Integrand size = 102, antiderivative size = 30 \[ \int \frac {e^{\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}} \left (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2)))\right )}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx=e^{\frac {\frac {1}{2}-e^x}{-i \pi +x-\log (4-\log (2))}} \] Output:
exp((1/2-exp(x))/(x-ln(ln(2)-4)))
Time = 1.38 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}} \left (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2)))\right )}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx=e^{\frac {-1+2 e^x}{2 (i \pi -x+\log (4-\log (2)))}} \] Input:
Integrate[(E^((-1 + 2*E^x)/(-2*x + 2*(I*Pi + Log[4 - Log[2]])))*(-1 + E^x* (2 - 2*x) + 2*E^x*(I*Pi + Log[4 - Log[2]])))/(2*x^2 - 4*x*(I*Pi + Log[4 - Log[2]]) + 2*(I*Pi + Log[4 - Log[2]])^2),x]
Output:
E^((-1 + 2*E^x)/(2*(I*Pi - x + Log[4 - Log[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.
\(\displaystyle \int \frac {\left (e^x (2-2 x)+2 e^x (\log (4-\log (2))+i \pi )-1\right ) \exp \left (\frac {2 e^x-1}{-2 x+2 (\log (4-\log (2))+i \pi )}\right )}{2 x^2-4 x (\log (4-\log (2))+i \pi )+2 (\log (4-\log (2))+i \pi )^2} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 8 \int -\frac {\exp \left (-\frac {1-2 e^x}{2 (-x+\log (4-\log (2))+i \pi )}\right ) \left (-2 e^x (1-x)-2 e^x (i \pi +\log (4-\log (2)))+1\right )}{16 (-x+\log (4-\log (2))+i \pi )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{2} \int \frac {\exp \left (-\frac {1-2 e^x}{2 (-x+\log (4-\log (2))+i \pi )}\right ) \left (-2 e^x (1-x)-2 e^x (i \pi +\log (4-\log (2)))+1\right )}{(-x+\log (4-\log (2))+i \pi )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {2 \exp \left (x-\frac {1-2 e^x}{2 (-x+\log (4-\log (2))+i \pi )}\right ) (-x+\log (4-\log (2))+i \pi +1)}{(i x-i \log (4-\log (2))+\pi )^2}-\frac {\exp \left (-\frac {1-2 e^x}{2 (-x+\log (4-\log (2))+i \pi )}\right )}{(i x-i \log (4-\log (2))+\pi )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\int \frac {\exp \left (-\frac {1-2 e^x}{2 (-x+\log (4-\log (2))+i \pi )}\right )}{(i x-i \log (4-\log (2))+\pi )^2}dx-2 \int \frac {\exp \left (x-\frac {1-2 e^x}{2 (-x+\log (4-\log (2))+i \pi )}\right )}{(i x-i \log (4-\log (2))+\pi )^2}dx-2 i \int \frac {\exp \left (x-\frac {1-2 e^x}{2 (-x+\log (4-\log (2))+i \pi )}\right )}{i x-i \log (4-\log (2))+\pi }dx\right )\) |
Input:
Int[(E^((-1 + 2*E^x)/(-2*x + 2*(I*Pi + Log[4 - Log[2]])))*(-1 + E^x*(2 - 2 *x) + 2*E^x*(I*Pi + Log[4 - Log[2]])))/(2*x^2 - 4*x*(I*Pi + Log[4 - Log[2] ]) + 2*(I*Pi + Log[4 - Log[2]])^2),x]
Output:
$Aborted
Time = 2.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70
method | result | size |
risch | \({\mathrm e}^{-\frac {2 \,{\mathrm e}^{x}-1}{2 \left (x -\ln \left (\ln \left (2\right )-4\right )\right )}}\) | \(21\) |
parallelrisch | \({\mathrm e}^{\frac {2 \,{\mathrm e}^{x}-1}{2 \ln \left (\ln \left (2\right )-4\right )-2 x}}\) | \(21\) |
norman | \(\frac {\ln \left (\ln \left (2\right )-4\right ) {\mathrm e}^{\frac {2 \,{\mathrm e}^{x}-1}{2 \ln \left (\ln \left (2\right )-4\right )-2 x}}-x \,{\mathrm e}^{\frac {2 \,{\mathrm e}^{x}-1}{2 \ln \left (\ln \left (2\right )-4\right )-2 x}}}{\ln \left (\ln \left (2\right )-4\right )-x}\) | \(65\) |
Input:
int((2*exp(x)*ln(ln(2)-4)+(2-2*x)*exp(x)-1)*exp((2*exp(x)-1)/(2*ln(ln(2)-4 )-2*x))/(2*ln(ln(2)-4)^2-4*x*ln(ln(2)-4)+2*x^2),x,method=_RETURNVERBOSE)
Output:
exp(-1/2*(2*exp(x)-1)/(x-ln(ln(2)-4)))
