3.4.78 \(\int \frac {1}{x \sqrt [3]{1-x^3} (1+x^3)} \, dx\)

Optimal. Leaf size=137 \[ \frac {\log \left (x^3+1\right )}{6 \sqrt [3]{2}}+\frac {1}{2} \log \left (1-\sqrt [3]{1-x^3}\right )-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\log (x)}{2} \]

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Rubi [A]  time = 0.09, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {446, 86, 55, 618, 204, 31, 617} \begin {gather*} \frac {\log \left (x^3+1\right )}{6 \sqrt [3]{2}}+\frac {1}{2} \log \left (1-\sqrt [3]{1-x^3}\right )-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\log (x)}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), 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 86

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), In
t[(e + f*x)^p/(a + b*x), x], x] - Dist[d/(b*c - a*d), Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] &&  !IntegerQ[p]

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]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 133, normalized size = 0.97 \begin {gather*} \frac {1}{12} \left (2^{2/3} \log \left (x^3+1\right )+6 \log \left (1-\sqrt [3]{1-x^3}\right )-3\ 2^{2/3} \log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )+4 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )-2\ 2^{2/3} \sqrt {3} \tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )-6 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[3]*ArcTan[(1 + 2*(1 - x^3)^(1/3))/Sqrt[3]] - 2*2^(2/3)*Sqrt[3]*ArcTan[(1 + 2^(2/3)*(1 - x^3)^(1/3))/Sq
rt[3]] - 6*Log[x] + 2^(2/3)*Log[1 + x^3] + 6*Log[1 - (1 - x^3)^(1/3)] - 3*2^(2/3)*Log[2^(1/3) - (1 - x^3)^(1/3
)])/12

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IntegrateAlgebraic [A]  time = 0.20, size = 195, normalized size = 1.42 \begin {gather*} \frac {1}{3} \log \left (\sqrt [3]{1-x^3}-1\right )-\frac {\log \left (2^{2/3} \sqrt [3]{1-x^3}-2\right )}{3 \sqrt [3]{2}}-\frac {1}{6} \log \left (\left (1-x^3\right )^{2/3}+\sqrt [3]{1-x^3}+1\right )+\frac {\log \left (\sqrt [3]{2} \left (1-x^3\right )^{2/3}+2^{2/3} \sqrt [3]{1-x^3}+2\right )}{6 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^3}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 1.48, size = 410, normalized size = 2.99 \begin {gather*} \frac {1}{12} \cdot 2^{\frac {2}{3}} {\left (i \, \sqrt {3} \left (-1\right )^{\frac {1}{3}} - \left (-1\right )^{\frac {1}{3}}\right )} \log \left (\frac {1}{8} \, {\left (i \, \sqrt {3} \left (-1\right )^{\frac {1}{3}} - \left (-1\right )^{\frac {1}{3}}\right )}^{3} - \frac {3}{4} \cdot 2^{\frac {1}{3}} {\left (i \, \sqrt {3} \left (-1\right )^{\frac {1}{3}} - \left (-1\right )^{\frac {1}{3}}\right )}^{2} + 3 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{24} \, {\left (2^{\frac {2}{3}} {\left (i \, \sqrt {3} \left (-1\right )^{\frac {1}{3}} - \left (-1\right )^{\frac {1}{3}}\right )} - 2 \, \sqrt {\frac {3}{2}} \sqrt {-2^{\frac {1}{3}} {\left (i \, \sqrt {3} \left (-1\right )^{\frac {1}{3}} - \left (-1\right )^{\frac {1}{3}}\right )}^{2}}\right )} \log \left (\frac {3}{8} \cdot 2^{\frac {2}{3}} \sqrt {\frac {3}{2}} \sqrt {-2^{\frac {1}{3}} {\left (i \, \sqrt {3} \left (-1\right )^{\frac {1}{3}} - \left (-1\right )^{\frac {1}{3}}\right )}^{2}} {\left (i \, \sqrt {3} \left (-1\right )^{\frac {1}{3}} - \left (-1\right )^{\frac {1}{3}}\right )} + \frac {3}{8} \cdot 2^{\frac {1}{3}} {\left (i \, \sqrt {3} \left (-1\right )^{\frac {1}{3}} - \left (-1\right )^{\frac {1}{3}}\right )}^{2} + 3 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) - \frac {1}{24} \, {\left (2^{\frac {2}{3}} {\left (i \, \sqrt {3} \left (-1\right )^{\frac {1}{3}} - \left (-1\right )^{\frac {1}{3}}\right )} + 2 \, \sqrt {\frac {3}{2}} \sqrt {-2^{\frac {1}{3}} {\left (i \, \sqrt {3} \left (-1\right )^{\frac {1}{3}} - \left (-1\right )^{\frac {1}{3}}\right )}^{2}}\right )} \log \left (-\frac {3}{8} \cdot 2^{\frac {2}{3}} \sqrt {\frac {3}{2}} \sqrt {-2^{\frac {1}{3}} {\left (i \, \sqrt {3} \left (-1\right )^{\frac {1}{3}} - \left (-1\right )^{\frac {1}{3}}\right )}^{2}} {\left (i \, \sqrt {3} \left (-1\right )^{\frac {1}{3}} - \left (-1\right )^{\frac {1}{3}}\right )} + \frac {3}{8} \cdot 2^{\frac {1}{3}} {\left (i \, \sqrt {3} \left (-1\right )^{\frac {1}{3}} - \left (-1\right )^{\frac {1}{3}}\right )}^{2} + 3 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) + \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + \frac {1}{3} \, \log \left (-\frac {1}{24} \, {\left (i \, \sqrt {3} \left (-1\right )^{\frac {1}{3}} - \left (-1\right )^{\frac {1}{3}}\right )}^{3} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - \frac {4}{3}\right ) - \frac {1}{6} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*2^(2/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))*log(1/8*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))^3 - 3/4*2^(1/3)*(
I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))^2 + 3*(-x^3 + 1)^(1/3) + 1) - 1/24*(2^(2/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1
/3)) - 2*sqrt(3/2)*sqrt(-2^(1/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))^2))*log(3/8*2^(2/3)*sqrt(3/2)*sqrt(-2^(1/
3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))^2)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3)) + 3/8*2^(1/3)*(I*sqrt(3)*(-1)^(1
/3) - (-1)^(1/3))^2 + 3*(-x^3 + 1)^(1/3)) - 1/24*(2^(2/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3)) + 2*sqrt(3/2)*sq
rt(-2^(1/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))^2))*log(-3/8*2^(2/3)*sqrt(3/2)*sqrt(-2^(1/3)*(I*sqrt(3)*(-1)^(
1/3) - (-1)^(1/3))^2)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3)) + 3/8*2^(1/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))^2
+ 3*(-x^3 + 1)^(1/3)) + 1/3*sqrt(3)*arctan(2/3*sqrt(3)*(-x^3 + 1)^(1/3) + 1/3*sqrt(3)) + 1/3*log(-1/24*(I*sqrt
(3)*(-1)^(1/3) - (-1)^(1/3))^3 + (-x^3 + 1)^(1/3) - 4/3) - 1/6*log((-x^3 + 1)^(2/3) + (-x^3 + 1)^(1/3) + 1)

