Integrand size = 54, antiderivative size = 28 \[ \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\frac {e^{2 e^{-x}}}{\log ^2\left (\frac {15}{\frac {x}{4}+e^2 x}\right )} \] Output:
exp(1/exp(x))^2/ln(3/(1/5*x*exp(1)^2+1/20*x))^2
Time = 1.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\frac {e^{2 e^{-x}}}{\log ^2\left (\frac {60}{x+4 e^2 x}\right )} \] Input:
Integrate[(E^(2/E^x - x)*(2*E^x - 2*x*Log[60/(x + 4*E^2*x)]))/(x*Log[60/(x + 4*E^2*x)]^3),x]
Output:
E^(2/E^x)/Log[60/(x + 4*E^2*x)]^2
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.
\(\displaystyle \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{4 e^2 x+x}\right )\right )}{x \log ^3\left (\frac {60}{4 e^2 x+x}\right )} \, dx\) |
\(\Big \downarrow \) 2894 |
\(\displaystyle \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{4 e^2 x+x}\right )\right )}{x \log ^3\left (\frac {60}{\left (1+4 e^2\right ) x}\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 e^{2 e^{-x}-x} \left (e^x-x \log \left (\frac {60}{4 e^2 x+x}\right )\right )}{x \log ^3\left (\frac {60}{\left (1+4 e^2\right ) x}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {e^{2 e^{-x}-x} \left (e^x-x \log \left (\frac {60}{\left (1+4 e^2\right ) x}\right )\right )}{x \log ^3\left (\frac {60}{\left (1+4 e^2\right ) x}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 \int \left (\frac {e^{2 e^{-x}}}{x \log ^3\left (\frac {60}{\left (1+4 e^2\right ) x}\right )}-\frac {e^{2 e^{-x}-x}}{\log ^2\left (\frac {60}{\left (1+4 e^2\right ) x}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\int \frac {e^{2 e^{-x}}}{x \log ^3\left (\frac {60}{\left (1+4 e^2\right ) x}\right )}dx-\int \frac {e^{2 e^{-x}-x}}{\log ^2\left (\frac {60}{\left (1+4 e^2\right ) x}\right )}dx\right )\) |
Input:
Int[(E^(2/E^x - x)*(2*E^x - 2*x*Log[60/(x + 4*E^2*x)]))/(x*Log[60/(x + 4*E ^2*x)]^3),x]
Output:
$Aborted
Time = 11.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(\frac {{\mathrm e}^{2 \,{\mathrm e}^{-x}}}{{\ln \left (\frac {60}{x \left (4 \,{\mathrm e}^{2}+1\right )}\right )}^{2}}\) | \(27\) |
risch | \(-\frac {4 \,{\mathrm e}^{2 \,{\mathrm e}^{-x}}}{{\left (2 i \ln \left (5\right )+2 i \ln \left (3\right )+4 i \ln \left (2\right )-2 i \ln \left (4 \,{\mathrm e}^{2}+1\right )-2 i \ln \left (x \right )\right )}^{2}}\) | \(43\) |
Input:
int((-2*x*ln(60/(4*x*exp(1)^2+x))+2*exp(x))*exp(1/exp(x))^2/x/exp(x)/ln(60 /(4*x*exp(1)^2+x))^3,x,method=_RETURNVERBOSE)
Output:
exp(1/exp(x))^2/ln(60/x/(4*exp(1)^2+1))^2
Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\frac {e^{\left (-{\left (x e^{x} - 2\right )} e^{\left (-x\right )} + x\right )}}{\log \left (\frac {60}{4 \, x e^{2} + x}\right )^{2}} \] Input:
integrate((-2*x*log(60/(4*x*exp(1)^2+x))+2*exp(x))*exp(1/exp(x))^2/x/exp(x )/log(60/(4*x*exp(1)^2+x))^3,x, algorithm="fricas")
Output:
e^(-(x*e^x - 2)*e^(-x) + x)/log(60/(4*x*e^2 + x))^2
Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\frac {e^{2 e^{- x}}}{\log {\left (\frac {60}{x + 4 x e^{2}} \right )}^{2}} \] Input:
integrate((-2*x*ln(60/(4*x*exp(1)**2+x))+2*exp(x))*exp(1/exp(x))**2/x/exp( x)/ln(60/(4*x*exp(1)**2+x))**3,x)
Output:
exp(2*exp(-x))/log(60/(x + 4*x*exp(2)))**2
\[ \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\int { -\frac {2 \, {\left (x \log \left (\frac {60}{4 \, x e^{2} + x}\right ) - e^{x}\right )} e^{\left (-x + 2 \, e^{\left (-x\right )}\right )}}{x \log \left (\frac {60}{4 \, x e^{2} + x}\right )^{3}} \,d x } \] Input:
integrate((-2*x*log(60/(4*x*exp(1)^2+x))+2*exp(x))*exp(1/exp(x))^2/x/exp(x )/log(60/(4*x*exp(1)^2+x))^3,x, algorithm="maxima")
Output:
-2*integrate((x*log(60/(4*x*e^2 + x)) - e^x)*e^(-x + 2*e^(-x))/(x*log(60/( 4*x*e^2 + x))^3), x)
\[ \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\int { -\frac {2 \, {\left (x \log \left (\frac {60}{4 \, x e^{2} + x}\right ) - e^{x}\right )} e^{\left (-x + 2 \, e^{\left (-x\right )}\right )}}{x \log \left (\frac {60}{4 \, x e^{2} + x}\right )^{3}} \,d x } \] Input:
integrate((-2*x*log(60/(4*x*exp(1)^2+x))+2*exp(x))*exp(1/exp(x))^2/x/exp(x )/log(60/(4*x*exp(1)^2+x))^3,x, algorithm="giac")
Output:
integrate(-2*(x*log(60/(4*x*e^2 + x)) - e^x)*e^(-x + 2*e^(-x))/(x*log(60/( 4*x*e^2 + x))^3), x)
Time = 4.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^{-x}}}{{\ln \left (\frac {60}{x+4\,x\,{\mathrm {e}}^2}\right )}^2} \] Input:
int((exp(2*exp(-x))*exp(-x)*(2*exp(x) - 2*x*log(60/(x + 4*x*exp(2)))))/(x* log(60/(x + 4*x*exp(2)))^3),x)
Output:
exp(2*exp(-x))/log(60/(x + 4*x*exp(2)))^2
Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\frac {e^{\frac {2}{e^{x}}}}{\mathrm {log}\left (\frac {60}{4 e^{2} x +x}\right )^{2}} \] Input:
int((-2*x*log(60/(4*x*exp(1)^2+x))+2*exp(x))*exp(1/exp(x))^2/x/exp(x)/log( 60/(4*x*exp(1)^2+x))^3,x)
Output:
e**(2/e**x)/log(60/(4*e**2*x + x))**2