Integrand size = 122, antiderivative size = 30 \[ \int \frac {\left (-x+6 x^2-3 x^3\right ) \log (2)+\left (-x+2 x^2-x^3\right ) \log (2) \log (4)+\left (\left (1+3 x-3 x^2\right ) \log (2)+\left (x-x^2\right ) \log (2) \log (4)\right ) \log (x)+\left (\left (x-x^2\right ) \log (2)-x \log (2) \log (x)\right ) \log \left (-x+x^2+x \log (x)\right )}{e^x \left (-2 x+2 x^2\right )+2 e^x x \log (x)} \, dx=\frac {1}{2} e^{-x} \log (2) \left (x (3+\log (4))+\log \left (-x+x^2+x \log (x)\right )\right ) \]
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {\left (-x+6 x^2-3 x^3\right ) \log (2)+\left (-x+2 x^2-x^3\right ) \log (2) \log (4)+\left (\left (1+3 x-3 x^2\right ) \log (2)+\left (x-x^2\right ) \log (2) \log (4)\right ) \log (x)+\left (\left (x-x^2\right ) \log (2)-x \log (2) \log (x)\right ) \log \left (-x+x^2+x \log (x)\right )}{e^x \left (-2 x+2 x^2\right )+2 e^x x \log (x)} \, dx=\frac {1}{2} e^{-x} \log (2) (x (3+\log (4))+\log (x (-1+x+\log (x)))) \]
Integrate[((-x + 6*x^2 - 3*x^3)*Log[2] + (-x + 2*x^2 - x^3)*Log[2]*Log[4] + ((1 + 3*x - 3*x^2)*Log[2] + (x - x^2)*Log[2]*Log[4])*Log[x] + ((x - x^2) *Log[2] - x*Log[2]*Log[x])*Log[-x + x^2 + x*Log[x]])/(E^x*(-2*x + 2*x^2) + 2*E^x*x*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.
| \(\displaystyle \int \frac {\left (\left (-3 x^2+3 x+1\right ) \log (2)+\left (x-x^2\right ) \log (2) \log (4)\right ) \log (x)+\left (\left (x-x^2\right ) \log (2)-x \log (2) \log (x)\right ) \log \left (x^2-x+x \log (x)\right )+\left (-3 x^3+6 x^2-x\right ) \log (2)+\left (-x^3+2 x^2-x\right ) \log (2) \log (4)}{e^x \left (2 x^2-2 x\right )+2 e^x x \log (x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{-x} \log (2) \left (3 x^3 \left (1+\frac {2 \log (2)}{3}\right )+3 x^2 \left (1+\frac {2 \log (2)}{3}\right ) \log (x)+x^2 \log (x (x+\log (x)-1))-6 x^2 \left (1+\frac {2 \log (2)}{3}\right )-3 x \left (1+\frac {2 \log (2)}{3}\right ) \log (x)+x \log (x) \log (x (x+\log (x)-1))-x \log (x (x+\log (x)-1))+x (1+\log (4))-\log (x)\right )}{2 x (-x-\log (x)+1)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \log (2) \int \frac {e^{-x} \left ((3+\log (4)) x^3+(3+\log (4)) \log (x) x^2+\log (-x (-x-\log (x)+1)) x^2-2 (3+\log (4)) x^2-(3+\log (4)) \log (x) x+\log (x) \log (-x (-x-\log (x)+1)) x-\log (-x (-x-\log (x)+1)) x+(1+\log (4)) x-\log (x)\right )}{x (-x-\log (x)+1)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{2} \log (2) \int \left (-\frac {e^{-x} (3+\log (4)) x^2}{x+\log (x)-1}-\frac {e^{-x} (3+\log (4)) \log (x) x}{x+\log (x)-1}+\frac {2 e^{-x} (3+\log (4)) x}{x+\log (x)-1}-e^{-x} \log \left (x^2+\log (x) x-x\right )+\frac {e^{-x} (3+\log (4)) \log (x)}{x+\log (x)-1}+\frac {e^{-x} (-1-\log (4))}{x+\log (x)-1}+\frac {e^{-x} \log (x)}{(x+\log (x)-1) x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \log (2) \left (-\int e^{-x} \log \left (x^2+\log (x) x-x\right )dx+\int \frac {e^{-x}}{-x-\log (x)+1}dx+(3+\log (4)) \int \frac {e^{-x}}{x+\log (x)-1}dx-(1+\log (4)) \int \frac {e^{-x}}{x+\log (x)-1}dx+\int \frac {e^{-x}}{x (x+\log (x)-1)}dx+\operatorname {ExpIntegralEi}(-x)+e^{-x} x (3+\log (4))\right )\) |
