3.17.84 \(\int \frac {(-4+x^3) (-2+x^3) (-1+x^3)^{2/3}}{x^9 (-2+3 x^3)} \, dx\)

Optimal. Leaf size=144 \[ \frac {5 \log \left (\sqrt [3]{2} \sqrt [3]{x^3-1}+x\right )}{3\ 2^{2/3}}+\frac {5 \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{2} \sqrt [3]{x^3-1}-x}\right )}{2^{2/3} \sqrt {3}}-\frac {5 \log \left (-\sqrt [3]{2} \sqrt [3]{x^3-1} x+2^{2/3} \left (x^3-1\right )^{2/3}+x^2\right )}{6\ 2^{2/3}}+\frac {\left (x^3-1\right )^{2/3} \left (16 x^6+4 x^3+5\right )}{10 x^8} \]

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Rubi [C]  time = 0.61, antiderivative size = 141, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {6725, 271, 264, 277, 239, 430, 429} \begin {gather*} -\frac {15 x \left (x^3-1\right )^{2/3} F_1\left (\frac {1}{3};-\frac {2}{3},1;\frac {4}{3};x^3,\frac {3 x^3}{2}\right )}{2 \left (1-x^3\right )^{2/3}}+\frac {5}{2} \log \left (\sqrt [3]{x^3-1}-x\right )-\frac {5 \tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{x^3-1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\left (x^3-1\right )^{5/3}}{2 x^8}-\frac {9 \left (x^3-1\right )^{5/3}}{10 x^5}+\frac {5 \left (x^3-1\right )^{2/3}}{2 x^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[((-4 + x^3)*(-2 + x^3)*(-1 + x^3)^(2/3))/(x^9*(-2 + 3*x^3)),x]

[Out]

(5*(-1 + x^3)^(2/3))/(2*x^2) - (-1 + x^3)^(5/3)/(2*x^8) - (9*(-1 + x^3)^(5/3))/(10*x^5) - (15*x*(-1 + x^3)^(2/
3)*AppellF1[1/3, -2/3, 1, 4/3, x^3, (3*x^3)/2])/(2*(1 - x^3)^(2/3)) - (5*ArcTan[(1 + (2*x)/(-1 + x^3)^(1/3))/S
qrt[3]])/Sqrt[3] + (5*Log[-x + (-1 + x^3)^(1/3)])/2

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 264

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 271

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 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 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (-4+x^3\right ) \left (-2+x^3\right ) \left (-1+x^3\right )^{2/3}}{x^9 \left (-2+3 x^3\right )} \, dx &=\int \left (-\frac {4 \left (-1+x^3\right )^{2/3}}{x^9}-\frac {3 \left (-1+x^3\right )^{2/3}}{x^6}-\frac {5 \left (-1+x^3\right )^{2/3}}{x^3}+\frac {15 \left (-1+x^3\right )^{2/3}}{-2+3 x^3}\right ) \, dx\\ &=-\left (3 \int \frac {\left (-1+x^3\right )^{2/3}}{x^6} \, dx\right )-4 \int \frac {\left (-1+x^3\right )^{2/3}}{x^9} \, dx-5 \int \frac {\left (-1+x^3\right )^{2/3}}{x^3} \, dx+15 \int \frac {\left (-1+x^3\right )^{2/3}}{-2+3 x^3} \, dx\\ &=\frac {5 \left (-1+x^3\right )^{2/3}}{2 x^2}-\frac {\left (-1+x^3\right )^{5/3}}{2 x^8}-\frac {3 \left (-1+x^3\right )^{5/3}}{5 x^5}-\frac {3}{2} \int \frac {\left (-1+x^3\right )^{2/3}}{x^6} \, dx-5 \int \frac {1}{\sqrt [3]{-1+x^3}} \, dx+\frac {\left (15 \left (-1+x^3\right )^{2/3}\right ) \int \frac {\left (1-x^3\right )^{2/3}}{-2+3 x^3} \, dx}{\left (1-x^3\right )^{2/3}}\\ &=\frac {5 \left (-1+x^3\right )^{2/3}}{2 x^2}-\frac {\left (-1+x^3\right )^{5/3}}{2 x^8}-\frac {9 \left (-1+x^3\right )^{5/3}}{10 x^5}-\frac {15 x \left (-1+x^3\right )^{2/3} F_1\left (\frac {1}{3};-\frac {2}{3},1;\frac {4}{3};x^3,\frac {3 x^3}{2}\right )}{2 \left (1-x^3\right )^{2/3}}-\frac {5 \tan ^{-1}\left (\frac {1+\frac {2 x}{\sqrt [3]{-1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {5}{2} \log \left (-x+\sqrt [3]{-1+x^3}\right )\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 140, normalized size = 0.97 \begin {gather*} \frac {5 \left (2 \log \left (\frac {2^{2/3} x}{\sqrt [3]{1-x^3}}+2\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2^{2/3} x}{\sqrt [3]{1-x^3}}-1}{\sqrt {3}}\right )-\log \left (-\frac {2^{2/3} x}{\sqrt [3]{1-x^3}}+\frac {\sqrt [3]{2} x^2}{\left (1-x^3\right )^{2/3}}+2\right )\right )}{6\ 2^{2/3}}+\frac {\left (x^3-1\right )^{2/3} \left (16 x^6+4 x^3+5\right )}{10 x^8} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((-4 + x^3)*(-2 + x^3)*(-1 + x^3)^(2/3))/(x^9*(-2 + 3*x^3)),x]

