Integrand size = 112, antiderivative size = 32 \[ \int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx=\frac {x \left (5+\frac {1}{4} e^{e^{-x} x}+x-\log \left (2 x^2\right )\right )}{x+\log (x)} \] Output:
(x-ln(2*x^2)+1/4*exp(x/exp(x))+5)/(x+ln(x))*x
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx=\frac {x \left (e^{e^{-x} x}+4 (5+x)-4 \log \left (2 x^2\right )\right )}{4 (x+\log (x))} \] Input:
Integrate[(E^x*(-20 - 12*x + 4*x^2) + E^x*(12 + 8*x)*Log[x] + E^(x/E^x)*(- E^x + x^2 - x^3 + (E^x + x - x^2)*Log[x]) + (4*E^x - 4*E^x*Log[x])*Log[2*x ^2])/(4*E^x*x^2 + 8*E^x*x*Log[x] + 4*E^x*Log[x]^2),x]
Output:
(x*(E^(x/E^x) + 4*(5 + x) - 4*Log[2*x^2]))/(4*(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.
| \(\displaystyle \int \frac {e^x \left (4 x^2-12 x-20\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )+e^{e^{-x} x} \left (-x^3+x^2+\left (-x^2+x+e^x\right ) \log (x)-e^x\right )+e^x (8 x+12) \log (x)}{4 e^x x^2+4 e^x \log ^2(x)+8 e^x x \log (x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{-x} \left (e^x \left (4 x^2-12 x-20\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )+e^{e^{-x} x} \left (-x^3+x^2+\left (-x^2+x+e^x\right ) \log (x)-e^x\right )+e^x (8 x+12) \log (x)\right )}{4 (x+\log (x))^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int -\frac {e^{-x} \left (4 e^x \left (-x^2+3 x+5\right )-4 e^x (2 x+3) \log (x)+e^{e^{-x} x} \left (x^3-x^2+e^x-\left (-x^2+x+e^x\right ) \log (x)\right )-4 \left (e^x-e^x \log (x)\right ) \log \left (2 x^2\right )\right )}{(x+\log (x))^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{4} \int \frac {e^{-x} \left (4 e^x \left (-x^2+3 x+5\right )-4 e^x (2 x+3) \log (x)+e^{e^{-x} x} \left (x^3-x^2+e^x-\left (-x^2+x+e^x\right ) \log (x)\right )-4 \left (e^x-e^x \log (x)\right ) \log \left (2 x^2\right )\right )}{(x+\log (x))^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {1}{4} \int \left (\frac {e^{e^{-x} x-x} \left (x^3+\log (x) x^2-x^2-\log (x) x+e^x-e^x \log (x)\right )}{(x+\log (x))^2}-\frac {4 \left (x^2+2 \log (x) x-3 x+3 \log (x)-\log (x) \log \left (2 x^2\right )+\log \left (2 x^2\right )-5\right )}{(x+\log (x))^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} \left (-4 \int \frac {x^2}{(x+\log (x))^2}dx-\int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} x^2}{x+\log (x)}dx+4 \int \frac {\log \left (2 x^2\right )}{(x+\log (x))^2}dx-4 \int \frac {\log (x) \log \left (2 x^2\right )}{(x+\log (x))^2}dx-20 \int \frac {1}{(x+\log (x))^2}dx-\int \frac {e^{e^{-x} x}}{(x+\log (x))^2}dx-24 \int \frac {x}{(x+\log (x))^2}dx-\int \frac {e^{e^{-x} x} x}{(x+\log (x))^2}dx+12 \int \frac {1}{x+\log (x)}dx+\int \frac {e^{e^{-x} x}}{x+\log (x)}dx+8 \int \frac {x}{x+\log (x)}dx+\int \frac {e^{-e^{-x} \left (-1+e^x\right ) x} x}{x+\log (x)}dx\right )\) |
Input:
Int[(E^x*(-20 - 12*x + 4*x^2) + E^x*(12 + 8*x)*Log[x] + E^(x/E^x)*(-E^x + x^2 - x^3 + (E^x + x - x^2)*Log[x]) + (4*E^x - 4*E^x*Log[x])*Log[2*x^2])/( 4*E^x*x^2 + 8*E^x*x*Log[x] + 4*E^x*Log[x]^2),x]
