\(\int \frac {-11552 x^3-4864 x^3 \log (x^2)-512 x^3 \log ^2(x^2)+e^{-\frac {1}{19+4 \log (x^2)}} (-353-152 \log (x^2)-16 \log ^2(x^2))}{361 x^2+152 x^2 \log (x^2)+16 x^2 \log ^2(x^2)} \, dx\) [683]

Optimal result

Integrand size = 86, antiderivative size = 28 \[ \int \frac {-11552 x^3-4864 x^3 \log \left (x^2\right )-512 x^3 \log ^2\left (x^2\right )+e^{-\frac {1}{19+4 \log \left (x^2\right )}} \left (-353-152 \log \left (x^2\right )-16 \log ^2\left (x^2\right )\right )}{361 x^2+152 x^2 \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx=2+\frac {e^{\frac {x}{x-4 x \left (5+\log \left (x^2\right )\right )}}}{x}-16 x^2 \] Output:

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-11552 x^3-4864 x^3 \log \left (x^2\right )-512 x^3 \log ^2\left (x^2\right )+e^{-\frac {1}{19+4 \log \left (x^2\right )}} \left (-353-152 \log \left (x^2\right )-16 \log ^2\left (x^2\right )\right )}{361 x^2+152 x^2 \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx=\frac {e^{-\frac {1}{19+4 \log \left (x^2\right )}}}{x}-16 x^2 \] Input:

Integrate[(-11552*x^3 - 4864*x^3*Log[x^2] - 512*x^3*Log[x^2]^2 + (-353 - 1 
52*Log[x^2] - 16*Log[x^2]^2)/E^(19 + 4*Log[x^2])^(-1))/(361*x^2 + 152*x^2* 
Log[x^2] + 16*x^2*Log[x^2]^2),x]
 

Output:

1/(E^(19 + 4*Log[x^2])^(-1)*x) - 16*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.

\(\displaystyle \int \frac {-11552 x^3+e^{-\frac {1}{4 \log \left (x^2\right )+19}} \left (-16 \log ^2\left (x^2\right )-152 \log \left (x^2\right )-353\right )-512 x^3 \log ^2\left (x^2\right )-4864 x^3 \log \left (x^2\right )}{361 x^2+16 x^2 \log ^2\left (x^2\right )+152 x^2 \log \left (x^2\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-11552 x^3+e^{-\frac {1}{4 \log \left (x^2\right )+19}} \left (-16 \log ^2\left (x^2\right )-152 \log \left (x^2\right )-353\right )-512 x^3 \log ^2\left (x^2\right )-4864 x^3 \log \left (x^2\right )}{x^2 \left (4 \log \left (x^2\right )+19\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {e^{\frac {1}{-4 \log \left (x^2\right )-19}} \left (16 \log ^2\left (x^2\right )+152 \log \left (x^2\right )+353\right )}{x^2 \left (4 \log \left (x^2\right )+19\right )^2}-32 x\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\int \frac {e^{\frac {1}{-4 \log \left (x^2\right )-19}}}{x^2}dx+8 \int \frac {e^{\frac {1}{-4 \log \left (x^2\right )-19}}}{x^2 \left (4 \log \left (x^2\right )+19\right )^2}dx-16 x^2\)

Input:

Int[(-11552*x^3 - 4864*x^3*Log[x^2] - 512*x^3*Log[x^2]^2 + (-353 - 152*Log 
[x^2] - 16*Log[x^2]^2)/E^(19 + 4*Log[x^2])^(-1))/(361*x^2 + 152*x^2*Log[x^ 
2] + 16*x^2*Log[x^2]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86

