\(\int \frac {e^{\frac {1}{16} (80+x^2)} (-32 x+8 x^2) \log (x)+(e^{\frac {1}{16} (80+x^2)} (-32+40 x-8 x^2) \log (-1+x)+e^{\frac {1}{16} (80+x^2)} (-8 x+12 x^2-5 x^3+x^4) \log (-1+x) \log (x)) \log (\log (-1+x))}{(-5 x+5 x^2) \log (-1+x) \log ^2(x)} \, dx\) [934]

Optimal result

Integrand size = 114, antiderivative size = 27 \[ \int \frac {e^{\frac {1}{16} \left (80+x^2\right )} \left (-32 x+8 x^2\right ) \log (x)+\left (e^{\frac {1}{16} \left (80+x^2\right )} \left (-32+40 x-8 x^2\right ) \log (-1+x)+e^{\frac {1}{16} \left (80+x^2\right )} \left (-8 x+12 x^2-5 x^3+x^4\right ) \log (-1+x) \log (x)\right ) \log (\log (-1+x))}{\left (-5 x+5 x^2\right ) \log (-1+x) \log ^2(x)} \, dx=\frac {8 e^{5+\frac {x^2}{16}} (-4+x) \log (\log (-1+x))}{5 \log (x)} \] Output:

8/5*ln(ln(-1+x))*exp(1/16*x^2+5)/ln(x)*(-4+x)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1}{16} \left (80+x^2\right )} \left (-32 x+8 x^2\right ) \log (x)+\left (e^{\frac {1}{16} \left (80+x^2\right )} \left (-32+40 x-8 x^2\right ) \log (-1+x)+e^{\frac {1}{16} \left (80+x^2\right )} \left (-8 x+12 x^2-5 x^3+x^4\right ) \log (-1+x) \log (x)\right ) \log (\log (-1+x))}{\left (-5 x+5 x^2\right ) \log (-1+x) \log ^2(x)} \, dx=\frac {8 e^{5+\frac {x^2}{16}} (-4+x) \log (\log (-1+x))}{5 \log (x)} \] Input:

Integrate[(E^((80 + x^2)/16)*(-32*x + 8*x^2)*Log[x] + (E^((80 + x^2)/16)*( 
-32 + 40*x - 8*x^2)*Log[-1 + x] + E^((80 + x^2)/16)*(-8*x + 12*x^2 - 5*x^3 
 + x^4)*Log[-1 + x]*Log[x])*Log[Log[-1 + x]])/((-5*x + 5*x^2)*Log[-1 + x]* 
Log[x]^2),x]
 

Output:

(8*E^(5 + x^2/16)*(-4 + x)*Log[Log[-1 + x]])/(5*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.

