Integrand size = 87, antiderivative size = 24 \[ \int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx=3+5 e^{-x}+2 x+\frac {3 x}{\log (-x+\log (x))} \] Output:
2*x+3+3*x/ln(ln(x)-x)+5/exp(x)
Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx=5 e^{-x}+2 x+\frac {3 x}{\log (-x+\log (x))} \] Input:
Integrate[(E^x*(-3 + 3*x) + (-3*E^x*x + 3*E^x*Log[x])*Log[-x + Log[x]] + ( 5*x - 2*E^x*x + (-5 + 2*E^x)*Log[x])*Log[-x + Log[x]]^2)/((-(E^x*x) + E^x* Log[x])*Log[-x + Log[x]]^2),x]
Output:
5/E^x + 2*x + (3*x)/Log[-x + Log[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.
\(\displaystyle \int \frac {e^x (3 x-3)+\left (-2 e^x x+5 x+\left (2 e^x-5\right ) \log (x)\right ) \log ^2(\log (x)-x)+\left (3 e^x \log (x)-3 e^x x\right ) \log (\log (x)-x)}{\left (e^x \log (x)-e^x x\right ) \log ^2(\log (x)-x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (-5 e^{-x}-\frac {3 (x-1)}{(x-\log (x)) \log ^2(\log (x)-x)}+\frac {3}{\log (\log (x)-x)}+2\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \int \frac {1}{(x-\log (x)) \log ^2(\log (x)-x)}dx-3 \int \frac {x}{(x-\log (x)) \log ^2(\log (x)-x)}dx+3 \int \frac {1}{\log (\log (x)-x)}dx+2 x+5 e^{-x}\) |
Input:
Int[(E^x*(-3 + 3*x) + (-3*E^x*x + 3*E^x*Log[x])*Log[-x + Log[x]] + (5*x - 2*E^x*x + (-5 + 2*E^x)*Log[x])*Log[-x + Log[x]]^2)/((-(E^x*x) + E^x*Log[x] )*Log[-x + Log[x]]^2),x]
Output:
$Aborted
Time = 0.43 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08
method | result | size |
risch | \(\left (2 \,{\mathrm e}^{x} x +5\right ) {\mathrm e}^{-x}+\frac {3 x}{\ln \left (\ln \left (x \right )-x \right )}\) | \(26\) |
default | \(5 \,{\mathrm e}^{-x}+\frac {3 x +2 \ln \left (\ln \left (x \right )-x \right ) x}{\ln \left (\ln \left (x \right )-x \right )}\) | \(32\) |
parts | \(5 \,{\mathrm e}^{-x}+\frac {3 x +2 \ln \left (\ln \left (x \right )-x \right ) x}{\ln \left (\ln \left (x \right )-x \right )}\) | \(32\) |
parallelrisch | \(\frac {\left (4 \,{\mathrm e}^{x} \ln \left (\ln \left (x \right )-x \right ) x +6 \,{\mathrm e}^{x} x +10 \ln \left (\ln \left (x \right )-x \right )\right ) {\mathrm e}^{-x}}{2 \ln \left (\ln \left (x \right )-x \right )}\) | \(43\) |
Input:
int((((2*exp(x)-5)*ln(x)-2*exp(x)*x+5*x)*ln(ln(x)-x)^2+(3*exp(x)*ln(x)-3*e xp(x)*x)*ln(ln(x)-x)+(-3+3*x)*exp(x))/(exp(x)*ln(x)-exp(x)*x)/ln(ln(x)-x)^ 2,x,method=_RETURNVERBOSE)
Output:
(2*exp(x)*x+5)/exp(x)+3*x/ln(ln(x)-x)
Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx=\frac {{\left (3 \, x e^{x} + {\left (2 \, x e^{x} + 5\right )} \log \left (-x + \log \left (x\right )\right )\right )} e^{\left (-x\right )}}{\log \left (-x + \log \left (x\right )\right )} \] Input:
integrate((((2*exp(x)-5)*log(x)-2*exp(x)*x+5*x)*log(log(x)-x)^2+(3*exp(x)* log(x)-3*exp(x)*x)*log(log(x)-x)+(-3+3*x)*exp(x))/(exp(x)*log(x)-exp(x)*x) /log(log(x)-x)^2,x, algorithm="fricas")
