Integrand size = 215, antiderivative size = 25 \[ \int \frac {1-x+e^x \left (2 x-2 x^2\right )+e^{2 x} \left (-2 x+2 x^3\right )+\left (-2 x+2 x^2+e^x \left (4 x-2 x^2-2 x^3\right )\right ) \log (x)+\left (-2 x+2 x^2\right ) \log ^2(x)+\left (x-e^{2 x} x^2+2 e^x x^2 \log (x)-x^2 \log ^2(x)\right ) \log \left (-x+e^{2 x} x^2-2 e^x x^2 \log (x)+x^2 \log ^2(x)\right )}{-x+2 x^2-x^3+e^{2 x} \left (x^2-2 x^3+x^4\right )+e^x \left (-2 x^2+4 x^3-2 x^4\right ) \log (x)+\left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {\log \left (-x+x^2 \left (e^x-\log (x)\right )^2\right )}{-1+x} \]
Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {1-x+e^x \left (2 x-2 x^2\right )+e^{2 x} \left (-2 x+2 x^3\right )+\left (-2 x+2 x^2+e^x \left (4 x-2 x^2-2 x^3\right )\right ) \log (x)+\left (-2 x+2 x^2\right ) \log ^2(x)+\left (x-e^{2 x} x^2+2 e^x x^2 \log (x)-x^2 \log ^2(x)\right ) \log \left (-x+e^{2 x} x^2-2 e^x x^2 \log (x)+x^2 \log ^2(x)\right )}{-x+2 x^2-x^3+e^{2 x} \left (x^2-2 x^3+x^4\right )+e^x \left (-2 x^2+4 x^3-2 x^4\right ) \log (x)+\left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {\log \left (x \left (-1+e^{2 x} x-2 e^x x \log (x)+x \log ^2(x)\right )\right )}{-1+x} \]
Integrate[(1 - x + E^x*(2*x - 2*x^2) + E^(2*x)*(-2*x + 2*x^3) + (-2*x + 2* x^2 + E^x*(4*x - 2*x^2 - 2*x^3))*Log[x] + (-2*x + 2*x^2)*Log[x]^2 + (x - E ^(2*x)*x^2 + 2*E^x*x^2*Log[x] - x^2*Log[x]^2)*Log[-x + E^(2*x)*x^2 - 2*E^x *x^2*Log[x] + x^2*Log[x]^2])/(-x + 2*x^2 - x^3 + E^(2*x)*(x^2 - 2*x^3 + x^ 4) + E^x*(-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.
| \(\displaystyle \int \frac {e^{2 x} \left (2 x^3-2 x\right )+e^x \left (2 x-2 x^2\right )+\left (2 x^2-2 x\right ) \log ^2(x)+\left (-e^{2 x} x^2-x^2 \log ^2(x)+2 e^x x^2 \log (x)+x\right ) \log \left (e^{2 x} x^2+x^2 \log ^2(x)-2 e^x x^2 \log (x)-x\right )+\left (2 x^2+e^x \left (-2 x^3-2 x^2+4 x\right )-2 x\right ) \log (x)-x+1}{-x^3+2 x^2+e^{2 x} \left (x^4-2 x^3+x^2\right )+\left (x^4-2 x^3+x^2\right ) \log ^2(x)+e^x \left (-2 x^4+4 x^3-2 x^2\right ) \log (x)-x} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-e^{2 x} \left (2 x^3-2 x\right )-e^x \left (2 x-2 x^2\right )-\left (\left (2 x^2-2 x\right ) \log ^2(x)\right )-\left (-e^{2 x} x^2-x^2 \log ^2(x)+2 e^x x^2 \log (x)+x\right ) \log \left (e^{2 x} x^2+x^2 \log ^2(x)-2 e^x x^2 \log (x)-x\right )-\left (2 x^2+e^x \left (-2 x^3-2 x^2+4 x\right )-2 x\right ) \log (x)+x-1}{(1-x)^2 x \left (-e^{2 x} x-x \log ^2(x)+2 e^x x \log (x)+1\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {-2 x^2 \log ^2(x)+2 e^x x^2 \log (x)-2 e^x x+2 x+2 x \log (x)+1}{(x-1) x \left (e^{2 x} x+x \log ^2(x)-2 e^x x \log (x)-1\right )}+\frac {2 x^2-x \log \left (x \left (e^{2 x} x+x \log ^2(x)-2 e^x x \log (x)-1\right )\right )-2}{(x-1)^2 x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\int \frac {\log \left (e^{2 x} x^2+\log ^2(x) x^2-2 e^x \log (x) x^2-x\right )}{(1-x)^2}dx+3 \int \frac {1}{(x-1) \left (x \log ^2(x)-2 e^x x \log (x)+e^{2 x} x-1\right )}dx-2 \int \frac {e^x}{(x-1) \left (x \log ^2(x)-2 e^x x \log (x)+e^{2 x} x-1\right )}dx-\int \frac {1}{x \left (x \log ^2(x)-2 e^x x \log (x)+e^{2 x} x-1\right )}dx+2 \int \frac {e^x \log (x)}{x \log ^2(x)-2 e^x x \log (x)+e^{2 x} x-1}dx+2 \int \frac {\log (x)}{(x-1) \left (x \log ^2(x)-2 e^x x \log (x)+e^{2 x} x-1\right )}dx+2 \int \frac {e^x \log (x)}{(x-1) \left (x \log ^2(x)-2 e^x x \log (x)+e^{2 x} x-1\right )}dx-2 \int \frac {\log ^2(x)}{x \log ^2(x)-2 e^x x \log (x)+e^{2 x} x-1}dx-2 \int \frac {\log ^2(x)}{(x-1) \left (x \log ^2(x)-2 e^x x \log (x)+e^{2 x} x-1\right )}dx+4 \log (1-x)-2 \log (x)\) |
Int[(1 - x + E^x*(2*x - 2*x^2) + E^(2*x)*(-2*x + 2*x^3) + (-2*x + 2*x^2 + E^x*(4*x - 2*x^2 - 2*x^3))*Log[x] + (-2*x + 2*x^2)*Log[x]^2 + (x - E^(2*x) *x^2 + 2*E^x*x^2*Log[x] - x^2*Log[x]^2)*Log[-x + E^(2*x)*x^2 - 2*E^x*x^2*L og[x] + x^2*Log[x]^2])/(-x + 2*x^2 - x^3 + E^(2*x)*(x^2 - 2*x^3 + x^4) + E ^x*(-2*x^2 + 4*x^3 - 2*x^4)*Log[x] + (x^2 - 2*x^3 + x^4)*Log[x]^2),x]
3.4.16.3.1 Defintions of rubi rules used
Time = 40.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48
| method | result | size |
| parallelrisch | \(\frac {\ln \left (x^{2} \ln \left (x \right )^{2}-2 x^{2} {\mathrm e}^{x} \ln \left (x \right )+{\mathrm e}^{2 x} x^{2}-x \right )}{-1+x}\) | \(37\) |
| risch | \(\frac {\ln \left (-1+\left (\ln \left (x \right )^{2}-2 \,{\mathrm e}^{x} \ln \left (x \right )+{\mathrm e}^{2 x}\right ) x \right )}{-1+x}+\frac {-i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (1-\left (\ln \left (x \right )^{2}-2 \,{\mathrm e}^{x} \ln \left (x \right )+{\mathrm e}^{2 x}\right ) x \right )\right ) \operatorname {csgn}\left (i x \left (1-\left (\ln \left (x \right )^{2}-2 \,{\mathrm e}^{x} \ln \left (x \right )+{\mathrm e}^{2 x}\right ) x \right )\right )+i \pi \,\operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (i x \left (1-\left (\ln \left (x \right )^{2}-2 \,{\mathrm e}^{x} \ln \left (x \right )+{\mathrm e}^{2 x}\right ) x \right )\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (1-\left (\ln \left (x \right )^{2}-2 \,{\mathrm e}^{x} \ln \left (x \right )+{\mathrm e}^{2 x}\right ) x \right )\right ) {\operatorname {csgn}\left (i x \left (1-\left (\ln \left (x \right )^{2}-2 \,{\mathrm e}^{x} \ln \left (x \right )+{\mathrm e}^{2 x}\right ) x \right )\right )}^{2}+i \pi {\operatorname {csgn}\left (i x \left (1-\left (\ln \left (x \right )^{2}-2 \,{\mathrm e}^{x} \ln \left (x \right )+{\mathrm e}^{2 x}\right ) x \right )\right )}^{3}+2 \ln \left (x \right )}{-2+2 x}\) | \(220\) |
