\(\int \frac {-2-13 x+x^3-x^4+(2 x+x^2) \log (x)+(-5 x+x^2-x^3+x \log (x)) \log (e^{-x} (-60+12 x-12 x^2+12 \log (x)))}{(-5 x+x^2-x^3+x \log (x)) \log ^2(e^{-x} (-60+12 x-12 x^2+12 \log (x)))} \, dx\) [2609]

Optimal result

Integrand size = 103, antiderivative size = 24 \[ \int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx=\frac {2+x}{\log \left (12 e^{-x} \left (-5+x-x^2+\log (x)\right )\right )} \] Output:

(2+x)/ln(12*(ln(x)+x-x^2-5)/exp(x))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx=\frac {2+x}{\log \left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \] Input:

Integrate[(-2 - 13*x + x^3 - x^4 + (2*x + x^2)*Log[x] + (-5*x + x^2 - x^3 
+ x*Log[x])*Log[(-60 + 12*x - 12*x^2 + 12*Log[x])/E^x])/((-5*x + x^2 - x^3 
 + x*Log[x])*Log[(-60 + 12*x - 12*x^2 + 12*Log[x])/E^x]^2),x]
 

Output:

(2 + x)/Log[(-12*(5 - x + x^2 - Log[x]))/E^x]
 

Rubi [F]

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.

\(\displaystyle \int \frac {-x^4+x^3+\left (x^2+2 x\right ) \log (x)+\left (-x^3+x^2-5 x+x \log (x)\right ) \log \left (e^{-x} \left (-12 x^2+12 x+12 \log (x)-60\right )\right )-13 x-2}{\left (-x^3+x^2-5 x+x \log (x)\right ) \log ^2\left (e^{-x} \left (-12 x^2+12 x+12 \log (x)-60\right )\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {x^4-x^3-\left (x^2+2 x\right ) \log (x)-\left (-x^3+x^2-5 x+x \log (x)\right ) \log \left (e^{-x} \left (-12 x^2+12 x+12 \log (x)-60\right )\right )+13 x+2}{\left (x^3-x^2+5 x-x \log (x)\right ) \log ^2\left (-12 e^{-x} \left (x^2-x-\log (x)+5\right )\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {x^2}{\left (x^2-x-\log (x)+5\right ) \log ^2\left (-12 e^{-x} \left (x^2-x-\log (x)+5\right )\right )}-\frac {(x+2) \log (x)}{\left (x^2-x-\log (x)+5\right ) \log ^2\left (-12 e^{-x} \left (x^2-x-\log (x)+5\right )\right )}+\frac {13}{\left (x^2-x-\log (x)+5\right ) \log ^2\left (-12 e^{-x} \left (x^2-x-\log (x)+5\right )\right )}+\frac {2}{x \left (x^2-x-\log (x)+5\right ) \log ^2\left (-12 e^{-x} \left (x^2-x-\log (x)+5\right )\right )}+\frac {1}{\log \left (-12 e^{-x} \left (x^2-x-\log (x)+5\right )\right )}+\frac {x^3}{\left (x^2-x-\log (x)+5\right ) \log ^2\left (-12 e^{-x} \left (x^2-x-\log (x)+5\right )\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 13 \int \frac {1}{\left (x^2-x-\log (x)+5\right ) \log ^2\left (-12 e^{-x} \left (x^2-x-\log (x)+5\right )\right )}dx+2 \int \frac {1}{x \left (x^2-x-\log (x)+5\right ) \log ^2\left (-12 e^{-x} \left (x^2-x-\log (x)+5\right )\right )}dx-\int \frac {x^2}{\left (x^2-x-\log (x)+5\right ) \log ^2\left (-12 e^{-x} \left (x^2-x-\log (x)+5\right )\right )}dx-2 \int \frac {\log (x)}{\left (x^2-x-\log (x)+5\right ) \log ^2\left (-12 e^{-x} \left (x^2-x-\log (x)+5\right )\right )}dx-\int \frac {x \log (x)}{\left (x^2-x-\log (x)+5\right ) \log ^2\left (-12 e^{-x} \left (x^2-x-\log (x)+5\right )\right )}dx+\int \frac {1}{\log \left (-12 e^{-x} \left (x^2-x-\log (x)+5\right )\right )}dx+\int \frac {x^3}{\left (x^2-x-\log (x)+5\right ) \log ^2\left (-12 e^{-x} \left (x^2-x-\log (x)+5\right )\right )}dx\)

Input:

Int[(-2 - 13*x + x^3 - x^4 + (2*x + x^2)*Log[x] + (-5*x + x^2 - x^3 + x*Lo 
g[x])*Log[(-60 + 12*x - 12*x^2 + 12*Log[x])/E^x])/((-5*x + x^2 - x^3 + x*L 
og[x])*Log[(-60 + 12*x - 12*x^2 + 12*Log[x])/E^x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.43 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12

method	result	size
parallelrisch	\(-\frac {-4 x -8}{4 \ln \left (12 \left (\ln \left (x \right )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )}\)	\(27\)
risch	\(-\frac {2 i \left (2+x \right )}{-2 \pi {\operatorname {csgn}\left (i \left (\ln \left (x \right )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left (\ln \left (x \right )+x -x^{2}-5\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i \left (\ln \left (x \right )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )-\pi \,\operatorname {csgn}\left (i \left (\ln \left (x \right )+x -x^{2}-5\right )\right ) {\operatorname {csgn}\left (i \left (\ln \left (x \right )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )}^{2}+\pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) {\operatorname {csgn}\left (i \left (\ln \left (x \right )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )}^{2}-\pi {\operatorname {csgn}\left (i \left (\ln \left (x \right )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )}^{3}+2 \pi -4 i \ln \left (2\right )-2 i \ln \left (3\right )-2 i \ln \left (-\ln \left (x \right )-x +x^{2}+5\right )+2 i \ln \left ({\mathrm e}^{x}\right )}\)	\(201\)

