3.4.87 \(\int \frac {x^{11}}{(1-x^3)^{2/3} (1+x^3)} \, dx\)

Optimal. Leaf size=125 \[ -\frac {1}{7} \left (1-x^3\right )^{7/3}+\frac {1}{4} \left (1-x^3\right )^{4/3}-\sqrt [3]{1-x^3}+\frac {\log \left (x^3+1\right )}{6\ 2^{2/3}}-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2\ 2^{2/3}}+\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}} \]

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Rubi [A]  time = 0.09, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {446, 88, 57, 617, 204, 31} \begin {gather*} -\frac {1}{7} \left (1-x^3\right )^{7/3}+\frac {1}{4} \left (1-x^3\right )^{4/3}-\sqrt [3]{1-x^3}+\frac {\log \left (x^3+1\right )}{6\ 2^{2/3}}-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2\ 2^{2/3}}+\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^11/((1 - x^3)^(2/3)*(1 + x^3)),x]

[Out]

-(1 - x^3)^(1/3) + (1 - x^3)^(4/3)/4 - (1 - x^3)^(7/3)/7 + ArcTan[(1 + 2^(2/3)*(1 - x^3)^(1/3))/Sqrt[3]]/(2^(2
/3)*Sqrt[3]) + Log[1 + x^3]/(6*2^(2/3)) - Log[2^(1/3) - (1 - x^3)^(1/3)]/(2*2^(2/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 446

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {x^{11}}{\left (1-x^3\right )^{2/3} \left (1+x^3\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^3}{(1-x)^{2/3} (1+x)} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {1}{(1-x)^{2/3}}-\sqrt [3]{1-x}+(1-x)^{4/3}-\frac {1}{(1-x)^{2/3} (1+x)}\right ) \, dx,x,x^3\right )\\ &=-\sqrt [3]{1-x^3}+\frac {1}{4} \left (1-x^3\right )^{4/3}-\frac {1}{7} \left (1-x^3\right )^{7/3}-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{(1-x)^{2/3} (1+x)} \, dx,x,x^3\right )\\ &=-\sqrt [3]{1-x^3}+\frac {1}{4} \left (1-x^3\right )^{4/3}-\frac {1}{7} \left (1-x^3\right )^{7/3}+\frac {\log \left (1+x^3\right )}{6\ 2^{2/3}}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{2}-x} \, dx,x,\sqrt [3]{1-x^3}\right )}{2\ 2^{2/3}}+\frac {\operatorname {Subst}\left (\int \frac {1}{2^{2/3}+\sqrt [3]{2} x+x^2} \, dx,x,\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}\\ &=-\sqrt [3]{1-x^3}+\frac {1}{4} \left (1-x^3\right )^{4/3}-\frac {1}{7} \left (1-x^3\right )^{7/3}+\frac {\log \left (1+x^3\right )}{6\ 2^{2/3}}-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2\ 2^{2/3}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2^{2/3} \sqrt [3]{1-x^3}\right )}{2^{2/3}}\\ &=-\sqrt [3]{1-x^3}+\frac {1}{4} \left (1-x^3\right )^{4/3}-\frac {1}{7} \left (1-x^3\right )^{7/3}+\frac {\tan ^{-1}\left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}+\frac {\log \left (1+x^3\right )}{6\ 2^{2/3}}-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2\ 2^{2/3}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 151, normalized size = 1.21 \begin {gather*} \frac {1}{84} \left (3 \sqrt [3]{1-x^3} x^3-75 \sqrt [3]{1-x^3}-14 \sqrt [3]{2} \log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )+7 \sqrt [3]{2} \log \left (\left (1-x^3\right )^{2/3}+\sqrt [3]{2-2 x^3}+2^{2/3}\right )+14 \sqrt [3]{2} \sqrt {3} \tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )-12 \sqrt [3]{1-x^3} x^6\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^11/((1 - x^3)^(2/3)*(1 + x^3)),x]

[Out]

(-75*(1 - x^3)^(1/3) + 3*x^3*(1 - x^3)^(1/3) - 12*x^6*(1 - x^3)^(1/3) + 14*2^(1/3)*Sqrt[3]*ArcTan[(1 + 2^(2/3)
*(1 - x^3)^(1/3))/Sqrt[3]] - 14*2^(1/3)*Log[2^(1/3) - (1 - x^3)^(1/3)] + 7*2^(1/3)*Log[2^(2/3) + (2 - 2*x^3)^(
1/3) + (1 - x^3)^(2/3)])/84

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IntegrateAlgebraic [A]  time = 0.14, size = 141, normalized size = 1.13 \begin {gather*} -\frac {\log \left (2^{2/3} \sqrt [3]{1-x^3}-2\right )}{3\ 2^{2/3}}+\frac {\log \left (\sqrt [3]{2} \left (1-x^3\right )^{2/3}+2^{2/3} \sqrt [3]{1-x^3}+2\right )}{6\ 2^{2/3}}+\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}+\frac {1}{28} \sqrt [3]{1-x^3} \left (-4 x^6+x^3-25\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^11/((1 - x^3)^(2/3)*(1 + x^3)),x]

