (x^{3}+1\right )^{\frac {2}{3}}}{60 x^{8}}\)	\(106\)
trager	\(-\frac {\left (17 x^{6}+2 x^{3}-5\right ) \left (x^{3}+1\right )^{\frac {2}{3}}}{20 x^{8}}-\frac {\ln \left (317 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{3}-555 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x +2358 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}} x^{2}-2120 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{3}-1803 x \left (x^{3}+1\right )^{\frac {2}{3}}-555 x^{2} \left (x^{3}+1\right )^{\frac {1}{3}}+2675 x^{3}-317 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2}-99 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+1070\right )}{3}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (-535 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{3}+555 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x +1803 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}} x^{2}-1823 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{3}-2358 x \left (x^{3}+1\right )^{\frac {2}{3}}+555 x^{2} \left (x^{3}+1\right )^{\frac {1}{3}}+1268 x^{3}+535 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2}-1922 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+951\right )}{3}\)	\(287\)

[In]

int((x^3+1)^(2/3)*(x^6+x^3-2)/x^9,x,method=_RETURNVERBOSE)

[Out]

-1/20*(17*x^9+19*x^6-3*x^3-5)/x^8/(x^3+1)^(1/3)+x*hypergeom([1/3,1/3],[4/3],-x^3)

Fricas [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.11 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (-2+x^3+x^6\right )}{x^9} \, dx=\frac {20 \, \sqrt {3} x^{8} \arctan \left (-\frac {25382 \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {1}{3}} x^{2} - 13720 \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {2}{3}} x + \sqrt {3} {\left (5831 \, x^{3} + 7200\right )}}{58653 \, x^{3} + 8000}\right ) - 10 \, x^{8} \log \left (3 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} x^{2} - 3 \, {\left (x^{3} + 1\right )}^{\frac {2}{3}} x + 1\right ) - 3 \, {\left (17 \, x^{6} + 2 \, x^{3} - 5\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{60 \, x^{8}} \]

[In]

integrate((x^3+1)^(2/3)*(x^6+x^3-2)/x^9,x, algorithm="fricas")

[Out]

1/60*(20*sqrt(3)*x^8*arctan(-(25382*sqrt(3)*(x^3 + 1)^(1/3)*x^2 - 13720*sqrt(3)*(x^3 + 1)^(2/3)*x + sqrt(3)*(5
831*x^3 + 7200))/(58653*x^3 + 8000)) - 10*x^8*log(3*(x^3 + 1)^(1/3)*x^2 - 3*(x^3 + 1)^(2/3)*x + 1) - 3*(17*x^6
+ 2*x^3 - 5)*(x^3 + 1)^(2/3))/x^8

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.85 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.67 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (-2+x^3+x^6\right )}{x^9} \, dx=\frac {\left (1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{3 \Gamma \left (- \frac {2}{3}\right )} - \frac {2 \left (x^{3} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{3 x^{2} \Gamma \left (- \frac {2}{3}\right )} + \frac {\Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x^{2} \Gamma \left (\frac {1}{3}\right )} + \frac {\left (1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{3 x^{3} \Gamma \left (- \frac {2}{3}\right )} + \frac {4 \left (x^{3} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{9 x^{5} \Gamma \left (- \frac {2}{3}\right )} + \frac {10 \left (x^{3} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{9 x^{8} \Gamma \left (- \frac {2}{3}\right )} \]

[In]

integrate((x**3+1)**(2/3)*(x**6+x**3-2)/x**9,x)

[Out]

(1 + x**(-3))**(2/3)*gamma(-5/3)/(3*gamma(-2/3)) - 2*(x**3 + 1)**(2/3)*gamma(-8/3)/(3*x**2*gamma(-2/3)) + gamm
a(-2/3)*hyper((-2/3, -2/3), (1/3,), x**3*exp_polar(I*pi))/(3*x**2*gamma(1/3)) + (1 + x**(-3))**(2/3)*gamma(-5/
3)/(3*x**3*gamma(-2/3)) + 4*(x**3 + 1)**(2/3)*gamma(-8/3)/(9*x**5*gamma(-2/3)) + 10*(x**3 + 1)**(2/3)*gamma(-8
/3)/(9*x**8*gamma(-2/3))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (-2+x^3+x^6\right )}{x^9} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) - \frac {{\left (x^{3} + 1\right )}^{\frac {2}{3}}}{2 \, x^{2}} - \frac {3 \, {\left (x^{3} + 1\right )}^{\frac {5}{3}}}{5 \, x^{5}} + \frac {{\left (x^{3} + 1\right )}^{\frac {8}{3}}}{4 \, x^{8}} + \frac {1}{6} \, \log \left (\frac {{\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) - \frac {1}{3} \, \log \left (\frac {{\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x} - 1\right ) \]

[In]

integrate((x^3+1)^(2/3)*(x^6+x^3-2)/x^9,x, algorithm="maxima")

[Out]

-1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 + 1)^(1/3)/x + 1)) - 1/2*(x^3 + 1)^(2/3)/x^2 - 3/5*(x^3 + 1)^(5/3)/x^5
+ 1/4*(x^3 + 1)^(8/3)/x^8 + 1/6*log((x^3 + 1)^(1/3)/x + (x^3 + 1)^(2/3)/x^2 + 1) - 1/3*log((x^3 + 1)^(1/3)/x
- 1)

Giac [F]

\[ \int \frac {\left (1+x^3\right )^{2/3} \left (-2+x^3+x^6\right )}{x^9} \, dx=\int { \frac {{\left (x^{6} + x^{3} - 2\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{x^{9}} \,d x } \]

[In]

integrate((x^3+1)^(2/3)*(x^6+x^3-2)/x^9,x, algorithm="giac")

[Out]

integrate((x^6 + x^3 - 2)*(x^3 + 1)^(2/3)/x^9, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (1+x^3\right )^{2/3} \left (-2+x^3+x^6\right )}{x^9} \, dx=\int \frac {{\left (x^3+1\right )}^{2/3}\,\left (x^6+x^3-2\right )}{x^9} \,d x \]

[In]

int(((x^3 + 1)^(2/3)*(x^3 + x^6 - 2))/x^9,x)

[Out]

int(((x^3 + 1)^(2/3)*(x^3 + x^6 - 2))/x^9, x)