\(\displaystyle \int \frac {e^{\frac {1}{16} \left (x^2+80\right )} \left (8 x^2-32 x\right ) \log (x)+\left (e^{\frac {1}{16} \left (x^2+80\right )} \left (-8 x^2+40 x-32\right ) \log (x-1)+e^{\frac {1}{16} \left (x^2+80\right )} \left (x^4-5 x^3+12 x^2-8 x\right ) \log (x) \log (x-1)\right ) \log (\log (x-1))}{\left (5 x^2-5 x\right ) \log (x-1) \log ^2(x)} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {e^{\frac {1}{16} \left (x^2+80\right )} \left (8 x^2-32 x\right ) \log (x)+\left (e^{\frac {1}{16} \left (x^2+80\right )} \left (-8 x^2+40 x-32\right ) \log (x-1)+e^{\frac {1}{16} \left (x^2+80\right )} \left (x^4-5 x^3+12 x^2-8 x\right ) \log (x) \log (x-1)\right ) \log (\log (x-1))}{x (5 x-5) \log (x-1) \log ^2(x)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {e^{\frac {x^2}{16}+5} \left (x^4 \log (x-1) \log (x) \log (\log (x-1))-5 x^3 \log (x-1) \log (x) \log (\log (x-1))+8 x^2 \log (x)-8 x^2 \log (x-1) \log (\log (x-1))+12 x^2 \log (x-1) \log (x) \log (\log (x-1))-32 x \log (x)+40 x \log (x-1) \log (\log (x-1))-8 x \log (x-1) \log (x) \log (\log (x-1))-32 \log (x-1) \log (\log (x-1))\right )}{5 (x-1) \log (x-1) \log ^2(x)}-\frac {e^{\frac {x^2}{16}+5} \left (x^4 \log (x-1) \log (x) \log (\log (x-1))-5 x^3 \log (x-1) \log (x) \log (\log (x-1))+8 x^2 \log (x)-8 x^2 \log (x-1) \log (\log (x-1))+12 x^2 \log (x-1) \log (x) \log (\log (x-1))-32 x \log (x)+40 x \log (x-1) \log (\log (x-1))-8 x \log (x-1) \log (x) \log (\log (x-1))-32 \log (x-1) \log (\log (x-1))\right )}{5 x \log (x-1) \log ^2(x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {8}{5} \int \frac {e^{\frac {x^2}{16}+5} \log (\log (x-1))}{\log ^2(x)}dx+\frac {32}{5} \int \frac {e^{\frac {x^2}{16}+5} \log (\log (x-1))}{x \log ^2(x)}dx+\frac {8}{5} \int \frac {e^{\frac {x^2}{16}+5}}{\log (x-1) \log (x)}dx-\frac {24}{5} \int \frac {e^{\frac {x^2}{16}+5}}{(x-1) \log (x-1) \log (x)}dx+\frac {8}{5} \int \frac {e^{\frac {x^2}{16}+5} \log (\log (x-1))}{\log (x)}dx-\frac {4}{5} \int \frac {e^{\frac {x^2}{16}+5} x \log (\log (x-1))}{\log (x)}dx+\frac {1}{5} \int \frac {e^{\frac {x^2}{16}+5} x^2 \log (\log (x-1))}{\log (x)}dx\)

Input:

Int[(E^((80 + x^2)/16)*(-32*x + 8*x^2)*Log[x] + (E^((80 + x^2)/16)*(-32 + 
40*x - 8*x^2)*Log[-1 + x] + E^((80 + x^2)/16)*(-8*x + 12*x^2 - 5*x^3 + x^4 
)*Log[-1 + x]*Log[x])*Log[Log[-1 + x]])/((-5*x + 5*x^2)*Log[-1 + x]*Log[x] 
^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 35.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85

method	result	size
risch	\(\frac {8 \ln \left (\ln \left (-1+x \right )\right ) {\mathrm e}^{\frac {x^{2}}{16}+5} \left (x -4\right )}{5 \ln \left (x \right )}\)	\(23\)
parallelrisch	\(\frac {16 \,{\mathrm e}^{\frac {x^{2}}{16}+5} \ln \left (\ln \left (-1+x \right )\right ) x -64 \,{\mathrm e}^{\frac {x^{2}}{16}+5} \ln \left (\ln \left (-1+x \right )\right )}{10 \ln \left (x \right )}\)	\(39\)

Input:

int((((x^4-5*x^3+12*x^2-8*x)*exp(1/16*x^2+5)*ln(-1+x)*ln(x)+(-8*x^2+40*x-3 
2)*exp(1/16*x^2+5)*ln(-1+x))*ln(ln(-1+x))+(8*x^2-32*x)*exp(1/16*x^2+5)*ln( 
x))/(5*x^2-5*x)/ln(-1+x)/ln(x)^2,x,method=_RETURNVERBOSE)
 

Output:

8/5*ln(ln(-1+x))*exp(1/16*x^2+5)/ln(x)*(x-4)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\frac {1}{16} \left (80+x^2\right )} \left (-32 x+8 x^2\right ) \log (x)+\left (e^{\frac {1}{16} \left (80+x^2\right )} \left (-32+40 x-8 x^2\right ) \log (-1+x)+e^{\frac {1}{16} \left (80+x^2\right )} \left (-8 x+12 x^2-5 x^3+x^4\right ) \log (-1+x) \log (x)\right ) \log (\log (-1+x))}{\left (-5 x+5 x^2\right ) \log (-1+x) \log ^2(x)} \, dx=\frac {8 \, {\left (x - 4\right )} e^{\left (\frac {1}{16} \, x^{2} + 5\right )} \log \left (\log \left (x - 1\right )\right )}{5 \, \log \left (x\right )} \] Input:

integrate((((x^4-5*x^3+12*x^2-8*x)*exp(1/16*x^2+5)*log(-1+x)*log(x)+(-8*x^ 
2+40*x-32)*exp(1/16*x^2+5)*log(-1+x))*log(log(-1+x))+(8*x^2-32*x)*exp(1/16 
*x^2+5)*log(x))/(5*x^2-5*x)/log(-1+x)/log(x)^2,x, algorithm="fricas")
 

