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

Optimal. Leaf size=154 \[ \frac {1}{2 \sqrt [3]{1-x^3}}+\frac {\log \left (x^3+1\right )}{12 \sqrt [3]{2}}+\frac {1}{2} \log \left (1-\sqrt [3]{1-x^3}\right )-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}-\frac {\log (x)}{2} \]

[Out]

1/2/(-x^3+1)^(1/3)-1/2*ln(x)+1/24*ln(x^3+1)*2^(2/3)+1/2*ln(1-(-x^3+1)^(1/3))-1/8*ln(2^(1/3)-(-x^3+1)^(1/3))*2^
(2/3)+1/3*arctan(1/3*(1+2*(-x^3+1)^(1/3))*3^(1/2))*3^(1/2)-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 = 0.11, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {446, 85, 156, 55, 618, 204, 31, 617} \[ \frac {1}{2 \sqrt [3]{1-x^3}}+\frac {\log \left (x^3+1\right )}{12 \sqrt [3]{2}}+\frac {1}{2} \log \left (1-\sqrt [3]{1-x^3}\right )-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}-\frac {\log (x)}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(2*(1 - x^3)^(1/3)) + 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*2^(1/3)*Sqrt[3]) - Log[x]/2 + Log[1 + x^3]/(12*2^(1/3)) + Log[1 - (1 - x^3)^(1/3)]/2 - 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 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 85

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

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

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Mathematica [C]  time = 0.02, size = 54, normalized size = 0.35 \[ \frac {2 \, _2F_1\left (-\frac {1}{3},1;\frac {2}{3};1-x^3\right )-\, _2F_1\left (-\frac {1}{3},1;\frac {2}{3};\frac {1}{2} \left (1-x^3\right )\right )}{2 \sqrt [3]{1-x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.78, size = 226, normalized size = 1.47 \[ -\frac {2 \, \sqrt {6} 2^{\frac {1}{6}} \left (-1\right )^{\frac {1}{3}} {\left (x^{3} - 1\right )} \arctan \left (\frac {1}{6} \cdot 2^{\frac {1}{6}} {\left (2 \, \sqrt {6} \left (-1\right )^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - \sqrt {6} 2^{\frac {1}{3}}\right )}\right ) + 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{3} - 1\right )} \log \left (2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) - 2 \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{3} - 1\right )} \log \left (-2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) - 8 \, \sqrt {3} {\left (x^{3} - 1\right )} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + 4 \, {\left (x^{3} - 1\right )} \log \left ({\left (-x^{3} + 1\right )}^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) - 8 \, {\left (x^{3} - 1\right )} \log \left ({\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 1\right ) + 12 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{24 \, {\left (x^{3} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 160, normalized size = 1.04 \[ -\frac {1}{12} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) + \frac {1}{24} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) - \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left ({\left | -2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} \right |}\right ) + \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}} - \frac {1}{6} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left | {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(-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))) + 1/3*sqrt(3
)*arctan(1/3*sqrt(3)*(2*(-x^3 + 1)^(1/3) + 1)) + 1/2/(-x^3 + 1)^(1/3) - 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 = 2.72, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-x^{3}+1\right )^{\frac {4}{3}} \left (x^{3}+1\right ) x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{3} + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {4}{3}} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.40, size = 253, normalized size = 1.64 \[ \frac {\ln \left (\frac {17}{4}-\frac {17\,{\left (1-x^3\right )}^{1/3}}{4}\right )}{3}+\ln \left (\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,\left (1458\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2-\frac {459\,{\left (1-x^3\right )}^{1/3}}{4}\right )-\frac {63}{4}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,\left (1458\,{\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2-\frac {459\,{\left (1-x^3\right )}^{1/3}}{4}\right )+\frac {63}{4}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\frac {2^{2/3}\,\ln \left (\frac {2^{2/3}\,\left (\frac {81\,2^{1/3}}{4}-\frac {459\,{\left (1-x^3\right )}^{1/3}}{4}\right )}{12}+\frac {63}{4}\right )}{12}+\frac {1}{2\,{\left (1-x^3\right )}^{1/3}}+\frac {{\left (-1\right )}^{1/3}\,2^{2/3}\,\ln \left (\frac {{\left (-1\right )}^{1/3}\,2^{2/3}\,\left (\frac {81\,{\left (-1\right )}^{2/3}\,2^{1/3}}{4}-\frac {459\,{\left (1-x^3\right )}^{1/3}}{4}\right )}{12}-\frac {63}{4}\right )}{12}-\frac {{\left (-1\right )}^{1/3}\,2^{2/3}\,\ln \left (\frac {{\left (-1\right )}^{1/3}\,2^{2/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (\frac {459\,{\left (1-x^3\right )}^{1/3}}{4}-\frac {81\,{\left (-1\right )}^{2/3}\,2^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )}{24}-\frac {63}{4}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{24} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(17/4 - (17*(1 - x^3)^(1/3))/4)/3 + log(((3^(1/2)*1i)/6 - 1/6)*(1458*((3^(1/2)*1i)/6 - 1/6)^2 - (459*(1 - x
^3)^(1/3))/4) - 63/4)*((3^(1/2)*1i)/6 - 1/6) - log(((3^(1/2)*1i)/6 + 1/6)*(1458*((3^(1/2)*1i)/6 + 1/6)^2 - (45
9*(1 - x^3)^(1/3))/4) + 63/4)*((3^(1/2)*1i)/6 + 1/6) - (2^(2/3)*log((2^(2/3)*((81*2^(1/3))/4 - (459*(1 - x^3)^
(1/3))/4))/12 + 63/4))/12 + 1/(2*(1 - x^3)^(1/3)) + ((-1)^(1/3)*2^(2/3)*log(((-1)^(1/3)*2^(2/3)*((81*(-1)^(2/3
)*2^(1/3))/4 - (459*(1 - x^3)^(1/3))/4))/12 - 63/4))/12 - ((-1)^(1/3)*2^(2/3)*log(((-1)^(1/3)*2^(2/3)*(3^(1/2)
*1i + 1)*((459*(1 - x^3)^(1/3))/4 - (81*(-1)^(2/3)*2^(1/3)*(3^(1/2)*1i + 1)^2)/16))/24 - 63/4)*(3^(1/2)*1i + 1
))/24

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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