Integrand size = 200, antiderivative size = 26 \[ \int \frac {4 x-5 x^2-2 x^3+3 x^4+e^{\frac {1}{-x+x^2}} \left (x-3 x^2+2 x^3-x^4\right )+\left (x^2-2 x^3+x^4\right ) \log (x)}{4+8 x-12 x^2-16 x^3+16 x^4+e^{\frac {1}{-x+x^2}} \left (-4 x+12 x^3-8 x^4\right )+e^{\frac {2}{-x+x^2}} \left (x^2-2 x^3+x^4\right )+\left (4 x-12 x^3+8 x^4+e^{\frac {1}{-x+x^2}} \left (-2 x^2+4 x^3-2 x^4\right )\right ) \log (x)+\left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {x}{4-e^{\frac {1}{-x+x^2}}+\frac {2}{x}+\log (x)} \]
Time = 1.40 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {4 x-5 x^2-2 x^3+3 x^4+e^{\frac {1}{-x+x^2}} \left (x-3 x^2+2 x^3-x^4\right )+\left (x^2-2 x^3+x^4\right ) \log (x)}{4+8 x-12 x^2-16 x^3+16 x^4+e^{\frac {1}{-x+x^2}} \left (-4 x+12 x^3-8 x^4\right )+e^{\frac {2}{-x+x^2}} \left (x^2-2 x^3+x^4\right )+\left (4 x-12 x^3+8 x^4+e^{\frac {1}{-x+x^2}} \left (-2 x^2+4 x^3-2 x^4\right )\right ) \log (x)+\left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {x^2}{2-\left (-4+e^{\frac {1}{(-1+x) x}}\right ) x+x \log (x)} \]
Integrate[(4*x - 5*x^2 - 2*x^3 + 3*x^4 + E^(-x + x^2)^(-1)*(x - 3*x^2 + 2* x^3 - x^4) + (x^2 - 2*x^3 + x^4)*Log[x])/(4 + 8*x - 12*x^2 - 16*x^3 + 16*x ^4 + E^(-x + x^2)^(-1)*(-4*x + 12*x^3 - 8*x^4) + E^(2/(-x + x^2))*(x^2 - 2 *x^3 + x^4) + (4*x - 12*x^3 + 8*x^4 + E^(-x + x^2)^(-1)*(-2*x^2 + 4*x^3 - 2*x^4))*Log[x] + (x^2 - 2*x^3 + x^4)*Log[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.
| \(\displaystyle \int \frac {3 x^4-2 x^3-5 x^2+e^{\frac {1}{x^2-x}} \left (-x^4+2 x^3-3 x^2+x\right )+\left (x^4-2 x^3+x^2\right ) \log (x)+4 x}{16 x^4-16 x^3-12 x^2+e^{\frac {1}{x^2-x}} \left (-8 x^4+12 x^3-4 x\right )+e^{\frac {2}{x^2-x}} \left (x^4-2 x^3+x^2\right )+\left (x^4-2 x^3+x^2\right ) \log ^2(x)+\left (8 x^4-12 x^3+e^{\frac {1}{x^2-x}} \left (-2 x^4+4 x^3-2 x^2\right )+4 x\right ) \log (x)+8 x+4} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {x \left (-e^{\frac {1}{x^2-x}} \left (x^3-2 x^2+3 x-1\right )+(3 x+4) (x-1)^2+x (x-1)^2 \log (x)\right )}{(1-x)^2 \left (-\left (\left (e^{\frac {1}{x^2-x}}-4\right ) x\right )+x \log (x)+2\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {x^3-2 x^2+3 x-1}{(x-1)^2 \left (e^{\frac {1}{x^2-x}} x-4 x-x \log (x)-2\right )}-\frac {x^4-4 x^3+13 x^2+2 x^2 \log (x)-2 