3.98.80 \(\int \frac {-e^{40+e^4-x^2 \log ^2(3 x)}+e^{40+e^4-x^2 \log ^2(3 x)} (-2 x^2 \log (x) \log (3 x)-2 x^2 \log (x) \log ^2(3 x)) \log (\log (x))}{x \log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=24 \[ \frac {e^{40+e^4-x^2 \log ^2(3 x)}}{\log (\log (x))} \]

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Rubi [B]  time = 1.13, antiderivative size = 52, normalized size of antiderivative = 2.17, number of steps used = 2, number of rules used = 2, integrand size = 81, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6688, 2288} \begin {gather*} \frac {x e^{-x^2 \log ^2(3 x)+e^4+40} \log (3 x) (\log (3 x)+1)}{\left (x \log ^2(3 x)+x \log (3 x)\right ) \log (\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^(40 + E^4 - x^2*Log[3*x]^2) + E^(40 + E^4 - x^2*Log[3*x]^2)*(-2*x^2*Log[x]*Log[3*x] - 2*x^2*Log[x]*Log
[3*x]^2)*Log[Log[x]])/(x*Log[x]*Log[Log[x]]^2),x]

[Out]

(E^(40 + E^4 - x^2*Log[3*x]^2)*x*Log[3*x]*(1 + Log[3*x]))/((x*Log[3*x] + x*Log[3*x]^2)*Log[Log[x]])

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{40 \left (1+\frac {e^4}{40}\right )-x^2 \log ^2(3 x)} \left (-1-2 x^2 \log (x) \log (3 x) (1+\log (3 x)) \log (\log (x))\right )}{x \log (x) \log ^2(\log (x))} \, dx\\ &=\frac {e^{40+e^4-x^2 \log ^2(3 x)} x \log (3 x) (1+\log (3 x))}{\left (x \log (3 x)+x \log ^2(3 x)\right ) \log (\log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 24, normalized size = 1.00 \begin {gather*} \frac {e^{40+e^4-x^2 \log ^2(3 x)}}{\log (\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^(40 + E^4 - x^2*Log[3*x]^2) + E^(40 + E^4 - x^2*Log[3*x]^2)*(-2*x^2*Log[x]*Log[3*x] - 2*x^2*Log[
x]*Log[3*x]^2)*Log[Log[x]])/(x*Log[x]*Log[Log[x]]^2),x]

[Out]

E^(40 + E^4 - x^2*Log[3*x]^2)/Log[Log[x]]

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fricas [A]  time = 0.82, size = 38, normalized size = 1.58 \begin {gather*} \frac {e^{\left (-x^{2} \log \relax (3)^{2} - 2 \, x^{2} \log \relax (3) \log \relax (x) - x^{2} \log \relax (x)^{2} + e^{4} + 40\right )}}{\log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2*log(x)*log(3*x)^2-2*x^2*log(x)*log(3*x))*exp(-x^2*log(3*x)^2+exp(4)+40)*log(log(x))-exp(-x^
2*log(3*x)^2+exp(4)+40))/x/log(x)/log(log(x))^2,x, algorithm="fricas")

[Out]

e^(-x^2*log(3)^2 - 2*x^2*log(3)*log(x) - x^2*log(x)^2 + e^4 + 40)/log(log(x))

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giac [A]  time = 0.40, size = 38, normalized size = 1.58 \begin {gather*} \frac {e^{\left (-x^{2} \log \relax (3)^{2} - 2 \, x^{2} \log \relax (3) \log \relax (x) - x^{2} \log \relax (x)^{2} + e^{4} + 40\right )}}{\log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2*log(x)*log(3*x)^2-2*x^2*log(x)*log(3*x))*exp(-x^2*log(3*x)^2+exp(4)+40)*log(log(x))-exp(-x^
2*log(3*x)^2+exp(4)+40))/x/log(x)/log(log(x))^2,x, algorithm="giac")

[Out]

e^(-x^2*log(3)^2 - 2*x^2*log(3)*log(x) - x^2*log(x)^2 + e^4 + 40)/log(log(x))

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maple [A]  time = 0.09, size = 39, normalized size = 1.62

method	result	size
risch	\(\frac {x^{-2 x^{2} \ln \relax (3)} {\mathrm e}^{-x^{2} \ln \relax (x )^{2}+40-x^{2} \ln \relax (3)^{2}+{\mathrm e}^{4}}}{\ln \left (\ln \relax (x )\right )}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2*ln(x)*ln(3*x)^2-2*x^2*ln(x)*ln(3*x))*exp(-x^2*ln(3*x)^2+exp(4)+40)*ln(ln(x))-exp(-x^2*ln(3*x)^2+e
xp(4)+40))/x/ln(x)/ln(ln(x))^2,x,method=_RETURNVERBOSE)

[Out]

x^(-2*x^2*ln(3))*exp(-x^2*ln(x)^2+40-x^2*ln(3)^2+exp(4))/ln(ln(x))

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maxima [A]  time = 0.54, size = 38, normalized size = 1.58 \begin {gather*} \frac {e^{\left (-x^{2} \log \relax (3)^{2} - 2 \, x^{2} \log \relax (3) \log \relax (x) - x^{2} \log \relax (x)^{2} + e^{4} + 40\right )}}{\log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2*log(x)*log(3*x)^2-2*x^2*log(x)*log(3*x))*exp(-x^2*log(3*x)^2+exp(4)+40)*log(log(x))-exp(-x^
2*log(3*x)^2+exp(4)+40))/x/log(x)/log(log(x))^2,x, algorithm="maxima")

[Out]

e^(-x^2*log(3)^2 - 2*x^2*log(3)*log(x) - x^2*log(x)^2 + e^4 + 40)/log(log(x))

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mupad [B]  time = 5.83, size = 42, normalized size = 1.75 \begin {gather*} \frac {{\mathrm {e}}^{40}\,{\mathrm {e}}^{-x^2\,{\ln \relax (3)}^2}\,{\mathrm {e}}^{{\mathrm {e}}^4}\,{\mathrm {e}}^{-x^2\,{\ln \relax (x)}^2}}{x^{2\,x^2\,\ln \relax (3)}\,\ln \left (\ln \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(4) - x^2*log(3*x)^2 + 40) + log(log(x))*exp(exp(4) - x^2*log(3*x)^2 + 40)*(2*x^2*log(3*x)*log(x)
+ 2*x^2*log(3*x)^2*log(x)))/(x*log(log(x))^2*log(x)),x)

[Out]

(exp(40)*exp(-x^2*log(3)^2)*exp(exp(4))*exp(-x^2*log(x)^2))/(x^(2*x^2*log(3))*log(log(x)))

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sympy [A]  time = 0.50, size = 22, normalized size = 0.92 \begin {gather*} \frac {e^{- x^{2} \left (\log {\relax (x )} + \log {\relax (3 )}\right )^{2} + 40 + e^{4}}}{\log {\left (\log {\relax (x )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2*ln(x)*ln(3*x)**2-2*x**2*ln(x)*ln(3*x))*exp(-x**2*ln(3*x)**2+exp(4)+40)*ln(ln(x))-exp(-x**2
*ln(3*x)**2+exp(4)+40))/x/ln(x)/ln(ln(x))**2,x)

[Out]

exp(-x**2*(log(x) + log(3))**2 + 40 + exp(4))/log(log(x))

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