3.32.48 \(\int \frac {1}{\log ^2(\log (\log (4+\log ^2(3))))} \, dx\)

Optimal. Leaf size=13 \[ \frac {x}{\log ^2\left (\log \left (\log \left (4+\log ^2(3)\right )\right )\right )} \]

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Rubi [A]  time = 0.00, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {8} \begin {gather*} \frac {x}{\log ^2\left (\log \left (\log \left (4+\log ^2(3)\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/Log[Log[Log[4 + Log[3]^2]]]^2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {x}{\log ^2\left (\log \left (\log \left (4+\log ^2(3)\right )\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 1.00 \begin {gather*} \frac {x}{\log ^2\left (\log \left (\log \left (4+\log ^2(3)\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/Log[Log[Log[4 + Log[3]^2]]]^2

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fricas [A]  time = 0.68, size = 13, normalized size = 1.00 \begin {gather*} \frac {x}{\log \left (\log \left (\log \left (\log \relax (3)^{2} + 4\right )\right )\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/log(log(log(log(3)^2+4)))^2,x, algorithm="fricas")

[Out]

x/log(log(log(log(3)^2 + 4)))^2

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giac [A]  time = 0.34, size = 13, normalized size = 1.00 \begin {gather*} \frac {x}{\log \left (\log \left (\log \left (\log \relax (3)^{2} + 4\right )\right )\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/log(log(log(log(3)^2+4)))^2,x, algorithm="giac")

[Out]

x/log(log(log(log(3)^2 + 4)))^2

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maple [A]  time = 0.03, size = 14, normalized size = 1.08

method	result	size
default	\(\frac {x}{\ln \left (\ln \left (\ln \left (\ln \relax (3)^{2}+4\right )\right )\right )^{2}}\)	\(14\)
norman	\(\frac {x}{\ln \left (\ln \left (\ln \left (\ln \relax (3)^{2}+4\right )\right )\right )^{2}}\)	\(14\)
risch	\(\frac {x}{\ln \left (\ln \left (\ln \left (\ln \relax (3)^{2}+4\right )\right )\right )^{2}}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/ln(ln(ln(ln(3)^2+4)))^2,x,method=_RETURNVERBOSE)

[Out]

x/ln(ln(ln(ln(3)^2+4)))^2

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maxima [A]  time = 0.52, size = 13, normalized size = 1.00 \begin {gather*} \frac {x}{\log \left (\log \left (\log \left (\log \relax (3)^{2} + 4\right )\right )\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/log(log(log(log(3)^2+4)))^2,x, algorithm="maxima")

[Out]

x/log(log(log(log(3)^2 + 4)))^2

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mupad [B]  time = 0.00, size = 13, normalized size = 1.00 \begin {gather*} \frac {x}{{\ln \left (\ln \left (\ln \left ({\ln \relax (3)}^2+4\right )\right )\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/log(log(log(log(3)^2 + 4)))^2,x)

[Out]

x/log(log(log(log(3)^2 + 4)))^2

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sympy [A]  time = 0.05, size = 14, normalized size = 1.08 \begin {gather*} \frac {x}{\log {\left (\log {\left (\log {\left (\log {\relax (3 )}^{2} + 4 \right )} \right )} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/ln(ln(ln(ln(3)**2+4)))**2,x)

[Out]

x/log(log(log(log(3)**2 + 4)))**2

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