Integrand size = 76, antiderivative size = 30 \[ \int \frac {e^{-x} \left (20-28 x+17 x^2-6 x^3+x^4+e^{\frac {-4-x}{-2+x}} \left (-20+10 x-25 x^2+5 x^3\right )+\left (-4+4 x-x^2\right ) \log (4)\right )}{4-4 x+x^2} \, dx=e^{-x} \left (-5-x \left (5 e^{\frac {4+x}{2-x}}+x\right )+\log (4)\right ) \]
Time = 10.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-x} \left (20-28 x+17 x^2-6 x^3+x^4+e^{\frac {-4-x}{-2+x}} \left (-20+10 x-25 x^2+5 x^3\right )+\left (-4+4 x-x^2\right ) \log (4)\right )}{4-4 x+x^2} \, dx=e^{-1-\frac {6}{-2+x}-x} \left (-5 x+e^{\frac {4+x}{-2+x}} \left (-5-x^2+\log (4)\right )\right ) \]
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 ]
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 |
\(\displaystyle -5 \int e^{\frac {-x^2+x-4}{x-2}}dx-60 \int \frac {e^{\frac {-x^2+x-4}{x-2}}}{(x-2)^2}dx-30 \int \frac {e^{\frac {-x^2+x-4}{x-2}}}{x-2}dx+5 \int e^{\frac {-x^2+x-4}{x-2}} xdx-e^{-x} x^2-5 e^{-x}+e^{-x} \log (4)\) |
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]
3.12.9.3.1 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 = 0.95 (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 | \(\frac {\left (\left (2 \ln \left (2\right )-5\right ) x +2 x^{2}-x^{3}+10 x \,{\mathrm e}^{\frac {-4-x}{-2+x}}-5 x^{2} {\mathrm e}^{\frac {-4-x}{-2+x}}+10-4 \ln \left (2\right )\right ) {\mathrm e}^{-x}}{-2+x}\) | \(67\) |
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)
Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-x} \left (20-28 x+17 x^2-6 x^3+x^4+e^{\frac {-4-x}{-2+x}} \left (-20+10 x-25 x^2+5 x^3\right )+\left (-4+4 x-x^2\right ) \log (4)\right )}{4-4 x+x^2} \, dx=-{\left (x^{2} - 2 \, \log \left (2\right ) + 5\right )} e^{\left (-x\right )} - 5 \, x e^{\left (-x - \frac {x + 4}{x - 2}\right )} \]
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=\
Time = 4.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-x} \left (20-28 x+17 x^2-6 x^3+x^4+e^{\frac {-4-x}{-2+x}} \left (-20+10 x-25 x^2+5 x^3\right )+\left (-4+4 x-x^2\right ) \log (4)\right )}{4-4 x+x^2} \, dx=- 5 x e^{- x} e^{\frac {- x - 4}{x - 2}} + \left (- x^{2} - 5 + 2 \log {\left (2 \right )}\right ) e^{- x} \]
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)
\[ \int \frac {e^{-x} \left (20-28 x+17 x^2-6 x^3+x^4+e^{\frac {-4-x}{-2+x}} \left (-20+10 x-25 x^2+5 x^3\right )+\left (-4+4 x-x^2\right ) \log (4)\right )}{4-4 x+x^2} \, dx=\int { \frac {{\left (x^{4} - 6 \, x^{3} + 17 \, x^{2} + 5 \, {\left (x^{3} - 5 \, x^{2} + 2 \, x - 4\right )} e^{\left (-\frac {x + 4}{x - 2}\right )} - 2 \, {\left (x^{2} - 4 \, x + 4\right )} \log \left (2\right ) - 28 \, x + 20\right )} e^{\left (-x\right )}}{x^{2} - 4 \, x + 4} \,d x } \]
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=\
-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 = 0.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {e^{-x} \left (20-28 x+17 x^2-6 x^3+x^4+e^{\frac {-4-x}{-2+x}} \left (-20+10 x-25 x^2+5 x^3\right )+\left (-4+4 x-x^2\right ) \log (4)\right )}{4-4 x+x^2} \, dx=-x^{2} e^{\left (-x\right )} - 5 \, x e^{\left (-\frac {x^{2} + x}{x - 2} + 2\right )} + 2 \, e^{\left (-x\right )} \log \left (2\right ) - 5 \, e^{\left (-x\right )} \]
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=\
Time = 13.70 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.60 \[ \int \frac {e^{-x} \left (20-28 x+17 x^2-6 x^3+x^4+e^{\frac {-4-x}{-2+x}} \left (-20+10 x-25 x^2+5 x^3\right )+\left (-4+4 x-x^2\right ) \log (4)\right )}{4-4 x+x^2} \, dx=2\,{\mathrm {e}}^{-x}\,\ln \left (2\right )-5\,{\mathrm {e}}^{-x}-x^2\,{\mathrm {e}}^{-x}-5\,x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-\frac {x}{x-2}}\,{\mathrm {e}}^{-\frac {4}{x-2}} \]