\(\int e^{5+3 e^{-e^4+x} x-4 x^2} (-8 x+e^{-e^4+x} (3+3 x)) \, dx\) [7290]

Optimal result

Integrand size = 41, antiderivative size = 27 \[ \int e^{5+3 e^{-e^4+x} x-4 x^2} \left (-8 x+e^{-e^4+x} (3+3 x)\right ) \, dx=-5+e^{5-x^2-3 x \left (-e^{-e^4+x}+x\right )} \]

[Out]

exp(5-x*(3*x-3*exp(-exp(2)^2+x))-x^2)-5

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6838} \[ \int e^{5+3 e^{-e^4+x} x-4 x^2} \left (-8 x+e^{-e^4+x} (3+3 x)\right ) \, dx=e^{-4 x^2+3 e^{x-e^4} x+5} \]

[In]

Int[E^(5 + 3*E^(-E^4 + x)*x - 4*x^2)*(-8*x + E^(-E^4 + x)*(3 + 3*x)),x]

[Out]

E^(5 + 3*E^(-E^4 + x)*x - 4*x^2)

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = e^{5+3 e^{-e^4+x} x-4 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int e^{5+3 e^{-e^4+x} x-4 x^2} \left (-8 x+e^{-e^4+x} (3+3 x)\right ) \, dx=e^{5+3 e^{-e^4+x} x-4 x^2} \]

[In]

Integrate[E^(5 + 3*E^(-E^4 + x)*x - 4*x^2)*(-8*x + E^(-E^4 + x)*(3 + 3*x)),x]

[Out]

E^(5 + 3*E^(-E^4 + x)*x - 4*x^2)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70

method	result	size
risch	\({\mathrm e}^{3 \,{\mathrm e}^{x -{\mathrm e}^{4}} x -4 x^{2}+5}\)	\(19\)
norman	\({\mathrm e}^{3 \,{\mathrm e}^{x -{\mathrm e}^{4}} x -4 x^{2}+5}\)	\(21\)
parallelrisch	\({\mathrm e}^{3 \,{\mathrm e}^{x -{\mathrm e}^{4}} x -4 x^{2}+5}\)	\(21\)

[In]

int(((3*x+3)*exp(-exp(2)^2+x)-8*x)*exp(3*x*exp(-exp(2)^2+x)-4*x^2+5),x,method=_RETURNVERBOSE)

[Out]

exp(3*exp(x-exp(4))*x-4*x^2+5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int e^{5+3 e^{-e^4+x} x-4 x^2} \left (-8 x+e^{-e^4+x} (3+3 x)\right ) \, dx=e^{\left (-4 \, x^{2} + 3 \, x e^{\left (x - e^{4}\right )} + 5\right )} \]

[In]

integrate(((3*x+3)*exp(-exp(2)^2+x)-8*x)*exp(3*x*exp(-exp(2)^2+x)-4*x^2+5),x, algorithm="fricas")

[Out]

e^(-4*x^2 + 3*x*e^(x - e^4) + 5)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int e^{5+3 e^{-e^4+x} x-4 x^2} \left (-8 x+e^{-e^4+x} (3+3 x)\right ) \, dx=e^{- 4 x^{2} + 3 x e^{x - e^{4}} + 5} \]

[In]

integrate(((3*x+3)*exp(-exp(2)**2+x)-8*x)*exp(3*x*exp(-exp(2)**2+x)-4*x**2+5),x)

[Out]

exp(-4*x**2 + 3*x*exp(x - exp(4)) + 5)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int e^{5+3 e^{-e^4+x} x-4 x^2} \left (-8 x+e^{-e^4+x} (3+3 x)\right ) \, dx=e^{\left (-4 \, x^{2} + 3 \, x e^{\left (x - e^{4}\right )} + 5\right )} \]

[In]

integrate(((3*x+3)*exp(-exp(2)^2+x)-8*x)*exp(3*x*exp(-exp(2)^2+x)-4*x^2+5),x, algorithm="maxima")

[Out]

e^(-4*x^2 + 3*x*e^(x - e^4) + 5)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int e^{5+3 e^{-e^4+x} x-4 x^2} \left (-8 x+e^{-e^4+x} (3+3 x)\right ) \, dx=e^{\left (-4 \, x^{2} + 3 \, x e^{\left (x - e^{4}\right )} + 5\right )} \]

[In]

integrate(((3*x+3)*exp(-exp(2)^2+x)-8*x)*exp(3*x*exp(-exp(2)^2+x)-4*x^2+5),x, algorithm="giac")

[Out]

e^(-4*x^2 + 3*x*e^(x - e^4) + 5)

Mupad [B] (verification not implemented)

Time = 12.41 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int e^{5+3 e^{-e^4+x} x-4 x^2} \left (-8 x+e^{-e^4+x} (3+3 x)\right ) \, dx={\mathrm {e}}^{3\,x\,{\mathrm {e}}^{-{\mathrm {e}}^4}\,{\mathrm {e}}^x}\,{\mathrm {e}}^5\,{\mathrm {e}}^{-4\,x^2} \]

[In]

int(-exp(3*x*exp(x - exp(4)) - 4*x^2 + 5)*(8*x - exp(x - exp(4))*(3*x + 3)),x)

[Out]

exp(3*x*exp(-exp(4))*exp(x))*exp(5)*exp(-4*x^2)