Output:

8/5*(x - 4)*e^(1/16*x^2 + 5)*log(log(x - 1))/log(x)
 

Sympy [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\frac {1}{16} \left (80+x^2\right )} \left (-32 x+8 x^2\right ) \log (x)+\left (e^{\frac {1}{16} \left (80+x^2\right )} \left (-32+40 x-8 x^2\right ) \log (-1+x)+e^{\frac {1}{16} \left (80+x^2\right )} \left (-8 x+12 x^2-5 x^3+x^4\right ) \log (-1+x) \log (x)\right ) \log (\log (-1+x))}{\left (-5 x+5 x^2\right ) \log (-1+x) \log ^2(x)} \, dx=\frac {\left (8 x \log {\left (\log {\left (x - 1 \right )} \right )} - 32 \log {\left (\log {\left (x - 1 \right )} \right )}\right ) e^{\frac {x^{2}}{16} + 5}}{5 \log {\left (x \right )}} \] Input:

integrate((((x**4-5*x**3+12*x**2-8*x)*exp(1/16*x**2+5)*ln(-1+x)*ln(x)+(-8* 
x**2+40*x-32)*exp(1/16*x**2+5)*ln(-1+x))*ln(ln(-1+x))+(8*x**2-32*x)*exp(1/ 
16*x**2+5)*ln(x))/(5*x**2-5*x)/ln(-1+x)/ln(x)**2,x)
 

Output:

(8*x*log(log(x - 1)) - 32*log(log(x - 1)))*exp(x**2/16 + 5)/(5*log(x))
 

Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {1}{16} \left (80+x^2\right )} \left (-32 x+8 x^2\right ) \log (x)+\left (e^{\frac {1}{16} \left (80+x^2\right )} \left (-32+40 x-8 x^2\right ) \log (-1+x)+e^{\frac {1}{16} \left (80+x^2\right )} \left (-8 x+12 x^2-5 x^3+x^4\right ) \log (-1+x) \log (x)\right ) \log (\log (-1+x))}{\left (-5 x+5 x^2\right ) \log (-1+x) \log ^2(x)} \, dx=\frac {8 \, {\left (x e^{5} - 4 \, e^{5}\right )} e^{\left (\frac {1}{16} \, x^{2}\right )} \log \left (\log \left (x - 1\right )\right )}{5 \, \log \left (x\right )} \] Input:

integrate((((x^4-5*x^3+12*x^2-8*x)*exp(1/16*x^2+5)*log(-1+x)*log(x)+(-8*x^ 
2+40*x-32)*exp(1/16*x^2+5)*log(-1+x))*log(log(-1+x))+(8*x^2-32*x)*exp(1/16 
*x^2+5)*log(x))/(5*x^2-5*x)/log(-1+x)/log(x)^2,x, algorithm="maxima")
 

Output:

8/5*(x*e^5 - 4*e^5)*e^(1/16*x^2)*log(log(x - 1))/log(x)
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {e^{\frac {1}{16} \left (80+x^2\right )} \left (-32 x+8 x^2\right ) \log (x)+\left (e^{\frac {1}{16} \left (80+x^2\right )} \left (-32+40 x-8 x^2\right ) \log (-1+x)+e^{\frac {1}{16} \left (80+x^2\right )} \left (-8 x+12 x^2-5 x^3+x^4\right ) \log (-1+x) \log (x)\right ) \log (\log (-1+x))}{\left (-5 x+5 x^2\right ) \log (-1+x) \log ^2(x)} \, dx=\frac {8 \, {\left (x e^{\left (\frac {1}{16} \, x^{2} + 5\right )} \log \left (\log \left (x - 1\right )\right ) - 4 \, e^{\left (\frac {1}{16} \, x^{2} + 5\right )} \log \left (\log \left (x - 1\right )\right )\right )}}{5 \, \log \left (x\right )} \] Input:

integrate((((x^4-5*x^3+12*x^2-8*x)*exp(1/16*x^2+5)*log(-1+x)*log(x)+(-8*x^ 
2+40*x-32)*exp(1/16*x^2+5)*log(-1+x))*log(log(-1+x))+(8*x^2-32*x)*exp(1/16 
*x^2+5)*log(x))/(5*x^2-5*x)/log(-1+x)/log(x)^2,x, algorithm="giac")
 