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giac [A]  time = 0.20, size = 149, normalized size = 1.09 \begin {gather*} -\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{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}{12} \cdot 2^{\frac {2}{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 {2}{3}} \log \left ({\left | -2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} \right |}\right ) + \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{6} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left | {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 2.44, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (-x^{3}+1\right )^{\frac {1}{3}} \left (x^{3}+1\right ) x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.80, size = 256, normalized size = 1.87 \begin {gather*} \frac {\ln \left (6-6\,{\left (1-x^3\right )}^{1/3}\right )}{3}+\ln \left ({\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^3\,\left (1458\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2-135\,{\left (1-x^3\right )}^{1/3}\right )-{\left (1-x^3\right )}^{1/3}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (-{\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^3\,\left (1458\,{\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2-135\,{\left (1-x^3\right )}^{1/3}\right )-{\left (1-x^3\right )}^{1/3}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\frac {2^{2/3}\,\ln \left (\frac {3\,{\left (1-x^3\right )}^{1/3}}{2}-\frac {3\,2^{1/3}}{2}\right )}{6}+\frac {{\left (-1\right )}^{1/3}\,2^{2/3}\,\ln \left (\frac {3\,{\left (1-x^3\right )}^{1/3}}{2}-\frac {3\,{\left (-1\right )}^{2/3}\,2^{1/3}}{2}\right )}{6}-\frac {{\left (-1\right )}^{1/3}\,2^{2/3}\,\ln \left (-\frac {{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^3\,\left (135\,{\left (1-x^3\right )}^{1/3}-\frac {81\,{\left (-1\right )}^{2/3}\,2^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )}{432}-{\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(1/(x*(1 - x^3)^(1/3)*(x^3 + 1)),x)

[Out]

log(6 - 6*(1 - x^3)^(1/3))/3 + log(((3^(1/2)*1i)/6 - 1/6)^3*(1458*((3^(1/2)*1i)/6 - 1/6)^2 - 135*(1 - x^3)^(1/
3)) - (1 - x^3)^(1/3))*((3^(1/2)*1i)/6 - 1/6) - log(- ((3^(1/2)*1i)/6 + 1/6)^3*(1458*((3^(1/2)*1i)/6 + 1/6)^2
- 135*(1 - x^3)^(1/3)) - (1 - x^3)^(1/3))*((3^(1/2)*1i)/6 + 1/6) - (2^(2/3)*log((3*(1 - x^3)^(1/3))/2 - (3*2^(
1/3))/2))/6 + ((-1)^(1/3)*2^(2/3)*log((3*(1 - x^3)^(1/3))/2 - (3*(-1)^(2/3)*2^(1/3))/2))/6 - ((-1)^(1/3)*2^(2/
3)*log(- ((3^(1/2)*1i + 1)^3*(135*(1 - x^3)^(1/3) - (81*(-1)^(2/3)*2^(1/3)*(3^(1/2)*1i + 1)^2)/4))/432 - (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 {1}{x \sqrt [3]{- \left (x - 1\right ) \left (x^{2} + x + 1\right )} \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(1/x/(-x**3+1)**(1/3)/(x**3+1),x)

[Out]

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

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