Integrand size = 109, antiderivative size = 32 \begin {dmath*} \int \frac {5-5 e^x+5 x-5 x^2+\left (5+4 x^2+x^3+e^x (-5+6 x)\right ) \log (x)+\left (-x^3+e^x \left (x-x^2\right )\right ) \log ^2(x)+\left (5+(5-x) \log (x)-x \log ^2(x)\right ) \log \left (\frac {1}{2} (-5+x \log (x))\right )}{-5 x^2 \log ^2(x)+x^3 \log ^3(x)} \, dx=\frac {1-e^x+x-x^2+\log \left (\frac {1}{2} (-5+x \log (x))\right )}{x \log (x)} \end {dmath*}
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {5-5 e^x+5 x-5 x^2+\left (5+4 x^2+x^3+e^x (-5+6 x)\right ) \log (x)+\left (-x^3+e^x \left (x-x^2\right )\right ) \log ^2(x)+\left (5+(5-x) \log (x)-x \log ^2(x)\right ) \log \left (\frac {1}{2} (-5+x \log (x))\right )}{-5 x^2 \log ^2(x)+x^3 \log ^3(x)} \, dx=\frac {1-e^x+x-x^2+\log \left (\frac {1}{2} (-5+x \log (x))\right )}{x \log (x)} \end {dmath*}
Integrate[(5 - 5*E^x + 5*x - 5*x^2 + (5 + 4*x^2 + x^3 + E^x*(-5 + 6*x))*Lo g[x] + (-x^3 + E^x*(x - x^2))*Log[x]^2 + (5 + (5 - x)*Log[x] - x*Log[x]^2) *Log[(-5 + x*Log[x])/2])/(-5*x^2*Log[x]^2 + x^3*Log[x]^3),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.
\(\displaystyle \int \frac {-5 x^2+\left (e^x \left (x-x^2\right )-x^3\right ) \log ^2(x)+\left (x^3+4 x^2+e^x (6 x-5)+5\right ) \log (x)+5 x-5 e^x+\left (-x \log ^2(x)+(5-x) \log (x)+5\right ) \log \left (\frac {1}{2} (x \log (x)-5)\right )+5}{x^3 \log ^3(x)-5 x^2 \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {5 x^2-\left (e^x \left (x-x^2\right )-x^3\right ) \log ^2(x)-\left (x^3+4 x^2+e^x (6 x-5)+5\right ) \log (x)-5 x+5 e^x-\left (-x \log ^2(x)+(5-x) \log (x)+5\right ) \log \left (\frac {1}{2} (x \log (x)-5)\right )-5}{x^2 \log ^2(x) (5-x \log (x))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {e^x (x \log (x)-\log (x)-1)}{x^2 \log ^2(x)}-\frac {(\log (x)+1) \log \left (\frac {1}{2} (x \log (x)-5)\right )}{x^2 \log ^2(x)}+\frac {5}{x^2 \log ^2(x) (x \log (x)-5)}+\frac {5}{x^2 \log (x) (x \log (x)-5)}-\frac {5}{\log ^2(x) (x \log (x)-5)}+\frac {5}{x \log ^2(x) (x \log (x)-5)}+\frac {x}{\log (x) (x \log (x)-5)}-\frac {x}{x \log (x)-5}+\frac {4}{\log (x) (x \log (x)-5)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {\log \left (\frac {1}{2} (x \log (x)-5)\right )}{x^2 \log ^2(x)}dx-\int \frac {\log \left (\frac {1}{2} (x \log (x)-5)\right )}{x^2 \log (x)}dx+\frac {1}{5} \int \frac {1}{x \log (x)-5}dx+\int \frac {1}{x (x \log (x)-5)}dx-\frac {x}{\log (x)}-\frac {1}{5} \log (\log (x))+\frac {1}{\log (x)}-\frac {e^x}{x \log (x)}+\frac {1}{x \log (x)}\) |
Int[(5 - 5*E^x + 5*x - 5*x^2 + (5 + 4*x^2 + x^3 + E^x*(-5 + 6*x))*Log[x] + (-x^3 + E^x*(x - x^2))*Log[x]^2 + (5 + (5 - x)*Log[x] - x*Log[x]^2)*Log[( -5 + x*Log[x])/2])/(-5*x^2*Log[x]^2 + x^3*Log[x]^3),x]
3.14.23.3.1 Defintions of rubi rules used
Time = 1.35 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16
method | result | size |
risch | \(\frac {\ln \left (\frac {x \ln \left (x \right )}{2}-\frac {5}{2}\right )}{x \ln \left (x \right )}-\frac {x^{2}-x +{\mathrm e}^{x}-1}{x \ln \left (x \right )}\) | \(37\) |
parallelrisch | \(\frac {10-10 x^{2}+x \ln \left (x \right )+10 x -10 \,{\mathrm e}^{x}+10 \ln \left (\frac {x \ln \left (x \right )}{2}-\frac {5}{2}\right )}{10 x \ln \left (x \right )}\) | \(38\) |