Output:

8/5*(x*e^(1/16*x^2 + 5)*log(log(x - 1)) - 4*e^(1/16*x^2 + 5)*log(log(x - 1 
)))/log(x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1}{16} \left (80+x^2\right )} \left (-32 x+8 x^2\right ) \log (x)+\left (e^{\frac {1}{16} \left (80+x^2\right )} \left (-32+40 x-8 x^2\right ) \log (-1+x)+e^{\frac {1}{16} \left (80+x^2\right )} \left (-8 x+12 x^2-5 x^3+x^4\right ) \log (-1+x) \log (x)\right ) \log (\log (-1+x))}{\left (-5 x+5 x^2\right ) \log (-1+x) \log ^2(x)} \, dx=\int \frac {\ln \left (\ln \left (x-1\right )\right )\,\left (\ln \left (x-1\right )\,{\mathrm {e}}^{\frac {x^2}{16}+5}\,\left (8\,x^2-40\,x+32\right )+\ln \left (x-1\right )\,{\mathrm {e}}^{\frac {x^2}{16}+5}\,\ln \left (x\right )\,\left (-x^4+5\,x^3-12\,x^2+8\,x\right )\right )+{\mathrm {e}}^{\frac {x^2}{16}+5}\,\ln \left (x\right )\,\left (32\,x-8\,x^2\right )}{\ln \left (x-1\right )\,{\ln \left (x\right )}^2\,\left (5\,x-5\,x^2\right )} \,d x \] Input:

int((log(log(x - 1))*(log(x - 1)*exp(x^2/16 + 5)*(8*x^2 - 40*x + 32) + log 
(x - 1)*exp(x^2/16 + 5)*log(x)*(8*x - 12*x^2 + 5*x^3 - x^4)) + exp(x^2/16 
+ 5)*log(x)*(32*x - 8*x^2))/(log(x - 1)*log(x)^2*(5*x - 5*x^2)),x)
 

Output:

int((log(log(x - 1))*(log(x - 1)*exp(x^2/16 + 5)*(8*x^2 - 40*x + 32) + log 
(x - 1)*exp(x^2/16 + 5)*log(x)*(8*x - 12*x^2 + 5*x^3 - x^4)) + exp(x^2/16 
+ 5)*log(x)*(32*x - 8*x^2))/(log(x - 1)*log(x)^2*(5*x - 5*x^2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\frac {1}{16} \left (80+x^2\right )} \left (-32 x+8 x^2\right ) \log (x)+\left (e^{\frac {1}{16} \left (80+x^2\right )} \left (-32+40 x-8 x^2\right ) \log (-1+x)+e^{\frac {1}{16} \left (80+x^2\right )} \left (-8 x+12 x^2-5 x^3+x^4\right ) \log (-1+x) \log (x)\right ) \log (\log (-1+x))}{\left (-5 x+5 x^2\right ) \log (-1+x) \log ^2(x)} \, dx=\frac {8 e^{\frac {x^{2}}{16}} \mathrm {log}\left (\mathrm {log}\left (x -1\right )\right ) e^{5} \left (x -4\right )}{5 \,\mathrm {log}\left (x \right )} \] Input:

int((((x^4-5*x^3+12*x^2-8*x)*exp(1/16*x^2+5)*log(-1+x)*log(x)+(-8*x^2+40*x 
-32)*exp(1/16*x^2+5)*log(-1+x))*log(log(-1+x))+(8*x^2-32*x)*exp(1/16*x^2+5 
)*log(x))/(5*x^2-5*x)/log(-1+x)/log(x)^2,x)
 

Output:

(8*e**(x**2/16)*log(log(x - 1))*e**5*(x - 4))/(5*log(x))