3.7.39 \(\int \frac {x^{11}}{(1-x^3)^{4/3} (1+x^3)} \, dx\) [639]

Optimal. Leaf size=130 \[ \frac {1}{2 \sqrt [3]{1-x^3}}+\frac {1}{2} \left (1-x^3\right )^{2/3}-\frac {1}{5} \left (1-x^3\right )^{5/3}-\frac {\tan ^{-1}\left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}+\frac {\log \left (1+x^3\right )}{12 \sqrt [3]{2}}-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}} \]

[Out]

1/2/(-x^3+1)^(1/3)+1/2*(-x^3+1)^(2/3)-1/5*(-x^3+1)^(5/3)+1/24*ln(x^3+1)*2^(2/3)-1/8*ln(2^(1/3)-(-x^3+1)^(1/3))
*2^(2/3)-1/12*arctan(1/3*(1+2^(2/3)*(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {457, 89, 45, 797, 79, 57, 631, 210, 31} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}-\frac {1}{5} \left (1-x^3\right )^{5/3}+\frac {1}{2} \left (1-x^3\right )^{2/3}+\frac {1}{2 \sqrt [3]{1-x^3}}+\frac {\log \left (x^3+1\right )}{12 \sqrt [3]{2}}-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^11/((1 - x^3)^(4/3)*(1 + x^3)),x]

[Out]

1/(2*(1 - x^3)^(1/3)) + (1 - x^3)^(2/3)/2 - (1 - x^3)^(5/3)/5 - ArcTan[(1 + 2^(2/3)*(1 - x^3)^(1/3))/Sqrt[3]]/
(2*2^(1/3)*Sqrt[3]) + Log[1 + x^3]/(12*2^(1/3)) - 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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 57

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 89

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], (c + d*x)^n*((e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

Rule 210

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

Rule 457

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 631

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 797

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m +
p)*(f + g*x)*(a/d + (c/e)*x)^p, x] /; FreeQ[{a, c, d, e, f, g, m}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p
] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rubi steps

\begin {align*} \int \frac {x^{11}}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x^3}{(1-x)^{4/3} (1+x)} \, dx,x,x^3\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \left (-\frac {x}{\sqrt [3]{1-x}}-\frac {x}{\sqrt [3]{1-x} \left (-1+x^2\right )}\right ) \, dx,x,x^3\right )\\ &=-\left (\frac {1}{3} \text {Subst}\left (\int \frac {x}{\sqrt [3]{1-x}} \, dx,x,x^3\right )\right )-\frac {1}{3} \text {Subst}\left (\int \frac {x}{\sqrt [3]{1-x} \left (-1+x^2\right )} \, dx,x,x^3\right )\\ &=-\left (\frac {1}{3} \text {Subst}\left (\int \left (\frac {1}{\sqrt [3]{1-x}}-(1-x)^{2/3}\right ) \, dx,x,x^3\right )\right )-\frac {1}{3} \text {Subst}\left (\int \frac {x}{(-1-x) (1-x)^{4/3}} \, dx,x,x^3\right )\\ &=\frac {1}{2 \sqrt [3]{1-x^3}}+\frac {1}{2} \left (1-x^3\right )^{2/3}-\frac {1}{5} \left (1-x^3\right )^{5/3}+\frac {1}{6} \text {Subst}\left (\int \frac {1}{(-1-x) \sqrt [3]{1-x}} \, dx,x,x^3\right )\\ &=\frac {1}{2 \sqrt [3]{1-x^3}}+\frac {1}{2} \left (1-x^3\right )^{2/3}-\frac {1}{5} \left (1-x^3\right )^{5/3}+\frac {\log \left (1+x^3\right )}{12 \sqrt [3]{2}}-\frac {1}{4} \text {Subst}\left (\int \frac {1}{2^{2/3}+\sqrt [3]{2} x+x^2} \, dx,x,\sqrt [3]{1-x^3}\right )+\frac {\text {Subst}\left (\int \frac {1}{\sqrt [3]{2}-x} \, dx,x,\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}\\ &=\frac {1}{2 \sqrt [3]{1-x^3}}+\frac {1}{2} \left (1-x^3\right )^{2/3}-\frac {1}{5} \left (1-x^3\right )^{5/3}+\frac {\log \left (1+x^3\right )}{12 \sqrt [3]{2}}-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}+\frac {\text {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 {1}{2 \sqrt [3]{1-x^3}}+\frac {1}{2} \left (1-x^3\right )^{2/3}-\frac {1}{5} \left (1-x^3\right )^{5/3}-\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 \left (1+x^3\right )}{12 \sqrt [3]{2}}-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 137, normalized size = 1.05 \begin {gather*} \frac {1}{120} \left (-\frac {12 \left (-8+x^3+2 x^6\right )}{\sqrt [3]{1-x^3}}-10\ 2^{2/3} \sqrt {3} \tan ^{-1}\left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )-10\ 2^{2/3} \log \left (-2+2^{2/3} \sqrt [3]{1-x^3}\right )+5\ 2^{2/3} \log \left (2+2^{2/3} \sqrt [3]{1-x^3}+\sqrt [3]{2} \left (1-x^3\right )^{2/3}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^11/((1 - x^3)^(4/3)*(1 + x^3)),x]