[Out]

((-1 + x^3)^(2/3)*(5 + 4*x^3 + 16*x^6))/(10*x^8) + (5*(2*Sqrt[3]*ArcTan[(-1 + (2^(2/3)*x)/(1 - x^3)^(1/3))/Sqr
t[3]] - Log[2 + (2^(1/3)*x^2)/(1 - x^3)^(2/3) - (2^(2/3)*x)/(1 - x^3)^(1/3)] + 2*Log[2 + (2^(2/3)*x)/(1 - x^3)
^(1/3)]))/(6*2^(2/3))

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IntegrateAlgebraic [A]  time = 0.36, size = 144, normalized size = 1.00 \begin {gather*} \frac {5 \log \left (\sqrt [3]{2} \sqrt [3]{x^3-1}+x\right )}{3\ 2^{2/3}}+\frac {5 \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{2} \sqrt [3]{x^3-1}-x}\right )}{2^{2/3} \sqrt {3}}-\frac {5 \log \left (-\sqrt [3]{2} \sqrt [3]{x^3-1} x+2^{2/3} \left (x^3-1\right )^{2/3}+x^2\right )}{6\ 2^{2/3}}+\frac {\left (x^3-1\right )^{2/3} \left (16 x^6+4 x^3+5\right )}{10 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-4 + x^3)*(-2 + x^3)*(-1 + x^3)^(2/3))/(x^9*(-2 + 3*x^3)),x]

[Out]

((-1 + x^3)^(2/3)*(5 + 4*x^3 + 16*x^6))/(10*x^8) + (5*ArcTan[(Sqrt[3]*x)/(-x + 2*2^(1/3)*(-1 + x^3)^(1/3))])/(
2^(2/3)*Sqrt[3]) + (5*Log[x + 2^(1/3)*(-1 + x^3)^(1/3)])/(3*2^(2/3)) - (5*Log[x^2 - 2^(1/3)*x*(-1 + x^3)^(1/3)
+ 2^(2/3)*(-1 + x^3)^(2/3)])/(6*2^(2/3))

