Integrand size = 79, antiderivative size = 30 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\frac {2 e^{-\frac {5+\log (x)}{5-x}} \left (-2-e^x+2 x\right )}{x} \] Output:
2/x*(2*x-2-exp(x))/exp((5+ln(x))/(5-x))
Time = 5.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=-2 e^{\frac {5}{-5+x}} \left (2+e^x-2 x\right ) x^{\frac {6-x}{-5+x}} \] Input:
Integrate[(120 - 44*x - 12*x^2 + E^x*(60 - 62*x + 22*x^2 - 2*x^3) + (4*x + 2*E^x*x - 4*x^2)*Log[x])/(E^((-5 - Log[x])/(-5 + x))*(25*x^2 - 10*x^3 + x ^4)),x]
Output:
-2*E^(5/(-5 + x))*(2 + E^x - 2*x)*x^((6 - x)/(-5 + 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.
\(\displaystyle \int \frac {e^{-\frac {-\log (x)-5}{x-5}} \left (-12 x^2+\left (-4 x^2+2 e^x x+4 x\right ) \log (x)+e^x \left (-2 x^3+22 x^2-62 x+60\right )-44 x+120\right )}{x^4-10 x^3+25 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-\frac {-\log (x)-5}{x-5}} \left (-12 x^2+\left (-4 x^2+2 e^x x+4 x\right ) \log (x)+e^x \left (-2 x^3+22 x^2-62 x+60\right )-44 x+120\right )}{x^2 \left (x^2-10 x+25\right )}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int \frac {e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}-2} \left (-6 x^2-22 x+e^x \left (-x^3+11 x^2-31 x+30\right )+\left (-2 x^2+e^x x+2 x\right ) \log (x)+60\right )}{2 (5-x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}-2} \left (-6 x^2-22 x+e^x \left (-x^3+11 x^2-31 x+30\right )+\left (-2 x^2+e^x x+2 x\right ) \log (x)+60\right )}{(5-x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 \int \left (-\frac {e^{x-\frac {5}{5-x}} \left (x^3-11 x^2-\log (x) x+31 x-30\right ) x^{\frac {1}{x-5}-2}}{(x-5)^2}+\frac {60 e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}-2}}{(x-5)^2}+\frac {2 e^{-\frac {5}{5-x}} \log (x) x^{\frac {1}{x-5}-1}}{(x-5)^2}-\frac {22 e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}-1}}{(x-5)^2}-\frac {2 e^{-\frac {5}{5-x}} \log (x) x^{\frac {1}{x-5}}}{(x-5)^2}-\frac {6 e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}}}{(x-5)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (60 \int \frac {e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}-2}}{(x-5)^2}dx+30 \int \frac {e^{x-\frac {5}{5-x}} x^{\frac {1}{x-5}-2}}{(x-5)^2}dx-22 \int \frac {e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}-1}}{(x-5)^2}dx-31 \int \frac {e^{x-\frac {5}{5-x}} x^{\frac {1}{x-5}-1}}{(x-5)^2}dx-2 \int \frac {\int \frac {e^{\frac {5}{x-5}} x^{\frac {1}{x-5}-1}}{(x-5)^2}dx}{x}dx-\int \frac {\int \frac {e^{x+\frac {5}{x-5}} x^{\frac {1}{x-5}-1}}{(x-5)^2}dx}{x}dx+2 \log (x) \int \frac {e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}-1}}{(x-5)^2}dx+\log (x) \int \frac {e^{x-\frac {5}{5-x}} x^{\frac {1}{x-5}-1}}{(x-5)^2}dx-6 \int \frac {e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}}}{(x-5)^2}dx+11 \int \frac {e^{x-\frac {5}{5-x}} x^{\frac {1}{x-5}}}{(x-5)^2}dx-\int \frac {e^{x-\frac {5}{5-x}} x^{\frac {x-4}{x-5}}}{(5-x)^2}dx+2 \int \frac {\int \frac {e^{\frac {5}{x-5}} x^{\frac {1}{x-5}}}{(x-5)^2}dx}{x}dx-2 \log (x) \int \frac {e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}}}{(x-5)^2}dx\right )\) |
Input:
Int[(120 - 44*x - 12*x^2 + E^x*(60 - 62*x + 22*x^2 - 2*x^3) + (4*x + 2*E^x *x - 4*x^2)*Log[x])/(E^((-5 - Log[x])/(-5 + x))*(25*x^2 - 10*x^3 + x^4)),x ]
