Integrand size = 68, antiderivative size = 26 \begin {dmath*} \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 )} \end {dmath*}
Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \begin {dmath*} \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 )} \end {dmath*}
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]
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 |
\(\displaystyle -\frac {1}{2} \int \frac {e^{-2 x^2}}{(x-2)^2 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}dx-\int \frac {e^{-2 x^2}}{\log \left (\log \left (x^2\right )\right )}dx-\int \frac {e^{-2 x^2}}{(x-2)^3 \log \left (\log \left (x^2\right )\right )}dx-\frac {17}{4} \int \frac {e^{-2 x^2}}{(x-2)^2 \log \left (\log \left (x^2\right )\right )}dx-4 \int \frac {e^{-2 x^2}}{(x-2) \log \left (\log \left (x^2\right )\right )}dx\) |
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]
3.2.84.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Time = 9.32 (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 | \(\frac {x \,{\mathrm e}^{-2 x^{2}}}{4 \left (x^{2}-4 x +4\right ) \ln \left (2 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}\right )}\) | \(56\) |
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)
Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \begin {dmath*} \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 )} \end {dmath*}
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=\
Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \begin {dmath*} \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 )}} \end {dmath*}
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)
Time = 0.33 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \begin {dmath*} \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 )}} \end {dmath*}
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=\
Time = 0.35 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \begin {dmath*} \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 )}} \end {dmath*}
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=\
Time = 14.42 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \begin {dmath*} \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} \end {dmath*}