Integrand size = 76, antiderivative size = 30 \[ \int \frac {e^{-x} \left (20-28 x+17 x^2-6 x^3+x^4+e^{\frac {-4-x}{-2+x}} \left (-20+10 x-25 x^2+5 x^3\right )+\left (-4+4 x-x^2\right ) \log (4)\right )}{4-4 x+x^2} \, dx=e^{-x} \left (-5-x \left (5 e^{\frac {4+x}{2-x}}+x\right )+\log (4)\right ) \] Output:
(2*ln(2)-(x+5*exp((4+x)/(2-x)))*x-5)/exp(x)
Time = 10.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-x} \left (20-28 x+17 x^2-6 x^3+x^4+e^{\frac {-4-x}{-2+x}} \left (-20+10 x-25 x^2+5 x^3\right )+\left (-4+4 x-x^2\right ) \log (4)\right )}{4-4 x+x^2} \, dx=e^{-1-\frac {6}{-2+x}-x} \left (-5 x+e^{\frac {4+x}{-2+x}} \left (-5-x^2+\log (4)\right )\right ) \] Input:
Integrate[(20 - 28*x + 17*x^2 - 6*x^3 + x^4 + E^((-4 - x)/(-2 + x))*(-20 + 10*x - 25*x^2 + 5*x^3) + (-4 + 4*x - x^2)*Log[4])/(E^x*(4 - 4*x + x^2)),x ]
Output:
E^(-1 - 6/(-2 + x) - x)*(-5*x + E^((4 + x)/(-2 + x))*(-5 - x^2 + Log[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.
\(\displaystyle \int \frac {e^{-x} \left (x^4-6 x^3+17 x^2+\left (-x^2+4 x-4\right ) \log (4)+e^{\frac {-x-4}{x-2}} \left (5 x^3-25 x^2+10 x-20\right )-28 x+20\right )}{x^2-4 x+4} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int \frac {e^{-x} \left (x^4-6 x^3+17 x^2-28 x-5 e^{\frac {x+4}{2-x}} \left (-x^3+5 x^2-2 x+4\right )-\left (x^2-4 x+4\right ) \log (4)+20\right )}{4 (2-x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {e^{-x} \left (x^4-6 x^3+17 x^2-\left (x^2-4 x+4\right ) \log (4)-5 e^{\frac {x+4}{2-x}} \left (-x^3+5 x^2-2 x+4\right )-28 x+20\right )}{(2-x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{-x} x^4}{(x-2)^2}-\frac {6 e^{-x} x^3}{(x-2)^2}+\frac {17 e^{-x} x^2}{(x-2)^2}+\frac {5 e^{\frac {-x-4}{x-2}-x} \left (x^3-5 x^2+2 x-4\right )}{(2-x)^2}-\frac {28 e^{-x} x}{(x-2)^2}+\frac {20 e^{-x}}{(x-2)^2}-e^{-x} \log (4)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \int e^{\frac {-x^2+x-4}{x-2}}dx-60 \int \frac {e^{\frac {-x^2+x-4}{x-2}}}{(x-2)^2}dx-30 \int \frac {e^{\frac {-x^2+x-4}{x-2}}}{x-2}dx+5 \int e^{\frac {-x^2+x-4}{x-2}} xdx-e^{-x} x^2-5 e^{-x}+e^{-x} \log (4)\) |
Input:
Int[(20 - 28*x + 17*x^2 - 6*x^3 + x^4 + E^((-4 - x)/(-2 + x))*(-20 + 10*x - 25*x^2 + 5*x^3) + (-4 + 4*x - x^2)*Log[4])/(E^x*(4 - 4*x + x^2)),x]
Output:
$Aborted
Time = 1.74 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03
method | result | size |
