Integrand size = 96, antiderivative size = 31 \[ \int \frac {2 e^{\frac {e^4-x^2}{x}} x-x^2+\left (-3 x^2+e^{\frac {e^4-x^2}{x}} \left (-2 e^4+4 x-2 x^2\right )\right ) \log (x)}{\left (2 e^{\frac {e^4-x^2}{x}} x^2-x^3\right ) \log (x)} \, dx=\log \left (\frac {3}{2} \left (-e^{\frac {e^4}{x}-x}+\frac {x}{2}\right ) x^2 \log (x)\right ) \] Output:
ln(3/2*x^2*(1/2*x-exp(exp(4)/x-x))*ln(x))
Time = 0.37 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {2 e^{\frac {e^4-x^2}{x}} x-x^2+\left (-3 x^2+e^{\frac {e^4-x^2}{x}} \left (-2 e^4+4 x-2 x^2\right )\right ) \log (x)}{\left (2 e^{\frac {e^4-x^2}{x}} x^2-x^3\right ) \log (x)} \, dx=-x+2 \log (x)+\log \left (2 e^{\frac {e^4}{x}}-e^x x\right )+\log (\log (x)) \] Input:
Integrate[(2*E^((E^4 - x^2)/x)*x - x^2 + (-3*x^2 + E^((E^4 - x^2)/x)*(-2*E ^4 + 4*x - 2*x^2))*Log[x])/((2*E^((E^4 - x^2)/x)*x^2 - x^3)*Log[x]),x]
Output:
-x + 2*Log[x] + Log[2*E^(E^4/x) - E^x*x] + Log[Log[x]]
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.
\(\displaystyle \int \frac {-x^2+2 e^{\frac {e^4-x^2}{x}} x+\left (e^{\frac {e^4-x^2}{x}} \left (-2 x^2+4 x-2 e^4\right )-3 x^2\right ) \log (x)}{\left (2 e^{\frac {e^4-x^2}{x}} x^2-x^3\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^x \left (-x^2+2 e^{\frac {e^4-x^2}{x}} x+\left (e^{\frac {e^4-x^2}{x}} \left (-2 x^2+4 x-2 e^4\right )-3 x^2\right ) \log (x)\right )}{x^2 \left (2 e^{\frac {e^4}{x}}-e^x x\right ) \log (x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^x \left (x^2+x+e^4\right )}{x \left (e^x x-2 e^{\frac {e^4}{x}}\right )}-\frac {x^2 \log (x)-x-2 x \log (x)+e^4 \log (x)}{x^2 \log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {e^x}{2 e^{\frac {e^4}{x}}-e^x x}dx+\int \frac {e^{x+4}}{x \left (e^x x-2 e^{\frac {e^4}{x}}\right )}dx+\int \frac {e^x x}{e^x x-2 e^{\frac {e^4}{x}}}dx-x+\frac {e^4}{x}+2 \log (x)+\log (\log (x))\) |
Input:
Int[(2*E^((E^4 - x^2)/x)*x - x^2 + (-3*x^2 + E^((E^4 - x^2)/x)*(-2*E^4 + 4 *x - 2*x^2))*Log[x])/((2*E^((E^4 - x^2)/x)*x^2 - x^3)*Log[x]),x]
Output:
$Aborted
Time = 5.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87
method | result | size |
norman | \(2 \ln \left (x \right )+\ln \left (\ln \left (x \right )\right )+\ln \left (x -2 \,{\mathrm e}^{\frac {{\mathrm e}^{4}-x^{2}}{x}}\right )\) | \(27\) |
risch | \(2 \ln \left (x \right )+\ln \left (-\frac {x}{2}+{\mathrm e}^{\frac {{\mathrm e}^{4}-x^{2}}{x}}\right )+\ln \left (\ln \left (x \right )\right )\) | \(27\) |
parallelrisch | \(2 \ln \left (x \right )+\ln \left (\ln \left (x \right )\right )+\ln \left (x -2 \,{\mathrm e}^{\frac {{\mathrm e}^{4}-x^{2}}{x}}\right )\) | \(27\) |
Input:
int((((-2*exp(4)-2*x^2+4*x)*exp((exp(4)-x^2)/x)-3*x^2)*ln(x)+2*x*exp((exp( 4)-x^2)/x)-x^2)/(2*x^2*exp((exp(4)-x^2)/x)-x^3)/ln(x),x,method=_RETURNVERB OSE)
Output:
2*ln(x)+ln(ln(x))+ln(x-2*exp((exp(4)-x^2)/x))
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {2 e^{\frac {e^4-x^2}{x}} x-x^2+\left (-3 x^2+e^{\frac {e^4-x^2}{x}} \left (-2 e^4+4 x-2 x^2\right )\right ) \log (x)}{\left (2 e^{\frac {e^4-x^2}{x}} x^2-x^3\right ) \log (x)} \, dx=2 \, \log \left (x\right ) + \log \left (-x + 2 \, e^{\left (-\frac {x^{2} - e^{4}}{x}\right )}\right ) + \log \left (\log \left (x\right )\right ) \] Input:
