Integrand size = 126, antiderivative size = 30 \[ \int \frac {-3 x+3 x^3+\left (-x+x^2\right ) \log (3)+e^x \left (1+x-3 x^2+x^3\right ) \log (3)+\left (3 x-e^x x \log (3)+\left (x-2 x^2\right ) \log (3)\right ) \log (x)}{-3 x^3+3 x^4+e^x \left (x^2-x^3\right ) \log (3)+\left (-3 x^2+e^x x \log (3)+\left (x^2-x^3\right ) \log (3)\right ) \log (x)+x \log (3) \log ^2(x)} \, dx=\log \left (\frac {x-x^2+\log (x)}{-e^x+\frac {3 x}{\log (3)}-\log (x)}\right ) \]
\[ \int \frac {-3 x+3 x^3+\left (-x+x^2\right ) \log (3)+e^x \left (1+x-3 x^2+x^3\right ) \log (3)+\left (3 x-e^x x \log (3)+\left (x-2 x^2\right ) \log (3)\right ) \log (x)}{-3 x^3+3 x^4+e^x \left (x^2-x^3\right ) \log (3)+\left (-3 x^2+e^x x \log (3)+\left (x^2-x^3\right ) \log (3)\right ) \log (x)+x \log (3) \log ^2(x)} \, dx=\int \frac {-3 x+3 x^3+\left (-x+x^2\right ) \log (3)+e^x \left (1+x-3 x^2+x^3\right ) \log (3)+\left (3 x-e^x x \log (3)+\left (x-2 x^2\right ) \log (3)\right ) \log (x)}{-3 x^3+3 x^4+e^x \left (x^2-x^3\right ) \log (3)+\left (-3 x^2+e^x x \log (3)+\left (x^2-x^3\right ) \log (3)\right ) \log (x)+x \log (3) \log ^2(x)} \, dx \]
Integrate[(-3*x + 3*x^3 + (-x + x^2)*Log[3] + E^x*(1 + x - 3*x^2 + x^3)*Lo g[3] + (3*x - E^x*x*Log[3] + (x - 2*x^2)*Log[3])*Log[x])/(-3*x^3 + 3*x^4 + E^x*(x^2 - x^3)*Log[3] + (-3*x^2 + E^x*x*Log[3] + (x^2 - x^3)*Log[3])*Log [x] + x*Log[3]*Log[x]^2),x]
Integrate[(-3*x + 3*x^3 + (-x + x^2)*Log[3] + E^x*(1 + x - 3*x^2 + x^3)*Lo g[3] + (3*x - E^x*x*Log[3] + (x - 2*x^2)*Log[3])*Log[x])/(-3*x^3 + 3*x^4 + E^x*(x^2 - x^3)*Log[3] + (-3*x^2 + E^x*x*Log[3] + (x^2 - x^3)*Log[3])*Log [x] + x*Log[3]*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.
\(\displaystyle \int \frac {3 x^3+\left (\left (x-2 x^2\right ) \log (3)+3 x-e^x x \log (3)\right ) \log (x)+\left (x^2-x\right ) \log (3)+e^x \left (x^3-3 x^2+x+1\right ) \log (3)-3 x}{3 x^4-3 x^3+\left (-3 x^2+\left (x^2-x^3\right ) \log (3)+e^x x \log (3)\right ) \log (x)+e^x \left (x^2-x^3\right ) \log (3)+x \log (3) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-3 x^3-\left (\left (x-2 x^2\right ) \log (3)+3 x-e^x x \log (3)\right ) \log (x)-\left (x^2-x\right ) \log (3)-e^x \left (x^3-3 x^2+x+1\right ) \log (3)+3 x}{x \left (-x^2+x+\log (x)\right ) \left (3 x-\log (3) \log (x)-e^x \log (3)\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {3 x^2-3 x-x \log (3) \log (x)+\log (3)}{x \left (3 x-\log (3) \log (x)-e^x \log (3)\right )}+\frac {-x^3+3 x^2-x+x \log (x)-1}{x \left (x^2-x-\log (x)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -3 \int \frac {1}{3 x-\log (3) \log (x)-e^x \log (3)}dx+\log (3) \int \frac {1}{x \left (3 x-\log (3) \log (x)-e^x \log (3)\right )}dx+3 \int \frac {x}{3 x-\log (3) \log (x)-e^x \log (3)}dx+\log (3) \int \frac {\log (x)}{-3 x+\log (3) \log (x)+e^x \log (3)}dx+\log \left (-x^2+x+\log (x)\right )-x\) |