parallelrisch | \(\left (-5-x^{2}-5 x \,{\mathrm e}^{-\frac {4+x}{-2+x}}+2 \ln \left (2\right )\right ) {\mathrm e}^{-x}\) | \(31\) |
risch | \(\left (-x^{2}+2 \ln \left (2\right )-5\right ) {\mathrm e}^{-x}-5 x \,{\mathrm e}^{-\frac {x^{2}-x +4}{-2+x}}\) | \(37\) |
norman | \(\frac {\left (\left (-5+2 \ln \left (2\right )\right ) x +2 x^{2}-x^{3}+10 x \,{\mathrm e}^{\frac {-4-x}{-2+x}}-5 x^{2} {\mathrm e}^{\frac {-4-x}{-2+x}}+10-4 \ln \left (2\right )\right ) {\mathrm e}^{-x}}{-2+x}\) | \(67\) |
Input:
int(((5*x^3-25*x^2+10*x-20)*exp((-4-x)/(-2+x))+2*(-x^2+4*x-4)*ln(2)+x^4-6* x^3+17*x^2-28*x+20)/(x^2-4*x+4)/exp(x),x,method=_RETURNVERBOSE)
Output:
(-5-x^2-5*x*exp(-(4+x)/(-2+x))+2*ln(2))/exp(x)
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-x} \left (20-28 x+17 x^2-6 x^3+x^4+e^{\frac {-4-x}{-2+x}} \left (-20+10 x-25 x^2+5 x^3\right )+\left (-4+4 x-x^2\right ) \log (4)\right )}{4-4 x+x^2} \, dx=-{\left (x^{2} - 2 \, \log \left (2\right ) + 5\right )} e^{\left (-x\right )} - 5 \, x e^{\left (-x - \frac {x + 4}{x - 2}\right )} \] Input:
integrate(((5*x^3-25*x^2+10*x-20)*exp((-4-x)/(-2+x))+2*(-x^2+4*x-4)*log(2) +x^4-6*x^3+17*x^2-28*x+20)/(x^2-4*x+4)/exp(x),x, algorithm="fricas")
Output:
-(x^2 - 2*log(2) + 5)*e^(-x) - 5*x*e^(-x - (x + 4)/(x - 2))
Time = 4.51 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-x} \left (20-28 x+17 x^2-6 x^3+x^4+e^{\frac {-4-x}{-2+x}} \left (-20+10 x-25 x^2+5 x^3\right )+\left (-4+4 x-x^2\right ) \log (4)\right )}{4-4 x+x^2} \, dx=- 5 x e^{- x} e^{\frac {- x - 4}{x - 2}} + \left (- x^{2} - 5 + 2 \log {\left (2 \right )}\right ) e^{- x} \] Input:
integrate(((5*x**3-25*x**2+10*x-20)*exp((-4-x)/(-2+x))+2*(-x**2+4*x-4)*ln( 2)+x**4-6*x**3+17*x**2-28*x+20)/(x**2-4*x+4)/exp(x),x)
Output:
-5*x*exp(-x)*exp((-x - 4)/(x - 2)) + (-x**2 - 5 + 2*log(2))*exp(-x)
\[ \int \frac {e^{-x} \left (20-28 x+17 x^2-6 x^3+x^4+e^{\frac {-4-x}{-2+x}} \left (-20+10 x-25 x^2+5 x^3\right )+\left (-4+4 x-x^2\right ) \log (4)\right )}{4-4 x+x^2} \, dx=\int { \frac {{\left (x^{4} - 6 \, x^{3} + 17 \, x^{2} + 5 \, {\left (x^{3} - 5 \, x^{2} + 2 \, x - 4\right )} e^{\left (-\frac {x + 4}{x - 2}\right )} - 2 \, {\left (x^{2} - 4 \, x + 4\right )} \log \left (2\right ) - 28 \, x + 20\right )} e^{\left (-x\right )}}{x^{2} - 4 \, x + 4} \,d x } \] Input:
integrate(((5*x^3-25*x^2+10*x-20)*exp((-4-x)/(-2+x))+2*(-x^2+4*x-4)*log(2) +x^4-6*x^3+17*x^2-28*x+20)/(x^2-4*x+4)/exp(x),x, algorithm="maxima")
Output:
-8*(2*log(2) - 5)*integrate(e^(-x)/(x^3 - 6*x^2 + 12*x - 8), x) + 8*e^(-2) *exp_integral_e(2, x - 2)*log(2)/(x - 2) - (x^4*e - 4*x^3*e - x^2*(2*log(2 ) - 9)*e + 4*x*(2*log(2) - 5)*e + 5*(x^3 - 4*x^2 + 4*x)*e^(-6/(x - 2)))*e^ (-x)/(x^2*e - 4*x*e + 4*e) - 20*e^(-2)*exp_integral_e(2, x - 2)/(x - 2)
Time = 0.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {e^{-x} \left (20-28 x+17 x^2-6 x^3+x^4+e^{\frac {-4-x}{-2+x}} \left (-20+10 x-25 x^2+5 x^3\right )+\left (-4+4 x-x^2\right ) \log (4)\right )}{4-4 x+x^2} \, dx=-x^{2} e^{\left (-x\right )} - 5 \, x e^{\left (-\frac {x^{2} + x}{x - 2} + 2\right )} + 2 \, e^{\left (-x\right )} \log \left (2\right ) - 5 \, e^{\left (-x\right )} \] Input:
integrate(((5*x^3-25*x^2+10*x-20)*exp((-4-x)/(-2+x))+2*(-x^2+4*x-4)*log(2) +x^4-6*x^3+17*x^2-28*x+20)/(x^2-4*x+4)/exp(x),x, algorithm="giac")
Output:
-x^2*e^(-x) - 5*x*e^(-(x^2 + x)/(x - 2) + 2) + 2*e^(-x)*log(2) - 5*e^(-x)
Time = 2.90 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.60 \[ \int \frac {e^{-x} \left (20-28 x+17 x^2-6 x^3+x^4+e^{\frac {-4-x}{-2+x}} \left (-20+10 x-25 x^2+5 x^3\right )+\left (-4+4 x-x^2\right ) \log (4)\right )}{4-4 x+x^2} \, dx=2\,{\mathrm {e}}^{-x}\,\ln \left (2\right )-5\,{\mathrm {e}}^{-x}-x^2\,{\mathrm {e}}^{-x}-5\,x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-\frac {x}{x-2}}\,{\mathrm {e}}^{-\frac {4}{x-2}} \] Input:
int((exp(-x)*(exp(-(x + 4)/(x - 2))*(10*x - 25*x^2 + 5*x^3 - 20) - 28*x + 17*x^2 - 6*x^3 + x^4 - 2*log(2)*(x^2 - 4*x + 4) + 20))/(x^2 - 4*x + 4),x)
Output:
2*exp(-x)*log(2) - 5*exp(-x) - x^2*exp(-x) - 5*x*exp(-x)*exp(-x/(x - 2))*e xp(-4/(x - 2))
Time = 0.15 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.23 \[ \int \frac {e^{-x} \left (20-28 x+17 x^2-6 x^3+x^4+e^{\frac {-4-x}{-2+x}} \left (-20+10 x-25 x^2+5 x^3\right )+\left (-4+4 x-x^2\right ) \log (4)\right )}{4-4 x+x^2} \, dx=\frac {2 e^{\frac {6}{x -2}} \mathrm {log}\left (2\right ) e -e^{\frac {6}{x -2}} e \,x^{2}-5 e^{\frac {6}{x -2}} e -5 x}{e^{\frac {x^{2}-2 x +6}{x -2}} e} \] Input:
int(((5*x^3-25*x^2+10*x-20)*exp((-4-x)/(-2+x))+2*(-x^2+4*x-4)*log(2)+x^4-6 *x^3+17*x^2-28*x+20)/(x^2-4*x+4)/exp(x),x)
Output:
(2*e**(6/(x - 2))*log(2)*e - e**(6/(x - 2))*e*x**2 - 5*e**(6/(x - 2))*e - 5*x)/(e**((x**2 - 2*x + 6)/(x - 2))*e)