Integrand size = 69, antiderivative size = 24 \[ \int \frac {e^2 (-4+x)+2 e^2 \log ^2(x)}{8 x^2+\left (-32 x+8 x^2\right ) \log (x)+\left (32-32 x+2 x^2\right ) \log ^2(x)+(32-8 x) \log ^3(x)+8 \log ^4(x)} \, dx=\frac {e^2 x}{2 (2+\log (x)) (-x+2 \log (x))} \] Output:
exp(2)/(2*ln(x)-x)*x/(2*ln(x)+4)
Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^2 (-4+x)+2 e^2 \log ^2(x)}{8 x^2+\left (-32 x+8 x^2\right ) \log (x)+\left (32-32 x+2 x^2\right ) \log ^2(x)+(32-8 x) \log ^3(x)+8 \log ^4(x)} \, dx=-\frac {e^2 x}{2 (x-2 \log (x)) (2+\log (x))} \] Input:
Integrate[(E^2*(-4 + x) + 2*E^2*Log[x]^2)/(8*x^2 + (-32*x + 8*x^2)*Log[x] + (32 - 32*x + 2*x^2)*Log[x]^2 + (32 - 8*x)*Log[x]^3 + 8*Log[x]^4),x]
Output:
-1/2*(E^2*x)/((x - 2*Log[x])*(2 + 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.
\(\displaystyle \int \frac {e^2 (x-4)+2 e^2 \log ^2(x)}{8 x^2+\left (2 x^2-32 x+32\right ) \log ^2(x)+\left (8 x^2-32 x\right ) \log (x)+8 \log ^4(x)+(32-8 x) \log ^3(x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^2 \left (x+2 \log ^2(x)-4\right )}{2 (x-2 \log (x))^2 (\log (x)+2)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} e^2 \int -\frac {-2 \log ^2(x)-x+4}{(x-2 \log (x))^2 (\log (x)+2)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} e^2 \int \frac {-2 \log ^2(x)-x+4}{(x-2 \log (x))^2 (\log (x)+2)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{2} e^2 \int \left (-\frac {2 (x-2)}{(x+4) (x-2 \log (x))^2}+\frac {8}{(x+4)^2 (x-2 \log (x))}+\frac {4}{(x+4)^2 (\log (x)+2)}-\frac {1}{(x+4) (\log (x)+2)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{2} e^2 \left (-2 \int \frac {1}{(x-2 \log (x))^2}dx+12 \int \frac {1}{(x+4) (x-2 \log (x))^2}dx+8 \int \frac {1}{(x+4)^2 (x-2 \log (x))}dx-\int \frac {1}{(x+4) (\log (x)+2)^2}dx+4 \int \frac {1}{(x+4)^2 (\log (x)+2)}dx\right )\) |
Input:
Int[(E^2*(-4 + x) + 2*E^2*Log[x]^2)/(8*x^2 + (-32*x + 8*x^2)*Log[x] + (32 - 32*x + 2*x^2)*Log[x]^2 + (32 - 8*x)*Log[x]^3 + 8*Log[x]^4),x]
Output:
$Aborted
Time = 0.38 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83
method | result | size |
norman | \(-\frac {{\mathrm e}^{2} x}{2 \left (\ln \left (x \right )+2\right ) \left (x -2 \ln \left (x \right )\right )}\) | \(20\) |
parallelrisch | \(-\frac {{\mathrm e}^{2} x}{2 \left (\ln \left (x \right )+2\right ) \left (x -2 \ln \left (x \right )\right )}\) | \(20\) |
risch | \(-\frac {{\mathrm e}^{2} x}{2 \left (x \ln \left (x \right )-2 \ln \left (x \right )^{2}+2 x -4 \ln \left (x \right )\right )}\) | \(26\) |
default | \(\frac {{\mathrm e}^{2} x}{4 \ln \left (x \right )^{2}-2 x \ln \left (x \right )+8 \ln \left (x \right )-4 x}\) | \(27\) |
Input:
int((2*exp(2)*ln(x)^2+(x-4)*exp(2))/(8*ln(x)^4+(-8*x+32)*ln(x)^3+(2*x^2-32 *x+32)*ln(x)^2+(8*x^2-32*x)*ln(x)+8*x^2),x,method=_RETURNVERBOSE)
Output:
-1/2*exp(2)*x/(ln(x)+2)/(x-2*ln(x))
Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^2 (-4+x)+2 e^2 \log ^2(x)}{8 x^2+\left (-32 x+8 x^2\right ) \log (x)+\left (32-32 x+2 x^2\right ) \log ^2(x)+(32-8 x) \log ^3(x)+8 \log ^4(x)} \, dx=-\frac {x e^{2}}{2 \, {\left ({\left (x - 4\right )} \log \left (x\right ) - 2 \, \log \left (x\right )^{2} + 2 \, x\right )}} \] Input:
