Integrand size = 160, antiderivative size = 24 \[ \int \frac {e^{2+x} \left (1-2 x+x^2\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (1-2 x+x^2\right )+\left (e^{2+x} \left (x-2 x^2+x^3\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (x+2 x^2-4 x^3+2 x^4\right )\right ) \log (x)}{\left (e^{2+x} \left (x-2 x^2+x^3\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (x-2 x^2+x^3\right )\right ) \log (x)} \, dx=\log \left (\left (e^{2+x}+e^{4+\frac {1}{1-x}+x^2}\right ) \log (x)\right ) \]
Time = 1.50 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {e^{2+x} \left (1-2 x+x^2\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (1-2 x+x^2\right )+\left (e^{2+x} \left (x-2 x^2+x^3\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (x+2 x^2-4 x^3+2 x^4\right )\right ) \log (x)}{\left (e^{2+x} \left (x-2 x^2+x^3\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (x-2 x^2+x^3\right )\right ) \log (x)} \, dx=-\frac {1}{-1+x}+\log \left (e^{\frac {1}{-1+x}+x}+e^{2+x^2}\right )+\log (\log (x)) \]
Integrate[(E^(2 + x)*(1 - 2*x + x^2) + E^((-5 + 4*x - x^2 + x^3)/(-1 + x)) *(1 - 2*x + x^2) + (E^(2 + x)*(x - 2*x^2 + x^3) + E^((-5 + 4*x - x^2 + x^3 )/(-1 + x))*(x + 2*x^2 - 4*x^3 + 2*x^4))*Log[x])/((E^(2 + x)*(x - 2*x^2 + x^3) + E^((-5 + 4*x - x^2 + x^3)/(-1 + x))*(x - 2*x^2 + x^3))*Log[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.
\(\displaystyle \int \frac {e^{x+2} \left (x^2-2 x+1\right )+e^{\frac {x^3-x^2+4 x-5}{x-1}} \left (x^2-2 x+1\right )+\left (e^{x+2} \left (x^3-2 x^2+x\right )+e^{\frac {x^3-x^2+4 x-5}{x-1}} \left (2 x^4-4 x^3+2 x^2+x\right )\right ) \log (x)}{\left (e^{x+2} \left (x^3-2 x^2+x\right )+e^{\frac {x^3-x^2+4 x-5}{x-1}} \left (x^3-2 x^2+x\right )\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{\frac {2 x^2+x+3}{x-1}} x \left (-2 x^2+5 x-4\right )}{\left (e^{\frac {x \left (x^2+4\right )}{x-1}}+e^{\frac {2 x^2}{x-1}+\frac {x}{x-1}+\frac {3}{x-1}}\right ) (1-x)^2}+\frac {2 x^4 \log (x)-4 x^3 \log (x)+x^2+2 x^2 \log (x)-2 x+x \log (x)+1}{(x-1)^2 x \log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {e^{\frac {2 x^2+x+3}{x-1}}}{e^{\frac {x \left (x^2+4\right )}{x-1}}+e^{\frac {2 x^2}{x-1}+\frac {x}{x-1}+\frac {3}{x-1}}}dx-\int \frac {e^{\frac {2 x^2+x+3}{x-1}}}{\left (e^{\frac {x \left (x^2+4\right )}{x-1}}+e^{\frac {2 x^2}{x-1}+\frac {x}{x-1}+\frac {3}{x-1}}\right ) (x-1)^2}dx-2 \int \frac {e^{\frac {2 x^2+x+3}{x-1}} x}{e^{\frac {x \left (x^2+4\right )}{x-1}}+e^{\frac {2 x^2}{x-1}+\frac {x}{x-1}+\frac {3}{x-1}}}dx+x^2+\frac {1}{1-x}+\log (\log (x))\) |
Int[(E^(2 + x)*(1 - 2*x + x^2) + E^((-5 + 4*x - x^2 + x^3)/(-1 + x))*(1 - 2*x + x^2) + (E^(2 + x)*(x - 2*x^2 + x^3) + E^((-5 + 4*x - x^2 + x^3)/(-1 + x))*(x + 2*x^2 - 4*x^3 + 2*x^4))*Log[x])/((E^(2 + x)*(x - 2*x^2 + x^3) + E^((-5 + 4*x - x^2 + x^3)/(-1 + x))*(x - 2*x^2 + x^3))*Log[x]),x]
3.14.29.3.1 Defintions of rubi rules used
Time = 4.71 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29
method | result | size |
parallelrisch | \(\ln \left (\ln \left (x \right )\right )+\ln \left ({\mathrm e}^{\frac {x^{3}-x^{2}+4 x -5}{-1+x}}+{\mathrm e}^{2+x}\right )\) | \(31\) |
risch | \(\ln \left ({\mathrm e}^{\frac {x^{3}-x^{2}+4 x -5}{-1+x}}+{\mathrm e}^{2+x}\right )-4+\ln \left (\ln \left (x \right )\right )\) | \(32\) |
int((((2*x^4-4*x^3+2*x^2+x)*exp((x^3-x^2+4*x-5)/(-1+x))+(x^3-2*x^2+x)*exp( 2+x))*ln(x)+(x^2-2*x+1)*exp((x^3-x^2+4*x-5)/(-1+x))+(x^2-2*x+1)*exp(2+x))/ ((x^3-2*x^2+x)*exp((x^3-x^2+4*x-5)/(-1+x))+(x^3-2*x^2+x)*exp(2+x))/ln(x),x ,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {e^{2+x} \left (1-2 x+x^2\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (1-2 x+x^2\right )+\left (e^{2+x} \left (x-2 x^2+x^3\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (x+2 x^2-4 x^3+2 x^4\right )\right ) \log (x)}{\left (e^{2+x} \left (x-2 x^2+x^3\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (x-2 x^2+x^3\right )\right ) \log (x)} \, dx=\log \left (e^{\left (x + 2\right )} + e^{\left (\frac {x^{3} - x^{2} + 4 \, x - 5}{x - 1}\right )}\right ) + \log \left (\log \left (x\right )\right ) \]