[Out]

((-12*(-8 + x^3 + 2*x^6))/(1 - x^3)^(1/3) - 10*2^(2/3)*Sqrt[3]*ArcTan[(1 + 2^(2/3)*(1 - x^3)^(1/3))/Sqrt[3]] -
10*2^(2/3)*Log[-2 + 2^(2/3)*(1 - x^3)^(1/3)] + 5*2^(2/3)*Log[2 + 2^(2/3)*(1 - x^3)^(1/3) + 2^(1/3)*(1 - x^3)^
(2/3)])/120

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 6.07, size = 677, normalized size = 5.21

method	result	size
risch	\(\text {Expression too large to display}\)	\(677\)
trager	\(\text {Expression too large to display}\)	\(684\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/(-x^3+1)^(4/3)/(x^3+1),x,method=_RETURNVERBOSE)

[Out]

-1/10*(2*x^6+x^3-8)/(-x^3+1)^(1/3)-1/2*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*ln((15*RootOf(Root
Of(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*RootOf(_Z^3-4)^3*x^3+18*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+
36*_Z^2)^2*RootOf(_Z^3-4)^2*x^3-5*RootOf(_Z^3-4)*x^3-6*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*x^
3+21*(-x^3+1)^(1/3)*RootOf(_Z^3-4)^2+42*(-x^3+1)^(2/3)+35*RootOf(_Z^3-4)+42*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootO
f(_Z^3-4)+36*_Z^2))/(x+1)/(x^2-x+1))+1/12*ln((-12*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*RootOf(
_Z^3-4)^3*x^3+18*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)^2*RootOf(_Z^3-4)^2*x^3-12*RootOf(_Z^3-4)
*x^3+18*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*x^3+21*(-x^3+1)^(1/3)*RootOf(_Z^3-4)^2+42*(-x^3+1
)^(2/3)+28*RootOf(_Z^3-4)-42*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2))/(x+1)/(x^2-x+1))*RootOf(_Z^
3-4)+1/2*ln((-12*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*RootOf(_Z^3-4)^3*x^3+18*RootOf(RootOf(_Z
^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)^2*RootOf(_Z^3-4)^2*x^3-12*RootOf(_Z^3-4)*x^3+18*RootOf(RootOf(_Z^3-4)^2+6
*_Z*RootOf(_Z^3-4)+36*_Z^2)*x^3+21*(-x^3+1)^(1/3)*RootOf(_Z^3-4)^2+42*(-x^3+1)^(2/3)+28*RootOf(_Z^3-4)-42*Root
Of(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2))/(x+1)/(x^2-x+1))*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)
+36*_Z^2)

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Maxima [A]
time = 0.62, size = 119, normalized size = 0.92 \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}{5} \, {\left (-x^{3} + 1\right )}^{\frac {5}{3}} + \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 (-2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) + \frac {1}{2} \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + \frac {1}{2 \, {\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)^(4/3)/(x^3+1),x, algorithm="maxima")

