Integrand size = 79, antiderivative size = 27 \[ \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx=e^{-e^{-x} x}+\frac {e^{e^x}+x+\log (5)}{4 x^2} \] Output:
1/4*(x+ln(5)+exp(exp(x)))/x^2+1/exp(x/exp(x))
Time = 3.71 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx=\frac {1}{4} \left (4 e^{-e^{-x} x}+\frac {e^{e^x}}{x^2}+\frac {1}{x}+\frac {\log (25)}{2 x^2}\right ) \] Input:
Integrate[(E^(-x - x/E^x)*(-4*x^3 + 4*x^4 + E^(E^x + x/E^x)*(-2*E^x + E^(2 *x)*x) + E^(x + x/E^x)*(-x - 2*Log[5])))/(4*x^3),x]
Output:
(4/E^(x/E^x) + E^E^x/x^2 + x^(-1) + Log[25]/(2*x^2))/4
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.
\(\displaystyle \int \frac {e^{-e^{-x} x-x} \left (4 x^4-4 x^3+e^{e^{-x} x+e^x} \left (e^{2 x} x-2 e^x\right )+e^{e^{-x} x+x} (-x-2 \log (5))\right )}{4 x^3} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {e^{-e^{-x} x-x} \left (-4 x^4+4 x^3+e^{e^{-x} x+e^x} \left (2 e^x-e^{2 x} x\right )+e^{e^{-x} x+x} (x+2 \log (5))\right )}{x^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {e^{-e^{-x} x-x} \left (-4 x^4+4 x^3+e^{e^{-x} x+e^x} \left (2 e^x-e^{2 x} x\right )+e^{e^{-x} x+x} (x+\log (25))\right )}{x^3}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {1}{4} \int \frac {e^{-e^{-x} \left (1+e^x\right ) x} \left (-4 x^4+4 x^3+e^{e^{-x} x+e^x} \left (2 e^x-e^{2 x} x\right )+e^{e^{-x} x+x} (x+\log (25))\right )}{x^3}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle -\frac {1}{4} \int \frac {-4 e^{\left (-1-e^{-x}\right ) x} (x-1) x^3-e^{x+e^x} x+x+2 e^{e^x}+\log (25)}{x^3}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{4} \int \left (4 e^{-e^{-x} \left (1+e^x\right ) x} (1-x)-\frac {e^{x+e^x} x-x-2 e^{e^x}-\log (25)}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-2 \int \frac {e^{e^x}}{x^3}dx+\int \frac {e^{x+e^x}}{x^2}dx-4 \int e^{-e^{-x} \left (1+e^x\right ) x}dx+4 \int e^{-e^{-x} \left (1+e^x\right ) x} xdx+\frac {(x+\log (25))^2}{2 x^2 \log (25)}\right )\) |
Input:
Int[(E^(-x - x/E^x)*(-4*x^3 + 4*x^4 + E^(E^x + x/E^x)*(-2*E^x + E^(2*x)*x) + E^(x + x/E^x)*(-x - 2*Log[5])))/(4*x^3),x]
Output:
$Aborted
Time = 2.36 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {\ln \left (5\right )+x}{4 x^{2}}+\frac {{\mathrm e}^{{\mathrm e}^{x}}}{4 x^{2}}+{\mathrm e}^{-x \,{\mathrm e}^{-x}}\) | \(27\) |
parallelrisch | \(\frac {{\mathrm e}^{-x} \left ({\mathrm e}^{x} {\mathrm e}^{x \,{\mathrm e}^{-x}} \ln \left (5\right )+4 \,{\mathrm e}^{x} x^{2}+{\mathrm e}^{x} {\mathrm e}^{x \,{\mathrm e}^{-x}} x +{\mathrm e}^{{\mathrm e}^{x}} {\mathrm e}^{x} {\mathrm e}^{x \,{\mathrm e}^{-x}}\right ) {\mathrm e}^{-x \,{\mathrm e}^{-x}}}{4 x^{2}}\) | \(63\) |
Input:
int(1/4*((x*exp(x)^2-2*exp(x))*exp(x/exp(x))*exp(exp(x))+(-2*ln(5)-x)*exp( x)*exp(x/exp(x))+4*x^4-4*x^3)/x^3/exp(x)/exp(x/exp(x)),x,method=_RETURNVER BOSE)
Output:
1/4*(ln(5)+x)/x^2+1/4/x^2*exp(exp(x))+exp(-x*exp(-x))
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (21) = 42\).
