\(\int e^{-\frac {1}{125} e^{14+e^x}+x \log (2) \log (3)} (-\frac {1}{125} e^{14+e^x+x}+\log (2) \log (3)) \, dx\) [590]

Optimal result

Integrand size = 39, antiderivative size = 26 \[ \int e^{-\frac {1}{125} e^{14+e^x}+x \log (2) \log (3)} \left (-\frac {1}{125} e^{14+e^x+x}+\log (2) \log (3)\right ) \, dx=e^{-e^{-1+e^x+3 (5-\log (5))}+x \log (2) \log (3)} \]

[Out]

exp(-exp(exp(x)-3*ln(5)+14)+x*ln(2)*ln(3))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6838} \[ \int e^{-\frac {1}{125} e^{14+e^x}+x \log (2) \log (3)} \left (-\frac {1}{125} e^{14+e^x+x}+\log (2) \log (3)\right ) \, dx=e^{-\frac {1}{125} e^{e^x+14}} 2^{x \log (3)} \]

[In]

Int[E^(-1/125*E^(14 + E^x) + x*Log[2]*Log[3])*(-1/125*E^(14 + E^x + x) + Log[2]*Log[3]),x]

[Out]

2^(x*Log[3])/E^(E^(14 + E^x)/125)

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}& = 2^{x \log (3)} e^{-\frac {1}{125} e^{14+e^x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int e^{-\frac {1}{125} e^{14+e^x}+x \log (2) \log (3)} \left (-\frac {1}{125} e^{14+e^x+x}+\log (2) \log (3)\right ) \, dx=e^{-\frac {1}{125} e^{14+e^x}+x \log (2) \log (3)} \]

[In]

Integrate[E^(-1/125*E^(14 + E^x) + x*Log[2]*Log[3])*(-1/125*E^(14 + E^x + x) + Log[2]*Log[3]),x]

[Out]

E^(-1/125*E^(14 + E^x) + x*Log[2]*Log[3])

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62

method	result	size
risch	\(2^{x \ln \left (3\right )} {\mathrm e}^{-\frac {{\mathrm e}^{{\mathrm e}^{x}+14}}{125}}\)	\(16\)
derivativedivides	\({\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{x}-3 \ln \left (5\right )+14}+x \ln \left (2\right ) \ln \left (3\right )}\)	\(20\)
default	\({\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{x}-3 \ln \left (5\right )+14}+x \ln \left (2\right ) \ln \left (3\right )}\)	\(20\)
norman	\({\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{x}-3 \ln \left (5\right )+14}+x \ln \left (2\right ) \ln \left (3\right )}\)	\(20\)
parallelrisch	\({\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{x}-3 \ln \left (5\right )+14}+x \ln \left (2\right ) \ln \left (3\right )}\)	\(20\)

[In]

int((-exp(x)*exp(exp(x)-3*ln(5)+14)+ln(2)*ln(3))*exp(-exp(exp(x)-3*ln(5)+14)+x*ln(2)*ln(3)),x,method=_RETURNVE
RBOSE)

[Out]

2^(x*ln(3))*exp(-1/125*exp(exp(x)+14))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int e^{-\frac {1}{125} e^{14+e^x}+x \log (2) \log (3)} \left (-\frac {1}{125} e^{14+e^x+x}+\log (2) \log (3)\right ) \, dx=e^{\left ({\left (x e^{x} \log \left (3\right ) \log \left (2\right ) - e^{\left (x + e^{x} - 3 \, \log \left (5\right ) + 14\right )}\right )} e^{\left (-x\right )}\right )} \]

[In]

integrate((-exp(x)*exp(exp(x)-3*log(5)+14)+log(2)*log(3))*exp(-exp(exp(x)-3*log(5)+14)+x*log(2)*log(3)),x, alg
orithm="fricas")

[Out]

e^((x*e^x*log(3)*log(2) - e^(x + e^x - 3*log(5) + 14))*e^(-x))

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int e^{-\frac {1}{125} e^{14+e^x}+x \log (2) \log (3)} \left (-\frac {1}{125} e^{14+e^x+x}+\log (2) \log (3)\right ) \, dx=e^{x \log {\left (2 \right )} \log {\left (3 \right )} - \frac {e^{e^{x} + 14}}{125}} \]

[In]

integrate((-exp(x)*exp(exp(x)-3*ln(5)+14)+ln(2)*ln(3))*exp(-exp(exp(x)-3*ln(5)+14)+x*ln(2)*ln(3)),x)

[Out]

exp(x*log(2)*log(3) - exp(exp(x) + 14)/125)

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.58 \[ \int e^{-\frac {1}{125} e^{14+e^x}+x \log (2) \log (3)} \left (-\frac {1}{125} e^{14+e^x+x}+\log (2) \log (3)\right ) \, dx=e^{\left (x \log \left (3\right ) \log \left (2\right ) - \frac {1}{125} \, e^{\left (e^{x} + 14\right )}\right )} \]

[In]

integrate((-exp(x)*exp(exp(x)-3*log(5)+14)+log(2)*log(3))*exp(-exp(exp(x)-3*log(5)+14)+x*log(2)*log(3)),x, alg
orithm="maxima")

[Out]

e^(x*log(3)*log(2) - 1/125*e^(e^x + 14))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int e^{-\frac {1}{125} e^{14+e^x}+x \log (2) \log (3)} \left (-\frac {1}{125} e^{14+e^x+x}+\log (2) \log (3)\right ) \, dx=e^{\left (x \log \left (3\right ) \log \left (2\right ) - e^{\left (e^{x} - 3 \, \log \left (5\right ) + 14\right )}\right )} \]

[In]

integrate((-exp(x)*exp(exp(x)-3*log(5)+14)+log(2)*log(3))*exp(-exp(exp(x)-3*log(5)+14)+x*log(2)*log(3)),x, alg
orithm="giac")

[Out]

e^(x*log(3)*log(2) - e^(e^x - 3*log(5) + 14))

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.58 \[ \int e^{-\frac {1}{125} e^{14+e^x}+x \log (2) \log (3)} \left (-\frac {1}{125} e^{14+e^x+x}+\log (2) \log (3)\right ) \, dx=2^{x\,\ln \left (3\right )}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{14}}{125}} \]

[In]

int(exp(x*log(2)*log(3) - exp(exp(x) - 3*log(5) + 14))*(log(2)*log(3) - exp(exp(x) - 3*log(5) + 14)*exp(x)),x)

[Out]

2^(x*log(3))*exp(-(exp(exp(x))*exp(14))/125)