Integrand size = 90, antiderivative size = 26 \[ \int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx=e^{-5-e^{32 e^{-1+e^{5 x}}} x^2} x^x \] Output:
exp(x*ln(x))/exp(x^2*exp(16/exp(-exp(5*x)+1))^2+5)
Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx=e^{-5-e^{32 e^{-1+e^{5 x}}} x^2} x^x \] Input:
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]
Output:
E^(-5 - E^(32*E^(-1 + E^(5*x)))*x^2)*x^x
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int e^{-e^{32 e^{e^{5 x}-1}} x^2+e^{5 x}-6} \left (e^{1-e^{5 x}} x^x (\log (x)+1)+e^{32 e^{e^{5 x}-1}} \left (-160 e^{5 x} x^2-2 e^{1-e^{5 x}} x\right ) x^x\right ) \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (e^{-e^{32 e^{e^{5 x}-1}} x^2-5} x^x (\log (x)+1)-2 x^{x+1} \left (80 e^{5 x+e^{5 x}} x+e\right ) \exp \left (-e^{32 e^{e^{5 x}-1}} x^2+32 e^{e^{5 x}-1}-6\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \exp \left (-e^{32 e^{-1+e^{5 x}}} x^2+32 e^{-1+e^{5 x}}-5\right ) x^{x+1}dx-160 \int \exp \left (-e^{32 e^{-1+e^{5 x}}} x^2+5 x+32 e^{-1+e^{5 x}}+e^{5 x}-6\right ) x^{x+2}dx+\int e^{-e^{32 e^{-1+e^{5 x}}} x^2-5} x^xdx-\int \frac {\int e^{-e^{32 e^{-1+e^{5 x}}} x^2-5} x^xdx}{x}dx+\log (x) \int e^{-e^{32 e^{-1+e^{5 x}}} x^2-5} x^xdx\) |
Input:
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]
Output:
$Aborted
Time = 46.70 (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 | \({\mathrm e}^{x \ln \left (x \right )} {\mathrm e}^{-x^{2} {\mathrm e}^{32 \,{\mathrm e}^{{\mathrm e}^{5 x}-1}}-5}\) | \(32\) |
Input:
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)
Output:
x^x*exp(-x^2*exp(32*exp(exp(5*x)-1))-5)
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx=x^{x} e^{\left (-x^{2} e^{\left (32 \, e^{\left (e^{\left (5 \, x\right )} - 1\right )}\right )} - 5\right )} \] Input:
integrate(((-2*x*exp(-exp(5*x)+1)-160*x^2*exp(5*x))*exp(x*log(x))*exp(16/e xp(-exp(5*x)+1))^2+(1+log(x))*exp(-exp(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")
Output:
x^x*e^(-x^2*e^(32*e^(e^(5*x) - 1)) - 5)
Time = 28.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx=e^{x \log {\left (x \right )}} e^{- x^{2} e^{32 e^{e^{5 x} - 1}} - 5} \] Input:
integrate(((-2*x*exp(-exp(5*x)+1)-160*x**2*exp(5*x))*exp(x*ln(x))*exp(16/e xp(-exp(5*x)+1))**2+(1+ln(x))*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)
Output:
exp(x*log(x))*exp(-x**2*exp(32*exp(exp(5*x) - 1)) - 5)
Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx=e^{\left (-x^{2} e^{\left (32 \, e^{\left (e^{\left (5 \, x\right )} - 1\right )}\right )} + x \log \left (x\right ) - 5\right )} \] Input:
integrate(((-2*x*exp(-exp(5*x)+1)-160*x^2*exp(5*x))*exp(x*log(x))*exp(16/e xp(-exp(5*x)+1))^2+(1+log(x))*exp(-exp(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")
Output:
e^(-x^2*e^(32*e^(e^(5*x) - 1)) + x*log(x) - 5)
\[ \int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx=\int { -{\left (2 \, {\left (80 \, x^{2} e^{\left (5 \, x\right )} + x e^{\left (-e^{\left (5 \, x\right )} + 1\right )}\right )} x^{x} e^{\left (32 \, e^{\left (e^{\left (5 \, x\right )} - 1\right )}\right )} - x^{x} {\left (\log \left (x\right ) + 1\right )} e^{\left (-e^{\left (5 \, x\right )} + 1\right )}\right )} e^{\left (-x^{2} e^{\left (32 \, e^{\left (e^{\left (5 \, x\right )} - 1\right )}\right )} + e^{\left (5 \, x\right )} - 6\right )} \,d x } \] Input:
integrate(((-2*x*exp(-exp(5*x)+1)-160*x^2*exp(5*x))*exp(x*log(x))*exp(16/e xp(-exp(5*x)+1))^2+(1+log(x))*exp(-exp(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")
Output:
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)
Time = 2.77 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx=x^x\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{-x^2\,{\mathrm {e}}^{32\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{{\mathrm {e}}^{5\,x}}}} \] Input:
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)
Output:
x^x*exp(-5)*exp(-x^2*exp(32*exp(-1)*exp(exp(5*x))))
Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int e^{-6+e^{5 x}-e^{32 e^{-1+e^{5 x}}} x^2} \left (e^{32 e^{-1+e^{5 x}}} x^x \left (-2 e^{1-e^{5 x}} x-160 e^{5 x} x^2\right )+e^{1-e^{5 x}} x^x (1+\log (x))\right ) \, dx=\frac {x^{x}}{e^{e^{\frac {32 e^{e^{5 x}}}{e}} x^{2}} e^{5}} \] Input:
int(((-2*x*exp(-exp(5*x)+1)-160*x^2*exp(5*x))*exp(x*log(x))*exp(16/exp(-ex p(5*x)+1))^2+(1+log(x))*exp(-exp(5*x)+1)*exp(x*log(x)))/exp(-exp(5*x)+1)/e xp(x^2*exp(16/exp(-exp(5*x)+1))^2+5),x)
Output:
x**x/(e**(e**((32*e**(e**(5*x)))/e)*x**2)*e**5)