\(\int \frac {e^{-20+5 x \log (-1-e+e^{e^x+x})} (e^{e^x+x} (5 x+5 e^x x)+(-5-5 e+5 e^{e^x+x}) \log (-1-e+e^{e^x+x}))}{-1-e+e^{e^x+x}} \, dx\) [2163]

Optimal result

Integrand size = 82, antiderivative size = 20 \[ \int \frac {e^{-20+5 x \log \left (-1-e+e^{e^x+x}\right )} \left (e^{e^x+x} \left (5 x+5 e^x x\right )+\left (-5-5 e+5 e^{e^x+x}\right ) \log \left (-1-e+e^{e^x+x}\right )\right )}{-1-e+e^{e^x+x}} \, dx=e^{-20+5 x \log \left (-1-e+e^{e^x+x}\right )} \]

[Out]

exp(5*x*ln(exp(exp(x)+x)-exp(1)-1)-20)

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {6838} \[ \int \frac {e^{-20+5 x \log \left (-1-e+e^{e^x+x}\right )} \left (e^{e^x+x} \left (5 x+5 e^x x\right )+\left (-5-5 e+5 e^{e^x+x}\right ) \log \left (-1-e+e^{e^x+x}\right )\right )}{-1-e+e^{e^x+x}} \, dx=\frac {\left (e^{x+e^x}-1-e\right )^{5 x}}{e^{20}} \]

[In]

Int[(E^(-20 + 5*x*Log[-1 - E + E^(E^x + x)])*(E^(E^x + x)*(5*x + 5*E^x*x) + (-5 - 5*E + 5*E^(E^x + x))*Log[-1
- E + E^(E^x + x)]))/(-1 - E + E^(E^x + x)),x]

[Out]

(-1 - E + E^(E^x + x))^(5*x)/E^20

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}& = \frac {\left (-1-e+e^{e^x+x}\right )^{5 x}}{e^{20}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-20+5 x \log \left (-1-e+e^{e^x+x}\right )} \left (e^{e^x+x} \left (5 x+5 e^x x\right )+\left (-5-5 e+5 e^{e^x+x}\right ) \log \left (-1-e+e^{e^x+x}\right )\right )}{-1-e+e^{e^x+x}} \, dx=\frac {\left (-1-e+e^{e^x+x}\right )^{5 x}}{e^{20}} \]

[In]

Integrate[(E^(-20 + 5*x*Log[-1 - E + E^(E^x + x)])*(E^(E^x + x)*(5*x + 5*E^x*x) + (-5 - 5*E + 5*E^(E^x + x))*L
og[-1 - E + E^(E^x + x)]))/(-1 - E + E^(E^x + x)),x]

[Out]

(-1 - E + E^(E^x + x))^(5*x)/E^20

Maple [A] (verified)

Time = 3.54 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95

method	result	size
risch	\(\left ({\mathrm e}^{{\mathrm e}^{x}+x}-{\mathrm e}-1\right )^{5 x} {\mathrm e}^{-20}\)	\(19\)
parallelrisch	\({\mathrm e}^{5 x \ln \left ({\mathrm e}^{{\mathrm e}^{x}+x}-{\mathrm e}-1\right )-20}\)	\(19\)

[In]

int(((5*exp(exp(x)+x)-5*exp(1)-5)*ln(exp(exp(x)+x)-exp(1)-1)+(5*exp(x)*x+5*x)*exp(exp(x)+x))*exp(5*x*ln(exp(ex
p(x)+x)-exp(1)-1)-20)/(exp(exp(x)+x)-exp(1)-1),x,method=_RETURNVERBOSE)

[Out]

(exp(exp(x)+x)-exp(1)-1)^(5*x)*exp(-20)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-20+5 x \log \left (-1-e+e^{e^x+x}\right )} \left (e^{e^x+x} \left (5 x+5 e^x x\right )+\left (-5-5 e+5 e^{e^x+x}\right ) \log \left (-1-e+e^{e^x+x}\right )\right )}{-1-e+e^{e^x+x}} \, dx=e^{\left (5 \, x \log \left (-e + e^{\left (x + e^{x}\right )} - 1\right ) - 20\right )} \]