int(((-x^2*ln(x)^2+2*x^2*exp(x)*ln(x)-exp(x)^2*x^2+x)*ln(x^2*ln(x)^2-2*x^2 *exp(x)*ln(x)+exp(x)^2*x^2-x)+(2*x^2-2*x)*ln(x)^2+((-2*x^3-2*x^2+4*x)*exp( x)+2*x^2-2*x)*ln(x)+(2*x^3-2*x)*exp(x)^2+(-2*x^2+2*x)*exp(x)-x+1)/((x^4-2* x^3+x^2)*ln(x)^2+(-2*x^4+4*x^3-2*x^2)*exp(x)*ln(x)+(x^4-2*x^3+x^2)*exp(x)^ 2-x^3+2*x^2-x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {1-x+e^x \left (2 x-2 x^2\right )+e^{2 x} \left (-2 x+2 x^3\right )+\left (-2 x+2 x^2+e^x \left (4 x-2 x^2-2 x^3\right )\right ) \log (x)+\left (-2 x+2 x^2\right ) \log ^2(x)+\left (x-e^{2 x} x^2+2 e^x x^2 \log (x)-x^2 \log ^2(x)\right ) \log \left (-x+e^{2 x} x^2-2 e^x x^2 \log (x)+x^2 \log ^2(x)\right )}{-x+2 x^2-x^3+e^{2 x} \left (x^2-2 x^3+x^4\right )+e^x \left (-2 x^2+4 x^3-2 x^4\right ) \log (x)+\left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {\log \left (-2 \, x^{2} e^{x} \log \left (x\right ) + x^{2} \log \left (x\right )^{2} + x^{2} e^{\left (2 \, x\right )} - x\right )}{x - 1} \]
integrate(((-x^2*log(x)^2+2*x^2*exp(x)*log(x)-exp(x)^2*x^2+x)*log(x^2*log( x)^2-2*x^2*exp(x)*log(x)+exp(x)^2*x^2-x)+(2*x^2-2*x)*log(x)^2+((-2*x^3-2*x ^2+4*x)*exp(x)+2*x^2-2*x)*log(x)+(2*x^3-2*x)*exp(x)^2+(-2*x^2+2*x)*exp(x)- x+1)/((x^4-2*x^3+x^2)*log(x)^2+(-2*x^4+4*x^3-2*x^2)*exp(x)*log(x)+(x^4-2*x ^3+x^2)*exp(x)^2-x^3+2*x^2-x),x, algorithm=\
Time = 0.74 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {1-x+e^x \left (2 x-2 x^2\right )+e^{2 x} \left (-2 x+2 x^3\right )+\left (-2 x+2 x^2+e^x \left (4 x-2 x^2-2 x^3\right )\right ) \log (x)+\left (-2 x+2 x^2\right ) \log ^2(x)+\left (x-e^{2 x} x^2+2 e^x x^2 \log (x)-x^2 \log ^2(x)\right ) \log \left (-x+e^{2 x} x^2-2 e^x x^2 \log (x)+x^2 \log ^2(x)\right )}{-x+2 x^2-x^3+e^{2 x} \left (x^2-2 x^3+x^4\right )+e^x \left (-2 x^2+4 x^3-2 x^4\right ) \log (x)+\left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {\log {\left (x^{2} e^{2 x} - 2 x^{2} e^{x} \log {\left (x \right )} + x^{2} \log {\left (x \right )}^{2} - x \right )}}{x - 1} \]
integrate(((-x**2*ln(x)**2+2*x**2*exp(x)*ln(x)-exp(x)**2*x**2+x)*ln(x**2*l n(x)**2-2*x**2*exp(x)*ln(x)+exp(x)**2*x**2-x)+(2*x**2-2*x)*ln(x)**2+((-2*x **3-2*x**2+4*x)*exp(x)+2*x**2-2*x)*ln(x)+(2*x**3-2*x)*exp(x)**2+(-2*x**2+2 *x)*exp(x)-x+1)/((x**4-2*x**3+x**2)*ln(x)**2+(-2*x**4+4*x**3-2*x**2)*exp(x )*ln(x)+(x**4-2*x**3+x**2)*exp(x)**2-x**3+2*x**2-x),x)
Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {1-x+e^x \left (2 x-2 x^2\right )+e^{2 x} \left (-2 x+2 x^3\right )+\left (-2 x+2 x^2+e^x \left (4 x-2 x^2-2 x^3\right )\right ) \log (x)+\left (-2 x+2 x^2\right ) \log ^2(x)+\left (x-e^{2 x} x^2+2 e^x x^2 \log (x)-x^2 \log ^2(x)\right ) \log \left (-x+e^{2 x} x^2-2 e^x x^2 \log (x)+x^2 \log ^2(x)\right )}{-x+2 x^2-x^3+e^{2 x} \left (x^2-2 x^3+x^4\right )+e^x \left (-2 x^2+4 x^3-2 x^4\right ) \log (x)+\left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {\log \left (-2 \, x e^{x} \log \left (x\right ) + x \log \left (x\right )^{2} + x e^{\left (2 \, x\right )} - 1\right ) + \log \left (x\right )}{x - 1} \]