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fricas [B]  time = 2.80, size = 267, normalized size = 1.85 \begin {gather*} \frac {100 \cdot 4^{\frac {1}{6}} \sqrt {3} x^{8} \arctan \left (\frac {4^{\frac {1}{6}} {\left (12 \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (3 \, x^{4} - 2 \, x\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}} - 4^{\frac {1}{3}} \sqrt {3} {\left (27 \, x^{9} - 72 \, x^{6} + 36 \, x^{3} + 8\right )} - 12 \, \sqrt {3} {\left (9 \, x^{8} - 6 \, x^{5} - 4 \, x^{2}\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}}\right )}}{6 \, {\left (27 \, x^{9} - 36 \, x^{3} + 8\right )}}\right ) + 50 \cdot 4^{\frac {2}{3}} x^{8} \log \left (\frac {6 \cdot 4^{\frac {1}{3}} {\left (x^{3} - 1\right )}^{\frac {1}{3}} x^{2} + 4^{\frac {2}{3}} {\left (3 \, x^{3} - 2\right )} + 12 \, {\left (x^{3} - 1\right )}^{\frac {2}{3}} x}{3 \, x^{3} - 2}\right ) - 25 \cdot 4^{\frac {2}{3}} x^{8} \log \left (\frac {6 \cdot 4^{\frac {2}{3}} {\left (x^{3} - 1\right )}^{\frac {2}{3}} x - 4^{\frac {1}{3}} {\left (9 \, x^{6} - 6 \, x^{3} - 4\right )} + 6 \, {\left (3 \, x^{5} - 4 \, x^{2}\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{9 \, x^{6} - 12 \, x^{3} + 4}\right ) + 36 \, {\left (16 \, x^{6} + 4 \, x^{3} + 5\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{360 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-4)*(x^3-2)*(x^3-1)^(2/3)/x^9/(3*x^3-2),x, algorithm="fricas")

[Out]

1/360*(100*4^(1/6)*sqrt(3)*x^8*arctan(1/6*4^(1/6)*(12*4^(2/3)*sqrt(3)*(3*x^4 - 2*x)*(x^3 - 1)^(2/3) - 4^(1/3)*
sqrt(3)*(27*x^9 - 72*x^6 + 36*x^3 + 8) - 12*sqrt(3)*(9*x^8 - 6*x^5 - 4*x^2)*(x^3 - 1)^(1/3))/(27*x^9 - 36*x^3
+ 8)) + 50*4^(2/3)*x^8*log((6*4^(1/3)*(x^3 - 1)^(1/3)*x^2 + 4^(2/3)*(3*x^3 - 2) + 12*(x^3 - 1)^(2/3)*x)/(3*x^3
- 2)) - 25*4^(2/3)*x^8*log((6*4^(2/3)*(x^3 - 1)^(2/3)*x - 4^(1/3)*(9*x^6 - 6*x^3 - 4) + 6*(3*x^5 - 4*x^2)*(x^
3 - 1)^(1/3))/(9*x^6 - 12*x^3 + 4)) + 36*(16*x^6 + 4*x^3 + 5)*(x^3 - 1)^(2/3))/x^8

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{3} - 1\right )}^{\frac {2}{3}} {\left (x^{3} - 2\right )} {\left (x^{3} - 4\right )}}{{\left (3 \, x^{3} - 2\right )} x^{9}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-4)*(x^3-2)*(x^3-1)^(2/3)/x^9/(3*x^3-2),x, algorithm="giac")

[Out]

integrate((x^3 - 1)^(2/3)*(x^3 - 2)*(x^3 - 4)/((3*x^3 - 2)*x^9), x)