[Out]

((1 - x^3)^(1/3)*(-25 + x^3 - 4*x^6))/28 + ArcTan[1/Sqrt[3] + (2^(2/3)*(1 - x^3)^(1/3))/Sqrt[3]]/(2^(2/3)*Sqrt
[3]) - Log[-2 + 2^(2/3)*(1 - x^3)^(1/3)]/(3*2^(2/3)) + Log[2 + 2^(2/3)*(1 - x^3)^(1/3) + 2^(1/3)*(1 - x^3)^(2/
3)]/(6*2^(2/3))

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fricas [A]  time = 0.45, size = 142, normalized size = 1.14 \begin {gather*} -\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} \left (-1\right )^{\frac {1}{3}} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} {\left (4^{\frac {2}{3}} \sqrt {3} \left (-1\right )^{\frac {2}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 4^{\frac {1}{3}} \sqrt {3}\right )}\right ) - \frac {1}{24} \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 2 \cdot 4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{12} \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) - \frac {1}{28} \, {\left (4 \, x^{6} - x^{3} + 25\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(-x^3+1)^(2/3)/(x^3+1),x, algorithm="fricas")

[Out]

-1/6*4^(1/6)*sqrt(3)*(-1)^(1/3)*arctan(1/6*4^(1/6)*(4^(2/3)*sqrt(3)*(-1)^(2/3)*(-x^3 + 1)^(1/3) + 4^(1/3)*sqrt
(3))) - 1/24*4^(2/3)*(-1)^(1/3)*log(-4^(2/3)*(-1)^(1/3)*(-x^3 + 1)^(1/3) + 2*4^(1/3)*(-1)^(2/3) + 2*(-x^3 + 1)
^(2/3)) + 1/12*4^(2/3)*(-1)^(1/3)*log(4^(2/3)*(-1)^(1/3) + 2*(-x^3 + 1)^(1/3)) - 1/28*(4*x^6 - x^3 + 25)*(-x^3
+ 1)^(1/3)

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giac [A]  time = 0.18, size = 127, normalized size = 1.02 \begin {gather*} -\frac {1}{7} \, {\left (x^{3} - 1\right )}^{2} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + \frac {1}{6} \, \sqrt {3} 2^{\frac {1}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) + \frac {1}{4} \, {\left (-x^{3} + 1\right )}^{\frac {4}{3}} + \frac {1}{12} \cdot 2^{\frac {1}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) - \frac {1}{6} \cdot 2^{\frac {1}{3}} \log \left ({\left | -2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} \right |}\right ) - {\left (-x^{3} + 1\right )}^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(-x^3+1)^(2/3)/(x^3+1),x, algorithm="giac")

[Out]

-1/7*(x^3 - 1)^2*(-x^3 + 1)^(1/3) + 1/6*sqrt(3)*2^(1/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3) + 2*(-x^3 + 1)^(1/
3))) + 1/4*(-x^3 + 1)^(4/3) + 1/12*2^(1/3)*log(2^(2/3) + 2^(1/3)*(-x^3 + 1)^(1/3) + (-x^3 + 1)^(2/3)) - 1/6*2^
(1/3)*log(abs(-2^(1/3) + (-x^3 + 1)^(1/3))) - (-x^3 + 1)^(1/3)

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maple [C]  time = 7.45, size = 1579, normalized size = 12.63 \begin {gather*} \text {Expression too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/(-x^3+1)^(2/3)/(x^3+1),x)

[Out]

