\(\int \frac {(1-32 x+256 x^2) \log ^2(x)+e^{\frac {-16 x^3+(-64 x-16 x^2) \log (x)}{(-1+16 x) \log (x)}} (-16 x^2+256 x^3+(48 x^2-512 x^3) \log (x)+(64+32 x-256 x^2) \log ^2(x))}{(1-32 x+256 x^2) \log ^2(x)} \, dx\) [1802]

Optimal result

Integrand size = 106, antiderivative size = 27 \[ \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+e^{\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}} \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx=e^{\frac {4+x+\frac {x^2}{\log (x)}}{-1+\frac {1}{16 x}}}+x \] Output:

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

Mathematica [A] (verified)

Time = 3.17 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+e^{\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}} \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx=e^{-\frac {65}{16}-x-\frac {65}{16 (-1+16 x)}-\frac {16 x^3}{(-1+16 x) \log (x)}}+x \] Input:

Integrate[((1 - 32*x + 256*x^2)*Log[x]^2 + E^((-16*x^3 + (-64*x - 16*x^2)* 
Log[x])/((-1 + 16*x)*Log[x]))*(-16*x^2 + 256*x^3 + (48*x^2 - 512*x^3)*Log[ 
x] + (64 + 32*x - 256*x^2)*Log[x]^2))/((1 - 32*x + 256*x^2)*Log[x]^2),x]
 

Output:

E^(-65/16 - x - 65/(16*(-1 + 16*x)) - (16*x^3)/((-1 + 16*x)*Log[x])) + 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.

\(\displaystyle \int \frac {\left (256 x^3-16 x^2+\left (-256 x^2+32 x+64\right ) \log ^2(x)+\left (48 x^2-512 x^3\right ) \log (x)\right ) \exp \left (\frac {\left (-16 x^2-64 x\right ) \log (x)-16 x^3}{(16 x-1) \log (x)}\right )+\left (256 x^2-32 x+1\right ) \log ^2(x)}{\left (256 x^2-32 x+1\right ) \log ^2(x)} \, dx\)

\(\Big \downarrow \) 7277

\(\displaystyle 1024 \int \frac {\left (256 x^2-32 x+1\right ) \log ^2(x)-16 e^{\frac {16 x^3}{(1-16 x) \log (x)}} x^{\frac {16 x^2+64 x}{(1-16 x) \log (x)}} \left (-16 x^3+x^2-2 \left (-8 x^2+x+2\right ) \log ^2(x)-\left (3 x^2-32 x^3\right ) \log (x)\right )}{1024 (1-16 x)^2 \log ^2(x)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {\left (256 x^2-32 x+1\right ) \log ^2(x)-16 e^{\frac {16 x^3}{(1-16 x) \log (x)}} x^{\frac {16 \left (x^2+4 x\right )}{(1-16 x) \log (x)}} \left (-16 x^3+x^2-2 \left (-8 x^2+x+2\right ) \log ^2(x)-\left (3 x^2-32 x^3\right ) \log (x)\right )}{(1-16 x)^2 \log ^2(x)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {16 \left (16 x^3-32 x^3 \log (x)-x^2-16 x^2 \log ^2(x)+3 x^2 \log (x)+2 x \log ^2(x)+4 \log ^2(x)\right ) \exp \left (-\frac {16 x \left (x^2+x \log (x)+4 \log (x)\right )}{(16 x-1) \log (x)}\right )}{(1-16 x)^2 \log ^2(x)}+1\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{16} \int \frac {\exp \left (-\frac {16 x \left (x^2+\log (x) x+4 \log (x)\right )}{(16 x-1) \log (x)}\right )}{\log ^2(x)}dx+\int \frac {\exp \left (-\frac {16 x \left (x^2+\log (x) x+4 \log (x)\right )}{(16 x-1) \log (x)}\right ) x}{\log ^2(x)}dx+\frac {1}{16} \int \frac {\exp \left (-\frac {16 x \left (x^2+\log (x) x+4 \log (x)\right )}{(16 x-1) \log (x)}\right )}{(16 x-1) \log ^2(x)}dx-\int \exp \left (-\frac {16 x \left (x^2+\log (x) x+4 \log (x)\right )}{(16 x-1) \log (x)}\right )dx+65 \int \frac {\exp \left (-\frac {16 x \left (x^2+\log (x) x+4 \log (x)\right )}{(16 x-1) \log (x)}\right )}{(16 x-1)^2}dx-\frac {1}{16} \int \frac {\exp \left (-\frac {16 x \left (x^2+\log (x) x+4 \log (x)\right )}{(16 x-1) \log (x)}\right )}{\log (x)}dx-2 \int \frac {\exp \left (-\frac {16 x \left (x^2+\log (x) x+4 \log (x)\right )}{(16 x-1) \log (x)}\right ) x}{\log (x)}dx+\frac {1}{16} \int \frac {\exp \left (-\frac {16 x \left (x^2+\log (x) x+4 \log (x)\right )}{(16 x-1) \log (x)}\right )}{(16 x-1)^2 \log (x)}dx+x\)