x-x \log (x)-2}{(x-1)^2 \left (e^{\frac {1}{x^2-x}} x-4 x-x \log (x)-2\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -8 \int \frac {1}{\left (e^{\frac {1}{x^2-x}} x-\log (x) x-4 x-2\right )^2}dx-6 \int \frac {1}{(x-1)^2 \left (e^{\frac {1}{x^2-x}} x-\log (x) x-4 x-2\right )^2}dx-16 \int \frac {1}{(x-1) \left (e^{\frac {1}{x^2-x}} x-\log (x) x-4 x-2\right )^2}dx+2 \int \frac {x}{\left (e^{\frac {1}{x^2-x}} x-\log (x) x-4 x-2\right )^2}dx-\int \frac {x^2}{\left (e^{\frac {1}{x^2-x}} x-\log (x) x-4 x-2\right )^2}dx-\int \frac {1}{(x-1)^2 \left (e^{\frac {1}{x^2-x}} x-\log (x) x-4 x-2\right )}dx-2 \int \frac {1}{(x-1) \left (e^{\frac {1}{x^2-x}} x-\log (x) x-4 x-2\right )}dx-\int \frac {x}{e^{\frac {1}{x^2-x}} x-\log (x) x-4 x-2}dx-2 \int \frac {\log (x)}{\left (-e^{\frac {1}{x^2-x}} x+\log (x) x+4 x+2\right )^2}dx-\int \frac {\log (x)}{(x-1)^2 \left (-e^{\frac {1}{x^2-x}} x+\log (x) x+4 x+2\right )^2}dx-3 \int \frac {\log (x)}{(x-1) \left (-e^{\frac {1}{x^2-x}} x+\log (x) x+4 x+2\right )^2}dx\) |
Int[(4*x - 5*x^2 - 2*x^3 + 3*x^4 + E^(-x + x^2)^(-1)*(x - 3*x^2 + 2*x^3 - x^4) + (x^2 - 2*x^3 + x^4)*Log[x])/(4 + 8*x - 12*x^2 - 16*x^3 + 16*x^4 + E ^(-x + x^2)^(-1)*(-4*x + 12*x^3 - 8*x^4) + E^(2/(-x + x^2))*(x^2 - 2*x^3 + x^4) + (4*x - 12*x^3 + 8*x^4 + E^(-x + x^2)^(-1)*(-2*x^2 + 4*x^3 - 2*x^4) )*Log[x] + (x^2 - 2*x^3 + x^4)*Log[x]^2),x]
3.18.23.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.98 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12
| method | result | size |
| parallelrisch | \(\frac {x^{2}}{x \ln \left (x \right )-{\mathrm e}^{\frac {1}{x \left (-1+x \right )}} x +4 x +2}\) | \(29\) |
| risch | \(-\frac {x^{2}}{{\mathrm e}^{\frac {1}{x \left (-1+x \right )}} x -x \ln \left (x \right )-4 x -2}\) | \(30\) |
int(((x^4-2*x^3+x^2)*ln(x)+(-x^4+2*x^3-3*x^2+x)*exp(1/(x^2-x))+3*x^4-2*x^3 -5*x^2+4*x)/((x^4-2*x^3+x^2)*ln(x)^2+((-2*x^4+4*x^3-2*x^2)*exp(1/(x^2-x))+ 8*x^4-12*x^3+4*x)*ln(x)+(x^4-2*x^3+x^2)*exp(1/(x^2-x))^2+(-8*x^4+12*x^3-4* x)*exp(1/(x^2-x))+16*x^4-16*x^3-12*x^2+8*x+4),x,method=_RETURNVERBOSE)
