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

Optimal. Leaf size=175 \[ -\frac {1}{3 x^3 \sqrt [3]{1-x^3}}+\frac {5}{6 \sqrt [3]{1-x^3}}-\frac {\log \left (x^3+1\right )}{12 \sqrt [3]{2}}+\frac {1}{6} \log \left (1-\sqrt [3]{1-x^3}\right )+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}-\frac {\log (x)}{6} \]

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Rubi [A]  time = 0.12, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {446, 103, 156, 51, 55, 618, 204, 31, 617} \begin {gather*} -\frac {1}{3 x^3 \sqrt [3]{1-x^3}}+\frac {5}{6 \sqrt [3]{1-x^3}}-\frac {\log \left (x^3+1\right )}{12 \sqrt [3]{2}}+\frac {1}{6} \log \left (1-\sqrt [3]{1-x^3}\right )+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}-\frac {\log (x)}{6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

5/(6*(1 - x^3)^(1/3)) - 1/(3*x^3*(1 - x^3)^(1/3)) + ArcTan[(1 + 2*(1 - x^3)^(1/3))/Sqrt[3]]/(3*Sqrt[3]) + ArcT
an[(1 + 2^(2/3)*(1 - x^3)^(1/3))/Sqrt[3]]/(2*2^(1/3)*Sqrt[3]) - Log[x]/6 - Log[1 + x^3]/(12*2^(1/3)) + Log[1 -
(1 - x^3)^(1/3)]/6 + Log[2^(1/3) - (1 - x^3)^(1/3)]/(4*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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, 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 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

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

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^4 \left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{(1-x)^{4/3} x^2 (1+x)} \, dx,x,x^3\right )\\ &=-\frac {1}{3 x^3 \sqrt [3]{1-x^3}}-\frac {1}{3} \operatorname {Subst}\left (\int \frac {-\frac {1}{3}-\frac {4 x}{3}}{(1-x)^{4/3} x (1+x)} \, dx,x,x^3\right )\\ &=-\frac {1}{3 x^3 \sqrt [3]{1-x^3}}+\frac {1}{9} \operatorname {Subst}\left (\int \frac {1}{(1-x)^{4/3} x} \, dx,x,x^3\right )+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{(1-x)^{4/3} (1+x)} \, dx,x,x^3\right )\\ &=\frac {5}{6 \sqrt [3]{1-x^3}}-\frac {1}{3 x^3 \sqrt [3]{1-x^3}}+\frac {1}{9} \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} x} \, dx,x,x^3\right )+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} (1+x)} \, dx,x,x^3\right )\\ &=\frac {5}{6 \sqrt [3]{1-x^3}}-\frac {1}{3 x^3 \sqrt [3]{1-x^3}}-\frac {\log (x)}{6}-\frac {\log \left (1+x^3\right )}{12 \sqrt [3]{2}}-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sqrt [3]{1-x^3}\right )+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\sqrt [3]{1-x^3}\right )+\frac {1}{4} \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 )}{4 \sqrt [3]{2}}\\ &=\frac {5}{6 \sqrt [3]{1-x^3}}-\frac {1}{3 x^3 \sqrt [3]{1-x^3}}-\frac {\log (x)}{6}-\frac {\log \left (1+x^3\right )}{12 \sqrt [3]{2}}+\frac {1}{6} \log \left (1-\sqrt [3]{1-x^3}\right )+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1-x^3}\right )-\frac {\operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2^{2/3} \sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}\\ &=\frac {5}{6 \sqrt [3]{1-x^3}}-\frac {1}{3 x^3 \sqrt [3]{1-x^3}}+\frac {\tan ^{-1}\left (\frac {1+2 \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}-\frac {\log (x)}{6}-\frac {\log \left (1+x^3\right )}{12 \sqrt [3]{2}}+\frac {1}{6} \log \left (1-\sqrt [3]{1-x^3}\right )+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 64, normalized size = 0.37 \begin {gather*} \frac {3 x^3 \, _2F_1\left (-\frac {1}{3},1;\frac {2}{3};\frac {1}{2} \left (1-x^3\right )\right )+2 x^3 \, _2F_1\left (-\frac {1}{3},1;\frac {2}{3};1-x^3\right )-2}{6 x^3 \sqrt [3]{1-x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.30, size = 232, normalized size = 1.33 \begin {gather*} \frac {\left (1-x^3\right )^{2/3} \left (2-5 x^3\right )}{6 x^3 \left (x^3-1\right )}+\frac {1}{9} \log \left (\sqrt [3]{1-x^3}-1\right )+\frac {\log \left (2^{2/3} \sqrt [3]{1-x^3}-2\right )}{6 \sqrt [3]{2}}-\frac {1}{18} \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 )}{12 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^3}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 238, normalized size = 1.36 \begin {gather*} \frac {6 \, \sqrt {6} 2^{\frac {1}{6}} {\left (x^{6} - x^{3}\right )} \arctan \left (\frac {1}{6} \cdot 2^{\frac {1}{6}} {\left (\sqrt {6} 2^{\frac {1}{3}} + 2 \, \sqrt {6} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) - 3 \cdot 2^{\frac {2}{3}} {\left (x^{6} - x^{3}\right )} \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 ) + 6 \cdot 2^{\frac {2}{3}} {\left (x^{6} - x^{3}\right )} \log \left (-2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) + 8 \, \sqrt {3} {\left (x^{6} - x^{3}\right )} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - 4 \, {\left (x^{6} - x^{3}\right )} \log \left ({\left (-x^{3} + 1\right )}^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + 8 \, {\left (x^{6} - x^{3}\right )} \log \left ({\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 1\right ) - 12 \, {\left (5 \, x^{3} - 2\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{72 \, {\left (x^{6} - x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(-x^3+1)^(4/3)/(x^3+1),x, algorithm="fricas")