method	result	size
risch	\(-16 x^{2}+\frac {{\mathrm e}^{-\frac {1}{4 \ln \left (x^{2}\right )+19}}}{x}\)	\(24\)
parallelrisch	\(\frac {-77824 x^{3}+4864 \,{\mathrm e}^{-\frac {1}{4 \ln \left (x^{2}\right )+19}}}{4864 x}\)	\(27\)
default	\(-16 x^{2}+\frac {\left (4 \ln \left (x^{2}\right )-8 \ln \left (x \right )+19\right ) {\mathrm e}^{-\frac {1}{4 \ln \left (x^{2}\right )+19}}+8 \ln \left (x \right ) {\mathrm e}^{-\frac {1}{4 \ln \left (x^{2}\right )+19}}}{x \left (4 \ln \left (x^{2}\right )+19\right )}\)	\(65\)
parts	\(-16 x^{2}+\frac {\left (4 \ln \left (x^{2}\right )-8 \ln \left (x \right )+19\right ) {\mathrm e}^{-\frac {1}{4 \ln \left (x^{2}\right )+19}}+8 \ln \left (x \right ) {\mathrm e}^{-\frac {1}{4 \ln \left (x^{2}\right )+19}}}{x \left (4 \ln \left (x^{2}\right )+19\right )}\)	\(65\)

Input:

int(((-16*ln(x^2)^2-152*ln(x^2)-353)*exp(-1/(4*ln(x^2)+19))-512*x^3*ln(x^2 
)^2-4864*x^3*ln(x^2)-11552*x^3)/(16*x^2*ln(x^2)^2+152*x^2*ln(x^2)+361*x^2) 
,x,method=_RETURNVERBOSE)
 

Output:

-16*x^2+1/x*exp(-1/(4*ln(x^2)+19))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {-11552 x^3-4864 x^3 \log \left (x^2\right )-512 x^3 \log ^2\left (x^2\right )+e^{-\frac {1}{19+4 \log \left (x^2\right )}} \left (-353-152 \log \left (x^2\right )-16 \log ^2\left (x^2\right )\right )}{361 x^2+152 x^2 \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx=-\frac {16 \, x^{3} - e^{\left (-\frac {1}{4 \, \log \left (x^{2}\right ) + 19}\right )}}{x} \] Input:

integrate(((-16*log(x^2)^2-152*log(x^2)-353)*exp(-1/(4*log(x^2)+19))-512*x 
^3*log(x^2)^2-4864*x^3*log(x^2)-11552*x^3)/(16*x^2*log(x^2)^2+152*x^2*log( 
x^2)+361*x^2),x, algorithm="fricas")
 

Output:

-(16*x^3 - e^(-1/(4*log(x^2) + 19)))/x
 

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-11552 x^3-4864 x^3 \log \left (x^2\right )-512 x^3 \log ^2\left (x^2\right )+e^{-\frac {1}{19+4 \log \left (x^2\right )}} \left (-353-152 \log \left (x^2\right )-16 \log ^2\left (x^2\right )\right )}{361 x^2+152 x^2 \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx=- 16 x^{2} + \frac {e^{- \frac {1}{4 \log {\left (x^{2} \right )} + 19}}}{x} \] Input:

integrate(((-16*ln(x**2)**2-152*ln(x**2)-353)*exp(-1/(4*ln(x**2)+19))-512* 
x**3*ln(x**2)**2-4864*x**3*ln(x**2)-11552*x**3)/(16*x**2*ln(x**2)**2+152*x 
**2*ln(x**2)+361*x**2),x)
 

Output:

-16*x**2 + exp(-1/(4*log(x**2) + 19))/x
 

Maxima [F]

\[ \int \frac {-11552 x^3-4864 x^3 \log \left (x^2\right )-512 x^3 \log ^2\left (x^2\right )+e^{-\frac {1}{19+4 \log \left (x^2\right )}} \left (-353-152 \log \left (x^2\right )-16 \log ^2\left (x^2\right )\right )}{361 x^2+152 x^2 \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx=\int { -\frac {512 \, x^{3} \log \left (x^{2}\right )^{2} + 4864 \, x^{3} \log \left (x^{2}\right ) + 11552 \, x^{3} + {\left (16 \, \log \left (x^{2}\right )^{2} + 152 \, \log \left (x^{2}\right ) + 353\right )} e^{\left (-\frac {1}{4 \, \log \left (x^{2}\right ) + 19}\right )}}{16 \, x^{2} \log \left (x^{2}\right )^{2} + 152 \, x^{2} \log \left (x^{2}\right ) + 361 \, x^{2}} \,d x } \] Input:

integrate(((-16*log(x^2)^2-152*log(x^2)-353)*exp(-1/(4*log(x^2)+19))-512*x 
^3*log(x^2)^2-4864*x^3*log(x^2)-11552*x^3)/(16*x^2*log(x^2)^2+152*x^2*log( 
x^2)+361*x^2),x, algorithm="maxima")
 