1/28*(4*x^6-x^3+25)*(x^3-1)/(-x^3+1)^(2/3)+(1/6*RootOf(_Z^3+2)*ln((36*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3
+2)+36*_Z^2)^2*RootOf(_Z^3+2)^2*x^6-24*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)^3*x
^6-36*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)^2*RootOf(_Z^3+2)^2*x^3+24*RootOf(RootOf(_Z^3+2)^2+6
*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)^3*x^3+6*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*x^6-4*
RootOf(_Z^3+2)*x^6-144*(x^6-2*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^3-9*RootOf(_Z^3+2)^2*(x^6-2*x^3+1)^(1/3)*x^3+90*(x^6-2*x^3+1)^(2/3)*RootOf(_Z^3+2)^2*RootOf(RootOf(_Z^3+2)^
2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)-48*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*x^3+32*RootOf(_Z^3+2)*x
^3+144*(x^6-2*x^3+1)^(1/3)*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)+9*RootOf(_Z^3+2
)^2*(x^6-2*x^3+1)^(1/3)-48*(x^6-2*x^3+1)^(2/3)+42*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)-28*Root
Of(_Z^3+2))/(x-1)/(x+1)/(x^2+x+1)/(x^2-x+1))-1/6*ln(-(90*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)^
2*RootOf(_Z^3+2)^2*x^6+12*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)^3*x^6-90*RootOf(
RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)^2*RootOf(_Z^3+2)^2*x^3-12*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z
^3+2)+36*_Z^2)*RootOf(_Z^3+2)^3*x^3-45*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*x^6-6*RootOf(_Z^3+
2)*x^6+27*(x^6-2*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^3+12*RootO
f(_Z^3+2)^2*(x^6-2*x^3+1)^(1/3)*x^3+45*(x^6-2*x^3+1)^(2/3)*RootOf(_Z^3+2)^2*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootO
f(_Z^3+2)+36*_Z^2)+150*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*x^3+20*RootOf(_Z^3+2)*x^3-27*(x^6-
2*x^3+1)^(1/3)*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)-12*RootOf(_Z^3+2)^2*(x^6-2*
x^3+1)^(1/3)+9*(x^6-2*x^3+1)^(2/3)-105*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)-14*RootOf(_Z^3+2))
/(x-1)/(x+1)/(x^2+x+1)/(x^2-x+1))*RootOf(_Z^3+2)-ln(-(90*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)^
2*RootOf(_Z^3+2)^2*x^6+12*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)^3*x^6-90*RootOf(
RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)^2*RootOf(_Z^3+2)^2*x^3-12*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z
^3+2)+36*_Z^2)*RootOf(_Z^3+2)^3*x^3-45*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*x^6-6*RootOf(_Z^3+
2)*x^6+27*(x^6-2*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^3+12*RootO
f(_Z^3+2)^2*(x^6-2*x^3+1)^(1/3)*x^3+45*(x^6-2*x^3+1)^(2/3)*RootOf(_Z^3+2)^2*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootO
f(_Z^3+2)+36*_Z^2)+150*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*x^3+20*RootOf(_Z^3+2)*x^3-27*(x^6-
2*x^3+1)^(1/3)*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)-12*RootOf(_Z^3+2)^2*(x^6-2*
x^3+1)^(1/3)+9*(x^6-2*x^3+1)^(2/3)-105*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)-14*RootOf(_Z^3+2))
/(x-1)/(x+1)/(x^2+x+1)/(x^2-x+1))*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2))/(-x^3+1)^(2/3)*((x^3-1
)^2)^(1/3)

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maxima [A]  time = 1.52, size = 119, normalized size = 0.95 \begin {gather*} -\frac {1}{7} \, {\left (-x^{3} + 1\right )}^{\frac {7}{3}} + \frac {1}{6} \, \sqrt {3} 2^{\frac {1}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) + \frac {1}{4} \, {\left (-x^{3} + 1\right )}^{\frac {4}{3}} + \frac {1}{12} \cdot 2^{\frac {1}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) - \frac {1}{6} \cdot 2^{\frac {1}{3}} \log \left (-2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) - {\left (-x^{3} + 1\right )}^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(-x^3+1)^(2/3)/(x^3+1),x, algorithm="maxima")

[Out]

-1/7*(-x^3 + 1)^(7/3) + 1/6*sqrt(3)*2^(1/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3) + 2*(-x^3 + 1)^(1/3))) + 1/4*(
-x^3 + 1)^(4/3) + 1/12*2^(1/3)*log(2^(2/3) + 2^(1/3)*(-x^3 + 1)^(1/3) + (-x^3 + 1)^(2/3)) - 1/6*2^(1/3)*log(-2
^(1/3) + (-x^3 + 1)^(1/3)) - (-x^3 + 1)^(1/3)

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mupad [B]  time = 5.18, size = 135, normalized size = 1.08 \begin {gather*} \frac {{\left (1-x^3\right )}^{4/3}}{4}-{\left (1-x^3\right )}^{1/3}-\frac {2^{1/3}\,\ln \left (3\,2^{1/3}-3\,{\left (1-x^3\right )}^{1/3}\right )}{6}-\frac {{\left (1-x^3\right )}^{7/3}}{7}-\frac {2^{1/3}\,\ln \left (3\,{\left (1-x^3\right )}^{1/3}-\frac {3\,2^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{12}+\frac {2^{1/3}\,\ln \left (\frac {3\,2^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}+3\,{\left (1-x^3\right )}^{1/3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/((1 - x^3)^(2/3)*(x^3 + 1)),x)

[Out]

(1 - x^3)^(4/3)/4 - (1 - x^3)^(1/3) - (2^(1/3)*log(3*2^(1/3) - 3*(1 - x^3)^(1/3)))/6 - (1 - x^3)^(7/3)/7 - (2^
(1/3)*log(3*(1 - x^3)^(1/3) - (3*2^(1/3)*(3^(1/2)*1i - 1))/2)*(3^(1/2)*1i - 1))/12 + (2^(1/3)*log((3*2^(1/3)*(
3^(1/2)*1i + 1))/2 + 3*(1 - x^3)^(1/3))*(3^(1/2)*1i + 1))/12

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{11}}{\left (- \left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {2}{3}} \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**11/(-x**3+1)**(2/3)/(x**3+1),x)

[Out]

Integral(x**11/((-(x - 1)*(x**2 + x + 1))**(2/3)*(x + 1)*(x**2 - x + 1)), x)

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