3.92.84 \(\int \frac {e^{-2 x^2} (4-2 x+(-2-x+8 x^2-4 x^3) \log (x^2) \log (\log (x^2)))}{(-32+48 x-24 x^2+4 x^3) \log (x^2) \log ^2(\log (x^2))} \, dx\)

Optimal. Leaf size=26 \[ \frac {e^{-2 x^2} x}{4 (2-x)^2 \log \left (\log \left (x^2\right )\right )} \]

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Rubi [F]  time = 1.54, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[In]

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]

[Out]

-1/2*Defer[Int][1/(E^(2*x^2)*(-2 + x)^2*Log[x^2]*Log[Log[x^2]]^2), x] - Defer[Int][1/(E^(2*x^2)*Log[Log[x^2]])
, x] - Defer[Int][1/(E^(2*x^2)*(-2 + x)^3*Log[Log[x^2]]), x] - (17*Defer[Int][1/(E^(2*x^2)*(-2 + x)^2*Log[Log[
x^2]]), x])/4 - 4*Defer[Int][1/(E^(2*x^2)*(-2 + x)*Log[Log[x^2]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {e^{-2 x^2}}{2 (-2+x)^2 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}+\frac {e^{-2 x^2} \left (-2-x+8 x^2-4 x^3\right )}{4 (-2+x)^3 \log \left (\log \left (x^2\right )\right )}\right ) \, dx\\ &=\frac {1}{4} \int \frac {e^{-2 x^2} \left (-2-x+8 x^2-4 x^3\right )}{(-2+x)^3 \log \left (\log \left (x^2\right )\right )} \, dx-\frac {1}{2} \int \frac {e^{-2 x^2}}{(-2+x)^2 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx\\ &=\frac {1}{4} \int \left (-\frac {4 e^{-2 x^2}}{\log \left (\log \left (x^2\right )\right )}-\frac {4 e^{-2 x^2}}{(-2+x)^3 \log \left (\log \left (x^2\right )\right )}-\frac {17 e^{-2 x^2}}{(-2+x)^2 \log \left (\log \left (x^2\right )\right )}-\frac {16 e^{-2 x^2}}{(-2+x) \log \left (\log \left (x^2\right )\right )}\right ) \, dx-\frac {1}{2} \int \frac {e^{-2 x^2}}{(-2+x)^2 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx\\ &=-\left (\frac {1}{2} \int \frac {e^{-2 x^2}}{(-2+x)^2 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx\right )-4 \int \frac {e^{-2 x^2}}{(-2+x) \log \left (\log \left (x^2\right )\right )} \, dx-\frac {17}{4} \int \frac {e^{-2 x^2}}{(-2+x)^2 \log \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}}{(-2+x)^3 \log \left (\log \left (x^2\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 24, normalized size = 0.92 \begin {gather*} \frac {e^{-2 x^2} x}{4 (-2+x)^2 \log \left (\log \left (x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

x/(4*E^(2*x^2)*(-2 + x)^2*Log[Log[x^2]])

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fricas [A]  time = 0.71, size = 26, normalized size = 1.00 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

1/4*x*e^(-2*x^2)/((x^2 - 4*x + 4)*log(log(x^2)))

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giac [A]  time = 0.25, size = 36, normalized size = 1.38 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

1/4*x*e^(-2*x^2)/(x^2*log(log(x^2)) - 4*x*log(log(x^2)) + 4*log(log(x^2)))

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maple [C]  time = 0.13, size = 56, normalized size = 2.15

method	result	size
risch	\(\frac {x \,{\mathrm e}^{-2 x^{2}}}{4 \left (x^{2}-4 x +4\right ) \ln \left (2 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right )}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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,m
ethod=_RETURNVERBOSE)

[Out]

1/4*x*exp(-2*x^2)/(x^2-4*x+4)/ln(2*ln(x)-1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2)

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maxima [A]  time = 0.49, size = 39, normalized size = 1.50 \begin {gather*} \frac {x e^{\left (-2 \, x^{2}\right )}}{4 \, {\left (x^{2} \log \relax (2) - 4 \, x \log \relax (2) + {\left (x^{2} - 4 \, x + 4\right )} \log \left (\log \relax (x)\right ) + 4 \, \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

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))

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mupad [B]  time = 7.81, size = 21, normalized size = 0.81 \begin {gather*} \frac {x\,{\mathrm {e}}^{-2\,x^2}}{4\,\ln \left (\ln \left (x^2\right )\right )\,{\left (x-2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

(x*exp(-2*x^2))/(4*log(log(x^2))*(x - 2)^2)

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sympy [A]  time = 0.43, size = 37, normalized size = 1.42 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

x*exp(-2*x**2)/(4*x**2*log(log(x**2)) - 16*x*log(log(x**2)) + 16*log(log(x**2)))

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