Int[(-3*x + 3*x^3 + (-x + x^2)*Log[3] + E^x*(1 + x - 3*x^2 + x^3)*Log[3] + (3*x - E^x*x*Log[3] + (x - 2*x^2)*Log[3])*Log[x])/(-3*x^3 + 3*x^4 + E^x*( x^2 - x^3)*Log[3] + (-3*x^2 + E^x*x*Log[3] + (x^2 - x^3)*Log[3])*Log[x] + x*Log[3]*Log[x]^2),x]
3.25.40.3.1 Defintions of rubi rules used
Time = 0.48 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03
method | result | size |
parallelrisch | \(-\ln \left (-\frac {\ln \left (3\right ) \ln \left (x \right )}{3}-\frac {\ln \left (3\right ) {\mathrm e}^{x}}{3}+x \right )+\ln \left (x^{2}-x -\ln \left (x \right )\right )\) | \(31\) |
risch | \(\ln \left (\ln \left (x \right )+x -x^{2}\right )-\ln \left (\ln \left (x \right )+\frac {\ln \left (3\right ) {\mathrm e}^{x}-3 x}{\ln \left (3\right )}\right )\) | \(32\) |
int(((-x*ln(3)*exp(x)+(-2*x^2+x)*ln(3)+3*x)*ln(x)+(x^3-3*x^2+x+1)*ln(3)*ex p(x)+(x^2-x)*ln(3)+3*x^3-3*x)/(x*ln(3)*ln(x)^2+(x*ln(3)*exp(x)+(-x^3+x^2)* ln(3)-3*x^2)*ln(x)+(-x^3+x^2)*ln(3)*exp(x)+3*x^4-3*x^3),x,method=_RETURNVE RBOSE)
Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {-3 x+3 x^3+\left (-x+x^2\right ) \log (3)+e^x \left (1+x-3 x^2+x^3\right ) \log (3)+\left (3 x-e^x x \log (3)+\left (x-2 x^2\right ) \log (3)\right ) \log (x)}{-3 x^3+3 x^4+e^x \left (x^2-x^3\right ) \log (3)+\left (-3 x^2+e^x x \log (3)+\left (x^2-x^3\right ) \log (3)\right ) \log (x)+x \log (3) \log ^2(x)} \, dx=\log \left (-x^{2} + x + \log \left (x\right )\right ) - \log \left (e^{x} \log \left (3\right ) + \log \left (3\right ) \log \left (x\right ) - 3 \, x\right ) \]
integrate(((-x*log(3)*exp(x)+(-2*x^2+x)*log(3)+3*x)*log(x)+(x^3-3*x^2+x+1) *log(3)*exp(x)+(x^2-x)*log(3)+3*x^3-3*x)/(x*log(3)*log(x)^2+(x*log(3)*exp( x)+(-x^3+x^2)*log(3)-3*x^2)*log(x)+(-x^3+x^2)*log(3)*exp(x)+3*x^4-3*x^3),x , algorithm=\
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {-3 x+3 x^3+\left (-x+x^2\right ) \log (3)+e^x \left (1+x-3 x^2+x^3\right ) \log (3)+\left (3 x-e^x x \log (3)+\left (x-2 x^2\right ) \log (3)\right ) \log (x)}{-3 x^3+3 x^4+e^x \left (x^2-x^3\right ) \log (3)+\left (-3 x^2+e^x x \log (3)+\left (x^2-x^3\right ) \log (3)\right ) \log (x)+x \log (3) \log ^2(x)} \, dx=- \log {\left (\frac {- 3 x + \log {\left (3 \right )} \log {\left (x \right )}}{\log {\left (3 \right )}} + e^{x} \right )} + \log {\left (- x^{2} + x + \log {\left (x \right )} \right )} \]
integrate(((-x*ln(3)*exp(x)+(-2*x**2+x)*ln(3)+3*x)*ln(x)+(x**3-3*x**2+x+1) *ln(3)*exp(x)+(x**2-x)*ln(3)+3*x**3-3*x)/(x*ln(3)*ln(x)**2+(x*ln(3)*exp(x) +(-x**3+x**2)*ln(3)-3*x**2)*ln(x)+(-x**3+x**2)*ln(3)*exp(x)+3*x**4-3*x**3) ,x)
