\(\int \frac {(120+10 e^x-2 x+10 x^2+(2 x-10 e^x x-20 x^2) \log (x)+40 x \log ^2(x)) \log (\frac {-60-5 e^x+x-5 x^2+(120+20 x) \log (x)}{20 \log (x)})}{(-60 x-5 e^x x+x^2-5 x^3) \log (x)+(120 x+20 x^2) \log ^2(x)} \, dx\) [1141]

Optimal result

Integrand size = 109, antiderivative size = 32 \[ \int \frac {\left (120+10 e^x-2 x+10 x^2+\left (2 x-10 e^x x-20 x^2\right ) \log (x)+40 x \log ^2(x)\right ) \log \left (\frac {-60-5 e^x+x-5 x^2+(120+20 x) \log (x)}{20 \log (x)}\right )}{\left (-60 x-5 e^x x+x^2-5 x^3\right ) \log (x)+\left (120 x+20 x^2\right ) \log ^2(x)} \, dx=-1+\log ^2\left (6+x-\frac {3+\frac {1}{4} \left (e^x-\frac {x}{5}+x^2\right )}{\log (x)}\right ) \] Output:

ln(6-(3+1/4*x^2-1/20*x+1/4*exp(x))/ln(x)+x)^2-1
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {\left (120+10 e^x-2 x+10 x^2+\left (2 x-10 e^x x-20 x^2\right ) \log (x)+40 x \log ^2(x)\right ) \log \left (\frac {-60-5 e^x+x-5 x^2+(120+20 x) \log (x)}{20 \log (x)}\right )}{\left (-60 x-5 e^x x+x^2-5 x^3\right ) \log (x)+\left (120 x+20 x^2\right ) \log ^2(x)} \, dx=\log ^2\left (\frac {-60-5 e^x+x-5 x^2+20 (6+x) \log (x)}{20 \log (x)}\right ) \] Input:

Integrate[((120 + 10*E^x - 2*x + 10*x^2 + (2*x - 10*E^x*x - 20*x^2)*Log[x] 
 + 40*x*Log[x]^2)*Log[(-60 - 5*E^x + x - 5*x^2 + (120 + 20*x)*Log[x])/(20* 
Log[x])])/((-60*x - 5*E^x*x + x^2 - 5*x^3)*Log[x] + (120*x + 20*x^2)*Log[x 
]^2),x]
 

Output:

Log[(-60 - 5*E^x + x - 5*x^2 + 20*(6 + x)*Log[x])/(20*Log[x])]^2
 

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.

