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

Optimal. Leaf size=140 \[ -\frac {1}{6} \log \left (\left (x^3-1\right )^{2/3}-\sqrt [3]{x^3-1}+1\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3-1}+x}\right )}{\sqrt {3}}+\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{x^3-1}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2}{3} \tanh ^{-1}\left (\frac {x+1}{-2 \sqrt [3]{x^3-1}+x-1}\right )+\frac {1}{6} \log \left (\sqrt [3]{x^3-1} x+\left (x^3-1\right )^{2/3}+x^2\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 94, normalized size of antiderivative = 0.67, number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {1844, 239, 266, 56, 618, 204, 31} \begin {gather*} \frac {1}{2} \log \left (\sqrt [3]{x^3-1}+1\right )-\frac {1}{2} \log \left (\sqrt [3]{x^3-1}-x\right )+\frac {\tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{x^3-1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\tan ^{-1}\left (\frac {1-2 \sqrt [3]{x^3-1}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\log (x)}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + x)/(x*(-1 + x^3)^(1/3)),x]

[Out]

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

Rule 31

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

Rule 56

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, Simp
[Log[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] && Ne
gQ[(b*c - a*d)/b]

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 239

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + (2*Rt[b, 3]*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 266

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

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]

Rule 1844

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !IGtQ[m, 0]

Rubi steps

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

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Mathematica [C]  time = 0.06, size = 103, normalized size = 0.74 \begin {gather*} \frac {1}{6} \left (-3 \left (x^3-1\right )^{2/3} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};1-x^3\right )-2 \log \left (1-\frac {x}{\sqrt [3]{x^3-1}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{x^3-1}}+1}{\sqrt {3}}\right )+\log \left (\frac {x}{\sqrt [3]{x^3-1}}+\frac {x^2}{\left (x^3-1\right )^{2/3}}+1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[3]*ArcTan[(1 + (2*x)/(-1 + x^3)^(1/3))/Sqrt[3]] - 3*(-1 + x^3)^(2/3)*Hypergeometric2F1[2/3, 1, 5/3, 1
- x^3] - 2*Log[1 - x/(-1 + x^3)^(1/3)] + Log[1 + x^2/(-1 + x^3)^(2/3) + x/(-1 + x^3)^(1/3)])/6

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IntegrateAlgebraic [A]  time = 3.53, size = 140, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^3}}\right )}{\sqrt {3}}+\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-1+x^3}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2}{3} \tanh ^{-1}\left (\frac {1+x}{-1+x-2 \sqrt [3]{-1+x^3}}\right )-\frac {1}{6} \log \left (1-\sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right )+\frac {1}{6} \log \left (x^2+x \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.04, size = 196, normalized size = 1.40 \begin {gather*} \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {2 \, \sqrt {3} {\left (x^{4} + 2 \, x^{3} + x^{2} - x - 1\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}} - 2 \, \sqrt {3} {\left (x^{5} + x^{4} - x^{3} - 2 \, x^{2} - x\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}} + \sqrt {3} {\left (x^{5} + 2 \, x^{4} + 2 \, x^{3} - x^{2} - 2 \, x - 1\right )}}{3 \, {\left (2 \, x^{6} + 3 \, x^{5} - 4 \, x^{3} - 3 \, x^{2} + 1\right )}}\right ) + \frac {1}{3} \, \log \left (-x^{3} - x^{2} - {\left (x^{3} - 1\right )}^{\frac {2}{3}} {\left (x + 1\right )} - {\left (x^{3} - 1\right )}^{\frac {1}{3}} {\left (x^{2} + x\right )} + 1\right ) - \frac {1}{6} \, \log \left (-x^{3} + {\left (x^{3} - 1\right )}^{\frac {2}{3}} {\left (x + 1\right )} - {\left (x^{3} - 1\right )}^{\frac {1}{3}} {\left (x + 1\right )} + x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x - 1}{{\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/(x^3-1)^(1/3),x, algorithm="giac")

[Out]

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

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maple [C]  time = 8.23, size = 113, normalized size = 0.81

method	result	size
meijerg	\(-\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {1}{3}} \left (\frac {2 \pi \sqrt {3}\, x^{3} \hypergeom \left (\left [1, 1, \frac {4}{3}\right ], \left [2, 2\right ], x^{3}\right )}{9 \Gamma \left (\frac {2}{3}\right )}+\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+3 \ln \relax (x )+i \pi \right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}\right )}{6 \pi \mathrm {signum}\left (x^{3}-1\right )^{\frac {1}{3}}}+\frac {\left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {1}{3}} x \hypergeom \left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], x^{3}\right )}{\mathrm {signum}\left (x^{3}-1\right )^{\frac {1}{3}}}\)	\(113\)
trager	\(\text {Expression too large to display}\)	\(2269\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/Pi*3^(1/2)*GAMMA(2/3)/signum(x^3-1)^(1/3)*(-signum(x^3-1))^(1/3)*(2/9*Pi*3^(1/2)/GAMMA(2/3)*x^3*hypergeom
([1,1,4/3],[2,2],x^3)+2/3*(-1/6*Pi*3^(1/2)-3/2*ln(3)+3*ln(x)+I*Pi)*Pi*3^(1/2)/GAMMA(2/3))+1/signum(x^3-1)^(1/3
)*(-signum(x^3-1))^(1/3)*x*hypergeom([1/3,1/3],[4/3],x^3)

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maxima [A]  time = 0.41, size = 124, normalized size = 0.89 \begin {gather*} -\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}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + 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 (x^{3} - 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{6} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) - \frac {1}{3} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.24, size = 100, normalized size = 0.71 \begin {gather*} \frac {\ln \left ({\left (x^3-1\right )}^{1/3}+1\right )}{3}+\ln \left (9\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2+{\left (x^3-1\right )}^{1/3}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (9\,{\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2+{\left (x^3-1\right )}^{1/3}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\frac {x\,{\left (1-x^3\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ x^3\right )}{{\left (x^3-1\right )}^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 1)/(x*(x^3 - 1)^(1/3)),x)

[Out]

log((x^3 - 1)^(1/3) + 1)/3 + log(9*((3^(1/2)*1i)/6 - 1/6)^2 + (x^3 - 1)^(1/3))*((3^(1/2)*1i)/6 - 1/6) - log(9*
((3^(1/2)*1i)/6 + 1/6)^2 + (x^3 - 1)^(1/3))*((3^(1/2)*1i)/6 + 1/6) + (x*(1 - x^3)^(1/3)*hypergeom([1/3, 1/3],
4/3, x^3))/(x^3 - 1)^(1/3)

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sympy [C]  time = 2.42, size = 60, normalized size = 0.43 \begin {gather*} \frac {x e^{- \frac {i \pi }{3}} \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {\Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{3}}} \right )}}{3 x \Gamma \left (\frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(-I*pi/3)*gamma(1/3)*hyper((1/3, 1/3), (4/3,), x**3)/(3*gamma(4/3)) + gamma(1/3)*hyper((1/3, 1/3), (4/3,)
, exp_polar(2*I*pi)/x**3)/(3*x*gamma(4/3))

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