[In]

integrate(((5*exp(exp(x)+x)-5*exp(1)-5)*log(exp(exp(x)+x)-exp(1)-1)+(5*exp(x)*x+5*x)*exp(exp(x)+x))*exp(5*x*lo
g(exp(exp(x)+x)-exp(1)-1)-20)/(exp(exp(x)+x)-exp(1)-1),x, algorithm="fricas")

[Out]

e^(5*x*log(-e + e^(x + e^x) - 1) - 20)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-20+5 x \log \left (-1-e+e^{e^x+x}\right )} \left (e^{e^x+x} \left (5 x+5 e^x x\right )+\left (-5-5 e+5 e^{e^x+x}\right ) \log \left (-1-e+e^{e^x+x}\right )\right )}{-1-e+e^{e^x+x}} \, dx=\text {Timed out} \]

[In]

integrate(((5*exp(exp(x)+x)-5*exp(1)-5)*ln(exp(exp(x)+x)-exp(1)-1)+(5*exp(x)*x+5*x)*exp(exp(x)+x))*exp(5*x*ln(
exp(exp(x)+x)-exp(1)-1)-20)/(exp(exp(x)+x)-exp(1)-1),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-20+5 x \log \left (-1-e+e^{e^x+x}\right )} \left (e^{e^x+x} \left (5 x+5 e^x x\right )+\left (-5-5 e+5 e^{e^x+x}\right ) \log \left (-1-e+e^{e^x+x}\right )\right )}{-1-e+e^{e^x+x}} \, dx=e^{\left (5 \, x \log \left (-e + e^{\left (x + e^{x}\right )} - 1\right ) - 20\right )} \]

[In]

integrate(((5*exp(exp(x)+x)-5*exp(1)-5)*log(exp(exp(x)+x)-exp(1)-1)+(5*exp(x)*x+5*x)*exp(exp(x)+x))*exp(5*x*lo
g(exp(exp(x)+x)-exp(1)-1)-20)/(exp(exp(x)+x)-exp(1)-1),x, algorithm="maxima")

[Out]

e^(5*x*log(-e + e^(x + e^x) - 1) - 20)

Giac [A] (verification not implemented)

none

Time = 0.62 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-20+5 x \log \left (-1-e+e^{e^x+x}\right )} \left (e^{e^x+x} \left (5 x+5 e^x x\right )+\left (-5-5 e+5 e^{e^x+x}\right ) \log \left (-1-e+e^{e^x+x}\right )\right )}{-1-e+e^{e^x+x}} \, dx=e^{\left (5 \, x \log \left (-e + e^{\left (x + e^{x}\right )} - 1\right ) - 20\right )} \]

[In]

integrate(((5*exp(exp(x)+x)-5*exp(1)-5)*log(exp(exp(x)+x)-exp(1)-1)+(5*exp(x)*x+5*x)*exp(exp(x)+x))*exp(5*x*lo
g(exp(exp(x)+x)-exp(1)-1)-20)/(exp(exp(x)+x)-exp(1)-1),x, algorithm="giac")

[Out]

e^(5*x*log(-e + e^(x + e^x) - 1) - 20)

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-20+5 x \log \left (-1-e+e^{e^x+x}\right )} \left (e^{e^x+x} \left (5 x+5 e^x x\right )+\left (-5-5 e+5 e^{e^x+x}\right ) \log \left (-1-e+e^{e^x+x}\right )\right )}{-1-e+e^{e^x+x}} \, dx={\mathrm {e}}^{-20}\,{\left ({\mathrm {e}}^{x+{\mathrm {e}}^x}-\mathrm {e}-1\right )}^{5\,x} \]

[In]

int(-(exp(5*x*log(exp(x + exp(x)) - exp(1) - 1) - 20)*(exp(x + exp(x))*(5*x + 5*x*exp(x)) - log(exp(x + exp(x)
) - exp(1) - 1)*(5*exp(1) - 5*exp(x + exp(x)) + 5)))/(exp(1) - exp(x + exp(x)) + 1),x)

[Out]

exp(-20)*(exp(x + exp(x)) - exp(1) - 1)^(5*x)