\(\displaystyle \int \frac {\left (10 x^2+\left (-20 x^2-10 e^x x+2 x\right ) \log (x)-2 x+10 e^x+40 x \log ^2(x)+120\right ) \log \left (\frac {-5 x^2+x-5 e^x+(20 x+120) \log (x)-60}{20 \log (x)}\right )}{\left (20 x^2+120 x\right ) \log ^2(x)+\left (-5 x^3+x^2-5 e^x x-60 x\right ) \log (x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {2 \left (-5 x^2+10 x^2 \log (x)+x-5 e^x-20 x \log ^2(x)+5 e^x x \log (x)-x \log (x)-60\right ) \log \left (\frac {-5 x^2+x-5 e^x+(20 x+120) \log (x)-60}{20 \log (x)}\right )}{x \log (x) \left (5 x^2-x+5 e^x-20 x \log (x)-120 \log (x)+60\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \int -\frac {\left (-10 \log (x) x^2+5 x^2+20 \log ^2(x) x-5 e^x \log (x) x+\log (x) x-x+5 e^x+60\right ) \log \left (-\frac {5 x^2-x+5 e^x-20 (x+6) \log (x)+60}{20 \log (x)}\right )}{x \log (x) \left (5 x^2-20 \log (x) x-x+5 e^x-120 \log (x)+60\right )}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -2 \int \frac {\left (-10 \log (x) x^2+5 x^2+20 \log ^2(x) x-5 e^x \log (x) x+\log (x) x-x+5 e^x+60\right ) \log \left (-\frac {5 x^2-x+5 e^x-20 (x+6) \log (x)+60}{20 \log (x)}\right )}{x \log (x) \left (5 x^2-20 \log (x) x-x+5 e^x-120 \log (x)+60\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -2 \int \left (\frac {\left (5 x^3-20 \log (x) x^2-11 x^2-100 \log (x) x+81 x+120\right ) \log \left (\frac {-5 x^2+x-5 e^x+20 (x+6) \log (x)-60}{20 \log (x)}\right )}{x \left (5 x^2-20 \log (x) x-x+5 e^x-120 \log (x)+60\right )}-\frac {(x \log (x)-1) \log \left (\frac {-5 x^2+x-5 e^x+20 (x+6) \log (x)-60}{20 \log (x)}\right )}{x \log (x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 \left (-120 \int \frac {1}{5 x^2-20 \log (x) x-x+5 e^x-120 \log (x)+60}dx-81 \int \frac {x}{5 x^2-20 \log (x) x-x+5 e^x-120 \log (x)+60}dx+11 \int \frac {x^2}{5 x^2-20 \log (x) x-x+5 e^x-120 \log (x)+60}dx+100 \int \frac {x \log (x)}{5 x^2-20 \log (x) x-x+5 e^x-120 \log (x)+60}dx+20 \int \frac {x^2 \log (x)}{5 x^2-20 \log (x) x-x+5 e^x-120 \log (x)+60}dx+\int \frac {\log \left (\frac {-5 x^2+x-5 e^x+20 (x+6) \log (x)-60}{20 \log (x)}\right )}{x \log (x)}dx+81 \int \frac {\log \left (\frac {-5 x^2+x-5 e^x+20 (x+6) \log (x)-60}{20 \log (x)}\right )}{5 x^2-20 \log (x) x-x+5 e^x-120 \log (x)+60}dx+120 \int \frac {\log \left (\frac {-5 x^2+x-5 e^x+20 (x+6) \log (x)-60}{20 \log (x)}\right )}{x \left (5 x^2-20 \log (x) x-x+5 e^x-120 \log (x)+60\right )}dx-11 \int \frac {x \log \left (\frac {-5 x^2+x-5 e^x+20 (x+6) \log (x)-60}{20 \log (x)}\right )}{5 x^2-20 \log (x) x-x+5 e^x-120 \log (x)+60}dx+5 \int \frac {x^2 \log \left (\frac {-5 x^2+x-5 e^x+20 (x+6) \log (x)-60}{20 \log (x)}\right )}{5 x^2-20 \log (x) x-x+5 e^x-120 \log (x)+60}dx-20 \int \frac {x \log (x) \log \left (\frac {-5 x^2+x-5 e^x+20 (x+6) \log (x)-60}{20 \log (x)}\right )}{5 x^2-20 \log (x) x-x+5 e^x-120 \log (x)+60}dx+100 \int \frac {\log (x) \log \left (\frac {-5 x^2+x-5 e^x+20 (x+6) \log (x)-60}{20 \log (x)}\right )}{-5 x^2+20 \log (x) x+x-5 e^x+120 \log (x)-60}dx-5 \int \frac {x^3}{5 x^2-20 \log (x) x-x+5 e^x-120 \log (x)+60}dx-\operatorname {LogIntegral}(x)+\frac {x^2}{2}-x \log \left (-\frac {5 x^2-x+5 e^x-20 (x+6) \log (x)+60}{20 \log (x)}\right )\right )\)

Input:

Int[((120 + 10*E^x - 2*x + 10*x^2 + (2*x - 10*E^x*x - 20*x^2)*Log[x] + 40* 
x*Log[x]^2)*Log[(-60 - 5*E^x + x - 5*x^2 + (120 + 20*x)*Log[x])/(20*Log[x] 
)])/((-60*x - 5*E^x*x + x^2 - 5*x^3)*Log[x] + (120*x + 20*x^2)*Log[x]^2),x 
]
 

