\(\int \frac {e^{64 x^3} (-4+768 x^3)}{x^2} \, dx\) [7457]

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

Optimal result

Integrand size = 18, antiderivative size = 15 \[ \int \frac {e^{64 x^3} \left (-4+768 x^3\right )}{x^2} \, dx=4 \left (2+\frac {e^{64 x^3}}{x}\right ) \]

[Out]

8+4*exp(64*x^3)/x

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2326} \[ \int \frac {e^{64 x^3} \left (-4+768 x^3\right )}{x^2} \, dx=\frac {4 e^{64 x^3}}{x} \]

[In]

Int[(E^(64*x^3)*(-4 + 768*x^3))/x^2,x]

[Out]

(4*E^(64*x^3))/x

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {4 e^{64 x^3}}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {e^{64 x^3} \left (-4+768 x^3\right )}{x^2} \, dx=\frac {4 e^{64 x^3}}{x} \]

[In]

Integrate[(E^(64*x^3)*(-4 + 768*x^3))/x^2,x]

[Out]

(4*E^(64*x^3))/x

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80

method result size
gosper \(\frac {4 \,{\mathrm e}^{64 x^{3}}}{x}\) \(12\)
norman \(\frac {4 \,{\mathrm e}^{64 x^{3}}}{x}\) \(12\)
risch \(\frac {4 \,{\mathrm e}^{64 x^{3}}}{x}\) \(12\)
parallelrisch \(\frac {4 \,{\mathrm e}^{64 x^{3}}}{x}\) \(12\)
meijerg \(-16 \left (-1\right )^{\frac {1}{3}} \left (\frac {x^{2} \left (-1\right )^{\frac {2}{3}} \Gamma \left (\frac {2}{3}\right )}{\left (-x^{3}\right )^{\frac {2}{3}}}-\frac {\left (-1\right )^{\frac {2}{3}} x^{2} \Gamma \left (\frac {2}{3}, -64 x^{3}\right )}{\left (-x^{3}\right )^{\frac {2}{3}}}\right )-\frac {16 \left (-1\right )^{\frac {1}{3}} \left (-\frac {3 x^{2} \left (-1\right )^{\frac {2}{3}} \Gamma \left (\frac {2}{3}\right )}{\left (-x^{3}\right )^{\frac {2}{3}}}+\frac {3 \left (-1\right )^{\frac {2}{3}} {\mathrm e}^{64 x^{3}}}{4 x}+\frac {3 \left (-1\right )^{\frac {2}{3}} x^{2} \Gamma \left (\frac {2}{3}, -64 x^{3}\right )}{\left (-x^{3}\right )^{\frac {2}{3}}}\right )}{3}\) \(105\)

[In]

int((768*x^3-4)*exp(64*x^3)/x^2,x,method=_RETURNVERBOSE)

[Out]

4*exp(64*x^3)/x

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {e^{64 x^3} \left (-4+768 x^3\right )}{x^2} \, dx=\frac {4 \, e^{\left (64 \, x^{3}\right )}}{x} \]

[In]

integrate((768*x^3-4)*exp(64*x^3)/x^2,x, algorithm="fricas")

[Out]

4*e^(64*x^3)/x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.53 \[ \int \frac {e^{64 x^3} \left (-4+768 x^3\right )}{x^2} \, dx=\frac {4 e^{64 x^{3}}}{x} \]

[In]

integrate((768*x**3-4)*exp(64*x**3)/x**2,x)

[Out]

4*exp(64*x**3)/x

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.60 \[ \int \frac {e^{64 x^3} \left (-4+768 x^3\right )}{x^2} \, dx=-\frac {16 \, x^{2} \Gamma \left (\frac {2}{3}, -64 \, x^{3}\right )}{\left (-x^{3}\right )^{\frac {2}{3}}} + \frac {16 \, \left (-x^{3}\right )^{\frac {1}{3}} \Gamma \left (-\frac {1}{3}, -64 \, x^{3}\right )}{3 \, x} \]

[In]

integrate((768*x^3-4)*exp(64*x^3)/x^2,x, algorithm="maxima")

[Out]

-16*x^2*gamma(2/3, -64*x^3)/(-x^3)^(2/3) + 16/3*(-x^3)^(1/3)*gamma(-1/3, -64*x^3)/x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {e^{64 x^3} \left (-4+768 x^3\right )}{x^2} \, dx=\frac {4 \, e^{\left (64 \, x^{3}\right )}}{x} \]

[In]

integrate((768*x^3-4)*exp(64*x^3)/x^2,x, algorithm="giac")

[Out]

4*e^(64*x^3)/x

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {e^{64 x^3} \left (-4+768 x^3\right )}{x^2} \, dx=\frac {4\,{\mathrm {e}}^{64\,x^3}}{x} \]

[In]

int((exp(64*x^3)*(768*x^3 - 4))/x^2,x)

[Out]

(4*exp(64*x^3))/x