3.4.30 \(\int \frac {e^{\frac {5}{-4 x^2+x^2 \log (\log (x))}} (-5+40 \log (x)-10 \log (x) \log (\log (x)))}{16 x^3 \log (x)-8 x^3 \log (x) \log (\log (x))+x^3 \log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=14 \[ e^{\frac {5}{x^2 (-4+\log (\log (x)))}} \]

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Rubi [A]  time = 1.13, antiderivative size = 16, normalized size of antiderivative = 1.14, number of steps used = 3, number of rules used = 3, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {6688, 12, 6706} \begin {gather*} e^{-\frac {5}{x^2 (4-\log (\log (x)))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(5/(-4*x^2 + x^2*Log[Log[x]]))*(-5 + 40*Log[x] - 10*Log[x]*Log[Log[x]]))/(16*x^3*Log[x] - 8*x^3*Log[x]*
Log[Log[x]] + x^3*Log[x]*Log[Log[x]]^2),x]

[Out]

E^(-5/(x^2*(4 - Log[Log[x]])))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6688

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

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} &=\int \frac {5 e^{\frac {5}{x^2 (-4+\log (\log (x)))}} (-1-2 \log (x) (-4+\log (\log (x))))}{x^3 \log (x) (4-\log (\log (x)))^2} \, dx\\ &=5 \int \frac {e^{\frac {5}{x^2 (-4+\log (\log (x)))}} (-1-2 \log (x) (-4+\log (\log (x))))}{x^3 \log (x) (4-\log (\log (x)))^2} \, dx\\ &=e^{-\frac {5}{x^2 (4-\log (\log (x)))}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 14, normalized size = 1.00 \begin {gather*} e^{\frac {5}{x^2 (-4+\log (\log (x)))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5/(-4*x^2 + x^2*Log[Log[x]]))*(-5 + 40*Log[x] - 10*Log[x]*Log[Log[x]]))/(16*x^3*Log[x] - 8*x^3*L
og[x]*Log[Log[x]] + x^3*Log[x]*Log[Log[x]]^2),x]

[Out]

E^(5/(x^2*(-4 + Log[Log[x]])))

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fricas [A]  time = 0.64, size = 18, normalized size = 1.29 \begin {gather*} e^{\left (\frac {5}{x^{2} \log \left (\log \relax (x)\right ) - 4 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*log(x)*log(log(x))+40*log(x)-5)*exp(5/(x^2*log(log(x))-4*x^2))/(x^3*log(x)*log(log(x))^2-8*x^3*
log(x)*log(log(x))+16*x^3*log(x)),x, algorithm="fricas")

[Out]

e^(5/(x^2*log(log(x)) - 4*x^2))

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giac [A]  time = 0.63, size = 18, normalized size = 1.29 \begin {gather*} e^{\left (\frac {5}{x^{2} \log \left (\log \relax (x)\right ) - 4 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*log(x)*log(log(x))+40*log(x)-5)*exp(5/(x^2*log(log(x))-4*x^2))/(x^3*log(x)*log(log(x))^2-8*x^3*
log(x)*log(log(x))+16*x^3*log(x)),x, algorithm="giac")

[Out]

e^(5/(x^2*log(log(x)) - 4*x^2))

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maple [A]  time = 0.02, size = 14, normalized size = 1.00

method	result	size
risch	\({\mathrm e}^{\frac {5}{x^{2} \left (\ln \left (\ln \relax (x )\right )-4\right )}}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-10*ln(x)*ln(ln(x))+40*ln(x)-5)*exp(5/(x^2*ln(ln(x))-4*x^2))/(x^3*ln(x)*ln(ln(x))^2-8*x^3*ln(x)*ln(ln(x))
+16*x^3*ln(x)),x,method=_RETURNVERBOSE)

[Out]

exp(5/x^2/(ln(ln(x))-4))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*log(x)*log(log(x))+40*log(x)-5)*exp(5/(x^2*log(log(x))-4*x^2))/(x^3*log(x)*log(log(x))^2-8*x^3*
log(x)*log(log(x))+16*x^3*log(x)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 0.52, size = 13, normalized size = 0.93 \begin {gather*} {\mathrm {e}}^{\frac {5}{x^2\,\left (\ln \left (\ln \relax (x)\right )-4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(5/(x^2*log(log(x)) - 4*x^2))*(10*log(log(x))*log(x) - 40*log(x) + 5))/(16*x^3*log(x) - 8*x^3*log(log
(x))*log(x) + x^3*log(log(x))^2*log(x)),x)

[Out]

exp(5/(x^2*(log(log(x)) - 4)))

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sympy [A]  time = 0.82, size = 15, normalized size = 1.07 \begin {gather*} e^{\frac {5}{x^{2} \log {\left (\log {\relax (x )} \right )} - 4 x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*ln(x)*ln(ln(x))+40*ln(x)-5)*exp(5/(x**2*ln(ln(x))-4*x**2))/(x**3*ln(x)*ln(ln(x))**2-8*x**3*ln(x
)*ln(ln(x))+16*x**3*ln(x)),x)

[Out]

exp(5/(x**2*log(log(x)) - 4*x**2))

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