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maple [C]  time = 2.50, size = 564, normalized size = 3.92 \begin {gather*} \frac {16 x^{9}-12 x^{6}+x^{3}-5}{10 x^{8} \left (x^{3}-1\right )^{\frac {1}{3}}}+\frac {5 \RootOf \left (\textit {\_Z}^{3}-2\right ) \ln \left (-\frac {18 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+6 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+36 \textit {\_Z}^{2}\right )^{2} \RootOf \left (\textit {\_Z}^{3}-2\right )^{2} x^{3}-3 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+6 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+36 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}-2\right )^{3} x^{3}-18 \left (x^{3}-1\right )^{\frac {2}{3}} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+6 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+36 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}-2\right )^{2} x -18 \left (x^{3}-1\right )^{\frac {1}{3}} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+6 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+36 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}-2\right ) x^{2}-3 \RootOf \left (\textit {\_Z}^{3}-2\right )^{2} \left (x^{3}-1\right )^{\frac {1}{3}} x^{2}-6 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+6 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+36 \textit {\_Z}^{2}\right ) x^{3}+\RootOf \left (\textit {\_Z}^{3}-2\right ) x^{3}+12 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+6 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+36 \textit {\_Z}^{2}\right )-2 \RootOf \left (\textit {\_Z}^{3}-2\right )}{3 x^{3}-2}\right )}{6}+5 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+6 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+36 \textit {\_Z}^{2}\right ) \ln \left (-\frac {18 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+6 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+36 \textit {\_Z}^{2}\right )^{2} \RootOf \left (\textit {\_Z}^{3}-2\right )^{2} x^{3}-3 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+6 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+36 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}-2\right )^{3} x^{3}+18 \left (x^{3}-1\right )^{\frac {2}{3}} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+6 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+36 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}-2\right )^{2} x +3 \RootOf \left (\textit {\_Z}^{3}-2\right )^{2} \left (x^{3}-1\right )^{\frac {1}{3}} x^{2}+12 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+6 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+36 \textit {\_Z}^{2}\right ) x^{3}-2 \RootOf \left (\textit {\_Z}^{3}-2\right ) x^{3}+6 x \left (x^{3}-1\right )^{\frac {2}{3}}-12 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+6 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+36 \textit {\_Z}^{2}\right )+2 \RootOf \left (\textit {\_Z}^{3}-2\right )}{3 x^{3}-2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3-4)*(x^3-2)*(x^3-1)^(2/3)/x^9/(3*x^3-2),x)

[Out]

1/10*(16*x^9-12*x^6+x^3-5)/x^8/(x^3-1)^(1/3)+5/6*RootOf(_Z^3-2)*ln(-(18*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z
^3-2)+36*_Z^2)^2*RootOf(_Z^3-2)^2*x^3-3*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^3*
x^3-18*(x^3-1)^(2/3)*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^2*x-18*(x^3-1)^(1/3)*
RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)*x^2-3*RootOf(_Z^3-2)^2*(x^3-1)^(1/3)*x^2-6
*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*x^3+RootOf(_Z^3-2)*x^3+12*RootOf(RootOf(_Z^3-2)^2+6*_Z*R
ootOf(_Z^3-2)+36*_Z^2)-2*RootOf(_Z^3-2))/(3*x^3-2))+5*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*ln(
-(18*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)^2*RootOf(_Z^3-2)^2*x^3-3*RootOf(RootOf(_Z^3-2)^2+6*_
Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^3*x^3+18*(x^3-1)^(2/3)*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36
*_Z^2)*RootOf(_Z^3-2)^2*x+3*RootOf(_Z^3-2)^2*(x^3-1)^(1/3)*x^2+12*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+
36*_Z^2)*x^3-2*RootOf(_Z^3-2)*x^3+6*x*(x^3-1)^(2/3)-12*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)+2*
RootOf(_Z^3-2))/(3*x^3-2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{3} - 1\right )}^{\frac {2}{3}} {\left (x^{3} - 2\right )} {\left (x^{3} - 4\right )}}{{\left (3 \, x^{3} - 2\right )} x^{9}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-4)*(x^3-2)*(x^3-1)^(2/3)/x^9/(3*x^3-2),x, algorithm="maxima")

[Out]

integrate((x^3 - 1)^(2/3)*(x^3 - 2)*(x^3 - 4)/((3*x^3 - 2)*x^9), x)

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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^3-2\right )\,\left (x^3-4\right )}{x^9\,\left (3\,x^3-2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3 - 1)^(2/3)*(x^3 - 2)*(x^3 - 4))/(x^9*(3*x^3 - 2)),x)

[Out]

int(((x^3 - 1)^(2/3)*(x^3 - 2)*(x^3 - 4))/(x^9*(3*x^3 - 2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {2}{3}} \left (x^{3} - 4\right ) \left (x^{3} - 2\right )}{x^{9} \left (3 x^{3} - 2\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3-4)*(x**3-2)*(x**3-1)**(2/3)/x**9/(3*x**3-2),x)

[Out]

Integral(((x - 1)*(x**2 + x + 1))**(2/3)*(x**3 - 4)*(x**3 - 2)/(x**9*(3*x**3 - 2)), x)

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