Time = 0.34 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {-3 x+3 x^3+\left (-x+x^2\right ) \log (3)+e^x \left (1+x-3 x^2+x^3\right ) \log (3)+\left (3 x-e^x x \log (3)+\left (x-2 x^2\right ) \log (3)\right ) \log (x)}{-3 x^3+3 x^4+e^x \left (x^2-x^3\right ) \log (3)+\left (-3 x^2+e^x x \log (3)+\left (x^2-x^3\right ) \log (3)\right ) \log (x)+x \log (3) \log ^2(x)} \, dx=\log \left (-x^{2} + x + \log \left (x\right )\right ) - \log \left (\frac {e^{x} \log \left (3\right ) + \log \left (3\right ) \log \left (x\right ) - 3 \, x}{\log \left (3\right )}\right ) \]
integrate(((-x*log(3)*exp(x)+(-2*x^2+x)*log(3)+3*x)*log(x)+(x^3-3*x^2+x+1) *log(3)*exp(x)+(x^2-x)*log(3)+3*x^3-3*x)/(x*log(3)*log(x)^2+(x*log(3)*exp( x)+(-x^3+x^2)*log(3)-3*x^2)*log(x)+(-x^3+x^2)*log(3)*exp(x)+3*x^4-3*x^3),x , algorithm=\
Time = 0.65 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-3 x+3 x^3+\left (-x+x^2\right ) \log (3)+e^x \left (1+x-3 x^2+x^3\right ) \log (3)+\left (3 x-e^x x \log (3)+\left (x-2 x^2\right ) \log (3)\right ) \log (x)}{-3 x^3+3 x^4+e^x \left (x^2-x^3\right ) \log (3)+\left (-3 x^2+e^x x \log (3)+\left (x^2-x^3\right ) \log (3)\right ) \log (x)+x \log (3) \log ^2(x)} \, dx=\log \left (x^{2} - x - \log \left (x\right )\right ) - \log \left (-e^{x} \log \left (3\right ) - \log \left (3\right ) \log \left (x\right ) + 3 \, x\right ) \]
integrate(((-x*log(3)*exp(x)+(-2*x^2+x)*log(3)+3*x)*log(x)+(x^3-3*x^2+x+1) *log(3)*exp(x)+(x^2-x)*log(3)+3*x^3-3*x)/(x*log(3)*log(x)^2+(x*log(3)*exp( x)+(-x^3+x^2)*log(3)-3*x^2)*log(x)+(-x^3+x^2)*log(3)*exp(x)+3*x^4-3*x^3),x , algorithm=\
Timed out. \[ \int \frac {-3 x+3 x^3+\left (-x+x^2\right ) \log (3)+e^x \left (1+x-3 x^2+x^3\right ) \log (3)+\left (3 x-e^x x \log (3)+\left (x-2 x^2\right ) \log (3)\right ) \log (x)}{-3 x^3+3 x^4+e^x \left (x^2-x^3\right ) \log (3)+\left (-3 x^2+e^x x \log (3)+\left (x^2-x^3\right ) \log (3)\right ) \log (x)+x \log (3) \log ^2(x)} \, dx=\int \frac {\ln \left (x\right )\,\left (3\,x+\ln \left (3\right )\,\left (x-2\,x^2\right )-x\,{\mathrm {e}}^x\,\ln \left (3\right )\right )-3\,x+3\,x^3-\ln \left (3\right )\,\left (x-x^2\right )+{\mathrm {e}}^x\,\ln \left (3\right )\,\left (x^3-3\,x^2+x+1\right )}{\ln \left (x\right )\,\left (\ln \left (3\right )\,\left (x^2-x^3\right )-3\,x^2+x\,{\mathrm {e}}^x\,\ln \left (3\right )\right )-3\,x^3+3\,x^4+{\mathrm {e}}^x\,\ln \left (3\right )\,\left (x^2-x^3\right )+x\,\ln \left (3\right )\,{\ln \left (x\right )}^2} \,d x \]
int((log(x)*(3*x + log(3)*(x - 2*x^2) - x*exp(x)*log(3)) - 3*x + 3*x^3 - l og(3)*(x - x^2) + exp(x)*log(3)*(x - 3*x^2 + x^3 + 1))/(log(x)*(log(3)*(x^ 2 - x^3) - 3*x^2 + x*exp(x)*log(3)) - 3*x^3 + 3*x^4 + exp(x)*log(3)*(x^2 - x^3) + x*log(3)*log(x)^2),x)