Integrand size = 102, antiderivative size = 34 \[ \int \frac {e^{-2+x} \left (e^{\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}} \left (-4-4 x-x^2\right )+e^{2-x} \left (32+16 x+2 x^2\right ) (i \pi +\log (9-e))^2\right )}{\left (16+8 x+x^2\right ) (i \pi +\log (9-e))^2} \, dx=-e^{\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}}+2 x \] Output:
2*x-exp(x/(4+x)/exp(2-x)/ln(exp(1)-9)^2)
Time = 5.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-2+x} \left (e^{\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}} \left (-4-4 x-x^2\right )+e^{2-x} \left (32+16 x+2 x^2\right ) (i \pi +\log (9-e))^2\right )}{\left (16+8 x+x^2\right ) (i \pi +\log (9-e))^2} \, dx=-e^{-\frac {e^{-2+x} x}{(4+x) (\pi -i \log (9-e))^2}}+2 x \] Input:
Integrate[(E^(-2 + x)*(E^((E^(-2 + x)*x)/((4 + x)*(I*Pi + Log[9 - E])^2))* (-4 - 4*x - x^2) + E^(2 - x)*(32 + 16*x + 2*x^2)*(I*Pi + Log[9 - E])^2))/( (16 + 8*x + x^2)*(I*Pi + Log[9 - E])^2),x]
Output:
-E^(-((E^(-2 + x)*x)/((4 + x)*(Pi - I*Log[9 - E])^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.
| \(\displaystyle \int \frac {e^{x-2} \left (\left (-x^2-4 x-4\right ) e^{\frac {e^{x-2} x}{(x+4) (\log (9-e)+i \pi )^2}}+e^{2-x} \left (2 x^2+16 x+32\right ) (\log (9-e)+i \pi )^2\right )}{\left (x^2+8 x+16\right ) (\log (9-e)+i \pi )^2} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int -\frac {e^{x-2} \left (e^{\frac {e^{x-2} x}{(x+4) (i \pi +\log (9-e))^2}} \left (x^2+4 x+4\right )-2 e^{2-x} \left (x^2+8 x+16\right ) (i \pi +\log (9-e))^2\right )}{x^2+8 x+16}dx}{(\log (9-e)+i \pi )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {e^{x-2} \left (e^{\frac {e^{x-2} x}{(x+4) (i \pi +\log (9-e))^2}} \left (x^2+4 x+4\right )-2 e^{2-x} \left (x^2+8 x+16\right ) (i \pi +\log (9-e))^2\right )}{x^2+8 x+16}dx}{(\log (9-e)+i \pi )^2}\) |
| \(\Big \downarrow \) 2007 |
| \(\displaystyle -\frac {\int \frac {e^{x-2} \left (e^{\frac {e^{x-2} x}{(x+4) (i \pi +\log (9-e))^2}} \left (x^2+4 x+4\right )-2 e^{2-x} \left (x^2+8 x+16\right ) (i \pi +\log (9-e))^2\right )}{(x+4)^2}dx}{(\log (9-e)+i \pi )^2}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {\int \left (\frac {\exp \left (\frac {e^{x-2} x}{(x+4) (i \pi +\log (9-e))^2}+x-2\right ) (x+2)^2}{(x+4)^2}+2 (\pi -i \log (9-e))^2\right )dx}{(\log (9-e)+i \pi )^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\int \exp \left (\frac {e^{x-2} x}{(x+4) (i \pi +\log (9-e))^2}+x-2\right )dx+4 \int \frac {\exp \left (\frac {e^{x-2} x}{(x+4) (i \pi +\log (9-e))^2}+x-2\right )}{(x+4)^2}dx-4 \int \frac {\exp \left (\frac {e^{x-2} x}{(x+4) (i \pi +\log (9-e))^2}+x-2\right )}{x+4}dx+2 x (\pi -i \log (9-e))^2}{(\log (9-e)+i \pi )^2}\) |
Input:
Int[(E^(-2 + x)*(E^((E^(-2 + x)*x)/((4 + x)*(I*Pi + Log[9 - E])^2))*(-4 - 4*x - x^2) + E^(2 - x)*(32 + 16*x + 2*x^2)*(I*Pi + Log[9 - E])^2))/((16 + 8*x + x^2)*(I*Pi + Log[9 - E])^2),x]
Output:
$Aborted
Time = 0.72 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76
| method | result | size |
| risch | \(2 x -{\mathrm e}^{\frac {x \,{\mathrm e}^{-2+x}}{\left (4+x \right ) \ln \left ({\mathrm e}-9\right )^{2}}}\) | \(26\) |