Input:

int(((x*ln(x)-x^3+x^2-5*x)*ln((12*ln(x)-12*x^2+12*x-60)/exp(x))+(x^2+2*x)* 
ln(x)-x^4+x^3-13*x-2)/(x*ln(x)-x^3+x^2-5*x)/ln((12*ln(x)-12*x^2+12*x-60)/e 
xp(x))^2,x,method=_RETURNVERBOSE)
 

Output:

-1/4*(-4*x-8)/ln(12*(ln(x)+x-x^2-5)/exp(x))
 

Fricas [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx=\frac {x + 2}{\log \left (-12 \, {\left (x^{2} - x + 5\right )} e^{\left (-x\right )} + 12 \, e^{\left (-x\right )} \log \left (x\right )\right )} \] Input:

integrate(((x*log(x)-x^3+x^2-5*x)*log((12*log(x)-12*x^2+12*x-60)/exp(x))+( 
x^2+2*x)*log(x)-x^4+x^3-13*x-2)/(x*log(x)-x^3+x^2-5*x)/log((12*log(x)-12*x 
^2+12*x-60)/exp(x))^2,x, algorithm="fricas")
 

Output:

(x + 2)/log(-12*(x^2 - x + 5)*e^(-x) + 12*e^(-x)*log(x))
 

Sympy [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx=\frac {x + 2}{\log {\left (\left (- 12 x^{2} + 12 x + 12 \log {\left (x \right )} - 60\right ) e^{- x} \right )}} \] Input:

integrate(((x*ln(x)-x**3+x**2-5*x)*ln((12*ln(x)-12*x**2+12*x-60)/exp(x))+( 
x**2+2*x)*ln(x)-x**4+x**3-13*x-2)/(x*ln(x)-x**3+x**2-5*x)/ln((12*ln(x)-12* 
x**2+12*x-60)/exp(x))**2,x)
 

Output:

(x + 2)/log((-12*x**2 + 12*x + 12*log(x) - 60)*exp(-x))
 

Maxima [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx=-\frac {x + 2}{x - \log \left (3\right ) - 2 \, \log \left (2\right ) - \log \left (-x^{2} + x + \log \left (x\right ) - 5\right )} \] Input:

integrate(((x*log(x)-x^3+x^2-5*x)*log((12*log(x)-12*x^2+12*x-60)/exp(x))+( 
x^2+2*x)*log(x)-x^4+x^3-13*x-2)/(x*log(x)-x^3+x^2-5*x)/log((12*log(x)-12*x 
^2+12*x-60)/exp(x))^2,x, algorithm="maxima")
 

Output:

-(x + 2)/(x - log(3) - 2*log(2) - log(-x^2 + x + log(x) - 5))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx=-\frac {x + 2}{x - \log \left (12\right ) - \log \left (-x^{2} + x + \log \left (x\right ) - 5\right )} \] Input:

integrate(((x*log(x)-x^3+x^2-5*x)*log((12*log(x)-12*x^2+12*x-60)/exp(x))+( 
x^2+2*x)*log(x)-x^4+x^3-13*x-2)/(x*log(x)-x^3+x^2-5*x)/log((12*log(x)-12*x 
^2+12*x-60)/exp(x))^2,x, algorithm="giac")
 

Output:

-(x + 2)/(x - log(12) - log(-x^2 + x + log(x) - 5))
 

Mupad [B] (verification not implemented)

Time = 2.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx=\frac {x+2}{\ln \left ({\mathrm {e}}^{-x}\,\left (12\,x+12\,\ln \left (x\right )-12\,x^2-60\right )\right )} \] Input:

int((13*x + log(exp(-x)*(12*x + 12*log(x) - 12*x^2 - 60))*(5*x - x*log(x) 
- x^2 + x^3) - log(x)*(2*x + x^2) - x^3 + x^4 + 2)/(log(exp(-x)*(12*x + 12 
*log(x) - 12*x^2 - 60))^2*(5*x - x*log(x) - x^2 + x^3)),x)
 

Output:

(x + 2)/log(exp(-x)*(12*x + 12*log(x) - 12*x^2 - 60))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx=\frac {x +2}{\mathrm {log}\left (\frac {12 \,\mathrm {log}\left (x \right )-12 x^{2}+12 x -60}{e^{x}}\right )} \] Input:

int(((x*log(x)-x^3+x^2-5*x)*log((12*log(x)-12*x^2+12*x-60)/exp(x))+(x^2+2* 
x)*log(x)-x^4+x^3-13*x-2)/(x*log(x)-x^3+x^2-5*x)/log((12*log(x)-12*x^2+12* 
x-60)/exp(x))^2,x)
 

Output:

(x + 2)/log((12*log(x) - 12*x**2 + 12*x - 60)/e**x)