Integrand size = 102, antiderivative size = 30 \[ \int \frac {e^{\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}} \left (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2)))\right )}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx=e^{\frac {\frac {1}{2}-e^x}{-i \pi +x-\log (4-\log (2))}} \]
Time = 2.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}} \left (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2)))\right )}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx=e^{\frac {-1+2 e^x}{2 (i \pi -x+\log (4-\log (2)))}} \]
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]
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 |
\(\displaystyle \frac {1}{2} \left (\int \frac {\exp \left (-\frac {1-2 e^x}{2 (-x+\log (4-\log (2))+i \pi )}\right )}{(i x-i \log (4-\log (2))+\pi )^2}dx-2 \int \frac {\exp \left (x-\frac {1-2 e^x}{2 (-x+\log (4-\log (2))+i \pi )}\right )}{(i x-i \log (4-\log (2))+\pi )^2}dx-2 i \int \frac {\exp \left (x-\frac {1-2 e^x}{2 (-x+\log (4-\log (2))+i \pi )}\right )}{i x-i \log (4-\log (2))+\pi }dx\right )\) |
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]
3.11.68.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 2.60 (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 | \(\frac {\ln \left (\ln \left (2\right )-4\right ) {\mathrm e}^{\frac {2 \,{\mathrm e}^{x}-1}{2 \ln \left (\ln \left (2\right )-4\right )-2 x}}-x \,{\mathrm e}^{\frac {2 \,{\mathrm e}^{x}-1}{2 \ln \left (\ln \left (2\right )-4\right )-2 x}}}{\ln \left (\ln \left (2\right )-4\right )-x}\) | \(65\) |
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)
Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {e^{\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}} \left (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2)))\right )}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx=e^{\left (-\frac {2 \, e^{x} - 1}{2 \, {\left (x - \log \left (\log \left (2\right ) - 4\right )\right )}}\right )} \]
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 =\
Timed out. \[ \int \frac {e^{\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}} \left (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2)))\right )}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx=\text {Timed out} \]
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)
Time = 0.37 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}} \left (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2)))\right )}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx=e^{\left (-\frac {e^{x}}{x - \log \left (\log \left (2\right ) - 4\right )} + \frac {1}{2 \, {\left (x - \log \left (\log \left (2\right ) - 4\right )\right )}}\right )} \]
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 =\
Time = 0.34 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}} \left (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2)))\right )}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx=e^{\left (-\frac {e^{x}}{x - \log \left (\log \left (2\right ) - 4\right )} + \frac {1}{2 \, {\left (x - \log \left (\log \left (2\right ) - 4\right )\right )}}\right )} \]
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 =\
Time = 13.86 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}} \left (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2)))\right )}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx={\mathrm {e}}^{-\frac {{\mathrm {e}}^x}{x-\ln \left (\ln \left (2\right )-4\right )}}\,{\mathrm {e}}^{\frac {1}{2\,x-2\,\ln \left (\ln \left (2\right )-4\right )}} \]