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {e^{\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}} \left (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2)))\right )}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx=e^{\left (-\frac {2 \, e^{x} - 1}{2 \, {\left (x - \log \left (\log \left (2\right ) - 4\right )\right )}}\right )} \] Input:
integrate((2*exp(x)*log(log(2)-4)+(2-2*x)*exp(x)-1)*exp((2*exp(x)-1)/(2*lo g(log(2)-4)-2*x))/(2*log(log(2)-4)^2-4*x*log(log(2)-4)+2*x^2),x, algorithm ="fricas")
Output:
e^(-1/2*(2*e^x - 1)/(x - log(log(2) - 4)))
Timed out. \[ \int \frac {e^{\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}} \left (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2)))\right )}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx=\text {Timed out} \] Input:
integrate((2*exp(x)*ln(ln(2)-4)+(2-2*x)*exp(x)-1)*exp((2*exp(x)-1)/(2*ln(l n(2)-4)-2*x))/(2*ln(ln(2)-4)**2-4*x*ln(ln(2)-4)+2*x**2),x)
Output:
Timed out
Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}} \left (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2)))\right )}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx=e^{\left (-\frac {e^{x}}{x - \log \left (\log \left (2\right ) - 4\right )} + \frac {1}{2 \, {\left (x - \log \left (\log \left (2\right ) - 4\right )\right )}}\right )} \] Input:
integrate((2*exp(x)*log(log(2)-4)+(2-2*x)*exp(x)-1)*exp((2*exp(x)-1)/(2*lo g(log(2)-4)-2*x))/(2*log(log(2)-4)^2-4*x*log(log(2)-4)+2*x^2),x, algorithm ="maxima")
Output:
e^(-e^x/(x - log(log(2) - 4)) + 1/2/(x - log(log(2) - 4)))
Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}} \left (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2)))\right )}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx=e^{\left (-\frac {e^{x}}{x - \log \left (\log \left (2\right ) - 4\right )} + \frac {1}{2 \, {\left (x - \log \left (\log \left (2\right ) - 4\right )\right )}}\right )} \] Input:
integrate((2*exp(x)*log(log(2)-4)+(2-2*x)*exp(x)-1)*exp((2*exp(x)-1)/(2*lo g(log(2)-4)-2*x))/(2*log(log(2)-4)^2-4*x*log(log(2)-4)+2*x^2),x, algorithm ="giac")
Output:
e^(-e^x/(x - log(log(2) - 4)) + 1/2/(x - log(log(2) - 4)))
Time = 0.37 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}} \left (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2)))\right )}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx={\mathrm {e}}^{-\frac {{\mathrm {e}}^x}{x-\ln \left (\ln \left (2\right )-4\right )}}\,{\mathrm {e}}^{\frac {1}{2\,x-2\,\ln \left (\ln \left (2\right )-4\right )}} \] Input:
int(-(exp(-(2*exp(x) - 1)/(2*x - 2*log(log(2) - 4)))*(exp(x)*(2*x - 2) - 2 *log(log(2) - 4)*exp(x) + 1))/(2*log(log(2) - 4)^2 - 4*x*log(log(2) - 4) + 2*x^2),x)
Output:
exp(-exp(x)/(x - log(log(2) - 4)))*exp(1/(2*x - 2*log(log(2) - 4)))
Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17 \[ \int \frac {e^{\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}} \left (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2)))\right )}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx=\frac {e^{\frac {e^{x}}{\mathrm {log}\left (\mathrm {log}\left (2\right )-4\right )-x}}}{e^{\frac {1}{2 \,\mathrm {log}\left (\mathrm {log}\left (2\right )-4\right )-2 x}}} \] Input:
int((2*exp(x)*log(log(2)-4)+(2-2*x)*exp(x)-1)*exp((2*exp(x)-1)/(2*log(log( 2)-4)-2*x))/(2*log(log(2)-4)^2-4*x*log(log(2)-4)+2*x^2),x)
Output:
e**(e**x/(log(log(2) - 4) - x))/e**(1/(2*log(log(2) - 4) - 2*x))