integrate((((2*x^4-4*x^3+2*x^2+x)*exp((x^3-x^2+4*x-5)/(-1+x))+(x^3-2*x^2+x )*exp(2+x))*log(x)+(x^2-2*x+1)*exp((x^3-x^2+4*x-5)/(-1+x))+(x^2-2*x+1)*exp (2+x))/((x^3-2*x^2+x)*exp((x^3-x^2+4*x-5)/(-1+x))+(x^3-2*x^2+x)*exp(2+x))/ log(x),x, algorithm=\
Time = 0.42 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{2+x} \left (1-2 x+x^2\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (1-2 x+x^2\right )+\left (e^{2+x} \left (x-2 x^2+x^3\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (x+2 x^2-4 x^3+2 x^4\right )\right ) \log (x)}{\left (e^{2+x} \left (x-2 x^2+x^3\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (x-2 x^2+x^3\right )\right ) \log (x)} \, dx=\log {\left (e^{\frac {x^{3} - x^{2} + 4 x - 5}{x - 1}} + e^{x + 2} \right )} + \log {\left (\log {\left (x \right )} \right )} \]
integrate((((2*x**4-4*x**3+2*x**2+x)*exp((x**3-x**2+4*x-5)/(-1+x))+(x**3-2 *x**2+x)*exp(2+x))*ln(x)+(x**2-2*x+1)*exp((x**3-x**2+4*x-5)/(-1+x))+(x**2- 2*x+1)*exp(2+x))/((x**3-2*x**2+x)*exp((x**3-x**2+4*x-5)/(-1+x))+(x**3-2*x* *2+x)*exp(2+x))/ln(x),x)
Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {e^{2+x} \left (1-2 x+x^2\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (1-2 x+x^2\right )+\left (e^{2+x} \left (x-2 x^2+x^3\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (x+2 x^2-4 x^3+2 x^4\right )\right ) \log (x)}{\left (e^{2+x} \left (x-2 x^2+x^3\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (x-2 x^2+x^3\right )\right ) \log (x)} \, dx=\frac {x^{2} - x - 1}{x - 1} + \log \left ({\left (e^{\left (x^{2} + 2\right )} + e^{\left (x + \frac {1}{x - 1}\right )}\right )} e^{\left (-x\right )}\right ) + \log \left (\log \left (x\right )\right ) \]
integrate((((2*x^4-4*x^3+2*x^2+x)*exp((x^3-x^2+4*x-5)/(-1+x))+(x^3-2*x^2+x )*exp(2+x))*log(x)+(x^2-2*x+1)*exp((x^3-x^2+4*x-5)/(-1+x))+(x^2-2*x+1)*exp (2+x))/((x^3-2*x^2+x)*exp((x^3-x^2+4*x-5)/(-1+x))+(x^3-2*x^2+x)*exp(2+x))/ log(x),x, algorithm=\
Time = 0.46 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {e^{2+x} \left (1-2 x+x^2\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (1-2 x+x^2\right )+\left (e^{2+x} \left (x-2 x^2+x^3\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (x+2 x^2-4 x^3+2 x^4\right )\right ) \log (x)}{\left (e^{2+x} \left (x-2 x^2+x^3\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (x-2 x^2+x^3\right )\right ) \log (x)} \, dx=\log \left (e^{x} + e^{\left (\frac {x^{3} - x^{2} - x}{x - 1} + 3\right )}\right ) + \log \left (\log \left (x\right )\right ) \]
integrate((((2*x^4-4*x^3+2*x^2+x)*exp((x^3-x^2+4*x-5)/(-1+x))+(x^3-2*x^2+x )*exp(2+x))*log(x)+(x^2-2*x+1)*exp((x^3-x^2+4*x-5)/(-1+x))+(x^2-2*x+1)*exp (2+x))/((x^3-2*x^2+x)*exp((x^3-x^2+4*x-5)/(-1+x))+(x^3-2*x^2+x)*exp(2+x))/ log(x),x, algorithm=\
Time = 9.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.08 \[ \int \frac {e^{2+x} \left (1-2 x+x^2\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (1-2 x+x^2\right )+\left (e^{2+x} \left (x-2 x^2+x^3\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (x+2 x^2-4 x^3+2 x^4\right )\right ) \log (x)}{\left (e^{2+x} \left (x-2 x^2+x^3\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (x-2 x^2+x^3\right )\right ) \log (x)} \, dx=\ln \left (\ln \left (x\right )\right )+\ln \left ({\mathrm {e}}^2\,{\mathrm {e}}^x+{\mathrm {e}}^{\frac {4\,x}{x-1}}\,{\mathrm {e}}^{\frac {x^3}{x-1}}\,{\mathrm {e}}^{-\frac {x^2}{x-1}}\,{\mathrm {e}}^{-\frac {5}{x-1}}\right ) \]
int((log(x)*(exp((4*x - x^2 + x^3 - 5)/(x - 1))*(x + 2*x^2 - 4*x^3 + 2*x^4 ) + exp(x + 2)*(x - 2*x^2 + x^3)) + exp(x + 2)*(x^2 - 2*x + 1) + exp((4*x - x^2 + x^3 - 5)/(x - 1))*(x^2 - 2*x + 1))/(log(x)*(exp(x + 2)*(x - 2*x^2 + x^3) + exp((4*x - x^2 + x^3 - 5)/(x - 1))*(x - 2*x^2 + x^3))),x)