[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))) - 1/5*(-x^3 + 1)^(5/3) + 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(-2^(1/3) + (-x^3 + 1)^(
1/3)) + 1/2*(-x^3 + 1)^(2/3) + 1/2/(-x^3 + 1)^(1/3)

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Fricas [A]
time = 4.26, size = 159, normalized size = 1.22 \begin {gather*} -\frac {10 \, \sqrt {6} 2^{\frac {1}{6}} \left (-1\right )^{\frac {1}{3}} {\left (x^{3} - 1\right )} \arctan \left (\frac {1}{6} \cdot 2^{\frac {1}{6}} {\left (2 \, \sqrt {6} \left (-1\right )^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - \sqrt {6} 2^{\frac {1}{3}}\right )}\right ) + 5 \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{3} - 1\right )} \log \left (2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) - 10 \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{3} - 1\right )} \log \left (-2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) - 12 \, {\left (2 \, x^{6} + x^{3} - 8\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{120 \, {\left (x^{3} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(-x^3+1)^(4/3)/(x^3+1),x, algorithm="fricas")

[Out]

-1/120*(10*sqrt(6)*2^(1/6)*(-1)^(1/3)*(x^3 - 1)*arctan(1/6*2^(1/6)*(2*sqrt(6)*(-1)^(1/3)*(-x^3 + 1)^(1/3) - sq
rt(6)*2^(1/3))) + 5*2^(2/3)*(-1)^(1/3)*(x^3 - 1)*log(2^(1/3)*(-1)^(2/3)*(-x^3 + 1)^(1/3) - 2^(2/3)*(-1)^(1/3)
+ (-x^3 + 1)^(2/3)) - 10*2^(2/3)*(-1)^(1/3)*(x^3 - 1)*log(-2^(1/3)*(-1)^(2/3) + (-x^3 + 1)^(1/3)) - 12*(2*x^6
+ x^3 - 8)*(-x^3 + 1)^(2/3))/(x^3 - 1)

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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 {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(x**11/(-x**3+1)**(4/3)/(x**3+1),x)

[Out]

Integral(x**11/((-(x - 1)*(x**2 + x + 1))**(4/3)*(x + 1)*(x**2 - x + 1)), x)

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Giac [A]
time = 1.67, size = 120, normalized size = 0.92 \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}{5} \, {\left (-x^{3} + 1\right )}^{\frac {5}{3}} + \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}{2} \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + \frac {1}{2 \, {\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)^(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))) - 1/5*(-x^3 + 1)^(5/3) + 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))) + 1/2*(-x^3 + 1)^(2/3) + 1/2/(-x^3 + 1)^(1/3)

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Mupad [B]
time = 5.15, size = 139, normalized size = 1.07 \begin {gather*} \frac {1}{2\,{\left (1-x^3\right )}^{1/3}}-\frac {2^{2/3}\,\ln \left (\frac {{\left (1-x^3\right )}^{1/3}}{4}-\frac {2^{1/3}}{4}\right )}{12}+\frac {{\left (1-x^3\right )}^{2/3}}{2}-\frac {{\left (1-x^3\right )}^{5/3}}{5}-\frac {2^{2/3}\,\ln \left (\frac {{\left (1-x^3\right )}^{1/3}}{4}-\frac {2^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{24}+\frac {2^{2/3}\,\ln \left (\frac {{\left (1-x^3\right )}^{1/3}}{4}-\frac {2^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{24} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/((1 - x^3)^(4/3)*(x^3 + 1)),x)

[Out]

1/(2*(1 - x^3)^(1/3)) - (2^(2/3)*log((1 - x^3)^(1/3)/4 - 2^(1/3)/4))/12 + (1 - x^3)^(2/3)/2 - (1 - x^3)^(5/3)/
5 - (2^(2/3)*log((1 - x^3)^(1/3)/4 - (2^(1/3)*(3^(1/2)*1i - 1)^2)/16)*(3^(1/2)*1i - 1))/24 + (2^(2/3)*log((1 -
x^3)^(1/3)/4 - (2^(1/3)*(3^(1/2)*1i + 1)^2)/16)*(3^(1/2)*1i + 1))/24

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