Output:

$Aborted
 

Maple [F]

\[\int \frac {\left (40 x \ln \left (x \right )^{2}+\left (-10 \,{\mathrm e}^{x} x -20 x^{2}+2 x \right ) \ln \left (x \right )+10 \,{\mathrm e}^{x}+10 x^{2}-2 x +120\right ) \ln \left (\frac {\left (20 x +120\right ) \ln \left (x \right )-5 \,{\mathrm e}^{x}-5 x^{2}+x -60}{20 \ln \left (x \right )}\right )}{\left (20 x^{2}+120 x \right ) \ln \left (x \right )^{2}+\left (-5 \,{\mathrm e}^{x} x -5 x^{3}+x^{2}-60 x \right ) \ln \left (x \right )}d x\]

Input:

int((40*x*ln(x)^2+(-10*exp(x)*x-20*x^2+2*x)*ln(x)+10*exp(x)+10*x^2-2*x+120 
)*ln(1/20*((20*x+120)*ln(x)-5*exp(x)-5*x^2+x-60)/ln(x))/((20*x^2+120*x)*ln 
(x)^2+(-5*exp(x)*x-5*x^3+x^2-60*x)*ln(x)),x)
 

Output:

int((40*x*ln(x)^2+(-10*exp(x)*x-20*x^2+2*x)*ln(x)+10*exp(x)+10*x^2-2*x+120 
)*ln(1/20*((20*x+120)*ln(x)-5*exp(x)-5*x^2+x-60)/ln(x))/((20*x^2+120*x)*ln 
(x)^2+(-5*exp(x)*x-5*x^3+x^2-60*x)*ln(x)),x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {\left (120+10 e^x-2 x+10 x^2+\left (2 x-10 e^x x-20 x^2\right ) \log (x)+40 x \log ^2(x)\right ) \log \left (\frac {-60-5 e^x+x-5 x^2+(120+20 x) \log (x)}{20 \log (x)}\right )}{\left (-60 x-5 e^x x+x^2-5 x^3\right ) \log (x)+\left (120 x+20 x^2\right ) \log ^2(x)} \, dx=\log \left (-\frac {5 \, x^{2} - 20 \, {\left (x + 6\right )} \log \left (x\right ) - x + 5 \, e^{x} + 60}{20 \, \log \left (x\right )}\right )^{2} \] Input:

integrate((40*x*log(x)^2+(-10*exp(x)*x-20*x^2+2*x)*log(x)+10*exp(x)+10*x^2 
-2*x+120)*log(1/20*((20*x+120)*log(x)-5*exp(x)-5*x^2+x-60)/log(x))/((20*x^ 
2+120*x)*log(x)^2+(-5*exp(x)*x-5*x^3+x^2-60*x)*log(x)),x, algorithm="frica 
s")
 

Output:

log(-1/20*(5*x^2 - 20*(x + 6)*log(x) - x + 5*e^x + 60)/log(x))^2
 

Sympy [A] (verification not implemented)

Time = 4.00 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {\left (120+10 e^x-2 x+10 x^2+\left (2 x-10 e^x x-20 x^2\right ) \log (x)+40 x \log ^2(x)\right ) \log \left (\frac {-60-5 e^x+x-5 x^2+(120+20 x) \log (x)}{20 \log (x)}\right )}{\left (-60 x-5 e^x x+x^2-5 x^3\right ) \log (x)+\left (120 x+20 x^2\right ) \log ^2(x)} \, dx=\log {\left (\frac {- \frac {x^{2}}{4} + \frac {x}{20} + \frac {\left (20 x + 120\right ) \log {\left (x \right )}}{20} - \frac {e^{x}}{4} - 3}{\log {\left (x \right )}} \right )}^{2} \] Input:

integrate((40*x*ln(x)**2+(-10*exp(x)*x-20*x**2+2*x)*ln(x)+10*exp(x)+10*x** 
2-2*x+120)*ln(1/20*((20*x+120)*ln(x)-5*exp(x)-5*x**2+x-60)/ln(x))/((20*x** 
2+120*x)*ln(x)**2+(-5*exp(x)*x-5*x**3+x**2-60*x)*ln(x)),x)
 

Output:

log((-x**2/4 + x/20 + (20*x + 120)*log(x)/20 - exp(x)/4 - 3)/log(x))**2
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (28) = 56\).

