\(\int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} (e^{32 e^{-1+e^{5 x}}} x^x (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2)+e^{1-e^{5 x}} x^x (1+\log (x))) \, dx\) [6723]

Optimal result

Integrand size = 90, antiderivative size = 26 \[ \int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx=e^{-5-e^{32 e^{-1+e^{5 x}}} x^2} x^x \]

[Out]

exp(x*ln(x))/exp(x^2*exp(16/exp(-exp(5*x)+1))^2+5)

Rubi [F]

\[ \int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx=\int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx \]

[In]

Int[E^(-6 + E^(5*x) - E^(32*E^(-1 + E^(5*x)))*x^2)*(E^(32*E^(-1 + E^(5*x)))*x^x*(-2*E^(1 - E^(5*x))*x - 160*E^
(5*x)*x^2) + E^(1 - E^(5*x))*x^x*(1 + Log[x])),x]

[Out]

Defer[Int][E^(-5 - E^(32*E^(-1 + E^(5*x)))*x^2)*x^x, x] + Log[x]*Defer[Int][E^(-5 - E^(32*E^(-1 + E^(5*x)))*x^
2)*x^x, x] - 2*Defer[Int][E^(-5 + 32*E^(-1 + E^(5*x)) - E^(32*E^(-1 + E^(5*x)))*x^2)*x^(1 + x), x] - 160*Defer
[Int][E^(-6 + 32*E^(-1 + E^(5*x)) + E^(5*x) + 5*x - E^(32*E^(-1 + E^(5*x)))*x^2)*x^(2 + x), x] - Defer[Int][De
fer[Int][E^(-5 - E^(32*E^(-1 + E^(5*x)))*x^2)*x^x, x]/x, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-2 \exp \left (-6+32 e^{-1+e^{5 x}}-e^{32 e^{-1+e^{5 x}}} x^2\right ) x^{1+x} \left (e+80 e^{e^{5 x}+5 x} x\right )+e^{-5-e^{32 e^{-1+e^{5 x}}} x^2} x^x (1+\log (x))\right ) \, dx \\ & = -\left (2 \int \exp \left (-6+32 e^{-1+e^{5 x}}-e^{32 e^{-1+e^{5 x}}} x^2\right ) x^{1+x} \left (e+80 e^{e^{5 x}+5 x} x\right ) \, dx\right )+\int e^{-5-e^{32 e^{-1+e^{5 x}}} x^2} x^x (1+\log (x)) \, dx \\ & = -\left (2 \int \left (\exp \left (-5+32 e^{-1+e^{5 x}}-e^{32 e^{-1+e^{5 x}}} x^2\right ) x^{1+x}+80 \exp \left (-6+32 e^{-1+e^{5 x}}+e^{5 x}+5 x-e^{32 e^{-1+e^{5 x}}} x^2\right ) x^{2+x}\right ) \, dx\right )+\int \left (e^{-5-e^{32 e^{-1+e^{5 x}}} x^2} x^x+e^{-5-e^{32 e^{-1+e^{5 x}}} x^2} x^x \log (x)\right ) \, dx \\ & = -\left (2 \int \exp \left (-5+32 e^{-1+e^{5 x}}-e^{32 e^{-1+e^{5 x}}} x^2\right ) x^{1+x} \, dx\right )-160 \int \exp \left (-6+32 e^{-1+e^{5 x}}+e^{5 x}+5 x-e^{32 e^{-1+e^{5 x}}} x^2\right ) x^{2+x} \, dx+\int e^{-5-e^{32 e^{-1+e^{5 x}}} x^2} x^x \, dx+\int e^{-5-e^{32 e^{-1+e^{5 x}}} x^2} x^x \log (x) \, dx \\ & = -\left (2 \int \exp \left (-5+32 e^{-1+e^{5 x}}-e^{32 e^{-1+e^{5 x}}} x^2\right ) x^{1+x} \, dx\right )-160 \int \exp \left (-6+32 e^{-1+e^{5 x}}+e^{5 x}+5 x-e^{32 e^{-1+e^{5 x}}} x^2\right ) x^{2+x} \, dx+\log (x) \int e^{-5-e^{32 e^{-1+e^{5 x}}} x^2} x^x \, dx+\int e^{-5-e^{32 e^{-1+e^{5 x}}} x^2} x^x \, dx-\int \frac {\int e^{-5-e^{32 e^{-1+e^{5 x}}} x^2} x^x \, dx}{x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx=e^{-5-e^{32 e^{-1+e^{5 x}}} x^2} x^x \]

[In]

Integrate[E^(-6 + E^(5*x) - E^(32*E^(-1 + E^(5*x)))*x^2)*(E^(32*E^(-1 + E^(5*x)))*x^x*(-2*E^(1 - E^(5*x))*x -
160*E^(5*x)*x^2) + E^(1 - E^(5*x))*x^x*(1 + Log[x])),x]

[Out]

E^(-5 - E^(32*E^(-1 + E^(5*x)))*x^2)*x^x

Maple [A] (verified)