int(((-x*ln(x)^2+ln(x)*(5-x)+5)*ln(1/2*x*ln(x)-5/2)+((-x^2+x)*exp(x)-x^3)* ln(x)^2+((6*x-5)*exp(x)+x^3+4*x^2+5)*ln(x)-5*exp(x)-5*x^2+5*x+5)/(x^3*ln(x )^3-5*x^2*ln(x)^2),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \begin {dmath*} \int \frac {5-5 e^x+5 x-5 x^2+\left (5+4 x^2+x^3+e^x (-5+6 x)\right ) \log (x)+\left (-x^3+e^x \left (x-x^2\right )\right ) \log ^2(x)+\left (5+(5-x) \log (x)-x \log ^2(x)\right ) \log \left (\frac {1}{2} (-5+x \log (x))\right )}{-5 x^2 \log ^2(x)+x^3 \log ^3(x)} \, dx=-\frac {x^{2} - x + e^{x} - \log \left (\frac {1}{2} \, x \log \left (x\right ) - \frac {5}{2}\right ) - 1}{x \log \left (x\right )} \end {dmath*}
integrate(((-x*log(x)^2+log(x)*(5-x)+5)*log(1/2*x*log(x)-5/2)+((-x^2+x)*ex p(x)-x^3)*log(x)^2+((6*x-5)*exp(x)+x^3+4*x^2+5)*log(x)-5*exp(x)-5*x^2+5*x+ 5)/(x^3*log(x)^3-5*x^2*log(x)^2),x, algorithm=\
Exception generated. \begin {dmath*} \int \frac {5-5 e^x+5 x-5 x^2+\left (5+4 x^2+x^3+e^x (-5+6 x)\right ) \log (x)+\left (-x^3+e^x \left (x-x^2\right )\right ) \log ^2(x)+\left (5+(5-x) \log (x)-x \log ^2(x)\right ) \log \left (\frac {1}{2} (-5+x \log (x))\right )}{-5 x^2 \log ^2(x)+x^3 \log ^3(x)} \, dx=\text {Exception raised: TypeError} \end {dmath*}
integrate(((-x*ln(x)**2+ln(x)*(5-x)+5)*ln(1/2*x*ln(x)-5/2)+((-x**2+x)*exp( x)-x**3)*ln(x)**2+((6*x-5)*exp(x)+x**3+4*x**2+5)*ln(x)-5*exp(x)-5*x**2+5*x +5)/(x**3*ln(x)**3-5*x**2*ln(x)**2),x)
Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \begin {dmath*} \int \frac {5-5 e^x+5 x-5 x^2+\left (5+4 x^2+x^3+e^x (-5+6 x)\right ) \log (x)+\left (-x^3+e^x \left (x-x^2\right )\right ) \log ^2(x)+\left (5+(5-x) \log (x)-x \log ^2(x)\right ) \log \left (\frac {1}{2} (-5+x \log (x))\right )}{-5 x^2 \log ^2(x)+x^3 \log ^3(x)} \, dx=-\frac {x^{2} - x + e^{x} + \log \left (2\right ) - \log \left (x \log \left (x\right ) - 5\right ) - 1}{x \log \left (x\right )} \end {dmath*}
integrate(((-x*log(x)^2+log(x)*(5-x)+5)*log(1/2*x*log(x)-5/2)+((-x^2+x)*ex p(x)-x^3)*log(x)^2+((6*x-5)*exp(x)+x^3+4*x^2+5)*log(x)-5*exp(x)-5*x^2+5*x+ 5)/(x^3*log(x)^3-5*x^2*log(x)^2),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \begin {dmath*} \int \frac {5-5 e^x+5 x-5 x^2+\left (5+4 x^2+x^3+e^x (-5+6 x)\right ) \log (x)+\left (-x^3+e^x \left (x-x^2\right )\right ) \log ^2(x)+\left (5+(5-x) \log (x)-x \log ^2(x)\right ) \log \left (\frac {1}{2} (-5+x \log (x))\right )}{-5 x^2 \log ^2(x)+x^3 \log ^3(x)} \, dx=-\frac {x^{2} - x + e^{x} + \log \left (2\right ) - \log \left (x \log \left (x\right ) - 5\right ) - 1}{x \log \left (x\right )} \end {dmath*}
integrate(((-x*log(x)^2+log(x)*(5-x)+5)*log(1/2*x*log(x)-5/2)+((-x^2+x)*ex p(x)-x^3)*log(x)^2+((6*x-5)*exp(x)+x^3+4*x^2+5)*log(x)-5*exp(x)-5*x^2+5*x+ 5)/(x^3*log(x)^3-5*x^2*log(x)^2),x, algorithm=\
Time = 17.49 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.47 \begin {dmath*} \int \frac {5-5 e^x+5 x-5 x^2+\left (5+4 x^2+x^3+e^x (-5+6 x)\right ) \log (x)+\left (-x^3+e^x \left (x-x^2\right )\right ) \log ^2(x)+\left (5+(5-x) \log (x)-x \log ^2(x)\right ) \log \left (\frac {1}{2} (-5+x \log (x))\right )}{-5 x^2 \log ^2(x)+x^3 \log ^3(x)} \, dx=\frac {1}{\ln \left (x\right )}-\frac {x}{\ln \left (x\right )}+\frac {1}{x\,\ln \left (x\right )}+\frac {\ln \left (\frac {x\,\ln \left (x\right )}{2}-\frac {5}{2}\right )}{x\,\ln \left (x\right )}-\frac {{\mathrm {e}}^x}{x\,\ln \left (x\right )} \end {dmath*}