integrate((2*exp(2)*log(x)^2+(-4+x)*exp(2))/(8*log(x)^4+(-8*x+32)*log(x)^3 +(2*x^2-32*x+32)*log(x)^2+(8*x^2-32*x)*log(x)+8*x^2),x, algorithm="fricas" )
Output:
-1/2*x*e^2/((x - 4)*log(x) - 2*log(x)^2 + 2*x)
Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^2 (-4+x)+2 e^2 \log ^2(x)}{8 x^2+\left (-32 x+8 x^2\right ) \log (x)+\left (32-32 x+2 x^2\right ) \log ^2(x)+(32-8 x) \log ^3(x)+8 \log ^4(x)} \, dx=\frac {x e^{2}}{- 4 x + \left (8 - 2 x\right ) \log {\left (x \right )} + 4 \log {\left (x \right )}^{2}} \] Input:
integrate((2*exp(2)*ln(x)**2+(-4+x)*exp(2))/(8*ln(x)**4+(-8*x+32)*ln(x)**3 +(2*x**2-32*x+32)*ln(x)**2+(8*x**2-32*x)*ln(x)+8*x**2),x)
Output:
x*exp(2)/(-4*x + (8 - 2*x)*log(x) + 4*log(x)**2)
Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^2 (-4+x)+2 e^2 \log ^2(x)}{8 x^2+\left (-32 x+8 x^2\right ) \log (x)+\left (32-32 x+2 x^2\right ) \log ^2(x)+(32-8 x) \log ^3(x)+8 \log ^4(x)} \, dx=-\frac {x e^{2}}{2 \, {\left ({\left (x - 4\right )} \log \left (x\right ) - 2 \, \log \left (x\right )^{2} + 2 \, x\right )}} \] Input:
integrate((2*exp(2)*log(x)^2+(-4+x)*exp(2))/(8*log(x)^4+(-8*x+32)*log(x)^3 +(2*x^2-32*x+32)*log(x)^2+(8*x^2-32*x)*log(x)+8*x^2),x, algorithm="maxima" )
Output:
-1/2*x*e^2/((x - 4)*log(x) - 2*log(x)^2 + 2*x)
Time = 0.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^2 (-4+x)+2 e^2 \log ^2(x)}{8 x^2+\left (-32 x+8 x^2\right ) \log (x)+\left (32-32 x+2 x^2\right ) \log ^2(x)+(32-8 x) \log ^3(x)+8 \log ^4(x)} \, dx=-\frac {x e^{2}}{2 \, {\left (x \log \left (x\right ) - 2 \, \log \left (x\right )^{2} + 2 \, x - 4 \, \log \left (x\right )\right )}} \] Input:
integrate((2*exp(2)*log(x)^2+(-4+x)*exp(2))/(8*log(x)^4+(-8*x+32)*log(x)^3 +(2*x^2-32*x+32)*log(x)^2+(8*x^2-32*x)*log(x)+8*x^2),x, algorithm="giac")
Output:
-1/2*x*e^2/(x*log(x) - 2*log(x)^2 + 2*x - 4*log(x))
Time = 0.35 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^2 (-4+x)+2 e^2 \log ^2(x)}{8 x^2+\left (-32 x+8 x^2\right ) \log (x)+\left (32-32 x+2 x^2\right ) \log ^2(x)+(32-8 x) \log ^3(x)+8 \log ^4(x)} \, dx=-\frac {x\,{\mathrm {e}}^2}{2\,\left (x-2\,\ln \left (x\right )\right )\,\left (\ln \left (x\right )+2\right )} \] Input:
int((2*exp(2)*log(x)^2 + exp(2)*(x - 4))/(log(x)^2*(2*x^2 - 32*x + 32) + 8 *log(x)^4 - log(x)*(32*x - 8*x^2) + 8*x^2 - log(x)^3*(8*x - 32)),x)
Output:
-(x*exp(2))/(2*(x - 2*log(x))*(log(x) + 2))
Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^2 (-4+x)+2 e^2 \log ^2(x)}{8 x^2+\left (-32 x+8 x^2\right ) \log (x)+\left (32-32 x+2 x^2\right ) \log ^2(x)+(32-8 x) \log ^3(x)+8 \log ^4(x)} \, dx=\frac {\mathrm {log}\left (x \right ) e^{2} \left (2 \,\mathrm {log}\left (x \right )-x +4\right )}{8 \mathrm {log}\left (x \right )^{2}-4 \,\mathrm {log}\left (x \right ) x +16 \,\mathrm {log}\left (x \right )-8 x} \] Input:
int((2*exp(2)*log(x)^2+(-4+x)*exp(2))/(8*log(x)^4+(-8*x+32)*log(x)^3+(2*x^ 2-32*x+32)*log(x)^2+(8*x^2-32*x)*log(x)+8*x^2),x)
Output:
(log(x)*e**2*(2*log(x) - x + 4))/(4*(2*log(x)**2 - log(x)*x + 4*log(x) - 2 *x))