\(\int \frac {(-1+x^3)^{2/3}}{x^6} \, dx\) [71]

   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 [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 16 \[ \int \frac {\left (-1+x^3\right )^{2/3}}{x^6} \, dx=\frac {\left (-1+x^3\right )^{5/3}}{5 x^5} \]

[Out]

1/5*(x^3-1)^(5/3)/x^5

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {270} \[ \int \frac {\left (-1+x^3\right )^{2/3}}{x^6} \, dx=\frac {\left (x^3-1\right )^{5/3}}{5 x^5} \]

[In]

Int[(-1 + x^3)^(2/3)/x^6,x]

[Out]

(-1 + x^3)^(5/3)/(5*x^5)

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (-1+x^3\right )^{5/3}}{5 x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-1+x^3\right )^{2/3}}{x^6} \, dx=\frac {\left (-1+x^3\right )^{5/3}}{5 x^5} \]

[In]

Integrate[(-1 + x^3)^(2/3)/x^6,x]

[Out]

(-1 + x^3)^(5/3)/(5*x^5)

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method result size
trager \(\frac {\left (x^{3}-1\right )^{\frac {5}{3}}}{5 x^{5}}\) \(13\)
pseudoelliptic \(\frac {\left (x^{3}-1\right )^{\frac {5}{3}}}{5 x^{5}}\) \(13\)
gosper \(\frac {\left (x -1\right ) \left (x^{2}+x +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}}}{5 x^{5}}\) \(22\)
risch \(\frac {x^{6}-2 x^{3}+1}{5 x^{5} \left (x^{3}-1\right )^{\frac {1}{3}}}\) \(23\)
meijerg \(-\frac {\operatorname {signum}\left (x^{3}-1\right )^{\frac {2}{3}} \left (-x^{3}+1\right )^{\frac {5}{3}}}{5 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {2}{3}} x^{5}}\) \(33\)

[In]

int((x^3-1)^(2/3)/x^6,x,method=_RETURNVERBOSE)

[Out]

1/5*(x^3-1)^(5/3)/x^5

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {\left (-1+x^3\right )^{2/3}}{x^6} \, dx=\frac {{\left (x^{3} - 1\right )}^{\frac {5}{3}}}{5 \, x^{5}} \]

[In]

integrate((x^3-1)^(2/3)/x^6,x, algorithm="fricas")

[Out]

1/5*(x^3 - 1)^(5/3)/x^5

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.51 (sec) , antiderivative size = 126, normalized size of antiderivative = 7.88 \[ \int \frac {\left (-1+x^3\right )^{2/3}}{x^6} \, dx=\begin {cases} \frac {\left (-1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} e^{- \frac {i \pi }{3}} \Gamma \left (- \frac {5}{3}\right )}{3 \Gamma \left (- \frac {2}{3}\right )} - \frac {\left (-1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} e^{- \frac {i \pi }{3}} \Gamma \left (- \frac {5}{3}\right )}{3 x^{3} \Gamma \left (- \frac {2}{3}\right )} & \text {for}\: \frac {1}{\left |{x^{3}}\right |} > 1 \\- \frac {\left (1 - \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{3 \Gamma \left (- \frac {2}{3}\right )} + \frac {\left (1 - \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{3 x^{3} \Gamma \left (- \frac {2}{3}\right )} & \text {otherwise} \end {cases} \]

[In]

integrate((x**3-1)**(2/3)/x**6,x)

[Out]

Piecewise(((-1 + x**(-3))**(2/3)*exp(-I*pi/3)*gamma(-5/3)/(3*gamma(-2/3)) - (-1 + x**(-3))**(2/3)*exp(-I*pi/3)
*gamma(-5/3)/(3*x**3*gamma(-2/3)), 1/Abs(x**3) > 1), (-(1 - 1/x**3)**(2/3)*gamma(-5/3)/(3*gamma(-2/3)) + (1 -
1/x**3)**(2/3)*gamma(-5/3)/(3*x**3*gamma(-2/3)), True))

Maxima [A] (verification not implemented)

none

Time = 0.16 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {\left (-1+x^3\right )^{2/3}}{x^6} \, dx=\frac {{\left (x^{3} - 1\right )}^{\frac {5}{3}}}{5 \, x^{5}} \]

[In]

integrate((x^3-1)^(2/3)/x^6,x, algorithm="maxima")

[Out]

1/5*(x^3 - 1)^(5/3)/x^5

Giac [F]

\[ \int \frac {\left (-1+x^3\right )^{2/3}}{x^6} \, dx=\int { \frac {{\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x^{6}} \,d x } \]

[In]

integrate((x^3-1)^(2/3)/x^6,x, algorithm="giac")

[Out]

integrate((x^3 - 1)^(2/3)/x^6, x)

Mupad [B] (verification not implemented)

Time = 5.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \frac {\left (-1+x^3\right )^{2/3}}{x^6} \, dx=-\frac {{\left (x^3-1\right )}^{2/3}-x^3\,{\left (x^3-1\right )}^{2/3}}{5\,x^5} \]

[In]

int((x^3 - 1)^(2/3)/x^6,x)

[Out]

-((x^3 - 1)^(2/3) - x^3*(x^3 - 1)^(2/3))/(5*x^5)