Time = 0.36 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {4 x-5 x^2-2 x^3+3 x^4+e^{\frac {1}{-x+x^2}} \left (x-3 x^2+2 x^3-x^4\right )+\left (x^2-2 x^3+x^4\right ) \log (x)}{4+8 x-12 x^2-16 x^3+16 x^4+e^{\frac {1}{-x+x^2}} \left (-4 x+12 x^3-8 x^4\right )+e^{\frac {2}{-x+x^2}} \left (x^2-2 x^3+x^4\right )+\left (4 x-12 x^3+8 x^4+e^{\frac {1}{-x+x^2}} \left (-2 x^2+4 x^3-2 x^4\right )\right ) \log (x)+\left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=-\frac {x^{2}}{x e^{\left (\frac {1}{x^{2} - x}\right )} - x \log \left (x\right ) - 4 \, x - 2} \]
integrate(((x^4-2*x^3+x^2)*log(x)+(-x^4+2*x^3-3*x^2+x)*exp(1/(x^2-x))+3*x^ 4-2*x^3-5*x^2+4*x)/((x^4-2*x^3+x^2)*log(x)^2+((-2*x^4+4*x^3-2*x^2)*exp(1/( x^2-x))+8*x^4-12*x^3+4*x)*log(x)+(x^4-2*x^3+x^2)*exp(1/(x^2-x))^2+(-8*x^4+ 12*x^3-4*x)*exp(1/(x^2-x))+16*x^4-16*x^3-12*x^2+8*x+4),x, algorithm=\
Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {4 x-5 x^2-2 x^3+3 x^4+e^{\frac {1}{-x+x^2}} \left (x-3 x^2+2 x^3-x^4\right )+\left (x^2-2 x^3+x^4\right ) \log (x)}{4+8 x-12 x^2-16 x^3+16 x^4+e^{\frac {1}{-x+x^2}} \left (-4 x+12 x^3-8 x^4\right )+e^{\frac {2}{-x+x^2}} \left (x^2-2 x^3+x^4\right )+\left (4 x-12 x^3+8 x^4+e^{\frac {1}{-x+x^2}} \left (-2 x^2+4 x^3-2 x^4\right )\right ) \log (x)+\left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=- \frac {x^{2}}{x e^{\frac {1}{x^{2} - x}} - x \log {\left (x \right )} - 4 x - 2} \]
integrate(((x**4-2*x**3+x**2)*ln(x)+(-x**4+2*x**3-3*x**2+x)*exp(1/(x**2-x) )+3*x**4-2*x**3-5*x**2+4*x)/((x**4-2*x**3+x**2)*ln(x)**2+((-2*x**4+4*x**3- 2*x**2)*exp(1/(x**2-x))+8*x**4-12*x**3+4*x)*ln(x)+(x**4-2*x**3+x**2)*exp(1 /(x**2-x))**2+(-8*x**4+12*x**3-4*x)*exp(1/(x**2-x))+16*x**4-16*x**3-12*x** 2+8*x+4),x)
Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {4 x-5 x^2-2 x^3+3 x^4+e^{\frac {1}{-x+x^2}} \left (x-3 x^2+2 x^3-x^4\right )+\left (x^2-2 x^3+x^4\right ) \log (x)}{4+8 x-12 x^2-16 x^3+16 x^4+e^{\frac {1}{-x+x^2}} \left (-4 x+12 x^3-8 x^4\right )+e^{\frac {2}{-x+x^2}} \left (x^2-2 x^3+x^4\right )+\left (4 x-12 x^3+8 x^4+e^{\frac {1}{-x+x^2}} \left (-2 x^2+4 x^3-2 x^4\right )\right ) \log (x)+\left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=-\frac {x^{2} e^{\frac {1}{x}}}{x e^{\left (\frac {1}{x - 1}\right )} - {\left (x \log \left (x\right ) + 4 \, x + 2\right )} e^{\frac {1}{x}}} \]
integrate(((x^4-2*x^3+x^2)*log(x)+(-x^4+2*x^3-3*x^2+x)*exp(1/(x^2-x))+3*x^ 4-2*x^3-5*x^2+4*x)/((x^4-2*x^3+x^2)*log(x)^2+((-2*x^4+4*x^3-2*x^2)*exp(1/( x^2-x))+8*x^4-12*x^3+4*x)*log(x)+(x^4-2*x^3+x^2)*exp(1/(x^2-x))^2+(-8*x^4+ 12*x^3-4*x)*exp(1/(x^2-x))+16*x^4-16*x^3-12*x^2+8*x+4),x, algorithm=\
