Integrand size = 78, antiderivative size = 19 \[ \int \frac {e^{-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}} \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx=e^{2 x-\frac {361}{8} e^{-e^x x} x} \] Output:
exp(x-361/16*x^2/exp(ln(x)+exp(x)*x))^2
Time = 1.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}} \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx=e^{2 x-\frac {361}{8} e^{-e^x x} x} \] Input:
Integrate[(E^(-(E^x*x) + (-361*x^2 + 16*E^(E^x*x)*x^2)/(8*E^(E^x*x)*x))*(- 361*x + 16*E^(E^x*x)*x + E^x*(361*x^2 + 361*x^3)))/(8*x),x]
Output:
E^(2*x - (361*x)/(8*E^(E^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.
\(\displaystyle \int \frac {\left (e^x \left (361 x^3+361 x^2\right )+16 e^{e^x x} x-361 x\right ) \exp \left (\frac {e^{-e^x x} \left (16 e^{e^x x} x^2-361 x^2\right )}{8 x}-e^x x\right )}{8 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \int -\frac {\exp \left (-e^x x-\frac {e^{-e^x x} \left (361 x^2-16 e^{e^x x} x^2\right )}{8 x}\right ) \left (-16 e^{e^x x} x+361 x-361 e^x \left (x^3+x^2\right )\right )}{x}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{8} \int \frac {\exp \left (-e^x x-\frac {e^{-e^x x} \left (361 x^2-16 e^{e^x x} x^2\right )}{8 x}\right ) \left (-16 e^{e^x x} x+361 x-361 e^x \left (x^3+x^2\right )\right )}{x}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {1}{8} \int \frac {\exp \left (-\frac {1}{8} e^{-e^x x} \left (361-16 e^{e^x x}+8 e^{e^x x+x}\right ) x\right ) \left (-16 e^{e^x x} x+361 x-361 e^x \left (x^3+x^2\right )\right )}{x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{8} \int \left (-361 \exp \left (-\frac {1}{8} e^{-e^x x} \left (361-16 e^{e^x x}+8 e^{e^x x+x}\right ) x\right ) \left (e^x x^2+e^x x-1\right )-16 \exp \left (e^x x-\frac {1}{8} e^{-e^x x} \left (361-16 e^{e^x x}+8 e^{e^x x+x}\right ) x\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{8} \left (361 \int \exp \left (-\frac {1}{8} e^{-e^x x} \left (361-24 e^{e^x x}+8 e^{e^x x+x}\right ) x\right ) x^2dx-361 \int \exp \left (-\frac {1}{8} e^{-e^x x} \left (361-16 e^{e^x x}+8 e^{e^x x+x}\right ) x\right )dx+361 \int \exp \left (-\frac {1}{8} e^{-e^x x} \left (361-24 e^{e^x x}+8 e^{e^x x+x}\right ) x\right ) xdx+16 \int e^{2 x-\frac {361}{8} e^{-e^x x} x}dx\right )\) |
Input:
Int[(E^(-(E^x*x) + (-361*x^2 + 16*E^(E^x*x)*x^2)/(8*E^(E^x*x)*x))*(-361*x + 16*E^(E^x*x)*x + E^x*(361*x^2 + 361*x^3)))/(8*x),x]
Output:
$Aborted
Time = 2.54 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05
method | result | size |
risch | \({\mathrm e}^{\frac {x \left (16 \,{\mathrm e}^{{\mathrm e}^{x} x}-361\right ) {\mathrm e}^{-{\mathrm e}^{x} x}}{8}}\) | \(20\) |
parallelrisch | \({\mathrm e}^{\frac {\left (16 x \,{\mathrm e}^{{\mathrm e}^{x} x}-361 x \right ) {\mathrm e}^{-{\mathrm e}^{x} x}}{8}}\) | \(31\) |
Input:
int(1/8*(16*exp(ln(x)+exp(x)*x)+(361*x^3+361*x^2)*exp(x)-361*x)*exp(1/16*( 16*x*exp(ln(x)+exp(x)*x)-361*x^2)/exp(ln(x)+exp(x)*x))^2/exp(ln(x)+exp(x)* x),x,method=_RETURNVERBOSE)
Output:
exp(1/8*x*(16*exp(exp(x)*x)-361)*exp(-exp(x)*x))
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (21) = 42\).
