\(\int \frac {-1+x}{x \sqrt [3]{-1+x^3}} \, dx\) [1972]

Optimal result

Integrand size = 16, antiderivative size = 140 \[ \int \frac {-1+x}{x \sqrt [3]{-1+x^3}} \, dx=\frac {\arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-1+x^3}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2}{3} \text {arctanh}\left (\frac {1+x}{-1+x-2 \sqrt [3]{-1+x^3}}\right )-\frac {1}{6} \log \left (1-\sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right )+\frac {1}{6} \log \left (x^2+x \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right ) \]

[Out]

1/3*arctan(3^(1/2)*x/(x+2*(x^3-1)^(1/3)))*3^(1/2)-1/3*arctan(-1/3*3^(1/2)+2/3*(x^3-1)^(1/3)*3^(1/2))*3^(1/2)-2
/3*arctanh((1+x)/(-1+x-2*(x^3-1)^(1/3)))-1/6*ln(1-(x^3-1)^(1/3)+(x^3-1)^(2/3))+1/6*ln(x^2+x*(x^3-1)^(1/3)+(x^3
-1)^(2/3))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.67, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {1858, 245, 272, 58, 632, 210, 31} \[ \int \frac {-1+x}{x \sqrt [3]{-1+x^3}} \, dx=\frac {\arctan \left (\frac {\frac {2 x}{\sqrt [3]{x^3-1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {1-2 \sqrt [3]{x^3-1}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \log \left (\sqrt [3]{x^3-1}+1\right )-\frac {1}{2} \log \left (\sqrt [3]{x^3-1}-x\right )-\frac {\log (x)}{2} \]

[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 58

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] && NegQ
[(b*c - a*d)/b]

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 245

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 272

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 632

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 1858

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*} \text {integral}& = \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 {\arctan \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} \text {Subst}\left (\int \frac {1}{\sqrt [3]{-1+x} x} \, dx,x,x^3\right ) \\ & = \frac {\arctan \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} \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\sqrt [3]{-1+x^3}\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\sqrt [3]{-1+x^3}\right ) \\ & = \frac {\arctan \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 )+\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \sqrt [3]{-1+x^3}\right ) \\ & = \frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{-1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \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*}

Mathematica [A] (verified)

Time = 4.29 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.98 \[ \int \frac {-1+x}{x \sqrt [3]{-1+x^3}} \, dx=\frac {1}{6} \left (2 \sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{-1+x^3}}{\sqrt {3}}\right )+2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^3}}\right )+4 \text {arctanh}\left (\frac {1+x}{1-x+2 \sqrt [3]{-1+x^3}}\right )-\log \left (1-\sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right )+\log \left (x^2+x \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right )\right ) \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 4.84 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.81

method	result	size
meijerg	\(-\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}} \left (\frac {2 \pi \sqrt {3}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {4}{3}\right ], \left [2, 2\right ], x^{3}\right )}{9 \Gamma \left (\frac {2}{3}\right )}+\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+3 \ln \left (x \right )+i \pi \right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}\right )}{6 \pi \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}}}+\frac {{\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}} x \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], x^{3}\right )}{\operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}}}\)	\(113\)
trager	\(\text {Expression too large to display}\)	\(2489\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.40 \[ \int \frac {-1+x}{x \sqrt [3]{-1+x^3}} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {2 \, \sqrt {3} {\left (x^{4} + 2 \, x^{3} + x^{2} - x - 1\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}} - 2 \, \sqrt {3} {\left (x^{5} + x^{4} - x^{3} - 2 \, x^{2} - x\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}} + \sqrt {3} {\left (x^{5} + 2 \, x^{4} + 2 \, x^{3} - x^{2} - 2 \, x - 1\right )}}{3 \, {\left (2 \, x^{6} + 3 \, x^{5} - 4 \, x^{3} - 3 \, x^{2} + 1\right )}}\right ) + \frac {1}{3} \, \log \left (-x^{3} - x^{2} - {\left (x^{3} - 1\right )}^{\frac {2}{3}} {\left (x + 1\right )} - {\left (x^{3} - 1\right )}^{\frac {1}{3}} {\left (x^{2} + x\right )} + 1\right ) - \frac {1}{6} \, \log \left (-x^{3} + {\left (x^{3} - 1\right )}^{\frac {2}{3}} {\left (x + 1\right )} - {\left (x^{3} - 1\right )}^{\frac {1}{3}} {\left (x + 1\right )} + x + 1\right ) \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.43 \[ \int \frac {-1+x}{x \sqrt [3]{-1+x^3}} \, dx=\frac {x e^{- \frac {i \pi }{3}} \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {\Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{3}}} \right )}}{3 x \Gamma \left (\frac {4}{3}\right )} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.89 \[ \int \frac {-1+x}{x \sqrt [3]{-1+x^3}} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}} - 1\right )}\right ) - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) - \frac {1}{6} \, \log \left ({\left (x^{3} - 1\right )}^{\frac {2}{3}} - {\left (x^{3} - 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left (x^{3} - 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{6} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) - \frac {1}{3} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} - 1\right ) \]

[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)

Giac [F]

\[ \int \frac {-1+x}{x \sqrt [3]{-1+x^3}} \, dx=\int { \frac {x - 1}{{\left (x^{3} - 1\right )}^{\frac {1}{3}} x} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.81 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.71 \[ \int \frac {-1+x}{x \sqrt [3]{-1+x^3}} \, dx=\frac {\ln \left ({\left (x^3-1\right )}^{1/3}+1\right )}{3}+\ln \left (9\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2+{\left (x^3-1\right )}^{1/3}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (9\,{\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2+{\left (x^3-1\right )}^{1/3}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\frac {x\,{\left (1-x^3\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ x^3\right )}{{\left (x^3-1\right )}^{1/3}} \]

[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)