3.16.79 \(\int \frac {(-1+x^3)^{2/3} (-2+x^3+x^6)}{x^6} \, dx\)

Optimal. Leaf size=108 \[ -\frac {1}{9} \log \left (\sqrt [3]{x^3-1}-x\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3-1}+x}\right )}{3 \sqrt {3}}+\frac {1}{18} \log \left (\sqrt [3]{x^3-1} x+\left (x^3-1\right )^{2/3}+x^2\right )+\frac {\left (x^3-1\right )^{2/3} \left (10 x^6-27 x^3+12\right )}{30 x^5} \]

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Rubi [A]  time = 0.04, antiderivative size = 95, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1478, 453, 277, 195, 239} \begin {gather*} -\frac {1}{6} x \left (x^3-1\right )^{2/3}-\frac {1}{6} \log \left (\sqrt [3]{x^3-1}-x\right )+\frac {\tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{x^3-1}}+1}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {2 \left (x^3-1\right )^{8/3}}{5 x^5}+\frac {\left (x^3-1\right )^{5/3}}{10 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(x*(-1 + x^3)^(2/3)) + (-1 + x^3)^(5/3)/(10*x^2) + (2*(-1 + x^3)^(8/3))/(5*x^5) + ArcTan[(1 + (2*x)/(-1 +
x^3)^(1/3))/Sqrt[3]]/(3*Sqrt[3]) - Log[-x + (-1 + x^3)^(1/3)]/6

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 239

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 277

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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 1478

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_.), x_Sym
bol] :> Int[(f*x)^m*(d + e*x^n)^(q + p)*(a/d + (c*x^n)/e)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, q}, x] && Eq
Q[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\left (-1+x^3\right )^{2/3} \left (-2+x^3+x^6\right )}{x^6} \, dx &=\int \frac {\left (-1+x^3\right )^{5/3} \left (2+x^3\right )}{x^6} \, dx\\ &=\frac {2 \left (-1+x^3\right )^{8/3}}{5 x^5}-\frac {1}{5} \int \frac {\left (-1+x^3\right )^{5/3}}{x^3} \, dx\\ &=\frac {\left (-1+x^3\right )^{5/3}}{10 x^2}+\frac {2 \left (-1+x^3\right )^{8/3}}{5 x^5}-\frac {1}{2} \int \left (-1+x^3\right )^{2/3} \, dx\\ &=-\frac {1}{6} x \left (-1+x^3\right )^{2/3}+\frac {\left (-1+x^3\right )^{5/3}}{10 x^2}+\frac {2 \left (-1+x^3\right )^{8/3}}{5 x^5}+\frac {1}{3} \int \frac {1}{\sqrt [3]{-1+x^3}} \, dx\\ &=-\frac {1}{6} x \left (-1+x^3\right )^{2/3}+\frac {\left (-1+x^3\right )^{5/3}}{10 x^2}+\frac {2 \left (-1+x^3\right )^{8/3}}{5 x^5}+\frac {\tan ^{-1}\left (\frac {1+\frac {2 x}{\sqrt [3]{-1+x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {1}{6} \log \left (-x+\sqrt [3]{-1+x^3}\right )\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 77, normalized size = 0.71 \begin {gather*} \frac {\left (x^3-1\right )^{2/3} \left (-5 x^3 \, _2F_1\left (-\frac {2}{3},-\frac {2}{3};\frac {1}{3};x^3\right )+10 x^6 \, _2F_1\left (-\frac {2}{3},\frac {1}{3};\frac {4}{3};x^3\right )+4 \left (1-x^3\right )^{5/3}\right )}{10 x^5 \left (1-x^3\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1 + x^3)^(2/3)*(4*(1 - x^3)^(5/3) - 5*x^3*Hypergeometric2F1[-2/3, -2/3, 1/3, x^3] + 10*x^6*Hypergeometric2F
1[-2/3, 1/3, 4/3, x^3]))/(10*x^5*(1 - x^3)^(2/3))

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IntegrateAlgebraic [A]  time = 0.22, size = 108, normalized size = 1.00 \begin {gather*} \frac {\left (-1+x^3\right )^{2/3} \left (12-27 x^3+10 x^6\right )}{30 x^5}+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^3}}\right )}{3 \sqrt {3}}-\frac {1}{9} \log \left (-x+\sqrt [3]{-1+x^3}\right )+\frac {1}{18} \log \left (x^2+x \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1 + x^3)^(2/3)*(12 - 27*x^3 + 10*x^6))/(30*x^5) + ArcTan[(Sqrt[3]*x)/(x + 2*(-1 + x^3)^(1/3))]/(3*Sqrt[3])
- Log[-x + (-1 + x^3)^(1/3)]/9 + Log[x^2 + x*(-1 + x^3)^(1/3) + (-1 + x^3)^(2/3)]/18

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fricas [A]  time = 1.09, size = 117, normalized size = 1.08 \begin {gather*} \frac {10 \, \sqrt {3} x^{5} \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 ) - 5 \, x^{5} \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 (10 \, x^{6} - 27 \, x^{3} + 12\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{90 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/90*(10*sqrt(3)*x^5*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)) - 5*x^5*log(-3*(x^3 - 1)^(1/3)*x^2 + 3*(x^3 - 1)^(2/3)*x + 1) + 3*(10*x^6
- 27*x^3 + 12)*(x^3 - 1)^(2/3))/x^5

