Integrand size = 100, antiderivative size = 32 \[ \int \frac {e^x x-2 x^2-\log \left (\frac {3}{2 x}\right )+\left (-1-8 x^2+2 x^3+e^x \left (2 x+x^2\right )+x \log \left (\frac {3}{2 x}\right )\right ) \log (x)+x \log ^2(x)}{\left (-e^x x^2+2 x^3+x \log \left (\frac {3}{2 x}\right )\right ) \log (x)+x \log ^2(x)} \, dx=x-\log \left (\log (x) \left (x \left (-e^x+2 x\right )+\log \left (\frac {3}{2 x}\right )+\log (x)\right )^2\right ) \]
Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {e^x x-2 x^2-\log \left (\frac {3}{2 x}\right )+\left (-1-8 x^2+2 x^3+e^x \left (2 x+x^2\right )+x \log \left (\frac {3}{2 x}\right )\right ) \log (x)+x \log ^2(x)}{\left (-e^x x^2+2 x^3+x \log \left (\frac {3}{2 x}\right )\right ) \log (x)+x \log ^2(x)} \, dx=x-\log (\log (x))-2 \log \left (-e^x x+2 x^2+\log \left (\frac {3}{2 x}\right )+\log (x)\right ) \]
Integrate[(E^x*x - 2*x^2 - Log[3/(2*x)] + (-1 - 8*x^2 + 2*x^3 + E^x*(2*x + x^2) + x*Log[3/(2*x)])*Log[x] + x*Log[x]^2)/((-(E^x*x^2) + 2*x^3 + x*Log[ 3/(2*x)])*Log[x] + x*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.
\(\displaystyle \int \frac {-2 x^2+\left (2 x^3-8 x^2+e^x \left (x^2+2 x\right )+x \log \left (\frac {3}{2 x}\right )-1\right ) \log (x)+e^x x+x \log ^2(x)-\log \left (\frac {3}{2 x}\right )}{\left (2 x^3-e^x x^2+x \log \left (\frac {3}{2 x}\right )\right ) \log (x)+x \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 x^2-\left (2 x^3-8 x^2+e^x \left (x^2+2 x\right )+x \log \left (\frac {3}{2 x}\right )-1\right ) \log (x)-e^x x-x \log ^2(x)+\log \left (\frac {3}{2 x}\right )}{x \left (-2 x^2+e^x x-\log \left (\frac {3}{2 x}\right )-\log (x)\right ) \log (x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (2 x^3-2 x^2+x \log \left (\frac {3}{2 x}\right )+x \log (x)+\log \left (\frac {3}{2 x}\right )+\log (x)\right )}{x \left (2 x^2-e^x x+\log \left (\frac {3}{2 x}\right )+\log (x)\right )}+\frac {-x \log (x)-2 \log (x)-1}{x \log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \int \frac {x}{2 x^2-e^x x+\log \left (\frac {3}{2 x}\right )+\log (x)}dx+4 \int \frac {x^2}{2 x^2-e^x x+\log \left (\frac {3}{2 x}\right )+\log (x)}dx+2 \int \frac {\log \left (\frac {3}{2 x}\right )}{2 x^2-e^x x+\log \left (\frac {3}{2 x}\right )+\log (x)}dx+2 \int \frac {\log \left (\frac {3}{2 x}\right )}{x \left (2 x^2-e^x x+\log \left (\frac {3}{2 x}\right )+\log (x)\right )}dx+2 \int \frac {\log (x)}{2 x^2-e^x x+\log \left (\frac {3}{2 x}\right )+\log (x)}dx+2 \int \frac {\log (x)}{x \left (2 x^2-e^x x+\log \left (\frac {3}{2 x}\right )+\log (x)\right )}dx-x-2 \log (x)-\log (\log (x))\) |
Int[(E^x*x - 2*x^2 - Log[3/(2*x)] + (-1 - 8*x^2 + 2*x^3 + E^x*(2*x + x^2) + x*Log[3/(2*x)])*Log[x] + x*Log[x]^2)/((-(E^x*x^2) + 2*x^3 + x*Log[3/(2*x )])*Log[x] + x*Log[x]^2),x]
3.5.74.3.1 Defintions of rubi rules used
Time = 1.34 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94
method | result | size |
default | \(x -\ln \left (\ln \left (x \right )\right )-2 \ln \left (-{\mathrm e}^{x} x +2 x^{2}+\ln \left (x \right )+\ln \left (\frac {3}{2 x}\right )\right )\) | \(30\) |
parallelrisch | \(x -\ln \left (\ln \left (x \right )\right )-2 \ln \left (x^{2}-\frac {{\mathrm e}^{x} x}{2}+\frac {\ln \left (x \right )}{2}+\frac {\ln \left (\frac {3}{2 x}\right )}{2}\right )\) | \(32\) |
risch | \(x -2 \ln \left (x \right )-2 \ln \left ({\mathrm e}^{x}+\frac {i \left (4 i x^{2}-2 i \ln \left (2\right )+2 i \ln \left (3\right )\right )}{2 x}\right )-\ln \left (\ln \left (x \right )\right )\) | \(41\) |