[Out]

1/72*(6*sqrt(6)*2^(1/6)*(x^6 - x^3)*arctan(1/6*2^(1/6)*(sqrt(6)*2^(1/3) + 2*sqrt(6)*(-x^3 + 1)^(1/3))) - 3*2^(
2/3)*(x^6 - x^3)*log(2^(2/3) + 2^(1/3)*(-x^3 + 1)^(1/3) + (-x^3 + 1)^(2/3)) + 6*2^(2/3)*(x^6 - x^3)*log(-2^(1/
3) + (-x^3 + 1)^(1/3)) + 8*sqrt(3)*(x^6 - x^3)*arctan(2/3*sqrt(3)*(-x^3 + 1)^(1/3) + 1/3*sqrt(3)) - 4*(x^6 - x
^3)*log((-x^3 + 1)^(2/3) + (-x^3 + 1)^(1/3) + 1) + 8*(x^6 - x^3)*log((-x^3 + 1)^(1/3) - 1) - 12*(5*x^3 - 2)*(-
x^3 + 1)^(2/3))/(x^6 - x^3)

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giac [A]  time = 0.21, size = 181, normalized size = 1.03 \begin {gather*} \frac {1}{12} \, \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}{24} \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}{12} \cdot 2^{\frac {2}{3}} \log \left ({\left | -2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} \right |}\right ) + \frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {5 \, x^{3} - 2}{6 \, {\left ({\left (-x^{3} + 1\right )}^{\frac {4}{3}} - {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}} - \frac {1}{18} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{9} \, \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^4/(-x^3+1)^(4/3)/(x^3+1),x, algorithm="giac")

[Out]

1/12*sqrt(3)*2^(2/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3) + 2*(-x^3 + 1)^(1/3))) - 1/24*2^(2/3)*log(2^(2/3) + 2
^(1/3)*(-x^3 + 1)^(1/3) + (-x^3 + 1)^(2/3)) + 1/12*2^(2/3)*log(abs(-2^(1/3) + (-x^3 + 1)^(1/3))) + 1/9*sqrt(3)
*arctan(1/3*sqrt(3)*(2*(-x^3 + 1)^(1/3) + 1)) - 1/6*(5*x^3 - 2)/((-x^3 + 1)^(4/3) - (-x^3 + 1)^(1/3)) - 1/18*l
og((-x^3 + 1)^(2/3) + (-x^3 + 1)^(1/3) + 1) + 1/9*log(abs((-x^3 + 1)^(1/3) - 1))

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maple [F]  time = 2.29, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (-x^{3}+1\right )^{\frac {4}{3}} \left (x^{3}+1\right ) x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(-x^3+1)^(4/3)/(x^3+1),x)

