\(\int e^{4 e^{-2 x^3}} (e^{12}-24 e^{12-2 x^3} x^3) \, dx\) [9733]

Optimal result

Integrand size = 30, antiderivative size = 15 \[ \int e^{4 e^{-2 x^3}} \left (e^{12}-24 e^{12-2 x^3} x^3\right ) \, dx=e^{12+4 e^{-2 x^3}} x \]

[Out]

x*exp(exp(-x^3)^2)^4*exp(3)^4

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {2326} \[ \int e^{4 e^{-2 x^3}} \left (e^{12}-24 e^{12-2 x^3} x^3\right ) \, dx=e^{4 e^{-2 x^3}+12} x \]

[In]

Int[E^(4/E^(2*x^3))*(E^12 - 24*E^(12 - 2*x^3)*x^3),x]

[Out]

E^(12 + 4/E^(2*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}& = e^{12+4 e^{-2 x^3}} x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int e^{4 e^{-2 x^3}} \left (e^{12}-24 e^{12-2 x^3} x^3\right ) \, dx=e^{4 \left (3+e^{-2 x^3}\right )} x \]

[In]

Integrate[E^(4/E^(2*x^3))*(E^12 - 24*E^(12 - 2*x^3)*x^3),x]

[Out]

E^(4*(3 + E^(-2*x^3)))*x

Maple [A] (verified)

Time = 1.43 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93

method	result	size
risch	\(x \,{\mathrm e}^{4 \,{\mathrm e}^{-2 x^{3}}+12}\)	\(14\)
parallelrisch	\(x \,{\mathrm e}^{4 \,{\mathrm e}^{-2 x^{3}}} {\mathrm e}^{12}\)	\(18\)

[In]

int((-24*x^3*exp(3)^4*exp(-x^3)^2+exp(3)^4)*exp(exp(-x^3)^2)^4,x,method=_RETURNVERBOSE)

[Out]

x*exp(4*exp(-2*x^3)+12)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int e^{4 e^{-2 x^3}} \left (e^{12}-24 e^{12-2 x^3} x^3\right ) \, dx=x e^{\left (4 \, e^{\left (-2 \, x^{3}\right )} + 12\right )} \]

[In]

integrate((-24*x^3*exp(3)^4*exp(-x^3)^2+exp(3)^4)*exp(exp(-x^3)^2)^4,x, algorithm="fricas")

[Out]

x*e^(4*e^(-2*x^3) + 12)

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int e^{4 e^{-2 x^3}} \left (e^{12}-24 e^{12-2 x^3} x^3\right ) \, dx=x e^{12} e^{4 e^{- 2 x^{3}}} \]

[In]

integrate((-24*x**3*exp(3)**4*exp(-x**3)**2+exp(3)**4)*exp(exp(-x**3)**2)**4,x)

[Out]

x*exp(12)*exp(4*exp(-2*x**3))

Maxima [F]

\[ \int e^{4 e^{-2 x^3}} \left (e^{12}-24 e^{12-2 x^3} x^3\right ) \, dx=\int { -{\left (24 \, x^{3} e^{\left (-2 \, x^{3} + 12\right )} - e^{12}\right )} e^{\left (4 \, e^{\left (-2 \, x^{3}\right )}\right )} \,d x } \]

[In]

integrate((-24*x^3*exp(3)^4*exp(-x^3)^2+exp(3)^4)*exp(exp(-x^3)^2)^4,x, algorithm="maxima")

[Out]

-integrate((24*x^3*e^(-2*x^3 + 12) - e^12)*e^(4*e^(-2*x^3)), x)

Giac [F]

\[ \int e^{4 e^{-2 x^3}} \left (e^{12}-24 e^{12-2 x^3} x^3\right ) \, dx=\int { -{\left (24 \, x^{3} e^{\left (-2 \, x^{3} + 12\right )} - e^{12}\right )} e^{\left (4 \, e^{\left (-2 \, x^{3}\right )}\right )} \,d x } \]

[In]

integrate((-24*x^3*exp(3)^4*exp(-x^3)^2+exp(3)^4)*exp(exp(-x^3)^2)^4,x, algorithm="giac")

[Out]

integrate(-(24*x^3*e^(-2*x^3 + 12) - e^12)*e^(4*e^(-2*x^3)), x)

Mupad [B] (verification not implemented)

Time = 15.41 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int e^{4 e^{-2 x^3}} \left (e^{12}-24 e^{12-2 x^3} x^3\right ) \, dx=x\,{\mathrm {e}}^{4\,{\mathrm {e}}^{-2\,x^3}}\,{\mathrm {e}}^{12} \]

[In]

int(exp(4*exp(-2*x^3))*(exp(12) - 24*x^3*exp(12)*exp(-2*x^3)),x)

[Out]

x*exp(4*exp(-2*x^3))*exp(12)