Time = 0.19 (sec) , antiderivative size = 125, normalized size of antiderivative = 3.91 \[ \int \frac {\left (120+10 e^x-2 x+10 x^2+\left (2 x-10 e^x x-20 x^2\right ) \log (x)+40 x \log ^2(x)\right ) \log \left (\frac {-60-5 e^x+x-5 x^2+(120+20 x) \log (x)}{20 \log (x)}\right )}{\left (-60 x-5 e^x x+x^2-5 x^3\right ) \log (x)+\left (120 x+20 x^2\right ) \log ^2(x)} \, dx=2 \, {\left (\log \left (5\right ) + \log \left (\log \left (x\right )\right )\right )} \log \left (5 \, x^{2} - 20 \, {\left (x + 6\right )} \log \left (x\right ) - x + 5 \, e^{x} + 60\right ) - \log \left (5 \, x^{2} - 20 \, {\left (x + 6\right )} \log \left (x\right ) - x + 5 \, e^{x} + 60\right )^{2} + 2 \, {\left (\log \left (x^{2} - 4 \, {\left (x + 6\right )} \log \left (x\right ) - \frac {1}{5} \, x + e^{x} + 12\right ) - \log \left (\log \left (x\right )\right )\right )} \log \left (-\frac {5 \, x^{2} - 20 \, {\left (x + 6\right )} \log \left (x\right ) - x + 5 \, e^{x} + 60}{20 \, \log \left (x\right )}\right ) - 2 \, \log \left (5\right ) \log \left (\log \left (x\right )\right ) - \log \left (\log \left (x\right )\right )^{2} \] Input:

integrate((40*x*log(x)^2+(-10*exp(x)*x-20*x^2+2*x)*log(x)+10*exp(x)+10*x^2 
-2*x+120)*log(1/20*((20*x+120)*log(x)-5*exp(x)-5*x^2+x-60)/log(x))/((20*x^ 
2+120*x)*log(x)^2+(-5*exp(x)*x-5*x^3+x^2-60*x)*log(x)),x, algorithm="maxim 
a")
                                                                                    
                                                                                    
 

Output:

2*(log(5) + log(log(x)))*log(5*x^2 - 20*(x + 6)*log(x) - x + 5*e^x + 60) - 
 log(5*x^2 - 20*(x + 6)*log(x) - x + 5*e^x + 60)^2 + 2*(log(x^2 - 4*(x + 6 
)*log(x) - 1/5*x + e^x + 12) - log(log(x)))*log(-1/20*(5*x^2 - 20*(x + 6)* 
log(x) - x + 5*e^x + 60)/log(x)) - 2*log(5)*log(log(x)) - log(log(x))^2
 

Giac [F]

\[ \int \frac {\left (120+10 e^x-2 x+10 x^2+\left (2 x-10 e^x x-20 x^2\right ) \log (x)+40 x \log ^2(x)\right ) \log \left (\frac {-60-5 e^x+x-5 x^2+(120+20 x) \log (x)}{20 \log (x)}\right )}{\left (-60 x-5 e^x x+x^2-5 x^3\right ) \log (x)+\left (120 x+20 x^2\right ) \log ^2(x)} \, dx=\int { \frac {2 \, {\left (20 \, x \log \left (x\right )^{2} + 5 \, x^{2} - {\left (10 \, x^{2} + 5 \, x e^{x} - x\right )} \log \left (x\right ) - x + 5 \, e^{x} + 60\right )} \log \left (-\frac {5 \, x^{2} - 20 \, {\left (x + 6\right )} \log \left (x\right ) - x + 5 \, e^{x} + 60}{20 \, \log \left (x\right )}\right )}{20 \, {\left (x^{2} + 6 \, x\right )} \log \left (x\right )^{2} - {\left (5 \, x^{3} - x^{2} + 5 \, x e^{x} + 60 \, x\right )} \log \left (x\right )} \,d x } \] Input:

