\(\int e^{-e^{2+e^4}-x+4 x \log (x)-8 e^2 x \log ^2(x)} (3+(4-16 e^2) \log (x)-8 e^2 \log ^2(x)) \, dx\) [2682]

Optimal result

Integrand size = 52, antiderivative size = 29 \[ \int e^{-e^{2+e^4}-x+4 x \log (x)-8 e^2 x \log ^2(x)} \left (3+\left (4-16 e^2\right ) \log (x)-8 e^2 \log ^2(x)\right ) \, dx=e^{-e^{2+e^4}-x+4 \log (x) \left (x-2 e^2 x \log (x)\right )} \]

[Out]

exp(4*(x-2*x*exp(2)*ln(x))*ln(x)-x-exp(2+exp(4)))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {6838} \[ \int e^{-e^{2+e^4}-x+4 x \log (x)-8 e^2 x \log ^2(x)} \left (3+\left (4-16 e^2\right ) \log (x)-8 e^2 \log ^2(x)\right ) \, dx=x^{4 x} e^{-x-8 e^2 x \log ^2(x)-e^{2+e^4}} \]

[In]

Int[E^(-E^(2 + E^4) - x + 4*x*Log[x] - 8*E^2*x*Log[x]^2)*(3 + (4 - 16*E^2)*Log[x] - 8*E^2*Log[x]^2),x]

[Out]

E^(-E^(2 + E^4) - x - 8*E^2*x*Log[x]^2)*x^(4*x)

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^{-e^{2+e^4}-x-8 e^2 x \log ^2(x)} x^{4 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int e^{-e^{2+e^4}-x+4 x \log (x)-8 e^2 x \log ^2(x)} \left (3+\left (4-16 e^2\right ) \log (x)-8 e^2 \log ^2(x)\right ) \, dx=e^{-e^{2+e^4}-x+4 x \log (x)-8 e^2 x \log ^2(x)} \]

[In]

Integrate[E^(-E^(2 + E^4) - x + 4*x*Log[x] - 8*E^2*x*Log[x]^2)*(3 + (4 - 16*E^2)*Log[x] - 8*E^2*Log[x]^2),x]

[Out]

E^(-E^(2 + E^4) - x + 4*x*Log[x] - 8*E^2*x*Log[x]^2)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93

method	result	size
norman	\({\mathrm e}^{-8 x \,{\mathrm e}^{2} \ln \left (x \right )^{2}+4 x \ln \left (x \right )-{\mathrm e}^{2+{\mathrm e}^{4}}-x}\)	\(27\)
parallelrisch	\({\mathrm e}^{-8 x \,{\mathrm e}^{2} \ln \left (x \right )^{2}+4 x \ln \left (x \right )-{\mathrm e}^{2+{\mathrm e}^{4}}-x}\)	\(27\)
risch	\(x^{4 x} {\mathrm e}^{-8 x \,{\mathrm e}^{2} \ln \left (x \right )^{2}-{\mathrm e}^{2+{\mathrm e}^{4}}-x}\)	\(28\)

[In]

int((-8*exp(2)*ln(x)^2+(-16*exp(2)+4)*ln(x)+3)*exp(-8*x*exp(2)*ln(x)^2+4*x*ln(x)-exp(2+exp(4))-x),x,method=_RE
TURNVERBOSE)

[Out]

exp(-8*x*exp(2)*ln(x)^2+4*x*ln(x)-exp(2+exp(4))-x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int e^{-e^{2+e^4}-x+4 x \log (x)-8 e^2 x \log ^2(x)} \left (3+\left (4-16 e^2\right ) \log (x)-8 e^2 \log ^2(x)\right ) \, dx=e^{\left (-8 \, x e^{2} \log \left (x\right )^{2} + 4 \, x \log \left (x\right ) - x - e^{\left (e^{4} + 2\right )}\right )} \]

[In]

integrate((-8*exp(2)*log(x)^2+(-16*exp(2)+4)*log(x)+3)*exp(-8*x*exp(2)*log(x)^2+4*x*log(x)-exp(2+exp(4))-x),x,
algorithm="fricas")

[Out]

e^(-8*x*e^2*log(x)^2 + 4*x*log(x) - x - e^(e^4 + 2))

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int e^{-e^{2+e^4}-x+4 x \log (x)-8 e^2 x \log ^2(x)} \left (3+\left (4-16 e^2\right ) \log (x)-8 e^2 \log ^2(x)\right ) \, dx=e^{- 8 x e^{2} \log {\left (x \right )}^{2} + 4 x \log {\left (x \right )} - x - e^{2 + e^{4}}} \]

[In]

integrate((-8*exp(2)*ln(x)**2+(-16*exp(2)+4)*ln(x)+3)*exp(-8*x*exp(2)*ln(x)**2+4*x*ln(x)-exp(2+exp(4))-x),x)

[Out]

exp(-8*x*exp(2)*log(x)**2 + 4*x*log(x) - x - exp(2 + exp(4)))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int e^{-e^{2+e^4}-x+4 x \log (x)-8 e^2 x \log ^2(x)} \left (3+\left (4-16 e^2\right ) \log (x)-8 e^2 \log ^2(x)\right ) \, dx=e^{\left (-8 \, x e^{2} \log \left (x\right )^{2} + 4 \, x \log \left (x\right ) - x - e^{\left (e^{4} + 2\right )}\right )} \]

[In]

integrate((-8*exp(2)*log(x)^2+(-16*exp(2)+4)*log(x)+3)*exp(-8*x*exp(2)*log(x)^2+4*x*log(x)-exp(2+exp(4))-x),x,
algorithm="maxima")

[Out]

e^(-8*x*e^2*log(x)^2 + 4*x*log(x) - x - e^(e^4 + 2))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int e^{-e^{2+e^4}-x+4 x \log (x)-8 e^2 x \log ^2(x)} \left (3+\left (4-16 e^2\right ) \log (x)-8 e^2 \log ^2(x)\right ) \, dx=e^{\left (-8 \, x e^{2} \log \left (x\right )^{2} + 4 \, x \log \left (x\right ) - x - e^{\left (e^{4} + 2\right )}\right )} \]

[In]

integrate((-8*exp(2)*log(x)^2+(-16*exp(2)+4)*log(x)+3)*exp(-8*x*exp(2)*log(x)^2+4*x*log(x)-exp(2+exp(4))-x),x,
algorithm="giac")

[Out]

e^(-8*x*e^2*log(x)^2 + 4*x*log(x) - x - e^(e^4 + 2))

Mupad [B] (verification not implemented)

Time = 9.36 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int e^{-e^{2+e^4}-x+4 x \log (x)-8 e^2 x \log ^2(x)} \left (3+\left (4-16 e^2\right ) \log (x)-8 e^2 \log ^2(x)\right ) \, dx=x^{4\,x}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-8\,x\,{\mathrm {e}}^2\,{\ln \left (x\right )}^2}\,{\mathrm {e}}^{-{\mathrm {e}}^2\,{\mathrm {e}}^{{\mathrm {e}}^4}} \]

[In]

int(-exp(4*x*log(x) - exp(exp(4) + 2) - x - 8*x*exp(2)*log(x)^2)*(8*exp(2)*log(x)^2 + log(x)*(16*exp(2) - 4) -
3),x)

[Out]

x^(4*x)*exp(-x)*exp(-8*x*exp(2)*log(x)^2)*exp(-exp(2)*exp(exp(4)))