[Out]

int(1/x^4/(-x^3+1)^(4/3)/(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 {4}{3}} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(-x^3+1)^(4/3)/(x^3+1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.02, size = 399, normalized size = 2.28 \begin {gather*} \frac {\ln \left (\frac {11\,{\left (1-x^3\right )}^{1/3}}{972}-\frac {11}{972}\right )}{9}+\frac {2^{2/3}\,\ln \left (\frac {2^{1/3}\,\left (\frac {2^{2/3}\,\left (\frac {81\,2^{1/3}}{4}-\frac {75\,{\left (1-x^3\right )}^{1/3}}{4}\right )}{12}-\frac {35}{12}\right )}{72}+\frac {{\left (1-x^3\right )}^{1/3}}{27}\right )}{12}+\ln \left ({\left (-\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )}^2\,\left (\left (-\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )\,\left (1458\,{\left (-\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )}^2-\frac {75\,{\left (1-x^3\right )}^{1/3}}{4}\right )-\frac {35}{12}\right )+\frac {{\left (1-x^3\right )}^{1/3}}{27}\right )\,\left (-\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )-\ln \left (\frac {{\left (1-x^3\right )}^{1/3}}{27}-{\left (\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )}^2\,\left (\left (\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )\,\left (1458\,{\left (\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )}^2-\frac {75\,{\left (1-x^3\right )}^{1/3}}{4}\right )+\frac {35}{12}\right )\right )\,\left (\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )+\frac {\frac {5\,x^3}{6}-\frac {1}{3}}{{\left (1-x^3\right )}^{1/3}-{\left (1-x^3\right )}^{4/3}}+\frac {2^{2/3}\,\ln \left (\frac {{\left (1-x^3\right )}^{1/3}}{27}+\frac {2^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (\frac {2^{2/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (\frac {81\,2^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}-\frac {75\,{\left (1-x^3\right )}^{1/3}}{4}\right )}{24}-\frac {35}{12}\right )}{288}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{24}-\frac {2^{2/3}\,\ln \left (\frac {{\left (1-x^3\right )}^{1/3}}{27}-\frac {2^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (\frac {2^{2/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (\frac {81\,2^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}-\frac {75\,{\left (1-x^3\right )}^{1/3}}{4}\right )}{24}+\frac {35}{12}\right )}{288}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{24} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((11*(1 - x^3)^(1/3))/972 - 11/972)/9 + (2^(2/3)*log((2^(1/3)*((2^(2/3)*((81*2^(1/3))/4 - (75*(1 - x^3)^(1/
3))/4))/12 - 35/12))/72 + (1 - x^3)^(1/3)/27))/12 + log(((3^(1/2)*1i)/18 - 1/18)^2*(((3^(1/2)*1i)/18 - 1/18)*(
1458*((3^(1/2)*1i)/18 - 1/18)^2 - (75*(1 - x^3)^(1/3))/4) - 35/12) + (1 - x^3)^(1/3)/27)*((3^(1/2)*1i)/18 - 1/
18) - log((1 - x^3)^(1/3)/27 - ((3^(1/2)*1i)/18 + 1/18)^2*(((3^(1/2)*1i)/18 + 1/18)*(1458*((3^(1/2)*1i)/18 + 1
/18)^2 - (75*(1 - x^3)^(1/3))/4) + 35/12))*((3^(1/2)*1i)/18 + 1/18) + ((5*x^3)/6 - 1/3)/((1 - x^3)^(1/3) - (1
- x^3)^(4/3)) + (2^(2/3)*log((1 - x^3)^(1/3)/27 + (2^(1/3)*(3^(1/2)*1i - 1)^2*((2^(2/3)*(3^(1/2)*1i - 1)*((81*
2^(1/3)*(3^(1/2)*1i - 1)^2)/16 - (75*(1 - x^3)^(1/3))/4))/24 - 35/12))/288)*(3^(1/2)*1i - 1))/24 - (2^(2/3)*lo
g((1 - x^3)^(1/3)/27 - (2^(1/3)*(3^(1/2)*1i + 1)^2*((2^(2/3)*(3^(1/2)*1i + 1)*((81*2^(1/3)*(3^(1/2)*1i + 1)^2)
/16 - (75*(1 - x^3)^(1/3))/4))/24 + 35/12))/288)*(3^(1/2)*1i + 1))/24

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

[Out]

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

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