Int[((-x + 6*x^2 - 3*x^3)*Log[2] + (-x + 2*x^2 - x^3)*Log[2]*Log[4] + ((1 + 3*x - 3*x^2)*Log[2] + (x - x^2)*Log[2]*Log[4])*Log[x] + ((x - x^2)*Log[2 ] - x*Log[2]*Log[x])*Log[-x + x^2 + x*Log[x]])/(E^x*(-2*x + 2*x^2) + 2*E^x *x*Log[x]),x]
3.10.22.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.88 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03
| method | result | size |
| parallelrisch | \(\frac {\left (2 x \ln \left (2\right )^{2}+3 x \ln \left (2\right )+\ln \left (2\right ) \ln \left (x \left (-1+\ln \left (x \right )+x \right )\right )\right ) {\mathrm e}^{-x}}{2}\) | \(31\) |
| risch | \(\frac {\ln \left (2\right ) {\mathrm e}^{-x} \ln \left (-1+\ln \left (x \right )+x \right )}{2}+\frac {\ln \left (2\right ) \left (i \pi \,\operatorname {csgn}\left (i \left (-1+\ln \left (x \right )+x \right )\right ) \operatorname {csgn}\left (i x \left (-1+\ln \left (x \right )+x \right )\right )^{2}-i \pi \,\operatorname {csgn}\left (i \left (-1+\ln \left (x \right )+x \right )\right ) \operatorname {csgn}\left (i x \left (-1+\ln \left (x \right )+x \right )\right ) \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (i x \left (-1+\ln \left (x \right )+x \right )\right )^{3}+i \pi \operatorname {csgn}\left (i x \left (-1+\ln \left (x \right )+x \right )\right )^{2} \operatorname {csgn}\left (i x \right )+4 x \ln \left (2\right )+6 x +2 \ln \left (x \right )\right ) {\mathrm e}^{-x}}{4}\) | \(127\) |
int(((-x*ln(2)*ln(x)+(-x^2+x)*ln(2))*ln(x*ln(x)+x^2-x)+(2*(-x^2+x)*ln(2)^2 +(-3*x^2+3*x+1)*ln(2))*ln(x)+2*(-x^3+2*x^2-x)*ln(2)^2+(-3*x^3+6*x^2-x)*ln( 2))/(2*x*exp(x)*ln(x)+(2*x^2-2*x)*exp(x)),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {\left (-x+6 x^2-3 x^3\right ) \log (2)+\left (-x+2 x^2-x^3\right ) \log (2) \log (4)+\left (\left (1+3 x-3 x^2\right ) \log (2)+\left (x-x^2\right ) \log (2) \log (4)\right ) \log (x)+\left (\left (x-x^2\right ) \log (2)-x \log (2) \log (x)\right ) \log \left (-x+x^2+x \log (x)\right )}{e^x \left (-2 x+2 x^2\right )+2 e^x x \log (x)} \, dx=\frac {1}{2} \, {\left (2 \, x \log \left (2\right )^{2} + 3 \, x \log \left (2\right ) + \log \left (2\right ) \log \left (x^{2} + x \log \left (x\right ) - x\right )\right )} e^{\left (-x\right )} \]
integrate(((-x*log(2)*log(x)+(-x^2+x)*log(2))*log(x*log(x)+x^2-x)+(2*(-x^2 +x)*log(2)^2+(-3*x^2+3*x+1)*log(2))*log(x)+2*(-x^3+2*x^2-x)*log(2)^2+(-3*x ^3+6*x^2-x)*log(2))/(2*x*exp(x)*log(x)+(2*x^2-2*x)*exp(x)),x, algorithm=\
Time = 12.78 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {\left (-x+6 x^2-3 x^3\right ) \log (2)+\left (-x+2 x^2-x^3\right ) \log (2) \log (4)+\left (\left (1+3 x-3 x^2\right ) \log (2)+\left (x-x^2\right ) \log (2) \log (4)\right ) \log (x)+\left (\left (x-x^2\right ) \log (2)-x \log (2) \log (x)\right ) \log \left (-x+x^2+x \log (x)\right )}{e^x \left (-2 x+2 x^2\right )+2 e^x x \log (x)} \, dx=\frac {\left (2 x \log {\left (2 \right )}^{2} + 3 x \log {\left (2 \right )} + \log {\left (2 \right )} \log {\left (x^{2} + x \log {\left (x \right )} - x \right )}\right ) e^{- x}}{2} \]
integrate(((-x*ln(2)*ln(x)+(-x**2+x)*ln(2))*ln(x*ln(x)+x**2-x)+(2*(-x**2+x )*ln(2)**2+(-3*x**2+3*x+1)*ln(2))*ln(x)+2*(-x**3+2*x**2-x)*ln(2)**2+(-3*x* *3+6*x**2-x)*ln(2))/(2*x*exp(x)*ln(x)+(2*x**2-2*x)*exp(x)),x)
Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {\left (-x+6 x^2-3 x^3\right ) \log (2)+\left (-x+2 x^2-x^3\right ) \log (2) \log (4)+\left (\left (1+3 x-3 x^2\right ) \log (2)+\left (x-x^2\right ) \log (2) \log (4)\right ) \log (x)+\left (\left (x-x^2\right ) \log (2)-x \log (2) \log (x)\right ) \log \left (-x+x^2+x \log (x)\right )}{e^x \left (-2 x+2 x^2\right )+2 e^x x \log (x)} \, dx=\frac {1}{2} \, {\left ({\left (2 \, \log \left (2\right )^{2} + 3 \, \log \left (2\right )\right )} x + \log \left (2\right ) \log \left (x + \log \left (x\right ) - 1\right ) + \log \left (2\right ) \log \left (x\right )\right )} e^{\left (-x\right )} \]
integrate(((-x*log(2)*log(x)+(-x^2+x)*log(2))*log(x*log(x)+x^2-x)+(2*(-x^2 +x)*log(2)^2+(-3*x^2+3*x+1)*log(2))*log(x)+2*(-x^3+2*x^2-x)*log(2)^2+(-3*x ^3+6*x^2-x)*log(2))/(2*x*exp(x)*log(x)+(2*x^2-2*x)*exp(x)),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.47 \[ \int \frac {\left (-x+6 x^2-3 x^3\right ) \log (2)+\left (-x+2 x^2-x^3\right ) \log (2) \log (4)+\left (\left (1+3 x-3 x^2\right ) \log (2)+\left (x-x^2\right ) \log (2) \log (4)\right ) \log (x)+\left (\left (x-x^2\right ) \log (2)-x \log (2) \log (x)\right ) \log \left (-x+x^2+x \log (x)\right )}{e^x \left (-2 x+2 x^2\right )+2 e^x x \log (x)} \, dx=x e^{\left (-x\right )} \log \left (2\right )^{2} + \frac {3}{2} \, x e^{\left (-x\right )} \log \left (2\right ) + \frac {1}{2} \, e^{\left (-x\right )} \log \left (2\right ) \log \left (x + \log \left (x\right ) - 1\right ) + \frac {1}{2} \, e^{\left (-x\right )} \log \left (2\right ) \log \left (x\right ) \]
integrate(((-x*log(2)*log(x)+(-x^2+x)*log(2))*log(x*log(x)+x^2-x)+(2*(-x^2 +x)*log(2)^2+(-3*x^2+3*x+1)*log(2))*log(x)+2*(-x^3+2*x^2-x)*log(2)^2+(-3*x ^3+6*x^2-x)*log(2))/(2*x*exp(x)*log(x)+(2*x^2-2*x)*exp(x)),x, algorithm=\
x*e^(-x)*log(2)^2 + 3/2*x*e^(-x)*log(2) + 1/2*e^(-x)*log(2)*log(x + log(x) - 1) + 1/2*e^(-x)*log(2)*log(x)
Timed out. \[ \int \frac {\left (-x+6 x^2-3 x^3\right ) \log (2)+\left (-x+2 x^2-x^3\right ) \log (2) \log (4)+\left (\left (1+3 x-3 x^2\right ) \log (2)+\left (x-x^2\right ) \log (2) \log (4)\right ) \log (x)+\left (\left (x-x^2\right ) \log (2)-x \log (2) \log (x)\right ) \log \left (-x+x^2+x \log (x)\right )}{e^x \left (-2 x+2 x^2\right )+2 e^x x \log (x)} \, dx=\int \frac {\ln \left (2\right )\,\left (3\,x^3-6\,x^2+x\right )+2\,{\ln \left (2\right )}^2\,\left (x^3-2\,x^2+x\right )-\ln \left (x\right )\,\left (\ln \left (2\right )\,\left (-3\,x^2+3\,x+1\right )+2\,{\ln \left (2\right )}^2\,\left (x-x^2\right )\right )-\ln \left (x\,\ln \left (x\right )-x+x^2\right )\,\left (\ln \left (2\right )\,\left (x-x^2\right )-x\,\ln \left (2\right )\,\ln \left (x\right )\right )}{{\mathrm {e}}^x\,\left (2\,x-2\,x^2\right )-2\,x\,{\mathrm {e}}^x\,\ln \left (x\right )} \,d x \]
int((log(2)*(x - 6*x^2 + 3*x^3) + 2*log(2)^2*(x - 2*x^2 + x^3) - log(x)*(l og(2)*(3*x - 3*x^2 + 1) + 2*log(2)^2*(x - x^2)) - log(x*log(x) - x + x^2)* (log(2)*(x - x^2) - x*log(2)*log(x)))/(exp(x)*(2*x - 2*x^2) - 2*x*exp(x)*l og(x)),x)