Output:
$Aborted
Time = 3.44 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.59
| method | result | size |
| parallelrisch | \(-\frac {48 x \ln \left (x \right )+240 \ln \left (x \right )-12 x \,{\mathrm e}^{x \,{\mathrm e}^{-x}}-48 \ln \left ({\mathrm e}^{x}\right ) x -48 \ln \left ({\mathrm e}^{x}\right ) \ln \left (x \right )+48 \ln \left (2 x^{2}\right ) x}{48 \left (x +\ln \left (x \right )\right )}\) | \(51\) |
| risch | \(-2 x +\frac {\left (i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+10+i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-2 \ln \left (2\right )+6 x \right ) x}{2 x +2 \ln \left (x \right )}+\frac {x \,{\mathrm e}^{x \,{\mathrm e}^{-x}}}{4 x +4 \ln \left (x \right )}\) | \(88\) |
Input:
int((((x+exp(x)-x^2)*ln(x)-exp(x)-x^3+x^2)*exp(x/exp(x))+(-4*exp(x)*ln(x)+ 4*exp(x))*ln(2*x^2)+(8*x+12)*exp(x)*ln(x)+(4*x^2-12*x-20)*exp(x))/(4*exp(x )*ln(x)^2+8*x*exp(x)*ln(x)+4*exp(x)*x^2),x,method=_RETURNVERBOSE)
Output:
-1/48*(48*x*ln(x)+240*ln(x)-12*exp(x/exp(x))*x-48*ln(exp(x))*x-48*ln(exp(x ))*ln(x)+48*ln(2*x^2)*x)/(x+ln(x))
Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx=\frac {4 \, x^{2} + x e^{\left (x e^{\left (-x\right )}\right )} - 4 \, x \log \left (2\right ) - 8 \, x \log \left (x\right ) + 20 \, x}{4 \, {\left (x + \log \left (x\right )\right )}} \] Input:
integrate((((x+exp(x)-x^2)*log(x)-exp(x)-x^3+x^2)*exp(x/exp(x))+(-4*exp(x) *log(x)+4*exp(x))*log(2*x^2)+(8*x+12)*exp(x)*log(x)+(4*x^2-12*x-20)*exp(x) )/(4*exp(x)*log(x)^2+8*x*exp(x)*log(x)+4*exp(x)*x^2),x, algorithm="fricas" )
Output:
1/4*(4*x^2 + x*e^(x*e^(-x)) - 4*x*log(2) - 8*x*log(x) + 20*x)/(x + log(x))
Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx=- 2 x + \frac {x e^{x e^{- x}}}{4 x + 4 \log {\left (x \right )}} + \frac {3 x^{2} - x \log {\left (2 \right )} + 5 x}{x + \log {\left (x \right )}} \] Input:
integrate((((x+exp(x)-x**2)*ln(x)-exp(x)-x**3+x**2)*exp(x/exp(x))+(-4*exp( x)*ln(x)+4*exp(x))*ln(2*x**2)+(8*x+12)*exp(x)*ln(x)+(4*x**2-12*x-20)*exp(x ))/(4*exp(x)*ln(x)**2+8*x*exp(x)*ln(x)+4*exp(x)*x**2),x)
Output:
-2*x + x*exp(x*exp(-x))/(4*x + 4*log(x)) + (3*x**2 - x*log(2) + 5*x)/(x + log(x))
\[ \int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx=\int { \frac {4 \, {\left (2 \, x + 3\right )} e^{x} \log \left (x\right ) - {\left (x^{3} - x^{2} + {\left (x^{2} - x - e^{x}\right )} \log \left (x\right ) + e^{x}\right )} e^{\left (x e^{\left (-x\right )}\right )} + 4 \, {\left (x^{2} - 3 \, x - 5\right )} e^{x} - 4 \, {\left (e^{x} \log \left (x\right ) - e^{x}\right )} \log \left (2 \, x^{2}\right )}{4 \, {\left (x^{2} e^{x} + 2 \, x e^{x} \log \left (x\right ) + e^{x} \log \left (x\right )^{2}\right )}} \,d x } \] Input:
integrate((((x+exp(x)-x^2)*log(x)-exp(x)-x^3+x^2)*exp(x/exp(x))+(-4*exp(x) *log(x)+4*exp(x))*log(2*x^2)+(8*x+12)*exp(x)*log(x)+(4*x^2-12*x-20)*exp(x) )/(4*exp(x)*log(x)^2+8*x*exp(x)*log(x)+4*exp(x)*x^2),x, algorithm="maxima" )
Output:
