Integrand size = 74, antiderivative size = 28 \[ \int \frac {e^{-x} \left (e^{5+\frac {4}{-3+x}} \left (-1440-1120 x+800 x^2-160 x^3\right )+e^5 \left (180+60 x-460 x^2+260 x^3-40 x^4\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {20 e^{5-x} \left (-1+8 e^{\frac {4}{-3+x}}+2 x\right )}{x} \] Output:
20*(2*x+8*exp(4/(-3+x))-1)/exp(x)*exp(5)/x
Time = 3.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (e^{5+\frac {4}{-3+x}} \left (-1440-1120 x+800 x^2-160 x^3\right )+e^5 \left (180+60 x-460 x^2+260 x^3-40 x^4\right )\right )}{9 x^2-6 x^3+x^4} \, dx=-\frac {20 e^{5-x} \left (1-8 e^{\frac {4}{-3+x}}-2 x\right )}{x} \] Input:
Integrate[(E^(5 + 4/(-3 + x))*(-1440 - 1120*x + 800*x^2 - 160*x^3) + E^5*( 180 + 60*x - 460*x^2 + 260*x^3 - 40*x^4))/(E^x*(9*x^2 - 6*x^3 + x^4)),x]
Output:
(-20*E^(5 - x)*(1 - 8*E^(4/(-3 + x)) - 2*x))/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.
\(\displaystyle \int \frac {e^{-x} \left (e^{\frac {4}{x-3}+5} \left (-160 x^3+800 x^2-1120 x-1440\right )+e^5 \left (-40 x^4+260 x^3-460 x^2+60 x+180\right )\right )}{x^4-6 x^3+9 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-x} \left (e^{\frac {4}{x-3}+5} \left (-160 x^3+800 x^2-1120 x-1440\right )+e^5 \left (-40 x^4+260 x^3-460 x^2+60 x+180\right )\right )}{x^2 \left (x^2-6 x+9\right )}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int -\frac {5 e^{-x} \left (8 e^{5-\frac {4}{3-x}} \left (x^3-5 x^2+7 x+9\right )-e^5 \left (-2 x^4+13 x^3-23 x^2+3 x+9\right )\right )}{(3-x)^2 x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -20 \int \frac {e^{-x} \left (8 e^{5-\frac {4}{3-x}} \left (x^3-5 x^2+7 x+9\right )-e^5 \left (-2 x^4+13 x^3-23 x^2+3 x+9\right )\right )}{(3-x)^2 x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -20 \int \left (\frac {e^{5-x} (x-1) (2 x+1)}{x^2}+\frac {8 e^{\frac {5 x-11}{x-3}-x} \left (x^3-5 x^2+7 x+9\right )}{(3-x)^2 x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -20 \left (\frac {32}{3} \int \frac {e^{\frac {-x^2+8 x-11}{x-3}}}{(x-3)^2}dx-\frac {32}{9} \int \frac {e^{\frac {-x^2+8 x-11}{x-3}}}{x-3}dx+8 \int \frac {e^{\frac {-x^2+8 x-11}{x-3}}}{x^2}dx+\frac {104}{9} \int \frac {e^{\frac {-x^2+8 x-11}{x-3}}}{x}dx-2 e^{5-x}+\frac {e^{5-x}}{x}\right )\) |
Input:
Int[(E^(5 + 4/(-3 + x))*(-1440 - 1120*x + 800*x^2 - 160*x^3) + E^5*(180 + 60*x - 460*x^2 + 260*x^3 - 40*x^4))/(E^x*(9*x^2 - 6*x^3 + x^4)),x]
Output:
$Aborted
Time = 0.63 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11
method | result | size |
parallelrisch | \(\frac {\left (40 x \,{\mathrm e}^{5}+160 \,{\mathrm e}^{5} {\mathrm e}^{\frac {4}{-3+x}}-20 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-x}}{x}\) | \(31\) |
risch | \(\frac {20 \left (-1+2 x \right ) {\mathrm e}^{5-x}}{x}+\frac {160 \,{\mathrm e}^{-\frac {x^{2}-8 x +11}{-3+x}}}{x}\) | \(39\) |