integrate((((-2*exp(4)-2*x^2+4*x)*exp((exp(4)-x^2)/x)-3*x^2)*log(x)+2*x*ex p((exp(4)-x^2)/x)-x^2)/(2*x^2*exp((exp(4)-x^2)/x)-x^3)/log(x),x, algorithm ="fricas")
Output:
2*log(x) + log(-x + 2*e^(-(x^2 - e^4)/x)) + log(log(x))
Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {2 e^{\frac {e^4-x^2}{x}} x-x^2+\left (-3 x^2+e^{\frac {e^4-x^2}{x}} \left (-2 e^4+4 x-2 x^2\right )\right ) \log (x)}{\left (2 e^{\frac {e^4-x^2}{x}} x^2-x^3\right ) \log (x)} \, dx=2 \log {\left (x \right )} + \log {\left (- \frac {x}{2} + e^{\frac {- x^{2} + e^{4}}{x}} \right )} + \log {\left (\log {\left (x \right )} \right )} \] Input:
integrate((((-2*exp(4)-2*x**2+4*x)*exp((exp(4)-x**2)/x)-3*x**2)*ln(x)+2*x* exp((exp(4)-x**2)/x)-x**2)/(2*x**2*exp((exp(4)-x**2)/x)-x**3)/ln(x),x)
Output:
2*log(x) + log(-x/2 + exp((-x**2 + exp(4))/x)) + log(log(x))
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {2 e^{\frac {e^4-x^2}{x}} x-x^2+\left (-3 x^2+e^{\frac {e^4-x^2}{x}} \left (-2 e^4+4 x-2 x^2\right )\right ) \log (x)}{\left (2 e^{\frac {e^4-x^2}{x}} x^2-x^3\right ) \log (x)} \, dx=-x + \log \left (-\frac {1}{2} \, x e^{x} + e^{\left (\frac {e^{4}}{x}\right )}\right ) + 2 \, \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \] Input:
integrate((((-2*exp(4)-2*x^2+4*x)*exp((exp(4)-x^2)/x)-3*x^2)*log(x)+2*x*ex p((exp(4)-x^2)/x)-x^2)/(2*x^2*exp((exp(4)-x^2)/x)-x^3)/log(x),x, algorithm ="maxima")
Output:
-x + log(-1/2*x*e^x + e^(e^4/x)) + 2*log(x) + log(log(x))
Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {2 e^{\frac {e^4-x^2}{x}} x-x^2+\left (-3 x^2+e^{\frac {e^4-x^2}{x}} \left (-2 e^4+4 x-2 x^2\right )\right ) \log (x)}{\left (2 e^{\frac {e^4-x^2}{x}} x^2-x^3\right ) \log (x)} \, dx=\log \left (x - 2 \, e^{\left (-\frac {x^{2} - e^{4}}{x}\right )}\right ) + 2 \, \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \] Input:
integrate((((-2*exp(4)-2*x^2+4*x)*exp((exp(4)-x^2)/x)-3*x^2)*log(x)+2*x*ex p((exp(4)-x^2)/x)-x^2)/(2*x^2*exp((exp(4)-x^2)/x)-x^3)/log(x),x, algorithm ="giac")
Output:
log(x - 2*e^(-(x^2 - e^4)/x)) + 2*log(x) + log(log(x))
Time = 3.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {2 e^{\frac {e^4-x^2}{x}} x-x^2+\left (-3 x^2+e^{\frac {e^4-x^2}{x}} \left (-2 e^4+4 x-2 x^2\right )\right ) \log (x)}{\left (2 e^{\frac {e^4-x^2}{x}} x^2-x^3\right ) \log (x)} \, dx=\ln \left (\ln \left (x\right )\right )+\ln \left (x-2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^4}{x}-x}\right )+2\,\ln \left (x\right ) \] Input:
int(-(log(x)*(exp((exp(4) - x^2)/x)*(2*exp(4) - 4*x + 2*x^2) + 3*x^2) - 2* x*exp((exp(4) - x^2)/x) + x^2)/(log(x)*(2*x^2*exp((exp(4) - x^2)/x) - x^3) ),x)
Output:
log(log(x)) + log(x - 2*exp(exp(4)/x - x)) + 2*log(x)
Time = 6.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {2 e^{\frac {e^4-x^2}{x}} x-x^2+\left (-3 x^2+e^{\frac {e^4-x^2}{x}} \left (-2 e^4+4 x-2 x^2\right )\right ) \log (x)}{\left (2 e^{\frac {e^4-x^2}{x}} x^2-x^3\right ) \log (x)} \, dx=\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+\mathrm {log}\left (2 e^{\frac {e^{4}}{x}}-e^{x} x \right )+2 \,\mathrm {log}\left (x \right )-x \] Input:
int((((-2*exp(4)-2*x^2+4*x)*exp((exp(4)-x^2)/x)-3*x^2)*log(x)+2*x*exp((exp (4)-x^2)/x)-x^2)/(2*x^2*exp((exp(4)-x^2)/x)-x^3)/log(x),x)
Output:
log(log(x)) + log(2*e**(e**4/x) - e**x*x) + 2*log(x) - x