\(\int \frac {1}{x \sqrt [3]{1-x^3}} \, dx\) [36]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [C] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 15, antiderivative size = 55 \[ \int \frac {1}{x \sqrt [3]{1-x^3}} \, dx=\frac {\arctan \left (\frac {1+2 \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\log (x)}{2}+\frac {1}{2} \log \left (1-\sqrt [3]{1-x^3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 57, 632, 210, 31} \[ \int \frac {1}{x \sqrt [3]{1-x^3}} \, dx=\frac {\arctan \left (\frac {2 \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \log \left (1-\sqrt [3]{1-x^3}\right )-\frac {\log (x)}{2} \]

[In]

Int[1/(x*(1 - x^3)^(1/3)),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.44 \[ \int \frac {1}{x \sqrt [3]{1-x^3}} \, dx=\frac {\arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (-1+\sqrt [3]{1-x^3}\right )-\frac {1}{6} \log \left (1+\sqrt [3]{1-x^3}+\left (1-x^3\right )^{2/3}\right ) \]

[In]

Integrate[1/(x*(1 - x^3)^(1/3)),x]

[Out]

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

Maple [A] (verified)

Time = 2.00 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.15

method result size
pseudoelliptic \(-\frac {\ln \left (\left (-x^{3}+1\right )^{\frac {2}{3}}+\left (-x^{3}+1\right )^{\frac {1}{3}}+1\right )}{6}+\frac {\arctan \left (\frac {\left (1+2 \left (-x^{3}+1\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}+\frac {\ln \left (\left (-x^{3}+1\right )^{\frac {1}{3}}-1\right )}{3}\) \(63\)
meijerg \(\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+3 \ln \left (x \right )+i \pi \right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}+\frac {2 \pi \sqrt {3}\, x^{3} {}_{3}^{}{\moversetsp {}{\mundersetsp {}{F_{2}^{}}}}\left (1,1,\frac {4}{3};2,2;x^{3}\right )}{9 \Gamma \left (\frac {2}{3}\right )}\right )}{6 \pi }\) \(65\)
trager \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\frac {-1438 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-9855 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+5502 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {2}{3}}+1477 x^{3}-14247 \left (-x^{3}+1\right )^{\frac {2}{3}}+5502 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {1}{3}}+11504 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-14247 \left (-x^{3}+1\right )^{\frac {1}{3}}+17006 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-2743}{x^{3}}\right )}{3}-\frac {\ln \left (-\frac {1438 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-6979 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+5502 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {2}{3}}-9894 x^{3}+19749 \left (-x^{3}+1\right )^{\frac {2}{3}}+5502 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {1}{3}}-11504 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}+19749 \left (-x^{3}+1\right )^{\frac {1}{3}}-6002 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+8245}{x^{3}}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )}{3}-\frac {\ln \left (-\frac {1438 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-6979 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+5502 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {2}{3}}-9894 x^{3}+19749 \left (-x^{3}+1\right )^{\frac {2}{3}}+5502 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {1}{3}}-11504 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}+19749 \left (-x^{3}+1\right )^{\frac {1}{3}}-6002 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+8245}{x^{3}}\right )}{3}\) \(372\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.16 \[ \int \frac {1}{x \sqrt [3]{1-x^3}} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{6} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.45 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.58 \[ \int \frac {1}{x \sqrt [3]{1-x^3}} \, dx=- \frac {e^{- \frac {i \pi }{3}} \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {1}{x^{3}}} \right )}}{3 x \Gamma \left (\frac {4}{3}\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x \sqrt [3]{1-x^3}} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{6} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x \sqrt [3]{1-x^3}} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{6} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left | {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.45 \[ \int \frac {1}{x \sqrt [3]{1-x^3}} \, dx=\frac {\ln \left ({\left (1-x^3\right )}^{1/3}-1\right )}{3}+\ln \left ({\left (1-x^3\right )}^{1/3}-9\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left ({\left (1-x^3\right )}^{1/3}-9\,{\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \]

[In]

int(1/(x*(1 - x^3)^(1/3)),x)

[Out]

log((1 - x^3)^(1/3) - 1)/3 + log((1 - x^3)^(1/3) - 9*((3^(1/2)*1i)/6 - 1/6)^2)*((3^(1/2)*1i)/6 - 1/6) - log((1
 - x^3)^(1/3) - 9*((3^(1/2)*1i)/6 + 1/6)^2)*((3^(1/2)*1i)/6 + 1/6)