Integrand size = 68, antiderivative size = 26 \[ \int \frac {e^{-2 x^2} \left (4-2 x+\left (-2-x+8 x^2-4 x^3\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right )}{\left (-32+48 x-24 x^2+4 x^3\right ) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\frac {e^{-2 x^2} x}{4 (2-x)^2 \log \left (\log \left (x^2\right )\right )} \] Output:
1/4*x/exp(x^2)^2/(2-x)^2/ln(ln(x^2))
Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-2 x^2} \left (4-2 x+\left (-2-x+8 x^2-4 x^3\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right )}{\left (-32+48 x-24 x^2+4 x^3\right ) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\frac {e^{-2 x^2} x}{4 (-2+x)^2 \log \left (\log \left (x^2\right )\right )} \] Input:
Integrate[(4 - 2*x + (-2 - x + 8*x^2 - 4*x^3)*Log[x^2]*Log[Log[x^2]])/(E^( 2*x^2)*(-32 + 48*x - 24*x^2 + 4*x^3)*Log[x^2]*Log[Log[x^2]]^2),x]
Output:
x/(4*E^(2*x^2)*(-2 + x)^2*Log[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.
\(\displaystyle \int \frac {e^{-2 x^2} \left (\left (-4 x^3+8 x^2-x-2\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )-2 x+4\right )}{\left (4 x^3-24 x^2+48 x-32\right ) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{-2 x^2} \left (\left (-4 x^3+8 x^2-x-2\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )-2 x+4\right )}{\left (2^{2/3} x-2\ 2^{2/3}\right )^3 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{-2 x^2} \left (-4 x^3+8 x^2-x-2\right )}{4 (x-2)^3 \log \left (\log \left (x^2\right )\right )}-\frac {e^{-2 x^2}}{2 (x-2)^2 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{2} \int \frac {e^{-2 x^2}}{(x-2)^2 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}dx-\int \frac {e^{-2 x^2}}{\log \left (\log \left (x^2\right )\right )}dx-\int \frac {e^{-2 x^2}}{(x-2)^3 \log \left (\log \left (x^2\right )\right )}dx-\frac {17}{4} \int \frac {e^{-2 x^2}}{(x-2)^2 \log \left (\log \left (x^2\right )\right )}dx-4 \int \frac {e^{-2 x^2}}{(x-2) \log \left (\log \left (x^2\right )\right )}dx\) |
Input:
Int[(4 - 2*x + (-2 - x + 8*x^2 - 4*x^3)*Log[x^2]*Log[Log[x^2]])/(E^(2*x^2) *(-32 + 48*x - 24*x^2 + 4*x^3)*Log[x^2]*Log[Log[x^2]]^2),x]
Output:
$Aborted
Time = 9.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(\frac {x \,{\mathrm e}^{-2 x^{2}}}{4 \ln \left (\ln \left (x^{2}\right )\right ) \left (x^{2}-4 x +4\right )}\) | \(27\) |
risch | \(\frac {x \,{\mathrm e}^{-2 x^{2}}}{4 \left (x^{2}-4 x +4\right ) \ln \left (2 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}\right )}\) | \(56\) |
Input:
int(((-4*x^3+8*x^2-x-2)*ln(x^2)*ln(ln(x^2))+4-2*x)/(4*x^3-24*x^2+48*x-32)/ exp(x^2)^2/ln(x^2)/ln(ln(x^2))^2,x,method=_RETURNVERBOSE)
Output:
1/4*x/exp(x^2)^2/ln(ln(x^2))/(x^2-4*x+4)
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 x^2} \left (4-2 x+\left (-2-x+8 x^2-4 x^3\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right )}{\left (-32+48 x-24 x^2+4 x^3\right ) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\frac {x e^{\left (-2 \, x^{2}\right )}}{4 \, {\left (x^{2} - 4 \, x + 4\right )} \log \left (\log \left (x^{2}\right )\right )} \] Input:
integrate(((-4*x^3+8*x^2-x-2)*log(x^2)*log(log(x^2))+4-2*x)/(4*x^3-24*x^2+ 48*x-32)/exp(x^2)^2/log(x^2)/log(log(x^2))^2,x, algorithm="fricas")
Output:
1/4*x*e^(-2*x^2)/((x^2 - 4*x + 4)*log(log(x^2)))
Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-2 x^2} \left (4-2 x+\left (-2-x+8 x^2-4 x^3\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right )}{\left (-32+48 x-24 x^2+4 x^3\right ) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\frac {x e^{- 2 x^{2}}}{4 x^{2} \log {\left (\log {\left (x^{2} \right )} \right )} - 16 x \log {\left (\log {\left (x^{2} \right )} \right )} + 16 \log {\left (\log {\left (x^{2} \right )} \right )}} \] Input:
integrate(((-4*x**3+8*x**2-x-2)*ln(x**2)*ln(ln(x**2))+4-2*x)/(4*x**3-24*x* *2+48*x-32)/exp(x**2)**2/ln(x**2)/ln(ln(x**2))**2,x)
Output:
x*exp(-2*x**2)/(4*x**2*log(log(x**2)) - 16*x*log(log(x**2)) + 16*log(log(x **2)))
Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {e^{-2 x^2} \left (4-2 x+\left (-2-x+8 x^2-4 x^3\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right )}{\left (-32+48 x-24 x^2+4 x^3\right ) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\frac {x e^{\left (-2 \, x^{2}\right )}}{4 \, {\left (x^{2} \log \left (2\right ) - 4 \, x \log \left (2\right ) + {\left (x^{2} - 4 \, x + 4\right )} \log \left (\log \left (x\right )\right ) + 4 \, \log \left (2\right )\right )}} \] Input:
integrate(((-4*x^3+8*x^2-x-2)*log(x^2)*log(log(x^2))+4-2*x)/(4*x^3-24*x^2+ 48*x-32)/exp(x^2)^2/log(x^2)/log(log(x^2))^2,x, algorithm="maxima")
Output:
1/4*x*e^(-2*x^2)/(x^2*log(2) - 4*x*log(2) + (x^2 - 4*x + 4)*log(log(x)) + 4*log(2))
Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {e^{-2 x^2} \left (4-2 x+\left (-2-x+8 x^2-4 x^3\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right )}{\left (-32+48 x-24 x^2+4 x^3\right ) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\frac {x e^{\left (-2 \, x^{2}\right )}}{4 \, {\left (x^{2} \log \left (\log \left (x^{2}\right )\right ) - 4 \, x \log \left (\log \left (x^{2}\right )\right ) + 4 \, \log \left (\log \left (x^{2}\right )\right )\right )}} \] Input:
integrate(((-4*x^3+8*x^2-x-2)*log(x^2)*log(log(x^2))+4-2*x)/(4*x^3-24*x^2+ 48*x-32)/exp(x^2)^2/log(x^2)/log(log(x^2))^2,x, algorithm="giac")
Output:
1/4*x*e^(-2*x^2)/(x^2*log(log(x^2)) - 4*x*log(log(x^2)) + 4*log(log(x^2)))
Time = 4.37 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-2 x^2} \left (4-2 x+\left (-2-x+8 x^2-4 x^3\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right )}{\left (-32+48 x-24 x^2+4 x^3\right ) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\frac {x\,{\mathrm {e}}^{-2\,x^2}}{4\,\ln \left (\ln \left (x^2\right )\right )\,{\left (x-2\right )}^2} \] Input:
int(-(exp(-2*x^2)*(2*x + log(x^2)*log(log(x^2))*(x - 8*x^2 + 4*x^3 + 2) - 4))/(log(x^2)*log(log(x^2))^2*(48*x - 24*x^2 + 4*x^3 - 32)),x)
Output:
(x*exp(-2*x^2))/(4*log(log(x^2))*(x - 2)^2)
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-2 x^2} \left (4-2 x+\left (-2-x+8 x^2-4 x^3\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right )}{\left (-32+48 x-24 x^2+4 x^3\right ) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\frac {x}{4 e^{2 x^{2}} \mathrm {log}\left (\mathrm {log}\left (x^{2}\right )\right ) \left (x^{2}-4 x +4\right )} \] Input:
int(((-4*x^3+8*x^2-x-2)*log(x^2)*log(log(x^2))+4-2*x)/(4*x^3-24*x^2+48*x-3 2)/exp(x^2)^2/log(x^2)/log(log(x^2))^2,x)
Output:
x/(4*e**(2*x**2)*log(log(x**2))*(x**2 - 4*x + 4))