| parallelrisch | \(\frac {-2 \ln \left ({\mathrm e}-9\right )^{2} \ln \left ({\mathrm e}^{2-x}\right )-{\mathrm e}^{\frac {x \,{\mathrm e}^{-2+x}}{\left (4+x \right ) \ln \left ({\mathrm e}-9\right )^{2}}} \ln \left ({\mathrm e}-9\right )^{2}}{\ln \left ({\mathrm e}-9\right )^{2}}\) | \(58\) |
| parts | \(2 x +\frac {\left (-4 \,{\mathrm e}^{2-x} \ln \left ({\mathrm e}-9\right ) {\mathrm e}^{\frac {x \,{\mathrm e}^{-2+x}}{\left (4+x \right ) \ln \left ({\mathrm e}-9\right )^{2}}}-\ln \left ({\mathrm e}-9\right ) x \,{\mathrm e}^{2-x} {\mathrm e}^{\frac {x \,{\mathrm e}^{-2+x}}{\left (4+x \right ) \ln \left ({\mathrm e}-9\right )^{2}}}\right ) {\mathrm e}^{-2+x}}{\left (4+x \right ) \ln \left ({\mathrm e}-9\right )}\) | \(100\) |
| norman | \(\frac {\left (-32 \,{\mathrm e}^{2-x} \ln \left ({\mathrm e}-9\right )-4 \,{\mathrm e}^{2-x} \ln \left ({\mathrm e}-9\right ) {\mathrm e}^{\frac {x \,{\mathrm e}^{-2+x}}{\left (4+x \right ) \ln \left ({\mathrm e}-9\right )^{2}}}+2 \ln \left ({\mathrm e}-9\right ) x^{2} {\mathrm e}^{2-x}-\ln \left ({\mathrm e}-9\right ) x \,{\mathrm e}^{2-x} {\mathrm e}^{\frac {x \,{\mathrm e}^{-2+x}}{\left (4+x \right ) \ln \left ({\mathrm e}-9\right )^{2}}}\right ) {\mathrm e}^{-2+x}}{\left (4+x \right ) \ln \left ({\mathrm e}-9\right )}\) | \(125\) |
Input:
int(((-x^2-4*x-4)*exp(x/(4+x)/exp(2-x)/ln(exp(1)-9)^2)+(2*x^2+16*x+32)*exp (2-x)*ln(exp(1)-9)^2)/(x^2+8*x+16)/exp(2-x)/ln(exp(1)-9)^2,x,method=_RETUR NVERBOSE)
Output:
2*x-exp(x/(4+x)*exp(-2+x)/ln(exp(1)-9)^2)
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74 \[ \int \frac {e^{-2+x} \left (e^{\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}} \left (-4-4 x-x^2\right )+e^{2-x} \left (32+16 x+2 x^2\right ) (i \pi +\log (9-e))^2\right )}{\left (16+8 x+x^2\right ) (i \pi +\log (9-e))^2} \, dx=2 \, x - e^{\left (\frac {x e^{\left (x - 2\right )}}{{\left (x + 4\right )} \log \left (e - 9\right )^{2}}\right )} \] Input:
integrate(((-x^2-4*x-4)*exp(x/(4+x)/exp(2-x)/log(exp(1)-9)^2)+(2*x^2+16*x+ 32)*exp(2-x)*log(exp(1)-9)^2)/(x^2+8*x+16)/exp(2-x)/log(exp(1)-9)^2,x, alg orithm="fricas")
Output:
2*x - e^(x*e^(x - 2)/((x + 4)*log(e - 9)^2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (26) = 52\).
Time = 1.00 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.50 \[ \int \frac {e^{-2+x} \left (e^{\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}} \left (-4-4 x-x^2\right )+e^{2-x} \left (32+16 x+2 x^2\right ) (i \pi +\log (9-e))^2\right )}{\left (16+8 x+x^2\right ) (i \pi +\log (9-e))^2} \, dx=2 x - e^{\frac {x e^{x}}{- \pi ^{2} x e^{2} + x e^{2} \log {\left (9 - e \right )}^{2} + 2 i \pi x e^{2} \log {\left (9 - e \right )} - 4 \pi ^{2} e^{2} + 4 e^{2} \log {\left (9 - e \right )}^{2} + 8 i \pi e^{2} \log {\left (9 - e \right )}}} \] Input:
integrate(((-x**2-4*x-4)*exp(x/(4+x)/exp(2-x)/ln(exp(1)-9)**2)+(2*x**2+16* x+32)*exp(2-x)*ln(exp(1)-9)**2)/(x**2+8*x+16)/exp(2-x)/ln(exp(1)-9)**2,x)
Output:
2*x - exp(x*exp(x)/(-pi**2*x*exp(2) + x*exp(2)*log(9 - E)**2 + 2*I*pi*x*ex p(2)*log(9 - E) - 4*pi**2*exp(2) + 4*exp(2)*log(9 - E)**2 + 8*I*pi*exp(2)* log(9 - E)))
Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (25) = 50\).
Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 3.53 \[ \int \frac {e^{-2+x} \left (e^{\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}} \left (-4-4 x-x^2\right )+e^{2-x} \left (32+16 x+2 x^2\right ) (i \pi +\log (9-e))^2\right )}{\left (16+8 x+x^2\right ) (i \pi +\log (9-e))^2} \, dx=\frac {2 \, {\left (x - \frac {16}{x + 4} - 8 \, \log \left (x + 4\right )\right )} \log \left (e - 9\right )^{2} + 16 \, {\left (\frac {4}{x + 4} + \log \left (x + 4\right )\right )} \log \left (e - 9\right )^{2} - e^{\left (-\frac {4 \, e^{x}}{x e^{2} \log \left (e - 9\right )^{2} + 4 \, e^{2} \log \left (e - 9\right )^{2}} + \frac {e^{\left (x - 2\right )}}{\log \left (e - 9\right )^{2}}\right )} \log \left (e - 9\right )^{2} - \frac {32 \, \log \left (e - 9\right )^{2}}{x + 4}}{\log \left (e - 9\right )^{2}} \] Input:
integrate(((-x^2-4*x-4)*exp(x/(4+x)/exp(2-x)/log(exp(1)-9)^2)+(2*x^2+16*x+ 32)*exp(2-x)*log(exp(1)-9)^2)/(x^2+8*x+16)/exp(2-x)/log(exp(1)-9)^2,x, alg orithm="maxima")
Output:
(2*(x - 16/(x + 4) - 8*log(x + 4))*log(e - 9)^2 + 16*(4/(x + 4) + log(x + 4))*log(e - 9)^2 - e^(-4*e^x/(x*e^2*log(e - 9)^2 + 4*e^2*log(e - 9)^2) + e ^(x - 2)/log(e - 9)^2)*log(e - 9)^2 - 32*log(e - 9)^2/(x + 4))/log(e - 9)^ 2
\[ \int \frac {e^{-2+x} \left (e^{\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}} \left (-4-4 x-x^2\right )+e^{2-x} \left (32+16 x+2 x^2\right ) (i \pi +\log (9-e))^2\right )}{\left (16+8 x+x^2\right ) (i \pi +\log (9-e))^2} \, dx=\int { \frac {{\left (2 \, {\left (x^{2} + 8 \, x + 16\right )} e^{\left (-x + 2\right )} \log \left (e - 9\right )^{2} - {\left (x^{2} + 4 \, x + 4\right )} e^{\left (\frac {x e^{\left (x - 2\right )}}{{\left (x + 4\right )} \log \left (e - 9\right )^{2}}\right )}\right )} e^{\left (x - 2\right )}}{{\left (x^{2} + 8 \, x + 16\right )} \log \left (e - 9\right )^{2}} \,d x } \] Input:
integrate(((-x^2-4*x-4)*exp(x/(4+x)/exp(2-x)/log(exp(1)-9)^2)+(2*x^2+16*x+ 32)*exp(2-x)*log(exp(1)-9)^2)/(x^2+8*x+16)/exp(2-x)/log(exp(1)-9)^2,x, alg orithm="giac")
Output:
undef
Time = 7.44 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2+x} \left (e^{\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}} \left (-4-4 x-x^2\right )+e^{2-x} \left (32+16 x+2 x^2\right ) (i \pi +\log (9-e))^2\right )}{\left (16+8 x+x^2\right ) (i \pi +\log (9-e))^2} \, dx=2\,x-{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}{x\,{\ln \left (\mathrm {e}-9\right )}^2+4\,{\ln \left (\mathrm {e}-9\right )}^2}} \] Input:
int(-(exp(x - 2)*(exp((x*exp(x - 2))/(log(exp(1) - 9)^2*(x + 4)))*(4*x + x ^2 + 4) - log(exp(1) - 9)^2*exp(2 - x)*(16*x + 2*x^2 + 32)))/(log(exp(1) - 9)^2*(8*x + x^2 + 16)),x)
Output:
2*x - exp((x*exp(-2)*exp(x))/(x*log(exp(1) - 9)^2 + 4*log(exp(1) - 9)^2))
Time = 0.17 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-2+x} \left (e^{\frac {e^{-2+x} x}{(4+x) (i \pi +\log (9-e))^2}} \left (-4-4 x-x^2\right )+e^{2-x} \left (32+16 x+2 x^2\right ) (i \pi +\log (9-e))^2\right )}{\left (16+8 x+x^2\right ) (i \pi +\log (9-e))^2} \, dx=-e^{\frac {e^{x} x}{\mathrm {log}\left (e -9\right )^{2} e^{2} x +4 \mathrm {log}\left (e -9\right )^{2} e^{2}}}+2 x \] Input:
int(((-x^2-4*x-4)*exp(x/(4+x)/exp(2-x)/log(exp(1)-9)^2)+(2*x^2+16*x+32)*ex p(2-x)*log(exp(1)-9)^2)/(x^2+8*x+16)/exp(2-x)/log(exp(1)-9)^2,x)
Output:
- e**((e**x*x)/(log(e - 9)**2*e**2*x + 4*log(e - 9)**2*e**2)) + 2*x