3.20.84 \(\int \frac {e^{-2 x-\log ^4(3)+2 x \log ^2(3) \log (\log (x))-x^2 \log ^2(\log (x))} (2 \log ^2(3)-2 \log (x)+(-2 x+2 \log ^2(3) \log (x)) \log (\log (x))-2 x \log (x) \log ^2(\log (x)))}{\log (x)} \, dx\)

Optimal. Leaf size=22 \[ e^{-2 x-\left (-\log ^2(3)+x \log (\log (x))\right )^2} \]

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Rubi [A]  time = 0.93, antiderivative size = 33, normalized size of antiderivative = 1.50, number of steps used = 1, number of rules used = 1, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6706} \begin {gather*} e^{-x^2 \log ^2(\log (x))-2 x-\log ^4(3)} \log ^{2 x \log ^2(3)}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-2*x - Log[3]^4 - x^2*Log[Log[x]]^2)*Log[x]^(2*x*Log[3]^2)

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{-2 x-\log ^4(3)-x^2 \log ^2(\log (x))} \log ^{2 x \log ^2(3)}(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.48, size = 33, normalized size = 1.50 \begin {gather*} e^{-2 x-\log ^4(3)-x^2 \log ^2(\log (x))} \log ^{2 x \log ^2(3)}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-2*x - Log[3]^4 + 2*x*Log[3]^2*Log[Log[x]] - x^2*Log[Log[x]]^2)*(2*Log[3]^2 - 2*Log[x] + (-2*x +
2*Log[3]^2*Log[x])*Log[Log[x]] - 2*x*Log[x]*Log[Log[x]]^2))/Log[x],x]

[Out]

E^(-2*x - Log[3]^4 - x^2*Log[Log[x]]^2)*Log[x]^(2*x*Log[3]^2)

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fricas [A]  time = 0.63, size = 31, normalized size = 1.41 \begin {gather*} e^{\left (-\log \relax (3)^{4} + 2 \, x \log \relax (3)^{2} \log \left (\log \relax (x)\right ) - x^{2} \log \left (\log \relax (x)\right )^{2} - 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)*log(log(x))^2+(2*log(3)^2*log(x)-2*x)*log(log(x))-2*log(x)+2*log(3)^2)/log(x)/exp(x^2*l
og(log(x))^2-2*x*log(3)^2*log(log(x))+log(3)^4+2*x),x, algorithm="fricas")

[Out]

e^(-log(3)^4 + 2*x*log(3)^2*log(log(x)) - x^2*log(log(x))^2 - 2*x)

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giac [A]  time = 0.41, size = 31, normalized size = 1.41 \begin {gather*} e^{\left (-\log \relax (3)^{4} + 2 \, x \log \relax (3)^{2} \log \left (\log \relax (x)\right ) - x^{2} \log \left (\log \relax (x)\right )^{2} - 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)*log(log(x))^2+(2*log(3)^2*log(x)-2*x)*log(log(x))-2*log(x)+2*log(3)^2)/log(x)/exp(x^2*l
og(log(x))^2-2*x*log(3)^2*log(log(x))+log(3)^4+2*x),x, algorithm="giac")

[Out]

e^(-log(3)^4 + 2*x*log(3)^2*log(log(x)) - x^2*log(log(x))^2 - 2*x)

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maple [A]  time = 0.04, size = 35, normalized size = 1.59

method	result	size
risch	\(\ln \relax (x )^{2 x \ln \relax (3)^{2}} {\mathrm e}^{-x^{2} \ln \left (\ln \relax (x )\right )^{2}-\ln \relax (3)^{4}-2 x}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*ln(x)*ln(ln(x))^2+(2*ln(3)^2*ln(x)-2*x)*ln(ln(x))-2*ln(x)+2*ln(3)^2)/ln(x)/exp(x^2*ln(ln(x))^2-2*x*l
n(3)^2*ln(ln(x))+ln(3)^4+2*x),x,method=_RETURNVERBOSE)

[Out]

1/(ln(x)^(-2*x*ln(3)^2))*exp(-x^2*ln(ln(x))^2-ln(3)^4-2*x)

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maxima [A]  time = 0.95, size = 31, normalized size = 1.41 \begin {gather*} e^{\left (-\log \relax (3)^{4} + 2 \, x \log \relax (3)^{2} \log \left (\log \relax (x)\right ) - x^{2} \log \left (\log \relax (x)\right )^{2} - 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)*log(log(x))^2+(2*log(3)^2*log(x)-2*x)*log(log(x))-2*log(x)+2*log(3)^2)/log(x)/exp(x^2*l
og(log(x))^2-2*x*log(3)^2*log(log(x))+log(3)^4+2*x),x, algorithm="maxima")

[Out]

e^(-log(3)^4 + 2*x*log(3)^2*log(log(x)) - x^2*log(log(x))^2 - 2*x)

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mupad [B]  time = 1.26, size = 33, normalized size = 1.50 \begin {gather*} {\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{-x^2\,{\ln \left (\ln \relax (x)\right )}^2}\,{\mathrm {e}}^{-{\ln \relax (3)}^4}\,{\ln \relax (x)}^{2\,x\,{\ln \relax (3)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x*log(log(x))*log(3)^2 - log(3)^4 - x^2*log(log(x))^2 - 2*x)*(2*log(x) - 2*log(3)^2 + log(log(x))*
(2*x - 2*log(3)^2*log(x)) + 2*x*log(log(x))^2*log(x)))/log(x),x)

[Out]

exp(-2*x)*exp(-x^2*log(log(x))^2)*exp(-log(3)^4)*log(x)^(2*x*log(3)^2)

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sympy [A]  time = 0.68, size = 32, normalized size = 1.45 \begin {gather*} e^{- x^{2} \log {\left (\log {\relax (x )} \right )}^{2} + 2 x \log {\relax (3 )}^{2} \log {\left (\log {\relax (x )} \right )} - 2 x - \log {\relax (3 )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*ln(x)*ln(ln(x))**2+(2*ln(3)**2*ln(x)-2*x)*ln(ln(x))-2*ln(x)+2*ln(3)**2)/ln(x)/exp(x**2*ln(ln(x
))**2-2*x*ln(3)**2*ln(ln(x))+ln(3)**4+2*x),x)

[Out]

exp(-x**2*log(log(x))**2 + 2*x*log(3)**2*log(log(x)) - 2*x - log(3)**4)

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