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.47 \[ \int \frac {e^{-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}} \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx=e^{\left (-\frac {1}{8} \, {\left (361 \, x^{2} + 8 \, {\left (x e^{x} - 2 \, x + \log \left (x\right )\right )} e^{\left (x e^{x} + \log \left (x\right )\right )}\right )} e^{\left (-x e^{x} - \log \left (x\right )\right )} + x e^{x} + \log \left (x\right )\right )} \] Input:
integrate(1/8*(16*exp(log(x)+exp(x)*x)+(361*x^3+361*x^2)*exp(x)-361*x)*exp (1/16*(16*x*exp(log(x)+exp(x)*x)-361*x^2)/exp(log(x)+exp(x)*x))^2/exp(log( x)+exp(x)*x),x, algorithm="fricas")
Output:
e^(-1/8*(361*x^2 + 8*(x*e^x - 2*x + log(x))*e^(x*e^x + log(x)))*e^(-x*e^x - log(x)) + x*e^x + log(x))
Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}} \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx=e^{\frac {2 \left (x^{2} e^{x e^{x}} - \frac {361 x^{2}}{16}\right ) e^{- x e^{x}}}{x}} \] Input:
integrate(1/8*(16*exp(ln(x)+exp(x)*x)+(361*x**3+361*x**2)*exp(x)-361*x)*ex p(1/16*(16*x*exp(ln(x)+exp(x)*x)-361*x**2)/exp(ln(x)+exp(x)*x))**2/exp(ln( x)+exp(x)*x),x)
Output:
exp(2*(x**2*exp(x*exp(x)) - 361*x**2/16)*exp(-x*exp(x))/x)
Time = 0.11 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {e^{-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}} \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx=e^{\left (-\frac {361}{8} \, x e^{\left (-x e^{x}\right )} + 2 \, x\right )} \] Input:
integrate(1/8*(16*exp(log(x)+exp(x)*x)+(361*x^3+361*x^2)*exp(x)-361*x)*exp (1/16*(16*x*exp(log(x)+exp(x)*x)-361*x^2)/exp(log(x)+exp(x)*x))^2/exp(log( x)+exp(x)*x),x, algorithm="maxima")
Output:
e^(-361/8*x*e^(-x*e^x) + 2*x)
\[ \int \frac {e^{-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}} \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx=\int { \frac {1}{8} \, {\left (361 \, {\left (x^{3} + x^{2}\right )} e^{x} - 361 \, x + 16 \, e^{\left (x e^{x} + \log \left (x\right )\right )}\right )} e^{\left (-\frac {1}{8} \, {\left (361 \, x^{2} - 16 \, x e^{\left (x e^{x} + \log \left (x\right )\right )}\right )} e^{\left (-x e^{x} - \log \left (x\right )\right )} - x e^{x} - \log \left (x\right )\right )} \,d x } \] Input:
integrate(1/8*(16*exp(log(x)+exp(x)*x)+(361*x^3+361*x^2)*exp(x)-361*x)*exp (1/16*(16*x*exp(log(x)+exp(x)*x)-361*x^2)/exp(log(x)+exp(x)*x))^2/exp(log( x)+exp(x)*x),x, algorithm="giac")
Output:
integrate(1/8*(361*(x^3 + x^2)*e^x - 361*x + 16*e^(x*e^x + log(x)))*e^(-1/ 8*(361*x^2 - 16*x*e^(x*e^x + log(x)))*e^(-x*e^x - log(x)) - x*e^x - log(x) ), x)
Time = 3.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {e^{-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}} \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx={\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-\frac {361\,x\,{\mathrm {e}}^{-x\,{\mathrm {e}}^x}}{8}} \] Input:
int((exp(2*exp(- log(x) - x*exp(x))*(x*exp(log(x) + x*exp(x)) - (361*x^2)/ 16))*exp(- log(x) - x*exp(x))*(16*exp(log(x) + x*exp(x)) - 361*x + exp(x)* (361*x^2 + 361*x^3)))/8,x)
Output:
exp(2*x)*exp(-(361*x*exp(-x*exp(x)))/8)
Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}} \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx=\frac {e^{2 x}}{e^{\frac {361 x}{8 e^{e^{x} x}}}} \] Input:
int(1/8*(16*exp(log(x)+exp(x)*x)+(361*x^3+361*x^2)*exp(x)-361*x)*exp(1/16* (16*x*exp(log(x)+exp(x)*x)-361*x^2)/exp(log(x)+exp(x)*x))^2/exp(log(x)+exp (x)*x),x)
Output:
e**(2*x)/e**((361*x)/(8*e**(e**x*x)))