\(\int \frac {-40 \log (2)+(x+e^x x) \log ^2(x)}{(40 x \log (2) \log (x)+(x+e^x x+x^2-8 x \log (2)) \log ^2(x)) \log (\frac {40 \log (2)+(1+e^x+x-8 \log (2)) \log (x)}{\log (x)})} \, dx\) [61]

Optimal result

Integrand size = 74, antiderivative size = 20 \[ \int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{\left (40 x \log (2) \log (x)+\left (x+e^x x+x^2-8 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx=\log \left (\log \left (1+e^x+x-\frac {8 \log (2) (-5+\log (x))}{\log (x)}\right )\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{\left (40 x \log (2) \log (x)+\left (x+e^x x+x^2-8 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx=\log \left (\log \left (1+e^x+x-8 \log (2)+\frac {40 \log (2)}{\log (x)}\right )\right ) \] Input:

Integrate[(-40*Log[2] + (x + E^x*x)*Log[x]^2)/((40*x*Log[2]*Log[x] + (x + 
E^x*x + x^2 - 8*x*Log[2])*Log[x]^2)*Log[(40*Log[2] + (1 + E^x + x - 8*Log[ 
2])*Log[x])/Log[x]]),x]
 

Output:

Log[Log[1 + E^x + x - 8*Log[2] + (40*Log[2])/Log[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.

\(\displaystyle \int \frac {\left (e^x x+x\right ) \log ^2(x)-40 \log (2)}{\left (\left (x^2+e^x x+x-8 x \log (2)\right ) \log ^2(x)+40 x \log (2) \log (x)\right ) \log \left (\frac {\left (x+e^x+1-8 \log (2)\right ) \log (x)+40 \log (2)}{\log (x)}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (e^x x+x\right ) \log ^2(x)-40 \log (2)}{x \log (x) \left (e^x \log (x)+x \log (x)+(1-8 \log (2)) \log (x)+40 \log (2)\right ) \log \left (\frac {\left (x+e^x+1-8 \log (2)\right ) \log (x)+40 \log (2)}{\log (x)}\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {-x^2 \log ^2(x)+8 x \log (2) \log ^2(x)-40 x \log (2) \log (x)-40 \log (2)}{x \log (x) \left (e^x \log (x)+x \log (x)+(1-8 \log (2)) \log (x)+40 \log (2)\right ) \log \left (x+e^x+\frac {40 \log (2)}{\log (x)}+1-8 \log (2)\right )}+\frac {1}{\log \left (x+e^x+\frac {40 \log (2)}{\log (x)}+1-8 \log (2)\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \int \frac {1}{\log \left (x+e^x+\frac {40 \log (2)}{\log (x)}-8 \log (2)+1\right )}dx+40 \log (2) \int \frac {1}{\left (-e^x \log (x)-x \log (x)-(1-8 \log (2)) \log (x)-40 \log (2)\right ) \log \left (x+e^x+\frac {40 \log (2)}{\log (x)}-8 \log (2)+1\right )}dx+40 \log (2) \int \frac {1}{x \log (x) \left (-e^x \log (x)-x \log (x)-(1-8 \log (2)) \log (x)-40 \log (2)\right ) \log \left (x+e^x+\frac {40 \log (2)}{\log (x)}-8 \log (2)+1\right )}dx+\int \frac {x \log (x)}{\left (-e^x \log (x)-x \log (x)-(1-8 \log (2)) \log (x)-40 \log (2)\right ) \log \left (x+e^x+\frac {40 \log (2)}{\log (x)}-8 \log (2)+1\right )}dx+8 \log (2) \int \frac {\log (x)}{\left (e^x \log (x)+x \log (x)+(1-8 \log (2)) \log (x)+40 \log (2)\right ) \log \left (x+e^x+\frac {40 \log (2)}{\log (x)}-8 \log (2)+1\right )}dx\)

Input:

Int[(-40*Log[2] + (x + E^x*x)*Log[x]^2)/((40*x*Log[2]*Log[x] + (x + E^x*x 
+ x^2 - 8*x*Log[2])*Log[x]^2)*Log[(40*Log[2] + (1 + E^x + x - 8*Log[2])*Lo 
g[x])/Log[x]]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 21.69 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25

method	result	size
parallelrisch	\(\ln \left (\ln \left (\frac {\left ({\mathrm e}^{x}-8 \ln \left (2\right )+x +1\right ) \ln \left (x \right )+40 \ln \left (2\right )}{\ln \left (x \right )}\right )\right )\)	\(25\)
risch	\(\ln \left (\ln \left (\left (\ln \left (x \right )-5\right ) \ln \left (2\right )-\frac {x \ln \left (x \right )}{8}-\frac {{\mathrm e}^{x} \ln \left (x \right )}{8}-\frac {\ln \left (x \right )}{8}\right )+\frac {i \left (-2 \pi {\operatorname {csgn}\left (\frac {i \left (-\left (\ln \left (x \right )-5\right ) \ln \left (2\right )+\frac {x \ln \left (x \right )}{8}+\frac {{\mathrm e}^{x} \ln \left (x \right )}{8}+\frac {\ln \left (x \right )}{8}\right )}{\ln \left (x \right )}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left (-\left (\ln \left (x \right )-5\right ) \ln \left (2\right )+\frac {x \ln \left (x \right )}{8}+\frac {{\mathrm e}^{x} \ln \left (x \right )}{8}+\frac {\ln \left (x \right )}{8}\right )\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i \left (-\left (\ln \left (x \right )-5\right ) \ln \left (2\right )+\frac {x \ln \left (x \right )}{8}+\frac {{\mathrm e}^{x} \ln \left (x \right )}{8}+\frac {\ln \left (x \right )}{8}\right )}{\ln \left (x \right )}\right )-\pi \,\operatorname {csgn}\left (i \left (-\left (\ln \left (x \right )-5\right ) \ln \left (2\right )+\frac {x \ln \left (x \right )}{8}+\frac {{\mathrm e}^{x} \ln \left (x \right )}{8}+\frac {\ln \left (x \right )}{8}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (-\left (\ln \left (x \right )-5\right ) \ln \left (2\right )+\frac {x \ln \left (x \right )}{8}+\frac {{\mathrm e}^{x} \ln \left (x \right )}{8}+\frac {\ln \left (x \right )}{8}\right )}{\ln \left (x \right )}\right )}^{2}+\pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) {\operatorname {csgn}\left (\frac {i \left (-\left (\ln \left (x \right )-5\right ) \ln \left (2\right )+\frac {x \ln \left (x \right )}{8}+\frac {{\mathrm e}^{x} \ln \left (x \right )}{8}+\frac {\ln \left (x \right )}{8}\right )}{\ln \left (x \right )}\right )}^{2}-\pi {\operatorname {csgn}\left (\frac {i \left (-\left (\ln \left (x \right )-5\right ) \ln \left (2\right )+\frac {x \ln \left (x \right )}{8}+\frac {{\mathrm e}^{x} \ln \left (x \right )}{8}+\frac {\ln \left (x \right )}{8}\right )}{\ln \left (x \right )}\right )}^{3}-6 i \ln \left (2\right )+2 i \ln \left (\ln \left (x \right )\right )+2 \pi \right )}{2}\right )\)	\(299\)

Input:

int(((exp(x)*x+x)*ln(x)^2-40*ln(2))/((exp(x)*x-8*x*ln(2)+x^2+x)*ln(x)^2+40 
*x*ln(2)*ln(x))/ln(((exp(x)-8*ln(2)+x+1)*ln(x)+40*ln(2))/ln(x)),x,method=_ 
RETURNVERBOSE)
 

Output:

ln(ln(((exp(x)-8*ln(2)+x+1)*ln(x)+40*ln(2))/ln(x)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{\left (40 x \log (2) \log (x)+\left (x+e^x x+x^2-8 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx=\log \left (\log \left (\frac {{\left (x + e^{x} - 8 \, \log \left (2\right ) + 1\right )} \log \left (x\right ) + 40 \, \log \left (2\right )}{\log \left (x\right )}\right )\right ) \] Input:

integrate(((exp(x)*x+x)*log(x)^2-40*log(2))/((exp(x)*x-8*x*log(2)+x^2+x)*l 
og(x)^2+40*x*log(2)*log(x))/log(((exp(x)-8*log(2)+x+1)*log(x)+40*log(2))/l 
og(x)),x, algorithm="fricas")
 

Output:

log(log(((x + e^x - 8*log(2) + 1)*log(x) + 40*log(2))/log(x)))
 

Sympy [A] (verification not implemented)