Input:

Int[((1 - 32*x + 256*x^2)*Log[x]^2 + E^((-16*x^3 + (-64*x - 16*x^2)*Log[x] 
)/((-1 + 16*x)*Log[x]))*(-16*x^2 + 256*x^3 + (48*x^2 - 512*x^3)*Log[x] + ( 
64 + 32*x - 256*x^2)*Log[x]^2))/((1 - 32*x + 256*x^2)*Log[x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11

method	result	size
risch	\(x +{\mathrm e}^{-\frac {16 x \left (x \ln \left (x \right )+x^{2}+4 \ln \left (x \right )\right )}{\left (16 x -1\right ) \ln \left (x \right )}}\)	\(30\)
parallelrisch	\(x +{\mathrm e}^{\frac {\left (-16 x^{2}-64 x \right ) \ln \left (x \right )-16 x^{3}}{\left (16 x -1\right ) \ln \left (x \right )}}+\frac {3}{32}\)	\(35\)

Input:

int((((-256*x^2+32*x+64)*ln(x)^2+(-512*x^3+48*x^2)*ln(x)+256*x^3-16*x^2)*e 
xp(((-16*x^2-64*x)*ln(x)-16*x^3)/(16*x-1)/ln(x))+(256*x^2-32*x+1)*ln(x)^2) 
/(256*x^2-32*x+1)/ln(x)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+e^{\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}} \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx=x + e^{\left (-\frac {16 \, {\left (x^{3} + {\left (x^{2} + 4 \, x\right )} \log \left (x\right )\right )}}{{\left (16 \, x - 1\right )} \log \left (x\right )}\right )} \] Input:

integrate((((-256*x^2+32*x+64)*log(x)^2+(-512*x^3+48*x^2)*log(x)+256*x^3-1 
6*x^2)*exp(((-16*x^2-64*x)*log(x)-16*x^3)/(16*x-1)/log(x))+(256*x^2-32*x+1 
)*log(x)^2)/(256*x^2-32*x+1)/log(x)^2,x, algorithm="fricas")
 

Output:

x + e^(-16*(x^3 + (x^2 + 4*x)*log(x))/((16*x - 1)*log(x)))
 

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+e^{\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}} \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx=x + e^{\frac {- 16 x^{3} + \left (- 16 x^{2} - 64 x\right ) \log {\left (x \right )}}{\left (16 x - 1\right ) \log {\left (x \right )}}} \] Input:

integrate((((-256*x**2+32*x+64)*ln(x)**2+(-512*x**3+48*x**2)*ln(x)+256*x** 
3-16*x**2)*exp(((-16*x**2-64*x)*ln(x)-16*x**3)/(16*x-1)/ln(x))+(256*x**2-3 
2*x+1)*ln(x)**2)/(256*x**2-32*x+1)/ln(x)**2,x)
 

Output:

x + exp((-16*x**3 + (-16*x**2 - 64*x)*log(x))/((16*x - 1)*log(x)))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (23) = 46\).

Time = 0.20 (sec) , antiderivative size = 89, normalized size of antiderivative = 3.30 \[ \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+e^{\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}} \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx={\left (x e^{\left (x + \frac {x^{2}}{\log \left (x\right )} + \frac {x}{16 \, \log \left (x\right )} + \frac {65}{16 \, {\left (16 \, x - 1\right )}} + \frac {1}{256 \, \log \left (x\right )} + \frac {65}{16}\right )} + e^{\left (-\frac {1}{256 \, {\left (16 \, x - 1\right )} \log \left (x\right )}\right )}\right )} e^{\left (-x - \frac {x^{2}}{\log \left (x\right )} - \frac {x}{16 \, \log \left (x\right )} - \frac {65}{16 \, {\left (16 \, x - 1\right )}} - \frac {1}{256 \, \log \left (x\right )} - \frac {65}{16}\right )} \] Input:

integrate((((-256*x^2+32*x+64)*log(x)^2+(-512*x^3+48*x^2)*log(x)+256*x^3-1 
6*x^2)*exp(((-16*x^2-64*x)*log(x)-16*x^3)/(16*x-1)/log(x))+(256*x^2-32*x+1 
)*log(x)^2)/(256*x^2-32*x+1)/log(x)^2,x, algorithm="maxima")
 