integrate(((-x^2*log(x)^2+2*x^2*exp(x)*log(x)-exp(x)^2*x^2+x)*log(x^2*log( x)^2-2*x^2*exp(x)*log(x)+exp(x)^2*x^2-x)+(2*x^2-2*x)*log(x)^2+((-2*x^3-2*x ^2+4*x)*exp(x)+2*x^2-2*x)*log(x)+(2*x^3-2*x)*exp(x)^2+(-2*x^2+2*x)*exp(x)- x+1)/((x^4-2*x^3+x^2)*log(x)^2+(-2*x^4+4*x^3-2*x^2)*exp(x)*log(x)+(x^4-2*x ^3+x^2)*exp(x)^2-x^3+2*x^2-x),x, algorithm=\
Time = 0.39 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {1-x+e^x \left (2 x-2 x^2\right )+e^{2 x} \left (-2 x+2 x^3\right )+\left (-2 x+2 x^2+e^x \left (4 x-2 x^2-2 x^3\right )\right ) \log (x)+\left (-2 x+2 x^2\right ) \log ^2(x)+\left (x-e^{2 x} x^2+2 e^x x^2 \log (x)-x^2 \log ^2(x)\right ) \log \left (-x+e^{2 x} x^2-2 e^x x^2 \log (x)+x^2 \log ^2(x)\right )}{-x+2 x^2-x^3+e^{2 x} \left (x^2-2 x^3+x^4\right )+e^x \left (-2 x^2+4 x^3-2 x^4\right ) \log (x)+\left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {\log \left (-2 \, x e^{x} \log \left (x\right ) + x \log \left (x\right )^{2} + x e^{\left (2 \, x\right )} - 1\right ) + \log \left (x\right )}{x - 1} \]
integrate(((-x^2*log(x)^2+2*x^2*exp(x)*log(x)-exp(x)^2*x^2+x)*log(x^2*log( x)^2-2*x^2*exp(x)*log(x)+exp(x)^2*x^2-x)+(2*x^2-2*x)*log(x)^2+((-2*x^3-2*x ^2+4*x)*exp(x)+2*x^2-2*x)*log(x)+(2*x^3-2*x)*exp(x)^2+(-2*x^2+2*x)*exp(x)- x+1)/((x^4-2*x^3+x^2)*log(x)^2+(-2*x^4+4*x^3-2*x^2)*exp(x)*log(x)+(x^4-2*x ^3+x^2)*exp(x)^2-x^3+2*x^2-x),x, algorithm=\
Time = 7.99 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {1-x+e^x \left (2 x-2 x^2\right )+e^{2 x} \left (-2 x+2 x^3\right )+\left (-2 x+2 x^2+e^x \left (4 x-2 x^2-2 x^3\right )\right ) \log (x)+\left (-2 x+2 x^2\right ) \log ^2(x)+\left (x-e^{2 x} x^2+2 e^x x^2 \log (x)-x^2 \log ^2(x)\right ) \log \left (-x+e^{2 x} x^2-2 e^x x^2 \log (x)+x^2 \log ^2(x)\right )}{-x+2 x^2-x^3+e^{2 x} \left (x^2-2 x^3+x^4\right )+e^x \left (-2 x^2+4 x^3-2 x^4\right ) \log (x)+\left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {\ln \left (x^2\,{\mathrm {e}}^{2\,x}-x+x^2\,{\ln \left (x\right )}^2-2\,x^2\,{\mathrm {e}}^x\,\ln \left (x\right )\right )}{x-1} \]
int((x + exp(2*x)*(2*x - 2*x^3) + log(x)^2*(2*x - 2*x^2) + log(x)*(2*x - 2 *x^2 + exp(x)*(2*x^2 - 4*x + 2*x^3)) - log(x^2*exp(2*x) - x + x^2*log(x)^2 - 2*x^2*exp(x)*log(x))*(x - x^2*exp(2*x) - x^2*log(x)^2 + 2*x^2*exp(x)*lo g(x)) - exp(x)*(2*x - 2*x^2) - 1)/(x - 2*x^2 + x^3 - exp(2*x)*(x^2 - 2*x^3 + x^4) - log(x)^2*(x^2 - 2*x^3 + x^4) + exp(x)*log(x)*(2*x^2 - 4*x^3 + 2* x^4)),x)