Integrand size = 97, antiderivative size = 28 \[ \int \frac {\left (-x-x^2\right ) \log (x)-x \log ^2(x)+e^{-x+x \log (x)} \left (x+x \log ^2(x)\right )+\left (e^{-x+x \log (x)}-x \log (x)\right ) \log \left (-e^{-x+x \log (x)}+x \log (x)\right )}{e^{-x+x \log (x)} x-x^2 \log (x)} \, dx=-e^3+x+\log (x) \log \left (-e^{-x+x \log (x)}+x \log (x)\right ) \]
Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {\left (-x-x^2\right ) \log (x)-x \log ^2(x)+e^{-x+x \log (x)} \left (x+x \log ^2(x)\right )+\left (e^{-x+x \log (x)}-x \log (x)\right ) \log \left (-e^{-x+x \log (x)}+x \log (x)\right )}{e^{-x+x \log (x)} x-x^2 \log (x)} \, dx=x+\log (x) \log \left (-e^{-x} x^x+x \log (x)\right ) \]
Integrate[((-x - x^2)*Log[x] - x*Log[x]^2 + E^(-x + x*Log[x])*(x + x*Log[x ]^2) + (E^(-x + x*Log[x]) - x*Log[x])*Log[-E^(-x + x*Log[x]) + x*Log[x]])/ (E^(-x + x*Log[x])*x - x^2*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.
\(\displaystyle \int \frac {\left (-x^2-x\right ) \log (x)-x \log ^2(x)+e^{x \log (x)-x} \left (x+x \log ^2(x)\right )+\left (e^{x \log (x)-x}-x \log (x)\right ) \log \left (x \log (x)-e^{x \log (x)-x}\right )}{x e^{x \log (x)-x}-x^2 \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^x \left (\left (-x^2-x\right ) \log (x)-x \log ^2(x)+e^{x \log (x)-x} \left (x+x \log ^2(x)\right )+\left (e^{x \log (x)-x}-x \log (x)\right ) \log \left (x \log (x)-e^{x \log (x)-x}\right )\right )}{x \left (x^x-e^x x \log (x)\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {x^x \log ^2(x)}{e^x x \log (x)-x^x}+\frac {e^x \log ^2(x)}{e^x x \log (x)-x^x}-\frac {x^x}{e^x x \log (x)-x^x}+\frac {e^x x \log (x)}{e^x x \log (x)-x^x}+\frac {e^x \log (x) \log \left (x \log (x)-e^{-x} x^x\right )}{e^x x \log (x)-x^x}+\frac {e^x \log (x)}{e^x x \log (x)-x^x}-\frac {x^{x-1} \log \left (x \log (x)-e^{-x} x^x\right )}{e^x x \log (x)-x^x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {e^x \log ^2(x)}{e^x x \log (x)-x^x}dx-\int \frac {x^x \log ^2(x)}{e^x x \log (x)-x^x}dx-\int \frac {x^x}{e^x x \log (x)-x^x}dx+\int \frac {e^x \log (x)}{e^x x \log (x)-x^x}dx+\int \frac {e^x x \log (x)}{e^x x \log (x)-x^x}dx+\int \frac {e^x \log (x) \log \left (x \log (x)-e^{-x} x^x\right )}{e^x x \log (x)-x^x}dx-\int \frac {x^{x-1} \log \left (x \log (x)-e^{-x} x^x\right )}{e^x x \log (x)-x^x}dx\) |
Int[((-x - x^2)*Log[x] - x*Log[x]^2 + E^(-x + x*Log[x])*(x + x*Log[x]^2) + (E^(-x + x*Log[x]) - x*Log[x])*Log[-E^(-x + x*Log[x]) + x*Log[x]])/(E^(-x + x*Log[x])*x - x^2*Log[x]),x]
3.2.1.3.1 Defintions of rubi rules used
Time = 0.38 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75
method | result | size |
risch | \(\ln \left (-x^{x} {\mathrm e}^{-x}+x \ln \left (x \right )\right ) \ln \left (x \right )+x\) | \(21\) |
