\(\int \frac {16+8 x^2+x^4+e^{5+x} (-16 x-8 x^3-x^5)+e^5 (-4 x+x^3) \log (2)+(e^x (-16 x-8 x^3-x^5)+(-4 x+x^3) \log (2)) \log (\frac {x}{4})}{e^5 (256 x+128 x^3+16 x^5)+(256 x+128 x^3+16 x^5) \log (\frac {x}{4})} \, dx\) [2990]

Optimal result

Integrand size = 122, antiderivative size = 35 \[ \int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx=\frac {1}{16} \left (2-e^x-\frac {\log (2)}{\frac {4}{x}+x}+\log \left (e^5+\log \left (\frac {x}{4}\right )\right )\right ) \] Output:

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

Mathematica [A] (verified)

Time = 16.66 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx=\frac {1}{16} \left (1-e^x-\frac {x \log (256)}{8 \left (4+x^2\right )}+\log \left (e^5+\log \left (\frac {x}{4}\right )\right )\right ) \] Input:

Integrate[(16 + 8*x^2 + x^4 + E^(5 + x)*(-16*x - 8*x^3 - x^5) + E^5*(-4*x 
+ x^3)*Log[2] + (E^x*(-16*x - 8*x^3 - x^5) + (-4*x + x^3)*Log[2])*Log[x/4] 
)/(E^5*(256*x + 128*x^3 + 16*x^5) + (256*x + 128*x^3 + 16*x^5)*Log[x/4]),x 
]
 

Output:

(1 - E^x - (x*Log[256])/(8*(4 + x^2)) + Log[E^5 + Log[x/4]])/16
 

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.

\(\displaystyle \int \frac {x^4+e^5 \left (x^3-4 x\right ) \log (2)+8 x^2+e^{x+5} \left (-x^5-8 x^3-16 x\right )+\left (\left (x^3-4 x\right ) \log (2)+e^x \left (-x^5-8 x^3-16 x\right )\right ) \log \left (\frac {x}{4}\right )+16}{e^5 \left (16 x^5+128 x^3+256 x\right )+\left (16 x^5+128 x^3+256 x\right ) \log \left (\frac {x}{4}\right )} \, dx\)

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(16 + 8*x^2 + x^4 + E^(5 + x)*(-16*x - 8*x^3 - x^5) + E^5*(-4*x + x^3) 
*Log[2] + (E^x*(-16*x - 8*x^3 - x^5) + (-4*x + x^3)*Log[2])*Log[x/4])/(E^5 
*(256*x + 128*x^3 + 16*x^5) + (256*x + 128*x^3 + 16*x^5)*Log[x/4]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 10.80 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03

method	result	size
risch	\(-\frac {{\mathrm e}^{x} x^{2}+x \ln \left (2\right )+4 \,{\mathrm e}^{x}}{16 \left (x^{2}+4\right )}+\frac {\ln \left ({\mathrm e}^{5}+\ln \left (\frac {x}{4}\right )\right )}{16}\)	\(36\)
parallelrisch	\(-\frac {{\mathrm e}^{x} x^{2}-\ln \left ({\mathrm e}^{5}+\ln \left (\frac {x}{4}\right )\right ) x^{2}+x \ln \left (2\right )+4 \,{\mathrm e}^{x}-4 \ln \left ({\mathrm e}^{5}+\ln \left (\frac {x}{4}\right )\right )}{16 \left (x^{2}+4\right )}\)	\(48\)

Input:

int((((-x^5-8*x^3-16*x)*exp(x)+(x^3-4*x)*ln(2))*ln(1/4*x)+(-x^5-8*x^3-16*x 
)*exp(5)*exp(x)+(x^3-4*x)*exp(5)*ln(2)+x^4+8*x^2+16)/((16*x^5+128*x^3+256* 
x)*ln(1/4*x)+(16*x^5+128*x^3+256*x)*exp(5)),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.29 \[ \int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx=-\frac {{\left (x e^{5} \log \left (2\right ) - {\left (x^{2} + 4\right )} e^{5} \log \left (e^{5} + \log \left (\frac {1}{4} \, x\right )\right ) + {\left (x^{2} + 4\right )} e^{\left (x + 5\right )}\right )} e^{\left (-5\right )}}{16 \, {\left (x^{2} + 4\right )}} \] Input:

integrate((((-x^5-8*x^3-16*x)*exp(x)+(x^3-4*x)*log(2))*log(1/4*x)+(-x^5-8* 
x^3-16*x)*exp(5)*exp(x)+(x^3-4*x)*exp(5)*log(2)+x^4+8*x^2+16)/((16*x^5+128 
*x^3+256*x)*log(1/4*x)+(16*x^5+128*x^3+256*x)*exp(5)),x, algorithm="fricas 
")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx=- \frac {x \log {\left (2 \right )}}{16 x^{2} + 64} - \frac {e^{x}}{16} + \frac {\log {\left (\log {\left (\frac {x}{4} \right )} + e^{5} \right )}}{16} \] Input:

integrate((((-x**5-8*x**3-16*x)*exp(x)+(x**3-4*x)*ln(2))*ln(1/4*x)+(-x**5- 
8*x**3-16*x)*exp(5)*exp(x)+(x**3-4*x)*exp(5)*ln(2)+x**4+8*x**2+16)/((16*x* 
*5+128*x**3+256*x)*ln(1/4*x)+(16*x**5+128*x**3+256*x)*exp(5)),x)
 

Output:

-x*log(2)/(16*x**2 + 64) - exp(x)/16 + log(log(x/4) + exp(5))/16
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx=-\frac {{\left (x^{2} + 4\right )} e^{x} + x \log \left (2\right )}{16 \, {\left (x^{2} + 4\right )}} + \frac {1}{16} \, \log \left (e^{5} - 2 \, \log \left (2\right ) + \log \left (x\right )\right ) \] Input:

integrate((((-x^5-8*x^3-16*x)*exp(x)+(x^3-4*x)*log(2))*log(1/4*x)+(-x^5-8* 
x^3-16*x)*exp(5)*exp(x)+(x^3-4*x)*exp(5)*log(2)+x^4+8*x^2+16)/((16*x^5+128 
*x^3+256*x)*log(1/4*x)+(16*x^5+128*x^3+256*x)*exp(5)),x, algorithm="maxima 
")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.34 \[ \int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx=-\frac {x^{2} e^{x} - x^{2} \log \left (e^{5} + \log \left (\frac {1}{4} \, x\right )\right ) + x \log \left (2\right ) + 4 \, e^{x} - 4 \, \log \left (e^{5} + \log \left (\frac {1}{4} \, x\right )\right )}{16 \, {\left (x^{2} + 4\right )}} \] Input:

integrate((((-x^5-8*x^3-16*x)*exp(x)+(x^3-4*x)*log(2))*log(1/4*x)+(-x^5-8* 
x^3-16*x)*exp(5)*exp(x)+(x^3-4*x)*exp(5)*log(2)+x^4+8*x^2+16)/((16*x^5+128 
*x^3+256*x)*log(1/4*x)+(16*x^5+128*x^3+256*x)*exp(5)),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 3.84 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx=\frac {\ln \left (\ln \left (\frac {x}{4}\right )+{\mathrm {e}}^5\right )}{16}-\frac {{\mathrm {e}}^x}{16}-\frac {x\,\ln \left (2\right )}{16\,x^2+64} \] Input:

int(-(log(x/4)*(log(2)*(4*x - x^3) + exp(x)*(16*x + 8*x^3 + x^5)) - 8*x^2 
- x^4 + exp(5)*log(2)*(4*x - x^3) + exp(5)*exp(x)*(16*x + 8*x^3 + x^5) - 1 
6)/(log(x/4)*(256*x + 128*x^3 + 16*x^5) + exp(5)*(256*x + 128*x^3 + 16*x^5 
)),x)
 

Output:

log(log(x/4) + exp(5))/16 - exp(x)/16 - (x*log(2))/(16*x^2 + 64)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.51 \[ \int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx=\frac {-e^{x} x^{2}-4 e^{x}+\mathrm {log}\left (\mathrm {log}\left (\frac {x}{4}\right )+e^{5}\right ) x^{2}+4 \,\mathrm {log}\left (\mathrm {log}\left (\frac {x}{4}\right )+e^{5}\right )-\mathrm {log}\left (2\right ) x}{16 x^{2}+64} \] Input:

int((((-x^5-8*x^3-16*x)*exp(x)+(x^3-4*x)*log(2))*log(1/4*x)+(-x^5-8*x^3-16 
*x)*exp(5)*exp(x)+(x^3-4*x)*exp(5)*log(2)+x^4+8*x^2+16)/((16*x^5+128*x^3+2 
56*x)*log(1/4*x)+(16*x^5+128*x^3+256*x)*exp(5)),x)
 

Output:

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