(x^2 - x*(log(2) - 5) - 2*x*log(x))/(x + log(x)) + 1/4*integrate(-(x^3 - x ^2 - (log(x) - 1)*e^x + (x^2 - x)*log(x))*e^(x*e^(-x) - x)/(x^2 + 2*x*log( x) + log(x)^2), x)
\[ \int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx=\int { \frac {4 \, {\left (2 \, x + 3\right )} e^{x} \log \left (x\right ) - {\left (x^{3} - x^{2} + {\left (x^{2} - x - e^{x}\right )} \log \left (x\right ) + e^{x}\right )} e^{\left (x e^{\left (-x\right )}\right )} + 4 \, {\left (x^{2} - 3 \, x - 5\right )} e^{x} - 4 \, {\left (e^{x} \log \left (x\right ) - e^{x}\right )} \log \left (2 \, x^{2}\right )}{4 \, {\left (x^{2} e^{x} + 2 \, x e^{x} \log \left (x\right ) + e^{x} \log \left (x\right )^{2}\right )}} \,d x } \] Input:
integrate((((x+exp(x)-x^2)*log(x)-exp(x)-x^3+x^2)*exp(x/exp(x))+(-4*exp(x) *log(x)+4*exp(x))*log(2*x^2)+(8*x+12)*exp(x)*log(x)+(4*x^2-12*x-20)*exp(x) )/(4*exp(x)*log(x)^2+8*x*exp(x)*log(x)+4*exp(x)*x^2),x, algorithm="giac")
Output:
integrate(1/4*(4*(2*x + 3)*e^x*log(x) - (x^3 - x^2 + (x^2 - x - e^x)*log(x ) + e^x)*e^(x*e^(-x)) + 4*(x^2 - 3*x - 5)*e^x - 4*(e^x*log(x) - e^x)*log(2 *x^2))/(x^2*e^x + 2*x*e^x*log(x) + e^x*log(x)^2), x)
Time = 6.95 (sec) , antiderivative size = 105, normalized size of antiderivative = 3.28 \[ \int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx=4\,x+\frac {\frac {x\,\left (3\,x-\ln \left (2\,x^2\right )+2\,\ln \left (x\right )-3\,x^2+5\right )}{x+1}-\frac {x\,\ln \left (x\right )\,\left (6\,x-\ln \left (2\,x^2\right )+2\,\ln \left (x\right )+5\right )}{x+1}}{x+\ln \left (x\right )}+\frac {\ln \left (2\,x^2\right )-2\,\ln \left (x\right )+1}{x+1}+\frac {x\,{\mathrm {e}}^{x\,{\mathrm {e}}^{-x}}}{4\,\left (x+\ln \left (x\right )\right )} \] Input:
int(-(exp(x*exp(-x))*(exp(x) - log(x)*(x + exp(x) - x^2) - x^2 + x^3) + ex p(x)*(12*x - 4*x^2 + 20) - log(2*x^2)*(4*exp(x) - 4*exp(x)*log(x)) - exp(x )*log(x)*(8*x + 12))/(4*x^2*exp(x) + 4*exp(x)*log(x)^2 + 8*x*exp(x)*log(x) ),x)
Output:
4*x + ((x*(3*x - log(2*x^2) + 2*log(x) - 3*x^2 + 5))/(x + 1) - (x*log(x)*( 6*x - log(2*x^2) + 2*log(x) + 5))/(x + 1))/(x + log(x)) + (log(2*x^2) - 2* log(x) + 1)/(x + 1) + (x*exp(x*exp(-x)))/(4*(x + log(x)))
Time = 0.17 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.66 \[ \int \frac {e^x \left (-20-12 x+4 x^2\right )+e^x (12+8 x) \log (x)+e^{e^{-x} x} \left (-e^x+x^2-x^3+\left (e^x+x-x^2\right ) \log (x)\right )+\left (4 e^x-4 e^x \log (x)\right ) \log \left (2 x^2\right )}{4 e^x x^2+8 e^x x \log (x)+4 e^x \log ^2(x)} \, dx=\frac {e^{\frac {x}{e^{x}}} x +4 \,\mathrm {log}\left (2 x^{2}\right ) \mathrm {log}\left (x \right )-8 \mathrm {log}\left (x \right )^{2}-8 \,\mathrm {log}\left (x \right ) x -20 \,\mathrm {log}\left (x \right )+4 x^{2}}{4 \,\mathrm {log}\left (x \right )+4 x} \] Input:
int((((x+exp(x)-x^2)*log(x)-exp(x)-x^3+x^2)*exp(x/exp(x))+(-4*exp(x)*log(x )+4*exp(x))*log(2*x^2)+(8*x+12)*exp(x)*log(x)+(4*x^2-12*x-20)*exp(x))/(4*e xp(x)*log(x)^2+8*x*exp(x)*log(x)+4*exp(x)*x^2),x)
Output:
(e**(x/e**x)*x + 4*log(2*x**2)*log(x) - 8*log(x)**2 - 8*log(x)*x - 20*log( x) + 4*x**2)/(4*(log(x) + x))