Integrand size = 109, antiderivative size = 30 \[ \int \frac {e^{\frac {3-e^x-5 x+x^2}{x}} \left (6-5 x-x^2-x^3+e^x \left (-2+x+x^2\right )+\left (-12+e^x (4-4 x)+4 x^2\right ) \log \left (x^2\right )\right )}{4 x^2+4 x^3+x^4+\left (-16 x^2-8 x^3\right ) \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx=\frac {e^{-5+\frac {3-e^x}{x}+x}}{-2-x+4 \log \left (x^2\right )} \] Output:
exp((-exp(x)+3)/x-5+x)/(4*ln(x^2)-2-x)
Time = 5.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\frac {3-e^x-5 x+x^2}{x}} \left (6-5 x-x^2-x^3+e^x \left (-2+x+x^2\right )+\left (-12+e^x (4-4 x)+4 x^2\right ) \log \left (x^2\right )\right )}{4 x^2+4 x^3+x^4+\left (-16 x^2-8 x^3\right ) \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx=\frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{-2-x+4 \log \left (x^2\right )} \] Input:
Integrate[(E^((3 - E^x - 5*x + x^2)/x)*(6 - 5*x - x^2 - x^3 + E^x*(-2 + x + x^2) + (-12 + E^x*(4 - 4*x) + 4*x^2)*Log[x^2]))/(4*x^2 + 4*x^3 + x^4 + ( -16*x^2 - 8*x^3)*Log[x^2] + 16*x^2*Log[x^2]^2),x]
Output:
E^(-5 + 3/x - E^x/x + x)/(-2 - x + 4*Log[x^2])
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.
\(\displaystyle \int \frac {e^{\frac {x^2-5 x-e^x+3}{x}} \left (-x^3-x^2+e^x \left (x^2+x-2\right )+\left (4 x^2+e^x (4-4 x)-12\right ) \log \left (x^2\right )-5 x+6\right )}{x^4+4 x^3+4 x^2+16 x^2 \log ^2\left (x^2\right )+\left (-8 x^3-16 x^2\right ) \log \left (x^2\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{x-\frac {e^x}{x}+\frac {3}{x}-5} \left (-x^3-x^2+e^x \left (x^2+x-2\right )+\left (4 x^2+e^x (4-4 x)-12\right ) \log \left (x^2\right )-5 x+6\right )}{x^2 \left (-4 \log \left (x^2\right )+x+2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{2 x-\frac {e^x}{x}+\frac {3}{x}-5} (x-1)}{x^2 \left (-4 \log \left (x^2\right )+x+2\right )}+\frac {4 e^{x-\frac {e^x}{x}+\frac {3}{x}-5} \log \left (x^2\right )}{\left (-4 \log \left (x^2\right )+x+2\right )^2}-\frac {12 e^{x-\frac {e^x}{x}+\frac {3}{x}-5} \log \left (x^2\right )}{x^2 \left (-4 \log \left (x^2\right )+x+2\right )^2}-\frac {e^{x-\frac {e^x}{x}+\frac {3}{x}-5}}{\left (-4 \log \left (x^2\right )+x+2\right )^2}-\frac {e^{x-\frac {e^x}{x}+\frac {3}{x}-5} x}{\left (-4 \log \left (x^2\right )+x+2\right )^2}-\frac {5 e^{x-\frac {e^x}{x}+\frac {3}{x}-5}}{x \left (-4 \log \left (x^2\right )+x+2\right )^2}+\frac {6 e^{x-\frac {e^x}{x}+\frac {3}{x}-5}}{x^2 \left (-4 \log \left (x^2\right )+x+2\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {e^{x-5-\frac {e^x}{x}+\frac {3}{x}}}{\left (x-4 \log \left (x^2\right )+2\right )^2}dx-8 \int \frac {e^{x-5-\frac {e^x}{x}+\frac {3}{x}}}{x \left (x-4 \log \left (x^2\right )+2\right )^2}dx-\int \frac {e^{x-5-\frac {e^x}{x}+\frac {3}{x}}}{x-4 \log \left (x^2\right )+2}dx+3 \int \frac {e^{x-5-\frac {e^x}{x}+\frac {3}{x}}}{x^2 \left (x-4 \log \left (x^2\right )+2\right )}dx-\int \frac {e^{2 x-5-\frac {e^x}{x}+\frac {3}{x}}}{x^2 \left (x-4 \log \left (x^2\right )+2\right )}dx+\int \frac {e^{2 x-5-\frac {e^x}{x}+\frac {3}{x}}}{x \left (x-4 \log \left (x^2\right )+2\right )}dx\) |
Input:
Int[(E^((3 - E^x - 5*x + x^2)/x)*(6 - 5*x - x^2 - x^3 + E^x*(-2 + x + x^2) + (-12 + E^x*(4 - 4*x) + 4*x^2)*Log[x^2]))/(4*x^2 + 4*x^3 + x^4 + (-16*x^ 2 - 8*x^3)*Log[x^2] + 16*x^2*Log[x^2]^2),x]
Output:
$Aborted
Time = 2.43 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07
method | result | size |
parallelrisch | \(-\frac {{\mathrm e}^{-\frac {-x^{2}+{\mathrm e}^{x}+5 x -3}{x}}}{x -4 \ln \left (x^{2}\right )+2}\) | \(32\) |
Input:
int((((4-4*x)*exp(x)+4*x^2-12)*ln(x^2)+(x^2+x-2)*exp(x)-x^3-x^2-5*x+6)*exp ((-exp(x)+x^2-5*x+3)/x)/(16*x^2*ln(x^2)^2+(-8*x^3-16*x^2)*ln(x^2)+x^4+4*x^ 3+4*x^2),x,method=_RETURNVERBOSE)
Output:
-exp(-(-x^2+exp(x)+5*x-3)/x)/(x-4*ln(x^2)+2)
Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {3-e^x-5 x+x^2}{x}} \left (6-5 x-x^2-x^3+e^x \left (-2+x+x^2\right )+\left (-12+e^x (4-4 x)+4 x^2\right ) \log \left (x^2\right )\right )}{4 x^2+4 x^3+x^4+\left (-16 x^2-8 x^3\right ) \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx=-\frac {e^{\left (\frac {x^{2} - 5 \, x - e^{x} + 3}{x}\right )}}{x - 4 \, \log \left (x^{2}\right ) + 2} \] Input:
integrate((((4-4*x)*exp(x)+4*x^2-12)*log(x^2)+(x^2+x-2)*exp(x)-x^3-x^2-5*x +6)*exp((-exp(x)+x^2-5*x+3)/x)/(16*x^2*log(x^2)^2+(-8*x^3-16*x^2)*log(x^2) +x^4+4*x^3+4*x^2),x, algorithm="fricas")
Output:
-e^((x^2 - 5*x - e^x + 3)/x)/(x - 4*log(x^2) + 2)
Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\frac {3-e^x-5 x+x^2}{x}} \left (6-5 x-x^2-x^3+e^x \left (-2+x+x^2\right )+\left (-12+e^x (4-4 x)+4 x^2\right ) \log \left (x^2\right )\right )}{4 x^2+4 x^3+x^4+\left (-16 x^2-8 x^3\right ) \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx=- \frac {e^{\frac {x^{2} - 5 x - e^{x} + 3}{x}}}{x - 4 \log {\left (x^{2} \right )} + 2} \] Input:
integrate((((4-4*x)*exp(x)+4*x**2-12)*ln(x**2)+(x**2+x-2)*exp(x)-x**3-x**2 -5*x+6)*exp((-exp(x)+x**2-5*x+3)/x)/(16*x**2*ln(x**2)**2+(-8*x**3-16*x**2) *ln(x**2)+x**4+4*x**3+4*x**2),x)
Output:
-exp((x**2 - 5*x - exp(x) + 3)/x)/(x - 4*log(x**2) + 2)
Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^{\frac {3-e^x-5 x+x^2}{x}} \left (6-5 x-x^2-x^3+e^x \left (-2+x+x^2\right )+\left (-12+e^x (4-4 x)+4 x^2\right ) \log \left (x^2\right )\right )}{4 x^2+4 x^3+x^4+\left (-16 x^2-8 x^3\right ) \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx=-\frac {e^{\left (x - \frac {e^{x}}{x} + \frac {3}{x}\right )}}{x e^{5} - 8 \, e^{5} \log \left (x\right ) + 2 \, e^{5}} \] Input:
integrate((((4-4*x)*exp(x)+4*x^2-12)*log(x^2)+(x^2+x-2)*exp(x)-x^3-x^2-5*x +6)*exp((-exp(x)+x^2-5*x+3)/x)/(16*x^2*log(x^2)^2+(-8*x^3-16*x^2)*log(x^2) +x^4+4*x^3+4*x^2),x, algorithm="maxima")
Output:
-e^(x - e^x/x + 3/x)/(x*e^5 - 8*e^5*log(x) + 2*e^5)
Exception generated. \[ \int \frac {e^{\frac {3-e^x-5 x+x^2}{x}} \left (6-5 x-x^2-x^3+e^x \left (-2+x+x^2\right )+\left (-12+e^x (4-4 x)+4 x^2\right ) \log \left (x^2\right )\right )}{4 x^2+4 x^3+x^4+\left (-16 x^2-8 x^3\right ) \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((((4-4*x)*exp(x)+4*x^2-12)*log(x^2)+(x^2+x-2)*exp(x)-x^3-x^2-5*x +6)*exp((-exp(x)+x^2-5*x+3)/x)/(16*x^2*log(x^2)^2+(-8*x^3-16*x^2)*log(x^2) +x^4+4*x^3+4*x^2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{-4,[0,1,9]%%%}+%%%{68,[0,1,8]%%%}+%%%{-320,[0,1,7]%%%}+%%% {256,[0,1
Time = 2.37 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {3-e^x-5 x+x^2}{x}} \left (6-5 x-x^2-x^3+e^x \left (-2+x+x^2\right )+\left (-12+e^x (4-4 x)+4 x^2\right ) \log \left (x^2\right )\right )}{4 x^2+4 x^3+x^4+\left (-16 x^2-8 x^3\right ) \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx=-\frac {{\mathrm {e}}^{-5}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^x}{x}}\,{\mathrm {e}}^{3/x}\,{\mathrm {e}}^x}{x-4\,\ln \left (x^2\right )+2} \] Input:
int(-(exp(-(5*x + exp(x) - x^2 - 3)/x)*(5*x + log(x^2)*(exp(x)*(4*x - 4) - 4*x^2 + 12) + x^2 + x^3 - exp(x)*(x + x^2 - 2) - 6))/(4*x^2 - log(x^2)*(1 6*x^2 + 8*x^3) + 4*x^3 + x^4 + 16*x^2*log(x^2)^2),x)
Output:
-(exp(-5)*exp(-exp(x)/x)*exp(3/x)*exp(x))/(x - 4*log(x^2) + 2)
Time = 5.29 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {e^{\frac {3-e^x-5 x+x^2}{x}} \left (6-5 x-x^2-x^3+e^x \left (-2+x+x^2\right )+\left (-12+e^x (4-4 x)+4 x^2\right ) \log \left (x^2\right )\right )}{4 x^2+4 x^3+x^4+\left (-16 x^2-8 x^3\right ) \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx=\frac {e^{\frac {x^{2}+3}{x}}}{e^{\frac {e^{x}}{x}} e^{5} \left (4 \,\mathrm {log}\left (x^{2}\right )-x -2\right )} \] Input:
int((((4-4*x)*exp(x)+4*x^2-12)*log(x^2)+(x^2+x-2)*exp(x)-x^3-x^2-5*x+6)*ex p((-exp(x)+x^2-5*x+3)/x)/(16*x^2*log(x^2)^2+(-8*x^3-16*x^2)*log(x^2)+x^4+4 *x^3+4*x^2),x)
Output:
e**((x**2 + 3)/x)/(e**(e**x/x)*e**5*(4*log(x**2) - x - 2))