Output:
(3*x*e^x + (2*x*e^x + 5)*log(-x + log(x)))*e^(-x)/log(-x + log(x))
Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx=2 x + \frac {3 x}{\log {\left (- x + \log {\left (x \right )} \right )}} + 5 e^{- x} \] Input:
integrate((((2*exp(x)-5)*ln(x)-2*exp(x)*x+5*x)*ln(ln(x)-x)**2+(3*exp(x)*ln (x)-3*exp(x)*x)*ln(ln(x)-x)+(-3+3*x)*exp(x))/(exp(x)*ln(x)-exp(x)*x)/ln(ln (x)-x)**2,x)
Output:
2*x + 3*x/log(-x + log(x)) + 5*exp(-x)
Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx=\frac {{\left (3 \, x e^{x} + {\left (2 \, x e^{x} + 5\right )} \log \left (-x + \log \left (x\right )\right )\right )} e^{\left (-x\right )}}{\log \left (-x + \log \left (x\right )\right )} \] Input:
integrate((((2*exp(x)-5)*log(x)-2*exp(x)*x+5*x)*log(log(x)-x)^2+(3*exp(x)* log(x)-3*exp(x)*x)*log(log(x)-x)+(-3+3*x)*exp(x))/(exp(x)*log(x)-exp(x)*x) /log(log(x)-x)^2,x, algorithm="maxima")
Output:
(3*x*e^x + (2*x*e^x + 5)*log(-x + log(x)))*e^(-x)/log(-x + log(x))
Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx=\frac {2 \, x \log \left (-x + \log \left (x\right )\right ) + 5 \, e^{\left (-x\right )} \log \left (-x + \log \left (x\right )\right ) + 3 \, x}{\log \left (-x + \log \left (x\right )\right )} \] Input:
integrate((((2*exp(x)-5)*log(x)-2*exp(x)*x+5*x)*log(log(x)-x)^2+(3*exp(x)* log(x)-3*exp(x)*x)*log(log(x)-x)+(-3+3*x)*exp(x))/(exp(x)*log(x)-exp(x)*x) /log(log(x)-x)^2,x, algorithm="giac")
Output:
(2*x*log(-x + log(x)) + 5*e^(-x)*log(-x + log(x)) + 3*x)/log(-x + log(x))
Time = 3.11 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.58 \[ \int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx=5\,x+5\,{\mathrm {e}}^{-x}-3\,\ln \left (x\right )+\frac {3}{x-1}-\frac {3\,x^2}{x-1}-\frac {3\,\ln \left (x\right )}{x-1}+\frac {3\,x}{\ln \left (\ln \left (x\right )-x\right )}+\frac {3\,x\,\ln \left (x\right )}{x-1} \] Input:
int((log(log(x) - x)^2*(5*x - 2*x*exp(x) + log(x)*(2*exp(x) - 5)) + log(lo g(x) - x)*(3*exp(x)*log(x) - 3*x*exp(x)) + exp(x)*(3*x - 3))/(log(log(x) - x)^2*(exp(x)*log(x) - x*exp(x))),x)
Output:
5*x + 5*exp(-x) - 3*log(x) + 3/(x - 1) - (3*x^2)/(x - 1) - (3*log(x))/(x - 1) + (3*x)/log(log(x) - x) + (3*x*log(x))/(x - 1)
Time = 0.16 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx=\frac {2 e^{x} \mathrm {log}\left (\mathrm {log}\left (x \right )-x \right ) x +3 e^{x} x +5 \,\mathrm {log}\left (\mathrm {log}\left (x \right )-x \right )}{e^{x} \mathrm {log}\left (\mathrm {log}\left (x \right )-x \right )} \] Input:
int((((2*exp(x)-5)*log(x)-2*exp(x)*x+5*x)*log(log(x)-x)^2+(3*exp(x)*log(x) -3*exp(x)*x)*log(log(x)-x)+(-3+3*x)*exp(x))/(exp(x)*log(x)-exp(x)*x)/log(l og(x)-x)^2,x)
Output:
(2*e**x*log(log(x) - x)*x + 3*e**x*x + 5*log(log(x) - x))/(e**x*log(log(x) - x))