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

Optimal result

Integrand size = 16, antiderivative size = 92 \[ \int \frac {1+x}{x \sqrt [3]{-1+x^3}} \, dx=-\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{-1+x^3}}{-2+2 x+\sqrt [3]{-1+x^3}}\right )-\log \left (1-x+\sqrt [3]{-1+x^3}\right )+\frac {1}{2} \log \left (1-2 x+x^2+(-1+x) \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.03, 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 = 0.47 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00 \[ \int \frac {1+x}{x \sqrt [3]{-1+x^3}} \, dx=-\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{-1+x^3}}{-2+2 x+\sqrt [3]{-1+x^3}}\right )-\log \left (1-x+\sqrt [3]{-1+x^3}\right )+\frac {1}{2} \log \left (1-2 x+x^2+(-1+x) \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right ) \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.94 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.23

method	result	size
meijerg	\(\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}} \left (\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+3 \ln \left (x \right )+i \pi \right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}+\frac {2 \pi \sqrt {3}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {4}{3}\right ], \left [2, 2\right ], x^{3}\right )}{9 \Gamma \left (\frac {2}{3}\right )}\right )}{6 \pi \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}}}+\frac {{\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}} x \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], x^{3}\right )}{\operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}}}\)	\(113\)
trager	\(-\ln \left (\frac {12 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}-30 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x +99 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}}-127 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x +16 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+12 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2}+127 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}}-97 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +28 \left (x^{3}-1\right )^{\frac {2}{3}}+99 \left (x^{3}-1\right )^{\frac {1}{3}} x -115 x^{2}+16 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-99 \left (x^{3}-1\right )^{\frac {1}{3}}+69 x -115}{x}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (\frac {46 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}-115 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x -99 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}}-28 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x +81 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+46 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2}+28 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}}+16 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +127 \left (x^{3}-1\right )^{\frac {2}{3}}-99 \left (x^{3}-1\right )^{\frac {1}{3}} x +18 x^{2}+81 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+99 \left (x^{3}-1\right )^{\frac {1}{3}}+12 x +18}{x}\right )\)	\(370\)

[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)*(2/3*(-1/6*Pi*3^(1/2)-3/2*ln(3)+3*ln(x)+I
*Pi)*Pi*3^(1/2)/GAMMA(2/3)+2/9*Pi*3^(1/2)/GAMMA(2/3)*x^3*hypergeom([1,1,4/3],[2,2],x^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 = 0.38 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95 \[ \int \frac {1+x}{x \sqrt [3]{-1+x^3}} \, dx=\sqrt {3} \arctan \left (-\frac {4 \, \sqrt {3} {\left (x^{3} - 1\right )}^{\frac {1}{3}} {\left (x - 1\right )} + \sqrt {3} {\left (x^{2} + x + 1\right )} - 2 \, \sqrt {3} {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{3 \, {\left (3 \, x^{2} - 5 \, x + 3\right )}}\right ) - \frac {1}{2} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}} {\left (x - 1\right )} + x - {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x}\right ) \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.39 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.65 \[ \int \frac {1+x}{x \sqrt [3]{-1+x^3}} \, dx=\frac {x e^{- \frac {i \pi }{3}} \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} - \frac {\Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{3}}} \right )}}{3 x \Gamma \left (\frac {4}{3}\right )} \]

[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 = 0.23 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.35 \[ \int \frac {1+x}{x \sqrt [3]{-1+x^3}} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}} - 1\right )}\right ) - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) + \frac {1}{6} \, \log \left ({\left (x^{3} - 1\right )}^{\frac {2}{3}} - {\left (x^{3} - 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{3} \, \log \left ({\left (x^{3} - 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{6} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) - \frac {1}{3} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} - 1\right ) \]

[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 = 0.46 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.09 \[ \int \frac {1+x}{x \sqrt [3]{-1+x^3}} \, dx=\frac {x\,{\left (1-x^3\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ x^3\right )}{{\left (x^3-1\right )}^{1/3}}-\ln \left (9\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2+{\left (x^3-1\right )}^{1/3}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\ln \left (9\,{\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2+{\left (x^3-1\right )}^{1/3}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\frac {\ln \left ({\left (x^3-1\right )}^{1/3}+1\right )}{3} \]

[In]

int((x + 1)/(x*(x^3 - 1)^(1/3)),x)

[Out]

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