Time = 0.11 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.11 \[ \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx=\frac {{\left (4 \, x^{2} e^{x} + {\left (x + \log \left (5\right )\right )} e^{\left ({\left (x e^{x} + x\right )} e^{\left (-x\right )}\right )} + e^{\left ({\left (x + e^{\left (2 \, x\right )}\right )} e^{\left (-x\right )} + x\right )}\right )} e^{\left (-{\left (x e^{x} + x\right )} e^{\left (-x\right )}\right )}}{4 \, x^{2}} \] Input:
integrate(1/4*((x*exp(x)^2-2*exp(x))*exp(x/exp(x))*exp(exp(x))+(-2*log(5)- x)*exp(x)*exp(x/exp(x))+4*x^4-4*x^3)/x^3/exp(x)/exp(x/exp(x)),x, algorithm ="fricas")
Output:
1/4*(4*x^2*e^x + (x + log(5))*e^((x*e^x + x)*e^(-x)) + e^((x + e^(2*x))*e^ (-x) + x))*e^(-(x*e^x + x)*e^(-x))/x^2
Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx=e^{- x e^{- x}} - \frac {- x - \log {\left (5 \right )}}{4 x^{2}} + \frac {e^{e^{x}}}{4 x^{2}} \] Input:
integrate(1/4*((x*exp(x)**2-2*exp(x))*exp(x/exp(x))*exp(exp(x))+(-2*ln(5)- x)*exp(x)*exp(x/exp(x))+4*x**4-4*x**3)/x**3/exp(x)/exp(x/exp(x)),x)
Output:
exp(-x*exp(-x)) - (-x - log(5))/(4*x**2) + exp(exp(x))/(4*x**2)
Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx=\frac {{\left (4 \, x^{2} + e^{\left (x e^{\left (-x\right )} + e^{x}\right )}\right )} e^{\left (-x e^{\left (-x\right )}\right )}}{4 \, x^{2}} + \frac {1}{4 \, x} + \frac {\log \left (5\right )}{4 \, x^{2}} \] Input:
integrate(1/4*((x*exp(x)^2-2*exp(x))*exp(x/exp(x))*exp(exp(x))+(-2*log(5)- x)*exp(x)*exp(x/exp(x))+4*x^4-4*x^3)/x^3/exp(x)/exp(x/exp(x)),x, algorithm ="maxima")
Output:
1/4*(4*x^2 + e^(x*e^(-x) + e^x))*e^(-x*e^(-x))/x^2 + 1/4/x + 1/4*log(5)/x^ 2
Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx=\frac {4 \, x^{2} e^{\left (-x e^{\left (-x\right )}\right )} + x + e^{\left (e^{x}\right )} + \log \left (5\right )}{4 \, x^{2}} \] Input:
integrate(1/4*((x*exp(x)^2-2*exp(x))*exp(x/exp(x))*exp(exp(x))+(-2*log(5)- x)*exp(x)*exp(x/exp(x))+4*x^4-4*x^3)/x^3/exp(x)/exp(x/exp(x)),x, algorithm ="giac")
Output:
1/4*(4*x^2*e^(-x*e^(-x)) + x + e^(e^x) + log(5))/x^2
Time = 2.64 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx={\mathrm {e}}^{-x\,{\mathrm {e}}^{-x}}+\frac {{\mathrm {e}}^{{\mathrm {e}}^x}}{4\,x^2}+\frac {x+\ln \left (5\right )}{4\,x^2} \] Input:
int(-(exp(-x)*exp(-x*exp(-x))*(x^3 - x^4 + (exp(x*exp(-x))*exp(x)*(x + 2*l og(5)))/4 + (exp(exp(x))*exp(x*exp(-x))*(2*exp(x) - x*exp(2*x)))/4))/x^3,x )
Output:
exp(-x*exp(-x)) + exp(exp(x))/(4*x^2) + (x + log(5))/(4*x^2)
Time = 0.18 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.22 \[ \int \frac {e^{-x-e^{-x} x} \left (-4 x^3+4 x^4+e^{e^x+e^{-x} x} \left (-2 e^x+e^{2 x} x\right )+e^{x+e^{-x} x} (-x-2 \log (5))\right )}{4 x^3} \, dx=\frac {e^{\frac {e^{2 x}+x}{e^{x}}}+e^{\frac {x}{e^{x}}} \mathrm {log}\left (5\right )+e^{\frac {x}{e^{x}}} x +4 x^{2}}{4 e^{\frac {x}{e^{x}}} x^{2}} \] Input:
int(1/4*((x*exp(x)^2-2*exp(x))*exp(x/exp(x))*exp(exp(x))+(-2*log(5)-x)*exp (x)*exp(x/exp(x))+4*x^4-4*x^3)/x^3/exp(x)/exp(x/exp(x)),x)
Output:
(e**((e**(2*x) + x)/e**x) + e**(x/e**x)*log(5) + e**(x/e**x)*x + 4*x**2)/( 4*e**(x/e**x)*x**2)