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{6} + x^{3} - 2\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 2.99, size = 61, normalized size = 0.56

method	result	size
risch	\(\frac {10 x^{9}-37 x^{6}+39 x^{3}-12}{30 x^{5} \left (x^{3}-1\right )^{\frac {1}{3}}}+\frac {\left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {1}{3}} x \hypergeom \left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], x^{3}\right )}{3 \mathrm {signum}\left (x^{3}-1\right )^{\frac {1}{3}}}\)	\(61\)
meijerg	\(\frac {\mathrm {signum}\left (x^{3}-1\right )^{\frac {2}{3}} x \hypergeom \left (\left [-\frac {2}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], x^{3}\right )}{\left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {2}{3}}}-\frac {\mathrm {signum}\left (x^{3}-1\right )^{\frac {2}{3}} \hypergeom \left (\left [-\frac {2}{3}, -\frac {2}{3}\right ], \left [\frac {1}{3}\right ], x^{3}\right )}{2 \left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {2}{3}} x^{2}}+\frac {2 \mathrm {signum}\left (x^{3}-1\right )^{\frac {2}{3}} \left (-x^{3}+1\right )^{\frac {5}{3}}}{5 \left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {2}{3}} x^{5}}\)	\(95\)
trager	\(\frac {\left (x^{3}-1\right )^{\frac {2}{3}} \left (10 x^{6}-27 x^{3}+12\right )}{30 x^{5}}+\frac {32 \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right ) \ln \left (11665390592 \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right )^{2} x^{3}[phone] \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x [phone] \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x^{2}[phone] \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right ) x^{3}-1508552373 x \left (x^{3}-1\right )^{\frac {2}{3}}-1508552373 x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}-1497160390 x^{3}-93323124736 \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right )^{2}[phone] \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right )[phone]\right )}{9}+\frac {\ln \left (11665390592 \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right )^{2} x^{3}-6206811648 \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x -6206811648 \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x^{2}-6571355104 \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right ) x^{3}-1314589509 x \left (x^{3}-1\right )^{\frac {2}{3}}-1314589509 x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}-1303197526 x^{3}-93323124736 \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right )^{2}-9037003232 \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right )[phone]\right )}{9}-\frac {32 \ln \left (11665390592 \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right )^{2} x^{3}-6206811648 \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x -6206811648 \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x^{2}-6571355104 \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right ) x^{3}-1314589509 x \left (x^{3}-1\right )^{\frac {2}{3}}-1314589509 x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}-1303197526 x^{3}-93323124736 \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right )^{2}-9037003232 \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right )[phone]\right ) \RootOf \left (1024 \textit {\_Z}^{2}-32 \textit {\_Z} +1\right )}{9}\)	\(462\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(10*x^9-37*x^6+39*x^3-12)/x^5/(x^3-1)^(1/3)+1/3/signum(x^3-1)^(1/3)*(-signum(x^3-1))^(1/3)*x*hypergeom([1
/3,1/3],[4/3],x^3)

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maxima [A]  time = 0.42, size = 118, normalized size = 1.09 \begin {gather*} -\frac {1}{9} \, \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 {{\left (x^{3} - 1\right )}^{\frac {2}{3}}}{3 \, x^{2} {\left (\frac {x^{3} - 1}{x^{3}} - 1\right )}} - \frac {2 \, {\left (x^{3} - 1\right )}^{\frac {5}{3}}}{5 \, x^{5}} + \frac {1}{18} \, \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}{9} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 - 1)^(1/3)/x + 1)) - 1/2*(x^3 - 1)^(2/3)/x^2 - 1/3*(x^3 - 1)^(2/3)/(x^
2*((x^3 - 1)/x^3 - 1)) - 2/5*(x^3 - 1)^(5/3)/x^5 + 1/18*log((x^3 - 1)^(1/3)/x + (x^3 - 1)^(2/3)/x^2 + 1) - 1/9
*log((x^3 - 1)^(1/3)/x - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x^3-1\right )}^{2/3}\,\left (x^6+x^3-2\right )}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 3.30, size = 199, normalized size = 1.84 \begin {gather*} - \frac {x e^{- \frac {i \pi }{3}} \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} - 2 \left (\begin {cases} \frac {\left (-1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} e^{- \frac {i \pi }{3}} \Gamma \left (- \frac {5}{3}\right )}{3 \Gamma \left (- \frac {2}{3}\right )} - \frac {\left (-1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} e^{- \frac {i \pi }{3}} \Gamma \left (- \frac {5}{3}\right )}{3 x^{3} \Gamma \left (- \frac {2}{3}\right )} & \text {for}\: \frac {1}{\left |{x^{3}}\right |} > 1 \\- \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 {\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 )} & \text {otherwise} \end {cases}\right ) + \frac {e^{\frac {2 i \pi }{3}} \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}} \right )}}{3 x^{2} \Gamma \left (\frac {1}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*exp(-I*pi/3)*gamma(1/3)*hyper((-2/3, 1/3), (4/3,), x**3)/(3*gamma(4/3)) - 2*Piecewise(((-1 + x**(-3))**(2/3
)*exp(-I*pi/3)*gamma(-5/3)/(3*gamma(-2/3)) - (-1 + x**(-3))**(2/3)*exp(-I*pi/3)*gamma(-5/3)/(3*x**3*gamma(-2/3
)), 1/Abs(x**3) > 1), (-(1 - 1/x**3)**(2/3)*gamma(-5/3)/(3*gamma(-2/3)) + (1 - 1/x**3)**(2/3)*gamma(-5/3)/(3*x
**3*gamma(-2/3)), True)) + exp(2*I*pi/3)*gamma(-2/3)*hyper((-2/3, -2/3), (1/3,), x**3)/(3*x**2*gamma(1/3))

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