3.20.89 \(\int \frac {-1-\log (\frac {\log (\frac {1}{4} \log ^2(1+e^{16}))}{x})}{x^2} \, dx\) [1989]

Optimal. Leaf size=22 \[ \frac {\log \left (\frac {\log \left (\frac {1}{4} \log ^2\left (1+e^{16}\right )\right )}{x}\right )}{x} \]

[Out]

ln(ln(1/4*ln(exp(16)+1)^2)/x)/x

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Rubi [A]
time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {2340} \begin {gather*} \frac {\log \left (\frac {\log \left (\frac {1}{4} \log ^2\left (1+e^{16}\right )\right )}{x}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - Log[Log[Log[1 + E^16]^2/4]/x])/x^2,x]

[Out]

Log[Log[Log[1 + E^16]^2/4]/x]/x

Rule 2340

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[b*(d*x)^(m + 1)*(Log[c*x^n]/(d
*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && EqQ[a*(m + 1) - b*n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\log \left (\frac {\log \left (\frac {1}{4} \log ^2\left (1+e^{16}\right )\right )}{x}\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 22, normalized size = 1.00 \begin {gather*} \frac {\log \left (\frac {\log \left (\frac {1}{4} \log ^2\left (1+e^{16}\right )\right )}{x}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - Log[Log[Log[1 + E^16]^2/4]/x])/x^2,x]

[Out]

Log[Log[Log[1 + E^16]^2/4]/x]/x

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Maple [A]
time = 0.07, size = 20, normalized size = 0.91

method	result	size
derivativedivides	\(\frac {\ln \left (\frac {\ln \left (\frac {\ln \left ({\mathrm e}^{16}+1\right )^{2}}{4}\right )}{x}\right )}{x}\)	\(20\)
default	\(\frac {\ln \left (\frac {\ln \left (\frac {\ln \left ({\mathrm e}^{16}+1\right )^{2}}{4}\right )}{x}\right )}{x}\)	\(20\)
norman	\(\frac {\ln \left (\frac {\ln \left (\frac {\ln \left ({\mathrm e}^{16}+1\right )^{2}}{4}\right )}{x}\right )}{x}\)	\(20\)
risch	\(\frac {\ln \left (\frac {-2 \ln \left (2\right )+2 \ln \left (\ln \left ({\mathrm e}^{16}+1\right )\right )}{x}\right )}{x}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(ln(1/4*ln(exp(16)+1)^2)/x)-1)/x^2,x,method=_RETURNVERBOSE)

[Out]

ln(ln(1/4*ln(exp(16)+1)^2)/x)/x

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (19) = 38\).
time = 0.26, size = 62, normalized size = 2.82 \begin {gather*} \frac {\frac {\log \left (\frac {1}{4} \, \log \left (e^{16} + 1\right )^{2}\right ) \log \left (\frac {\log \left (\frac {1}{4} \, \log \left (e^{16} + 1\right )^{2}\right )}{x}\right )}{x} - \frac {\log \left (\frac {1}{4} \, \log \left (e^{16} + 1\right )^{2}\right )}{x}}{\log \left (\frac {1}{4} \, \log \left (e^{16} + 1\right )^{2}\right )} + \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(log(1/4*log(exp(16)+1)^2)/x)-1)/x^2,x, algorithm="maxima")

[Out]

(log(1/4*log(e^16 + 1)^2)*log(log(1/4*log(e^16 + 1)^2)/x)/x - log(1/4*log(e^16 + 1)^2)/x)/log(1/4*log(e^16 + 1
)^2) + 1/x

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Fricas [A]
time = 0.34, size = 19, normalized size = 0.86 \begin {gather*} \frac {\log \left (\frac {\log \left (\frac {1}{4} \, \log \left (e^{16} + 1\right )^{2}\right )}{x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(log(1/4*log(exp(16)+1)^2)/x)-1)/x^2,x, algorithm="fricas")

[Out]

log(log(1/4*log(e^16 + 1)^2)/x)/x

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Sympy [A]
time = 0.05, size = 15, normalized size = 0.68 \begin {gather*} \frac {\log {\left (\frac {\log {\left (\frac {\log {\left (1 + e^{16} \right )}^{2}}{4} \right )}}{x} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(ln(1/4*ln(exp(16)+1)**2)/x)-1)/x**2,x)

[Out]

log(log(log(1 + exp(16))**2/4)/x)/x

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Giac [A]
time = 0.40, size = 19, normalized size = 0.86 \begin {gather*} \frac {\log \left (\frac {\log \left (\frac {1}{4} \, \log \left (e^{16} + 1\right )^{2}\right )}{x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(log(1/4*log(exp(16)+1)^2)/x)-1)/x^2,x, algorithm="giac")

[Out]

log(log(1/4*log(e^16 + 1)^2)/x)/x

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Mupad [B]
time = 1.20, size = 20, normalized size = 0.91 \begin {gather*} \frac {\ln \left (\frac {1}{x}\right )+\ln \left (\ln \left (\frac {{\ln \left ({\mathrm {e}}^{16}+1\right )}^2}{4}\right )\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(log(exp(16) + 1)^2/4)/x) + 1)/x^2,x)

[Out]

(log(1/x) + log(log(log(exp(16) + 1)^2/4)))/x

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