Output:
$Aborted
Time = 2.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87
method | result | size |
risch | \(\frac {2 \left (2 x -2-{\mathrm e}^{x}\right ) {\mathrm e}^{\frac {5+\ln \left (x \right )}{-5+x}}}{x}\) | \(26\) |
parallelrisch | \(\frac {\left (3000+600 x^{2}-300 \,{\mathrm e}^{x} x -3600 x +1500 \,{\mathrm e}^{x}\right ) {\mathrm e}^{\frac {5+\ln \left (x \right )}{-5+x}}}{150 x \left (-5+x \right )}\) | \(44\) |
Input:
int(((2*exp(x)*x-4*x^2+4*x)*ln(x)+(-2*x^3+22*x^2-62*x+60)*exp(x)-12*x^2-44 *x+120)/(x^4-10*x^3+25*x^2)/exp((-5-ln(x))/(-5+x)),x,method=_RETURNVERBOSE )
Output:
2/x*(2*x-2-exp(x))*exp((5+ln(x))/(-5+x))
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\frac {2 \, {\left (2 \, x - e^{x} - 2\right )} e^{\left (\frac {\log \left (x\right ) + 5}{x - 5}\right )}}{x} \] Input:
integrate(((2*exp(x)*x-4*x^2+4*x)*log(x)+(-2*x^3+22*x^2-62*x+60)*exp(x)-12 *x^2-44*x+120)/(x^4-10*x^3+25*x^2)/exp((-5-log(x))/(-5+x)),x, algorithm="f ricas")
Output:
2*(2*x - e^x - 2)*e^((log(x) + 5)/(x - 5))/x
Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\frac {\left (4 x - 2 e^{x} - 4\right ) e^{- \frac {- \log {\left (x \right )} - 5}{x - 5}}}{x} \] Input:
integrate(((2*exp(x)*x-4*x**2+4*x)*ln(x)+(-2*x**3+22*x**2-62*x+60)*exp(x)- 12*x**2-44*x+120)/(x**4-10*x**3+25*x**2)/exp((-5-ln(x))/(-5+x)),x)
Output:
(4*x - 2*exp(x) - 4)*exp(-(-log(x) - 5)/(x - 5))/x
Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\frac {2 \, {\left (2 \, x - e^{x} - 2\right )} e^{\left (\frac {\log \left (x\right )}{x - 5} + \frac {5}{x - 5}\right )}}{x} \] Input:
integrate(((2*exp(x)*x-4*x^2+4*x)*log(x)+(-2*x^3+22*x^2-62*x+60)*exp(x)-12 *x^2-44*x+120)/(x^4-10*x^3+25*x^2)/exp((-5-log(x))/(-5+x)),x, algorithm="m axima")
Output:
2*(2*x - e^x - 2)*e^(log(x)/(x - 5) + 5/(x - 5))/x
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (25) = 50\).
Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.77 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\frac {2 \, {\left (2 \, x e^{\left (\frac {x + \log \left (x\right )}{x - 5}\right )} - e^{\left (\frac {x^{2} - 4 \, x + \log \left (x\right )}{x - 5}\right )} - 2 \, e^{\left (\frac {x + \log \left (x\right )}{x - 5}\right )}\right )} e^{\left (-1\right )}}{x} \] Input:
integrate(((2*exp(x)*x-4*x^2+4*x)*log(x)+(-2*x^3+22*x^2-62*x+60)*exp(x)-12 *x^2-44*x+120)/(x^4-10*x^3+25*x^2)/exp((-5-log(x))/(-5+x)),x, algorithm="g iac")
Output:
2*(2*x*e^((x + log(x))/(x - 5)) - e^((x^2 - 4*x + log(x))/(x - 5)) - 2*e^( (x + log(x))/(x - 5)))*e^(-1)/x
Time = 3.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=-x^{\frac {1}{x-5}-1}\,{\mathrm {e}}^{\frac {5}{x-5}}\,\left (2\,{\mathrm {e}}^x-4\,x+4\right ) \] Input:
int(-(exp((log(x) + 5)/(x - 5))*(44*x - log(x)*(4*x + 2*x*exp(x) - 4*x^2) + 12*x^2 + exp(x)*(62*x - 22*x^2 + 2*x^3 - 60) - 120))/(25*x^2 - 10*x^3 + x^4),x)
Output:
-x^(1/(x - 5) - 1)*exp(5/(x - 5))*(2*exp(x) - 4*x + 4)
Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\frac {2 e^{\frac {\mathrm {log}\left (x \right )+5}{-5+x}} \left (-e^{x}+2 x -2\right )}{x} \] Input:
int(((2*exp(x)*x-4*x^2+4*x)*log(x)+(-2*x^3+22*x^2-62*x+60)*exp(x)-12*x^2-4 4*x+120)/(x^4-10*x^3+25*x^2)/exp((-5-log(x))/(-5+x)),x)
Output:
(2*e**((log(x) + 5)/(x - 5))*( - e**x + 2*x - 2))/x