Output:

-16*x^2 - integrate((64*log(x)^2 + 304*log(x) + 353)*e^(-1/(8*log(x) + 19) 
)/(64*x^2*log(x)^2 + 304*x^2*log(x) + 361*x^2), x)
 

Giac [F]

\[ \int \frac {-11552 x^3-4864 x^3 \log \left (x^2\right )-512 x^3 \log ^2\left (x^2\right )+e^{-\frac {1}{19+4 \log \left (x^2\right )}} \left (-353-152 \log \left (x^2\right )-16 \log ^2\left (x^2\right )\right )}{361 x^2+152 x^2 \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx=\int { -\frac {512 \, x^{3} \log \left (x^{2}\right )^{2} + 4864 \, x^{3} \log \left (x^{2}\right ) + 11552 \, x^{3} + {\left (16 \, \log \left (x^{2}\right )^{2} + 152 \, \log \left (x^{2}\right ) + 353\right )} e^{\left (-\frac {1}{4 \, \log \left (x^{2}\right ) + 19}\right )}}{16 \, x^{2} \log \left (x^{2}\right )^{2} + 152 \, x^{2} \log \left (x^{2}\right ) + 361 \, x^{2}} \,d x } \] Input:

integrate(((-16*log(x^2)^2-152*log(x^2)-353)*exp(-1/(4*log(x^2)+19))-512*x 
^3*log(x^2)^2-4864*x^3*log(x^2)-11552*x^3)/(16*x^2*log(x^2)^2+152*x^2*log( 
x^2)+361*x^2),x, algorithm="giac")
 

Output:

integrate(-(512*x^3*log(x^2)^2 + 4864*x^3*log(x^2) + 11552*x^3 + (16*log(x 
^2)^2 + 152*log(x^2) + 353)*e^(-1/(4*log(x^2) + 19)))/(16*x^2*log(x^2)^2 + 
 152*x^2*log(x^2) + 361*x^2), x)
 

Mupad [B] (verification not implemented)

Time = 4.46 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-11552 x^3-4864 x^3 \log \left (x^2\right )-512 x^3 \log ^2\left (x^2\right )+e^{-\frac {1}{19+4 \log \left (x^2\right )}} \left (-353-152 \log \left (x^2\right )-16 \log ^2\left (x^2\right )\right )}{361 x^2+152 x^2 \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx=\frac {{\mathrm {e}}^{-\frac {1}{\ln \left (x^8\right )+19}}}{x}-16\,x^2 \] Input:

int(-(4864*x^3*log(x^2) + exp(-1/(4*log(x^2) + 19))*(152*log(x^2) + 16*log 
(x^2)^2 + 353) + 11552*x^3 + 512*x^3*log(x^2)^2)/(152*x^2*log(x^2) + 361*x 
^2 + 16*x^2*log(x^2)^2),x)
 

Output:

exp(-1/(log(x^8) + 19))/x - 16*x^2
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {-11552 x^3-4864 x^3 \log \left (x^2\right )-512 x^3 \log ^2\left (x^2\right )+e^{-\frac {1}{19+4 \log \left (x^2\right )}} \left (-353-152 \log \left (x^2\right )-16 \log ^2\left (x^2\right )\right )}{361 x^2+152 x^2 \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx=\frac {-16 e^{\frac {1}{4 \,\mathrm {log}\left (x^{2}\right )+19}} x^{3}+1}{e^{\frac {1}{4 \,\mathrm {log}\left (x^{2}\right )+19}} x} \] Input:

int(((-16*log(x^2)^2-152*log(x^2)-353)*exp(-1/(4*log(x^2)+19))-512*x^3*log 
(x^2)^2-4864*x^3*log(x^2)-11552*x^3)/(16*x^2*log(x^2)^2+152*x^2*log(x^2)+3 
61*x^2),x)
 

Output:

( - 16*e**(1/(4*log(x**2) + 19))*x**3 + 1)/(e**(1/(4*log(x**2) + 19))*x)