Output:

(x*e^(x + x^2/log(x) + 1/16*x/log(x) + 65/16/(16*x - 1) + 1/256/log(x) + 6 
5/16) + e^(-1/256/((16*x - 1)*log(x))))*e^(-x - x^2/log(x) - 1/16*x/log(x) 
 - 65/16/(16*x - 1) - 1/256/log(x) - 65/16)
 

Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+e^{\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}} \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx=x + e^{\left (-\frac {16 \, {\left (x^{3} + x^{2} \log \left (x\right ) + 4 \, x \log \left (x\right )\right )}}{16 \, x \log \left (x\right ) - \log \left (x\right )}\right )} \] Input:

integrate((((-256*x^2+32*x+64)*log(x)^2+(-512*x^3+48*x^2)*log(x)+256*x^3-1 
6*x^2)*exp(((-16*x^2-64*x)*log(x)-16*x^3)/(16*x-1)/log(x))+(256*x^2-32*x+1 
)*log(x)^2)/(256*x^2-32*x+1)/log(x)^2,x, algorithm="giac")
 

Output:

x + e^(-16*(x^3 + x^2*log(x) + 4*x*log(x))/(16*x*log(x) - log(x)))
 

Mupad [B] (verification not implemented)

Time = 3.36 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+e^{\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}} \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx=x+{\mathrm {e}}^{\frac {64\,x\,\ln \left (x\right )}{\ln \left (x\right )-16\,x\,\ln \left (x\right )}}\,{\mathrm {e}}^{\frac {16\,x^2\,\ln \left (x\right )}{\ln \left (x\right )-16\,x\,\ln \left (x\right )}}\,{\mathrm {e}}^{\frac {16\,x^3}{\ln \left (x\right )-16\,x\,\ln \left (x\right )}} \] Input:

int((log(x)^2*(256*x^2 - 32*x + 1) + exp(-(log(x)*(64*x + 16*x^2) + 16*x^3 
)/(log(x)*(16*x - 1)))*(log(x)^2*(32*x - 256*x^2 + 64) + log(x)*(48*x^2 - 
512*x^3) - 16*x^2 + 256*x^3))/(log(x)^2*(256*x^2 - 32*x + 1)),x)
 

Output:

x + exp((64*x*log(x))/(log(x) - 16*x*log(x)))*exp((16*x^2*log(x))/(log(x) 
- 16*x*log(x)))*exp((16*x^3)/(log(x) - 16*x*log(x)))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.85 \[ \int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+e^{\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}} \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx=\frac {e^{\frac {16 \,\mathrm {log}\left (x \right ) x^{2}+4 \,\mathrm {log}\left (x \right )+16 x^{3}}{16 \,\mathrm {log}\left (x \right ) x -\mathrm {log}\left (x \right )}} e^{4} x +1}{e^{\frac {16 \,\mathrm {log}\left (x \right ) x^{2}+4 \,\mathrm {log}\left (x \right )+16 x^{3}}{16 \,\mathrm {log}\left (x \right ) x -\mathrm {log}\left (x \right )}} e^{4}} \] Input:

int((((-256*x^2+32*x+64)*log(x)^2+(-512*x^3+48*x^2)*log(x)+256*x^3-16*x^2) 
*exp(((-16*x^2-64*x)*log(x)-16*x^3)/(16*x-1)/log(x))+(256*x^2-32*x+1)*log( 
x)^2)/(256*x^2-32*x+1)/log(x)^2,x)
 

Output:

(e**((16*log(x)*x**2 + 4*log(x) + 16*x**3)/(16*log(x)*x - log(x)))*e**4*x 
+ 1)/(e**((16*log(x)*x**2 + 4*log(x) + 16*x**3)/(16*log(x)*x - log(x)))*e* 
*4)