Time = 0.36 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {4 x-5 x^2-2 x^3+3 x^4+e^{\frac {1}{-x+x^2}} \left (x-3 x^2+2 x^3-x^4\right )+\left (x^2-2 x^3+x^4\right ) \log (x)}{4+8 x-12 x^2-16 x^3+16 x^4+e^{\frac {1}{-x+x^2}} \left (-4 x+12 x^3-8 x^4\right )+e^{\frac {2}{-x+x^2}} \left (x^2-2 x^3+x^4\right )+\left (4 x-12 x^3+8 x^4+e^{\frac {1}{-x+x^2}} \left (-2 x^2+4 x^3-2 x^4\right )\right ) \log (x)+\left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=-\frac {x^{2}}{x e^{\left (\frac {1}{x^{2} - x}\right )} - x \log \left (x\right ) - 4 \, x - 2} \]
integrate(((x^4-2*x^3+x^2)*log(x)+(-x^4+2*x^3-3*x^2+x)*exp(1/(x^2-x))+3*x^ 4-2*x^3-5*x^2+4*x)/((x^4-2*x^3+x^2)*log(x)^2+((-2*x^4+4*x^3-2*x^2)*exp(1/( x^2-x))+8*x^4-12*x^3+4*x)*log(x)+(x^4-2*x^3+x^2)*exp(1/(x^2-x))^2+(-8*x^4+ 12*x^3-4*x)*exp(1/(x^2-x))+16*x^4-16*x^3-12*x^2+8*x+4),x, algorithm=\
Time = 13.29 (sec) , antiderivative size = 153, normalized size of antiderivative = 5.88 \[ \int \frac {4 x-5 x^2-2 x^3+3 x^4+e^{\frac {1}{-x+x^2}} \left (x-3 x^2+2 x^3-x^4\right )+\left (x^2-2 x^3+x^4\right ) \log (x)}{4+8 x-12 x^2-16 x^3+16 x^4+e^{\frac {1}{-x+x^2}} \left (-4 x+12 x^3-8 x^4\right )+e^{\frac {2}{-x+x^2}} \left (x^2-2 x^3+x^4\right )+\left (4 x-12 x^3+8 x^4+e^{\frac {1}{-x+x^2}} \left (-2 x^2+4 x^3-2 x^4\right )\right ) \log (x)+\left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=-\frac {x\,{\left (x^3-2\,x^2+x\right )}^2\,\left (-x^4+4\,x^3-13\,x^2+2\,x+2\right )+x\,\ln \left (x\right )\,\left (x-2\,x^2\right )\,{\left (x^3-2\,x^2+x\right )}^2}{{\left (x-1\right )}^2\,\left (4\,x-x\,{\mathrm {e}}^{-\frac {1}{x-x^2}}+x\,\ln \left (x\right )+2\right )\,\left (4\,x^3\,\ln \left (x\right )-x^2\,\ln \left (x\right )-2\,x-5\,x^4\,\ln \left (x\right )+2\,x^5\,\ln \left (x\right )+2\,x^2+15\,x^3-32\,x^4+22\,x^5-6\,x^6+x^7\right )} \]
int((4*x + exp(-1/(x - x^2))*(x - 3*x^2 + 2*x^3 - x^4) + log(x)*(x^2 - 2*x ^3 + x^4) - 5*x^2 - 2*x^3 + 3*x^4)/(8*x - exp(-1/(x - x^2))*(4*x - 12*x^3 + 8*x^4) + log(x)*(4*x - exp(-1/(x - x^2))*(2*x^2 - 4*x^3 + 2*x^4) - 12*x^ 3 + 8*x^4) + exp(-2/(x - x^2))*(x^2 - 2*x^3 + x^4) - 12*x^2 - 16*x^3 + 16* x^4 + log(x)^2*(x^2 - 2*x^3 + x^4) + 4),x)