Time = 29.65 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88

method	result	size
risch	\(x^{x} {\mathrm e}^{-x^{2} {\mathrm e}^{32 \,{\mathrm e}^{{\mathrm e}^{5 x}-1}}-5}\)	\(23\)
parallelrisch	\({\mathrm e}^{x \ln \left (x \right )} {\mathrm e}^{-x^{2} {\mathrm e}^{32 \,{\mathrm e}^{{\mathrm e}^{5 x}-1}}-5}\)	\(32\)

[In]

int(((-2*x*exp(-exp(5*x)+1)-160*x^2*exp(5*x))*exp(x*ln(x))*exp(16/exp(-exp(5*x)+1))^2+(ln(x)+1)*exp(-exp(5*x)+
1)*exp(x*ln(x)))/exp(-exp(5*x)+1)/exp(x^2*exp(16/exp(-exp(5*x)+1))^2+5),x,method=_RETURNVERBOSE)

[Out]

x^x*exp(-x^2*exp(32*exp(exp(5*x)-1))-5)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx=x^{x} e^{\left (-x^{2} e^{\left (32 \, e^{\left (e^{\left (5 \, x\right )} - 1\right )}\right )} - 5\right )} \]

[In]

integrate(((-2*x*exp(-exp(5*x)+1)-160*x^2*exp(5*x))*exp(x*log(x))*exp(16/exp(-exp(5*x)+1))^2+(log(x)+1)*exp(-e
xp(5*x)+1)*exp(x*log(x)))/exp(-exp(5*x)+1)/exp(x^2*exp(16/exp(-exp(5*x)+1))^2+5),x, algorithm="fricas")

[Out]

x^x*e^(-x^2*e^(32*e^(e^(5*x) - 1)) - 5)

Sympy [A] (verification not implemented)

Time = 28.48 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx=e^{x \log {\left (x \right )}} e^{- x^{2} e^{32 e^{e^{5 x} - 1}} - 5} \]

[In]

integrate(((-2*x*exp(-exp(5*x)+1)-160*x**2*exp(5*x))*exp(x*ln(x))*exp(16/exp(-exp(5*x)+1))**2+(ln(x)+1)*exp(-e
xp(5*x)+1)*exp(x*ln(x)))/exp(-exp(5*x)+1)/exp(x**2*exp(16/exp(-exp(5*x)+1))**2+5),x)

[Out]

exp(x*log(x))*exp(-x**2*exp(32*exp(exp(5*x) - 1)) - 5)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx=e^{\left (-x^{2} e^{\left (32 \, e^{\left (e^{\left (5 \, x\right )} - 1\right )}\right )} + x \log \left (x\right ) - 5\right )} \]

[In]

integrate(((-2*x*exp(-exp(5*x)+1)-160*x^2*exp(5*x))*exp(x*log(x))*exp(16/exp(-exp(5*x)+1))^2+(log(x)+1)*exp(-e
xp(5*x)+1)*exp(x*log(x)))/exp(-exp(5*x)+1)/exp(x^2*exp(16/exp(-exp(5*x)+1))^2+5),x, algorithm="maxima")

[Out]

e^(-x^2*e^(32*e^(e^(5*x) - 1)) + x*log(x) - 5)

Giac [F]

\[ \int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx=\int { -{\left (2 \, {\left (80 \, x^{2} e^{\left (5 \, x\right )} + x e^{\left (-e^{\left (5 \, x\right )} + 1\right )}\right )} x^{x} e^{\left (32 \, e^{\left (e^{\left (5 \, x\right )} - 1\right )}\right )} - x^{x} {\left (\log \left (x\right ) + 1\right )} e^{\left (-e^{\left (5 \, x\right )} + 1\right )}\right )} e^{\left (-x^{2} e^{\left (32 \, e^{\left (e^{\left (5 \, x\right )} - 1\right )}\right )} + e^{\left (5 \, x\right )} - 6\right )} \,d x } \]

[In]

integrate(((-2*x*exp(-exp(5*x)+1)-160*x^2*exp(5*x))*exp(x*log(x))*exp(16/exp(-exp(5*x)+1))^2+(log(x)+1)*exp(-e
xp(5*x)+1)*exp(x*log(x)))/exp(-exp(5*x)+1)/exp(x^2*exp(16/exp(-exp(5*x)+1))^2+5),x, algorithm="giac")

[Out]

integrate(-(2*(80*x^2*e^(5*x) + x*e^(-e^(5*x) + 1))*x^x*e^(32*e^(e^(5*x) - 1)) - x^x*(log(x) + 1)*e^(-e^(5*x)
+ 1))*e^(-x^2*e^(32*e^(e^(5*x) - 1)) + e^(5*x) - 6), x)

Mupad [B] (verification not implemented)

Time = 12.37 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx=x^x\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{-x^2\,{\mathrm {e}}^{32\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{{\mathrm {e}}^{5\,x}}}} \]

[In]

int(exp(exp(5*x) - 1)*exp(- x^2*exp(32*exp(exp(5*x) - 1)) - 5)*(exp(x*log(x))*exp(1 - exp(5*x))*(log(x) + 1) -
exp(x*log(x))*exp(32*exp(exp(5*x) - 1))*(2*x*exp(1 - exp(5*x)) + 160*x^2*exp(5*x))),x)

[Out]

x^x*exp(-5)*exp(-x^2*exp(32*exp(-1)*exp(exp(5*x))))