int((x*ln(x)^2+(x*ln(3/2/x)+(x^2+2*x)*exp(x)+2*x^3-8*x^2-1)*ln(x)-ln(3/2/x )+exp(x)*x-2*x^2)/(x*ln(x)^2+(x*ln(3/2/x)-exp(x)*x^2+2*x^3)*ln(x)),x,metho d=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.28 \[ \int \frac {e^x x-2 x^2-\log \left (\frac {3}{2 x}\right )+\left (-1-8 x^2+2 x^3+e^x \left (2 x+x^2\right )+x \log \left (\frac {3}{2 x}\right )\right ) \log (x)+x \log ^2(x)}{\left (-e^x x^2+2 x^3+x \log \left (\frac {3}{2 x}\right )\right ) \log (x)+x \log ^2(x)} \, dx=x - 2 \, \log \left (x\right ) - 2 \, \log \left (-\frac {2 \, x^{2} - x e^{x} + \log \left (\frac {3}{2}\right )}{x}\right ) - \log \left (-\log \left (\frac {3}{2}\right ) + \log \left (\frac {3}{2 \, x}\right )\right ) \]
integrate((x*log(x)^2+(x*log(3/2/x)+(x^2+2*x)*exp(x)+2*x^3-8*x^2-1)*log(x) -log(3/2/x)+exp(x)*x-2*x^2)/(x*log(x)^2+(x*log(3/2/x)-exp(x)*x^2+2*x^3)*lo g(x)),x, algorithm=\
Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {e^x x-2 x^2-\log \left (\frac {3}{2 x}\right )+\left (-1-8 x^2+2 x^3+e^x \left (2 x+x^2\right )+x \log \left (\frac {3}{2 x}\right )\right ) \log (x)+x \log ^2(x)}{\left (-e^x x^2+2 x^3+x \log \left (\frac {3}{2 x}\right )\right ) \log (x)+x \log ^2(x)} \, dx=x - 2 \log {\left (x \right )} - 2 \log {\left (e^{x} + \frac {- 2 x^{2} - \log {\left (3 \right )} + \log {\left (2 \right )}}{x} \right )} - \log {\left (\log {\left (x \right )} \right )} \]
integrate((x*ln(x)**2+(x*ln(3/2/x)+(x**2+2*x)*exp(x)+2*x**3-8*x**2-1)*ln(x )-ln(3/2/x)+exp(x)*x-2*x**2)/(x*ln(x)**2+(x*ln(3/2/x)-exp(x)*x**2+2*x**3)* ln(x)),x)
Time = 0.34 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {e^x x-2 x^2-\log \left (\frac {3}{2 x}\right )+\left (-1-8 x^2+2 x^3+e^x \left (2 x+x^2\right )+x \log \left (\frac {3}{2 x}\right )\right ) \log (x)+x \log ^2(x)}{\left (-e^x x^2+2 x^3+x \log \left (\frac {3}{2 x}\right )\right ) \log (x)+x \log ^2(x)} \, dx=x - 2 \, \log \left (x\right ) - 2 \, \log \left (-\frac {2 \, x^{2} - x e^{x} + \log \left (3\right ) - \log \left (2\right )}{x}\right ) - \log \left (\log \left (x\right )\right ) \]
integrate((x*log(x)^2+(x*log(3/2/x)+(x^2+2*x)*exp(x)+2*x^3-8*x^2-1)*log(x) -log(3/2/x)+exp(x)*x-2*x^2)/(x*log(x)^2+(x*log(3/2/x)-exp(x)*x^2+2*x^3)*lo g(x)),x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^x x-2 x^2-\log \left (\frac {3}{2 x}\right )+\left (-1-8 x^2+2 x^3+e^x \left (2 x+x^2\right )+x \log \left (\frac {3}{2 x}\right )\right ) \log (x)+x \log ^2(x)}{\left (-e^x x^2+2 x^3+x \log \left (\frac {3}{2 x}\right )\right ) \log (x)+x \log ^2(x)} \, dx=x - 2 \, \log \left (-2 \, x^{2} + x e^{x} - \log \left (3\right ) + \log \left (2\right )\right ) - \log \left (\log \left (x\right )\right ) \]
integrate((x*log(x)^2+(x*log(3/2/x)+(x^2+2*x)*exp(x)+2*x^3-8*x^2-1)*log(x) -log(3/2/x)+exp(x)*x-2*x^2)/(x*log(x)^2+(x*log(3/2/x)-exp(x)*x^2+2*x^3)*lo g(x)),x, algorithm=\
Time = 9.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {e^x x-2 x^2-\log \left (\frac {3}{2 x}\right )+\left (-1-8 x^2+2 x^3+e^x \left (2 x+x^2\right )+x \log \left (\frac {3}{2 x}\right )\right ) \log (x)+x \log ^2(x)}{\left (-e^x x^2+2 x^3+x \log \left (\frac {3}{2 x}\right )\right ) \log (x)+x \log ^2(x)} \, dx=x-\ln \left (\ln \left (x\right )\right )-2\,\ln \left (\frac {\ln \left (\frac {3}{2\,x}\right )+\ln \left (x\right )-x\,{\mathrm {e}}^x+2\,x^2}{x}\right )-2\,\ln \left (x\right ) \]