integrate((40*x*log(x)^2+(-10*exp(x)*x-20*x^2+2*x)*log(x)+10*exp(x)+10*x^2 
-2*x+120)*log(1/20*((20*x+120)*log(x)-5*exp(x)-5*x^2+x-60)/log(x))/((20*x^ 
2+120*x)*log(x)^2+(-5*exp(x)*x-5*x^3+x^2-60*x)*log(x)),x, algorithm="giac" 
)
 

Output:

integrate(2*(20*x*log(x)^2 + 5*x^2 - (10*x^2 + 5*x*e^x - x)*log(x) - x + 5 
*e^x + 60)*log(-1/20*(5*x^2 - 20*(x + 6)*log(x) - x + 5*e^x + 60)/log(x))/ 
(20*(x^2 + 6*x)*log(x)^2 - (5*x^3 - x^2 + 5*x*e^x + 60*x)*log(x)), x)
 

Mupad [B] (verification not implemented)

Time = 1.91 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {\left (120+10 e^x-2 x+10 x^2+\left (2 x-10 e^x x-20 x^2\right ) \log (x)+40 x \log ^2(x)\right ) \log \left (\frac {-60-5 e^x+x-5 x^2+(120+20 x) \log (x)}{20 \log (x)}\right )}{\left (-60 x-5 e^x x+x^2-5 x^3\right ) \log (x)+\left (120 x+20 x^2\right ) \log ^2(x)} \, dx={\ln \left (-\frac {\frac {{\mathrm {e}}^x}{4}-\frac {x}{20}-\frac {\ln \left (x\right )\,\left (20\,x+120\right )}{20}+\frac {x^2}{4}+3}{\ln \left (x\right )}\right )}^2 \] Input:

int((log(-(exp(x)/4 - x/20 - (log(x)*(20*x + 120))/20 + x^2/4 + 3)/log(x)) 
*(10*exp(x) - 2*x + 40*x*log(x)^2 - log(x)*(10*x*exp(x) - 2*x + 20*x^2) + 
10*x^2 + 120))/(log(x)^2*(120*x + 20*x^2) - log(x)*(60*x + 5*x*exp(x) - x^ 
2 + 5*x^3)),x)
 

Output:

log(-(exp(x)/4 - x/20 - (log(x)*(20*x + 120))/20 + x^2/4 + 3)/log(x))^2
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {\left (120+10 e^x-2 x+10 x^2+\left (2 x-10 e^x x-20 x^2\right ) \log (x)+40 x \log ^2(x)\right ) \log \left (\frac {-60-5 e^x+x-5 x^2+(120+20 x) \log (x)}{20 \log (x)}\right )}{\left (-60 x-5 e^x x+x^2-5 x^3\right ) \log (x)+\left (120 x+20 x^2\right ) \log ^2(x)} \, dx=\mathrm {log}\left (\frac {-5 e^{x}+20 \,\mathrm {log}\left (x \right ) x +120 \,\mathrm {log}\left (x \right )-5 x^{2}+x -60}{20 \,\mathrm {log}\left (x \right )}\right )^{2} \] Input:

int((40*x*log(x)^2+(-10*exp(x)*x-20*x^2+2*x)*log(x)+10*exp(x)+10*x^2-2*x+1 
20)*log(1/20*((20*x+120)*log(x)-5*exp(x)-5*x^2+x-60)/log(x))/((20*x^2+120* 
x)*log(x)^2+(-5*exp(x)*x-5*x^3+x^2-60*x)*log(x)),x)
 

Output:

log(( - 5*e**x + 20*log(x)*x + 120*log(x) - 5*x**2 + x - 60)/(20*log(x)))* 
*2