\(\int \frac {\sqrt [3]{-1+x^3} (-1+2 x^3)}{x^8} \, dx\) [316]

   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 = 20, antiderivative size = 28 \[ \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^8} \, dx=\frac {\sqrt [3]{-1+x^3} \left (4-15 x^3+11 x^6\right )}{28 x^7} \]

[Out]

1/28*(x^3-1)^(1/3)*(11*x^6-15*x^3+4)/x^7

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {464, 270} \[ \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^8} \, dx=\frac {11 \left (x^3-1\right )^{4/3}}{28 x^4}-\frac {\left (x^3-1\right )^{4/3}}{7 x^7} \]

[In]

Int[((-1 + x^3)^(1/3)*(-1 + 2*x^3))/x^8,x]

[Out]

-1/7*(-1 + x^3)^(4/3)/x^7 + (11*(-1 + x^3)^(4/3))/(28*x^4)

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]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (-1+x^3\right )^{4/3}}{7 x^7}+\frac {11}{7} \int \frac {\sqrt [3]{-1+x^3}}{x^5} \, dx \\ & = -\frac {\left (-1+x^3\right )^{4/3}}{7 x^7}+\frac {11 \left (-1+x^3\right )^{4/3}}{28 x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^8} \, dx=\frac {\left (-1+x^3\right )^{4/3} \left (-4+11 x^3\right )}{28 x^7} \]

[In]

Integrate[((-1 + x^3)^(1/3)*(-1 + 2*x^3))/x^8,x]

[Out]

((-1 + x^3)^(4/3)*(-4 + 11*x^3))/(28*x^7)

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71

method result size
pseudoelliptic \(\frac {\left (x^{3}-1\right )^{\frac {4}{3}} \left (11 x^{3}-4\right )}{28 x^{7}}\) \(20\)
trager \(\frac {\left (x^{3}-1\right )^{\frac {1}{3}} \left (11 x^{6}-15 x^{3}+4\right )}{28 x^{7}}\) \(25\)
gosper \(\frac {\left (x -1\right ) \left (x^{2}+x +1\right ) \left (11 x^{3}-4\right ) \left (x^{3}-1\right )^{\frac {1}{3}}}{28 x^{7}}\) \(29\)
risch \(\frac {11 x^{9}-26 x^{6}+19 x^{3}-4}{28 \left (x^{3}-1\right )^{\frac {2}{3}} x^{7}}\) \(30\)
meijerg \(-\frac {\operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}} \left (-x^{3}+1\right )^{\frac {4}{3}}}{2 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}} x^{4}}+\frac {\operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}} \left (-\frac {3}{4} x^{6}-\frac {1}{4} x^{3}+1\right ) \left (-x^{3}+1\right )^{\frac {1}{3}}}{7 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}} x^{7}}\) \(78\)

[In]

int((x^3-1)^(1/3)*(2*x^3-1)/x^8,x,method=_RETURNVERBOSE)

[Out]

1/28*(x^3-1)^(4/3)*(11*x^3-4)/x^7

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^8} \, dx=\frac {{\left (11 \, x^{6} - 15 \, x^{3} + 4\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{28 \, x^{7}} \]

[In]

integrate((x^3-1)^(1/3)*(2*x^3-1)/x^8,x, algorithm="fricas")

[Out]

