3.76.22 \(\int \frac {1+\log (-\frac {x}{\log (16)})+\log (-\frac {x}{\log (16)}) \log (x \log (-\frac {x}{\log (16)}))}{x \log (-\frac {x}{\log (16)}) \log (x \log (-\frac {x}{\log (16)}))} \, dx\)

Optimal. Leaf size=21 \[ \log \left (x \log \left (x-x \left (1-\log \left (-\frac {x}{\log (16)}\right )\right )\right )\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 14, normalized size of antiderivative = 0.67, number of steps used = 1, number of rules used = 1, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {6685} \begin {gather*} \log \left (x \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + Log[-(x/Log[16])] + Log[-(x/Log[16])]*Log[x*Log[-(x/Log[16])]])/(x*Log[-(x/Log[16])]*Log[x*Log[-(x/Lo
g[16])]]),x]

[Out]

Log[x*Log[x*Log[-(x/Log[16])]]]

Rule 6685

Int[(u_)/((w_)*(y_)), x_Symbol] :> With[{q = DerivativeDivides[y*w, u, x]}, Simp[q*Log[RemoveContent[y*w, x]],
x] /;  !FalseQ[q]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log \left (x \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 21, normalized size = 1.00 \begin {gather*} \log \left (-\frac {x}{\log (16)}\right )+\log \left (\log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + Log[-(x/Log[16])] + Log[-(x/Log[16])]*Log[x*Log[-(x/Log[16])]])/(x*Log[-(x/Log[16])]*Log[x*Log[
-(x/Log[16])]]),x]

[Out]

Log[-(x/Log[16])] + Log[Log[x*Log[-(x/Log[16])]]]

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fricas [A]  time = 1.00, size = 21, normalized size = 1.00 \begin {gather*} \log \left (-\frac {x}{4 \, \log \relax (2)}\right ) + \log \left (\log \left (x \log \left (-\frac {x}{4 \, \log \relax (2)}\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-1/4*x/log(2)) + log(log(x*log(-1/4*x/log(2))))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\log \left (x \log \left (-\frac {x}{4 \, \log \relax (2)}\right )\right ) \log \left (-\frac {x}{4 \, \log \relax (2)}\right ) + \log \left (-\frac {x}{4 \, \log \relax (2)}\right ) + 1}{x \log \left (x \log \left (-\frac {x}{4 \, \log \relax (2)}\right )\right ) \log \left (-\frac {x}{4 \, \log \relax (2)}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((log(x*log(-1/4*x/log(2)))*log(-1/4*x/log(2)) + log(-1/4*x/log(2)) + 1)/(x*log(x*log(-1/4*x/log(2)))
*log(-1/4*x/log(2))), x)

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maple [A]  time = 0.04, size = 22, normalized size = 1.05

method	result	size
norman	\(\ln \left (-\frac {x}{4 \ln \relax (2)}\right )+\ln \left (\ln \left (x \ln \left (-\frac {x}{4 \ln \relax (2)}\right )\right )\right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(-1/4*x/ln(2))*ln(x*ln(-1/4*x/ln(2)))+ln(-1/4*x/ln(2))+1)/x/ln(-1/4*x/ln(2))/ln(x*ln(-1/4*x/ln(2))),x,m
ethod=_RETURNVERBOSE)

[Out]

ln(-1/4*x/ln(2))+ln(ln(x*ln(-1/4*x/ln(2))))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, \log \relax (2) - \log \relax (x) + \log \left (-\log \relax (2)\right ) - 1}{x {\left (2 \, \log \relax (2) + \log \left (-\log \relax (2)\right )\right )} \log \relax (x) - x \log \relax (x)^{2} + {\left (x {\left (2 \, \log \relax (2) + \log \left (-\log \relax (2)\right )\right )} - x \log \relax (x)\right )} \log \left (-2 \, \log \relax (2) + \log \left (-x\right ) - \log \left (\log \relax (2)\right )\right )}\,{d x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((2*log(2) - log(x) + log(-log(2)) - 1)/(x*(2*log(2) + log(-log(2)))*log(x) - x*log(x)^2 + (x*(2*log(
2) + log(-log(2))) - x*log(x))*log(-2*log(2) + log(-x) - log(log(2)))), x) + log(x)

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mupad [B]  time = 5.69, size = 18, normalized size = 0.86 \begin {gather*} \ln \left (\ln \left (x\,\ln \left (-\frac {x}{4}\right )-x\,\ln \left (\ln \relax (2)\right )\right )\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(-x/(4*log(2))) + log(-x/(4*log(2)))*log(x*log(-x/(4*log(2)))) + 1)/(x*log(-x/(4*log(2)))*log(x*log(-x
/(4*log(2))))),x)

[Out]

log(log(x*log(-x/4) - x*log(log(2)))) + log(x)

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sympy [A]  time = 0.32, size = 17, normalized size = 0.81 \begin {gather*} \log {\relax (x )} + \log {\left (\log {\left (x \log {\left (- \frac {x}{4 \log {\relax (2 )}} \right )} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) + log(log(x*log(-x/(4*log(2)))))

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