norman | \(\frac {\left (-140 x \,{\mathrm e}^{5}+40 x^{2} {\mathrm e}^{5}-480 \,{\mathrm e}^{5} {\mathrm e}^{\frac {4}{-3+x}}+160 x \,{\mathrm e}^{5} {\mathrm e}^{\frac {4}{-3+x}}+60 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-x}}{x \left (-3+x \right )}\) | \(56\) |
orering | \(-\frac {2 \left (x^{5}-7 x^{4}+24 x^{3}-38 x^{2}+5 x +63\right ) \left (\left (-160 x^{3}+800 x^{2}-1120 x -1440\right ) {\mathrm e}^{5} {\mathrm e}^{\frac {4}{-3+x}}+\left (-40 x^{4}+260 x^{3}-460 x^{2}+60 x +180\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-x}}{\left (x^{5}-6 x^{4}+20 x^{3}-16 x^{2}-57 x +130\right ) \left (x^{4}-6 x^{3}+9 x^{2}\right )}-\frac {\left (x^{2}-2 x +7\right ) \left (-3+x \right )^{2} x \left (\frac {\left (\left (-480 x^{2}+1600 x -1120\right ) {\mathrm e}^{5} {\mathrm e}^{\frac {4}{-3+x}}-\frac {4 \left (-160 x^{3}+800 x^{2}-1120 x -1440\right ) {\mathrm e}^{5} {\mathrm e}^{\frac {4}{-3+x}}}{\left (-3+x \right )^{2}}+\left (-160 x^{3}+780 x^{2}-920 x +60\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-x}}{x^{4}-6 x^{3}+9 x^{2}}-\frac {\left (\left (-160 x^{3}+800 x^{2}-1120 x -1440\right ) {\mathrm e}^{5} {\mathrm e}^{\frac {4}{-3+x}}+\left (-40 x^{4}+260 x^{3}-460 x^{2}+60 x +180\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-x} \left (4 x^{3}-18 x^{2}+18 x \right )}{\left (x^{4}-6 x^{3}+9 x^{2}\right )^{2}}-\frac {\left (\left (-160 x^{3}+800 x^{2}-1120 x -1440\right ) {\mathrm e}^{5} {\mathrm e}^{\frac {4}{-3+x}}+\left (-40 x^{4}+260 x^{3}-460 x^{2}+60 x +180\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-x}}{x^{4}-6 x^{3}+9 x^{2}}\right )}{x^{5}-6 x^{4}+20 x^{3}-16 x^{2}-57 x +130}\) | \(415\) |
Input:
int(((-160*x^3+800*x^2-1120*x-1440)*exp(5)*exp(4/(-3+x))+(-40*x^4+260*x^3- 460*x^2+60*x+180)*exp(5))/(x^4-6*x^3+9*x^2)/exp(x),x,method=_RETURNVERBOSE )
Output:
(40*x*exp(5)+160*exp(5)*exp(4/(-3+x))-20*exp(5))/exp(x)/x
Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-x} \left (e^{5+\frac {4}{-3+x}} \left (-1440-1120 x+800 x^2-160 x^3\right )+e^5 \left (180+60 x-460 x^2+260 x^3-40 x^4\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {20 \, {\left ({\left (2 \, x - 1\right )} e^{\left (-x + 5\right )} + 8 \, e^{\left (-x + \frac {5 \, x - 11}{x - 3}\right )}\right )}}{x} \] Input:
integrate(((-160*x^3+800*x^2-1120*x-1440)*exp(5)*exp(4/(-3+x))+(-40*x^4+26 0*x^3-460*x^2+60*x+180)*exp(5))/(x^4-6*x^3+9*x^2)/exp(x),x, algorithm="fri cas")
Output:
20*((2*x - 1)*e^(-x + 5) + 8*e^(-x + (5*x - 11)/(x - 3)))/x
Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-x} \left (e^{5+\frac {4}{-3+x}} \left (-1440-1120 x+800 x^2-160 x^3\right )+e^5 \left (180+60 x-460 x^2+260 x^3-40 x^4\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {\left (40 x e^{5} - 20 e^{5}\right ) e^{- x}}{x} + \frac {160 e^{5} e^{- x} e^{\frac {4}{x - 3}}}{x} \] Input:
integrate(((-160*x**3+800*x**2-1120*x-1440)*exp(5)*exp(4/(-3+x))+(-40*x**4 +260*x**3-460*x**2+60*x+180)*exp(5))/(x**4-6*x**3+9*x**2)/exp(x),x)
Output:
(40*x*exp(5) - 20*exp(5))*exp(-x)/x + 160*exp(5)*exp(-x)*exp(4/(x - 3))/x
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^{-x} \left (e^{5+\frac {4}{-3+x}} \left (-1440-1120 x+800 x^2-160 x^3\right )+e^5 \left (180+60 x-460 x^2+260 x^3-40 x^4\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {20 \, {\left (2 \, x e^{5} - e^{5} + 8 \, e^{\left (\frac {4}{x - 3} + 5\right )}\right )} e^{\left (-x\right )}}{x} \] Input:
integrate(((-160*x^3+800*x^2-1120*x-1440)*exp(5)*exp(4/(-3+x))+(-40*x^4+26 0*x^3-460*x^2+60*x+180)*exp(5))/(x^4-6*x^3+9*x^2)/exp(x),x, algorithm="max ima")
Output:
20*(2*x*e^5 - e^5 + 8*e^(4/(x - 3) + 5))*e^(-x)/x
Time = 0.14 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {e^{-x} \left (e^{5+\frac {4}{-3+x}} \left (-1440-1120 x+800 x^2-160 x^3\right )+e^5 \left (180+60 x-460 x^2+260 x^3-40 x^4\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {20 \, {\left (2 \, x e^{\left (-x + 5\right )} - e^{\left (-x + 5\right )} + 8 \, e^{\left (-\frac {3 \, x^{2} - 13 \, x}{3 \, {\left (x - 3\right )}} + \frac {11}{3}\right )}\right )}}{x} \] Input:
integrate(((-160*x^3+800*x^2-1120*x-1440)*exp(5)*exp(4/(-3+x))+(-40*x^4+26 0*x^3-460*x^2+60*x+180)*exp(5))/(x^4-6*x^3+9*x^2)/exp(x),x, algorithm="gia c")
Output:
20*(2*x*e^(-x + 5) - e^(-x + 5) + 8*e^(-1/3*(3*x^2 - 13*x)/(x - 3) + 11/3) )/x
Time = 0.47 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {e^{-x} \left (e^{5+\frac {4}{-3+x}} \left (-1440-1120 x+800 x^2-160 x^3\right )+e^5 \left (180+60 x-460 x^2+260 x^3-40 x^4\right )\right )}{9 x^2-6 x^3+x^4} \, dx=40\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5-\frac {20\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5}{x}+\frac {160\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5\,{\mathrm {e}}^{\frac {4}{x-3}}}{x} \] Input:
int((exp(-x)*(exp(5)*(60*x - 460*x^2 + 260*x^3 - 40*x^4 + 180) - exp(5)*ex p(4/(x - 3))*(1120*x - 800*x^2 + 160*x^3 + 1440)))/(9*x^2 - 6*x^3 + x^4),x )
Output:
40*exp(-x)*exp(5) - (20*exp(-x)*exp(5))/x + (160*exp(-x)*exp(5)*exp(4/(x - 3)))/x
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-x} \left (e^{5+\frac {4}{-3+x}} \left (-1440-1120 x+800 x^2-160 x^3\right )+e^5 \left (180+60 x-460 x^2+260 x^3-40 x^4\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {20 e^{5} \left (8 e^{\frac {4}{x -3}}+2 x -1\right )}{e^{x} x} \] Input:
int(((-160*x^3+800*x^2-1120*x-1440)*exp(5)*exp(4/(-3+x))+(-40*x^4+260*x^3- 460*x^2+60*x+180)*exp(5))/(x^4-6*x^3+9*x^2)/exp(x),x)
Output:
(20*e**5*(8*e**(4/(x - 3)) + 2*x - 1))/(e**x*x)