Integrand size = 102, antiderivative size = 23 \[ \int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx=e^{x+\frac {x}{2-x}}+\log \left ((26-\log (x))^2\right ) \] Output:
exp(x+x/(2-x))+ln((26-ln(x))^2)
Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx=e^{-1-\frac {2}{-2+x}+x}+2 \log (26-\log (x)) \] Input:
Integrate[(8 - 8*x + 2*x^2 + E^((-3*x + x^2)/(-2 + x))*(-156*x + 104*x^2 - 26*x^3) + E^((-3*x + x^2)/(-2 + x))*(6*x - 4*x^2 + x^3)*Log[x])/(-104*x + 104*x^2 - 26*x^3 + (4*x - 4*x^2 + x^3)*Log[x]),x]
Output:
E^(-1 - 2/(-2 + x) + x) + 2*Log[26 - Log[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.
\(\displaystyle \int \frac {2 x^2+e^{\frac {x^2-3 x}{x-2}} \left (-26 x^3+104 x^2-156 x\right )+e^{\frac {x^2-3 x}{x-2}} \left (x^3-4 x^2+6 x\right ) \log (x)-8 x+8}{-26 x^3+104 x^2+\left (x^3-4 x^2+4 x\right ) \log (x)-104 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-2 x^2-e^{\frac {x^2-3 x}{x-2}} \left (-26 x^3+104 x^2-156 x\right )-e^{\frac {x^2-3 x}{x-2}} \left (x^3-4 x^2+6 x\right ) \log (x)+8 x-8}{(2-x)^2 x (26-\log (x))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{\frac {(x-3) x}{x-2}} \left (x^2-4 x+6\right )}{(x-2)^2}+\frac {2 x}{(x-2)^2 (\log (x)-26)}-\frac {8}{(x-2)^2 (\log (x)-26)}+\frac {8}{(x-2)^2 x (\log (x)-26)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -8 \int \frac {1}{(x-2)^2 (\log (x)-26)}dx+8 \int \frac {1}{(x-2)^2 x (\log (x)-26)}dx+2 \int \frac {x}{(x-2)^2 (\log (x)-26)}dx+e^{\frac {(3-x) x}{2-x}}\) |
Input:
Int[(8 - 8*x + 2*x^2 + E^((-3*x + x^2)/(-2 + x))*(-156*x + 104*x^2 - 26*x^ 3) + E^((-3*x + x^2)/(-2 + x))*(6*x - 4*x^2 + x^3)*Log[x])/(-104*x + 104*x ^2 - 26*x^3 + (4*x - 4*x^2 + x^3)*Log[x]),x]
Output:
$Aborted
Time = 0.43 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87
method | result | size |
risch | \({\mathrm e}^{\frac {x \left (-3+x \right )}{-2+x}}+2 \ln \left (-26+\ln \left (x \right )\right )\) | \(20\) |
parallelrisch | \({\mathrm e}^{\frac {x^{2}-3 x}{-2+x}}+2 \ln \left (-26+\ln \left (x \right )\right )\) | \(23\) |
default | \(2 \ln \left (-26+\ln \left (x \right )\right )+\frac {x \,{\mathrm e}^{\frac {x^{2}-3 x}{-2+x}}-2 \,{\mathrm e}^{\frac {x^{2}-3 x}{-2+x}}}{-2+x}\) | \(48\) |
norman | \(2 \ln \left (-26+\ln \left (x \right )\right )+\frac {x \,{\mathrm e}^{\frac {x^{2}-3 x}{-2+x}}-2 \,{\mathrm e}^{\frac {x^{2}-3 x}{-2+x}}}{-2+x}\) | \(48\) |
parts | \(2 \ln \left (-26+\ln \left (x \right )\right )+\frac {x \,{\mathrm e}^{\frac {x^{2}-3 x}{-2+x}}-2 \,{\mathrm e}^{\frac {x^{2}-3 x}{-2+x}}}{-2+x}\) | \(48\) |
Input:
int(((x^3-4*x^2+6*x)*exp((x^2-3*x)/(-2+x))*ln(x)+(-26*x^3+104*x^2-156*x)*e xp((x^2-3*x)/(-2+x))+2*x^2-8*x+8)/((x^3-4*x^2+4*x)*ln(x)-26*x^3+104*x^2-10 4*x),x,method=_RETURNVERBOSE)
Output:
exp(x*(-3+x)/(-2+x))+2*ln(-26+ln(x))
Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx=e^{\left (\frac {x^{2} - 3 \, x}{x - 2}\right )} + 2 \, \log \left (\log \left (x\right ) - 26\right ) \] Input:
integrate(((x^3-4*x^2+6*x)*exp((x^2-3*x)/(-2+x))*log(x)+(-26*x^3+104*x^2-1 56*x)*exp((x^2-3*x)/(-2+x))+2*x^2-8*x+8)/((x^3-4*x^2+4*x)*log(x)-26*x^3+10 4*x^2-104*x),x, algorithm="fricas")
Output:
e^((x^2 - 3*x)/(x - 2)) + 2*log(log(x) - 26)
Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx=e^{\frac {x^{2} - 3 x}{x - 2}} + 2 \log {\left (\log {\left (x \right )} - 26 \right )} \] Input:
integrate(((x**3-4*x**2+6*x)*exp((x**2-3*x)/(-2+x))*ln(x)+(-26*x**3+104*x* *2-156*x)*exp((x**2-3*x)/(-2+x))+2*x**2-8*x+8)/((x**3-4*x**2+4*x)*ln(x)-26 *x**3+104*x**2-104*x),x)
Output:
exp((x**2 - 3*x)/(x - 2)) + 2*log(log(x) - 26)
Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx=e^{\left (x - \frac {2}{x - 2} - 1\right )} + 2 \, \log \left (\log \left (x\right ) - 26\right ) \] Input:
integrate(((x^3-4*x^2+6*x)*exp((x^2-3*x)/(-2+x))*log(x)+(-26*x^3+104*x^2-1 56*x)*exp((x^2-3*x)/(-2+x))+2*x^2-8*x+8)/((x^3-4*x^2+4*x)*log(x)-26*x^3+10 4*x^2-104*x),x, algorithm="maxima")
Output:
e^(x - 2/(x - 2) - 1) + 2*log(log(x) - 26)
Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx=e^{\left (\frac {x^{2} - 3 \, x}{x - 2}\right )} + 2 \, \log \left (\log \left (x\right ) - 26\right ) \] Input:
integrate(((x^3-4*x^2+6*x)*exp((x^2-3*x)/(-2+x))*log(x)+(-26*x^3+104*x^2-1 56*x)*exp((x^2-3*x)/(-2+x))+2*x^2-8*x+8)/((x^3-4*x^2+4*x)*log(x)-26*x^3+10 4*x^2-104*x),x, algorithm="giac")
Output:
e^((x^2 - 3*x)/(x - 2)) + 2*log(log(x) - 26)
Time = 3.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx=2\,\ln \left (\ln \left (x\right )-26\right )+{\mathrm {e}}^{-\frac {3\,x}{x-2}}\,{\mathrm {e}}^{\frac {x^2}{x-2}} \] Input:
int(-(2*x^2 - exp(-(3*x - x^2)/(x - 2))*(156*x - 104*x^2 + 26*x^3) - 8*x + exp(-(3*x - x^2)/(x - 2))*log(x)*(6*x - 4*x^2 + x^3) + 8)/(104*x - log(x) *(4*x - 4*x^2 + x^3) - 104*x^2 + 26*x^3),x)
Output:
2*log(log(x) - 26) + exp(-(3*x)/(x - 2))*exp(x^2/(x - 2))
Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx=\frac {2 e^{\frac {2}{x -2}} \mathrm {log}\left (\mathrm {log}\left (x \right )-26\right ) e +e^{x}}{e^{\frac {2}{x -2}} e} \] Input:
int(((x^3-4*x^2+6*x)*exp((x^2-3*x)/(-2+x))*log(x)+(-26*x^3+104*x^2-156*x)* exp((x^2-3*x)/(-2+x))+2*x^2-8*x+8)/((x^3-4*x^2+4*x)*log(x)-26*x^3+104*x^2- 104*x),x)
Output:
(2*e**(2/(x - 2))*log(log(x) - 26)*e + e**x)/(e**(2/(x - 2))*e)