parallelrisch | \(\ln \left (x \right ) \ln \left (-{\mathrm e}^{\left (\ln \left (x \right )-1\right ) x}+x \ln \left (x \right )\right )+x\) | \(21\) |
int(((exp(x*ln(x)-x)-x*ln(x))*ln(-exp(x*ln(x)-x)+x*ln(x))+(x*ln(x)^2+x)*ex p(x*ln(x)-x)-x*ln(x)^2+(-x^2-x)*ln(x))/(x*exp(x*ln(x)-x)-x^2*ln(x)),x,meth od=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {\left (-x-x^2\right ) \log (x)-x \log ^2(x)+e^{-x+x \log (x)} \left (x+x \log ^2(x)\right )+\left (e^{-x+x \log (x)}-x \log (x)\right ) \log \left (-e^{-x+x \log (x)}+x \log (x)\right )}{e^{-x+x \log (x)} x-x^2 \log (x)} \, dx=\log \left (x \log \left (x\right ) - e^{\left (x \log \left (x\right ) - x\right )}\right ) \log \left (x\right ) + x \]
integrate(((exp(x*log(x)-x)-x*log(x))*log(-exp(x*log(x)-x)+x*log(x))+(x*lo g(x)^2+x)*exp(x*log(x)-x)-x*log(x)^2+(-x^2-x)*log(x))/(x*exp(x*log(x)-x)-x ^2*log(x)),x, algorithm=\
Time = 0.38 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {\left (-x-x^2\right ) \log (x)-x \log ^2(x)+e^{-x+x \log (x)} \left (x+x \log ^2(x)\right )+\left (e^{-x+x \log (x)}-x \log (x)\right ) \log \left (-e^{-x+x \log (x)}+x \log (x)\right )}{e^{-x+x \log (x)} x-x^2 \log (x)} \, dx=x + \log {\left (x \right )} \log {\left (x \log {\left (x \right )} - e^{x \log {\left (x \right )} - x} \right )} \]
integrate(((exp(x*ln(x)-x)-x*ln(x))*ln(-exp(x*ln(x)-x)+x*ln(x))+(x*ln(x)** 2+x)*exp(x*ln(x)-x)-x*ln(x)**2+(-x**2-x)*ln(x))/(x*exp(x*ln(x)-x)-x**2*ln( x)),x)
Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {\left (-x-x^2\right ) \log (x)-x \log ^2(x)+e^{-x+x \log (x)} \left (x+x \log ^2(x)\right )+\left (e^{-x+x \log (x)}-x \log (x)\right ) \log \left (-e^{-x+x \log (x)}+x \log (x)\right )}{e^{-x+x \log (x)} x-x^2 \log (x)} \, dx=-x \log \left (x\right ) + \log \left (x e^{x} \log \left (x\right ) - x^{x}\right ) \log \left (x\right ) + x \]
integrate(((exp(x*log(x)-x)-x*log(x))*log(-exp(x*log(x)-x)+x*log(x))+(x*lo g(x)^2+x)*exp(x*log(x)-x)-x*log(x)^2+(-x^2-x)*log(x))/(x*exp(x*log(x)-x)-x ^2*log(x)),x, algorithm=\
Timed out. \[ \int \frac {\left (-x-x^2\right ) \log (x)-x \log ^2(x)+e^{-x+x \log (x)} \left (x+x \log ^2(x)\right )+\left (e^{-x+x \log (x)}-x \log (x)\right ) \log \left (-e^{-x+x \log (x)}+x \log (x)\right )}{e^{-x+x \log (x)} x-x^2 \log (x)} \, dx=\text {Timed out} \]
integrate(((exp(x*log(x)-x)-x*log(x))*log(-exp(x*log(x)-x)+x*log(x))+(x*lo g(x)^2+x)*exp(x*log(x)-x)-x*log(x)^2+(-x^2-x)*log(x))/(x*exp(x*log(x)-x)-x ^2*log(x)),x, algorithm=\
Time = 9.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {\left (-x-x^2\right ) \log (x)-x \log ^2(x)+e^{-x+x \log (x)} \left (x+x \log ^2(x)\right )+\left (e^{-x+x \log (x)}-x \log (x)\right ) \log \left (-e^{-x+x \log (x)}+x \log (x)\right )}{e^{-x+x \log (x)} x-x^2 \log (x)} \, dx=x+\ln \left (x\,\ln \left (x\right )-x^x\,{\mathrm {e}}^{-x}\right )\,\ln \left (x\right ) \]