1/28*(11*x^6 - 15*x^3 + 4)*(x^3 - 1)^(1/3)/x^7

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.24 (sec) , antiderivative size = 425, normalized size of antiderivative = 15.18 \[ \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^8} \, dx=2 \left (\begin {cases} \frac {\sqrt [3]{-1 + \frac {1}{x^{3}}} e^{- \frac {2 i \pi }{3}} \Gamma \left (- \frac {4}{3}\right )}{3 \Gamma \left (- \frac {1}{3}\right )} - \frac {\sqrt [3]{-1 + \frac {1}{x^{3}}} e^{- \frac {2 i \pi }{3}} \Gamma \left (- \frac {4}{3}\right )}{3 x^{3} \Gamma \left (- \frac {1}{3}\right )} & \text {for}\: \frac {1}{\left |{x^{3}}\right |} > 1 \\- \frac {\sqrt [3]{1 - \frac {1}{x^{3}}} \Gamma \left (- \frac {4}{3}\right )}{3 \Gamma \left (- \frac {1}{3}\right )} + \frac {\sqrt [3]{1 - \frac {1}{x^{3}}} \Gamma \left (- \frac {4}{3}\right )}{3 x^{3} \Gamma \left (- \frac {1}{3}\right )} & \text {otherwise} \end {cases}\right ) - \begin {cases} \frac {3 x^{6} \sqrt [3]{-1 + \frac {1}{x^{3}}} e^{\frac {i \pi }{3}} \Gamma \left (- \frac {7}{3}\right )}{9 x^{6} \Gamma \left (- \frac {1}{3}\right ) - 9 x^{3} \Gamma \left (- \frac {1}{3}\right )} - \frac {2 x^{3} \sqrt [3]{-1 + \frac {1}{x^{3}}} e^{\frac {i \pi }{3}} \Gamma \left (- \frac {7}{3}\right )}{9 x^{6} \Gamma \left (- \frac {1}{3}\right ) - 9 x^{3} \Gamma \left (- \frac {1}{3}\right )} + \frac {4 \sqrt [3]{-1 + \frac {1}{x^{3}}} e^{\frac {i \pi }{3}} \Gamma \left (- \frac {7}{3}\right )}{9 x^{9} \Gamma \left (- \frac {1}{3}\right ) - 9 x^{6} \Gamma \left (- \frac {1}{3}\right )} - \frac {5 \sqrt [3]{-1 + \frac {1}{x^{3}}} e^{\frac {i \pi }{3}} \Gamma \left (- \frac {7}{3}\right )}{9 x^{6} \Gamma \left (- \frac {1}{3}\right ) - 9 x^{3} \Gamma \left (- \frac {1}{3}\right )} & \text {for}\: \frac {1}{\left |{x^{3}}\right |} > 1 \\\frac {\sqrt [3]{1 - \frac {1}{x^{3}}} \Gamma \left (- \frac {7}{3}\right )}{3 \Gamma \left (- \frac {1}{3}\right )} + \frac {\sqrt [3]{1 - \frac {1}{x^{3}}} \Gamma \left (- \frac {7}{3}\right )}{9 x^{3} \Gamma \left (- \frac {1}{3}\right )} - \frac {4 \sqrt [3]{1 - \frac {1}{x^{3}}} \Gamma \left (- \frac {7}{3}\right )}{9 x^{6} \Gamma \left (- \frac {1}{3}\right )} & \text {otherwise} \end {cases} \]

[In]

integrate((x**3-1)**(1/3)*(2*x**3-1)/x**8,x)

[Out]

2*Piecewise(((-1 + x**(-3))**(1/3)*exp(-2*I*pi/3)*gamma(-4/3)/(3*gamma(-1/3)) - (-1 + x**(-3))**(1/3)*exp(-2*I
*pi/3)*gamma(-4/3)/(3*x**3*gamma(-1/3)), 1/Abs(x**3) > 1), (-(1 - 1/x**3)**(1/3)*gamma(-4/3)/(3*gamma(-1/3)) +
 (1 - 1/x**3)**(1/3)*gamma(-4/3)/(3*x**3*gamma(-1/3)), True)) - Piecewise((3*x**6*(-1 + x**(-3))**(1/3)*exp(I*
pi/3)*gamma(-7/3)/(9*x**6*gamma(-1/3) - 9*x**3*gamma(-1/3)) - 2*x**3*(-1 + x**(-3))**(1/3)*exp(I*pi/3)*gamma(-
7/3)/(9*x**6*gamma(-1/3) - 9*x**3*gamma(-1/3)) + 4*(-1 + x**(-3))**(1/3)*exp(I*pi/3)*gamma(-7/3)/(9*x**9*gamma
(-1/3) - 9*x**6*gamma(-1/3)) - 5*(-1 + x**(-3))**(1/3)*exp(I*pi/3)*gamma(-7/3)/(9*x**6*gamma(-1/3) - 9*x**3*ga
mma(-1/3)), 1/Abs(x**3) > 1), ((1 - 1/x**3)**(1/3)*gamma(-7/3)/(3*gamma(-1/3)) + (1 - 1/x**3)**(1/3)*gamma(-7/
3)/(9*x**3*gamma(-1/3)) - 4*(1 - 1/x**3)**(1/3)*gamma(-7/3)/(9*x**6*gamma(-1/3)), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^8} \, dx=\frac {{\left (x^{3} - 1\right )}^{\frac {4}{3}}}{4 \, x^{4}} + \frac {{\left (x^{3} - 1\right )}^{\frac {7}{3}}}{7 \, x^{7}} \]

[In]

integrate((x^3-1)^(1/3)*(2*x^3-1)/x^8,x, algorithm="maxima")

[Out]

1/4*(x^3 - 1)^(4/3)/x^4 + 1/7*(x^3 - 1)^(7/3)/x^7

Giac [F]

\[ \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^8} \, dx=\int { \frac {{\left (2 \, x^{3} - 1\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x^{8}} \,d x } \]

[In]

integrate((x^3-1)^(1/3)*(2*x^3-1)/x^8,x, algorithm="giac")

[Out]

integrate((2*x^3 - 1)*(x^3 - 1)^(1/3)/x^8, x)

Mupad [B] (verification not implemented)

Time = 5.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^8} \, dx=\frac {4\,{\left (x^3-1\right )}^{1/3}-15\,x^3\,{\left (x^3-1\right )}^{1/3}+11\,x^6\,{\left (x^3-1\right )}^{1/3}}{28\,x^7} \]

[In]

int(((x^3 - 1)^(1/3)*(2*x^3 - 1))/x^8,x)

[Out]

(4*(x^3 - 1)^(1/3) - 15*x^3*(x^3 - 1)^(1/3) + 11*x^6*(x^3 - 1)^(1/3))/(28*x^7)