Time = 2.45 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{\left (40 x \log (2) \log (x)+\left (x+e^x x+x^2-8 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx=\log {\left (\log {\left (\frac {\left (x + e^{x} - 8 \log {\left (2 \right )} + 1\right ) \log {\left (x \right )} + 40 \log {\left (2 \right )}}{\log {\left (x \right )}} \right )} \right )} \] Input:

integrate(((exp(x)*x+x)*ln(x)**2-40*ln(2))/((exp(x)*x-8*x*ln(2)+x**2+x)*ln 
(x)**2+40*x*ln(2)*ln(x))/ln(((exp(x)-8*ln(2)+x+1)*ln(x)+40*ln(2))/ln(x)),x 
)
 

Output:

log(log(((x + exp(x) - 8*log(2) + 1)*log(x) + 40*log(2))/log(x)))
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{\left (40 x \log (2) \log (x)+\left (x+e^x x+x^2-8 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx=\log \left (\log \left ({\left (x + e^{x} - 8 \, \log \left (2\right ) + 1\right )} \log \left (x\right ) + 40 \, \log \left (2\right )\right ) - \log \left (\log \left (x\right )\right )\right ) \] Input:

integrate(((exp(x)*x+x)*log(x)^2-40*log(2))/((exp(x)*x-8*x*log(2)+x^2+x)*l 
og(x)^2+40*x*log(2)*log(x))/log(((exp(x)-8*log(2)+x+1)*log(x)+40*log(2))/l 
og(x)),x, algorithm="maxima")
 

Output:

log(log((x + e^x - 8*log(2) + 1)*log(x) + 40*log(2)) - log(log(x)))
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{\left (40 x \log (2) \log (x)+\left (x+e^x x+x^2-8 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx=\log \left (-\log \left (x \log \left (x\right ) + e^{x} \log \left (x\right ) - 8 \, \log \left (2\right ) \log \left (x\right ) + 40 \, \log \left (2\right ) + \log \left (x\right )\right ) + \log \left (\log \left (x\right )\right )\right ) \] Input:

integrate(((exp(x)*x+x)*log(x)^2-40*log(2))/((exp(x)*x-8*x*log(2)+x^2+x)*l 
og(x)^2+40*x*log(2)*log(x))/log(((exp(x)-8*log(2)+x+1)*log(x)+40*log(2))/l 
og(x)),x, algorithm="giac")
 

Output:

log(-log(x*log(x) + e^x*log(x) - 8*log(2)*log(x) + 40*log(2) + log(x)) + l 
og(log(x)))
 

Mupad [B] (verification not implemented)

Time = 3.38 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{\left (40 x \log (2) \log (x)+\left (x+e^x x+x^2-8 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx=\ln \left (\ln \left (\frac {40\,\ln \left (2\right )+\ln \left (x\right )\,\left (x-8\,\ln \left (2\right )+{\mathrm {e}}^x+1\right )}{\ln \left (x\right )}\right )\right ) \] Input:

int(-(40*log(2) - log(x)^2*(x + x*exp(x)))/(log((40*log(2) + log(x)*(x - 8 
*log(2) + exp(x) + 1))/log(x))*(log(x)^2*(x - 8*x*log(2) + x*exp(x) + x^2) 
 + 40*x*log(2)*log(x))),x)
 

Output:

log(log((40*log(2) + log(x)*(x - 8*log(2) + exp(x) + 1))/log(x)))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \frac {-40 \log (2)+\left (x+e^x x\right ) \log ^2(x)}{\left (40 x \log (2) \log (x)+\left (x+e^x x+x^2-8 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {40 \log (2)+\left (1+e^x+x-8 \log (2)\right ) \log (x)}{\log (x)}\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\frac {e^{x} \mathrm {log}\left (x \right )-8 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (2\right )+\mathrm {log}\left (x \right ) x +\mathrm {log}\left (x \right )+40 \,\mathrm {log}\left (2\right )}{\mathrm {log}\left (x \right )}\right )\right ) \] Input:

int(((exp(x)*x+x)*log(x)^2-40*log(2))/((exp(x)*x-8*x*log(2)+x^2+x)*log(x)^ 
2+40*x*log(2)*log(x))/log(((exp(x)-8*log(2)+x+1)*log(x)+40*log(2))/log(x)) 
,x)
 

Output:

log(log((e**x*log(x) - 8*log(x)*log(2) + log(x)*x + log(x) + 40*log(2))/lo 
g(x)))