3.101.89 \(\int \frac {-1-2 x}{2 x-2 x^2+x \log (\frac {4 \log (\log (\log (5)))}{5 x})} \, dx\)

Optimal. Leaf size=20 \[ -1+\log \left (2-2 x+\log \left (\frac {4 \log (\log (\log (5)))}{5 x}\right )\right ) \]

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Rubi [F]  time = 0.13, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-1-2 x}{2 x-2 x^2+x \log \left (\frac {4 \log (\log (\log (5)))}{5 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1 - 2*x)/(2*x - 2*x^2 + x*Log[(4*Log[Log[Log[5]]])/(5*x)]),x]

[Out]

2*Defer[Int][(-2 + 2*x - Log[(4*Log[Log[Log[5]]])/(5*x)])^(-1), x] + Defer[Int][1/(x*(-2 + 2*x - Log[(4*Log[Lo
g[Log[5]]])/(5*x)])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2}{-2+2 x-\log \left (\frac {4 \log (\log (\log (5)))}{5 x}\right )}+\frac {1}{x \left (-2+2 x-\log \left (\frac {4 \log (\log (\log (5)))}{5 x}\right )\right )}\right ) \, dx\\ &=2 \int \frac {1}{-2+2 x-\log \left (\frac {4 \log (\log (\log (5)))}{5 x}\right )} \, dx+\int \frac {1}{x \left (-2+2 x-\log \left (\frac {4 \log (\log (\log (5)))}{5 x}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 18, normalized size = 0.90 \begin {gather*} \log \left (2-2 x+\log \left (\frac {4 \log (\log (\log (5)))}{5 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 2*x)/(2*x - 2*x^2 + x*Log[(4*Log[Log[Log[5]]])/(5*x)]),x]

[Out]

Log[2 - 2*x + Log[(4*Log[Log[Log[5]]])/(5*x)]]

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fricas [A]  time = 0.55, size = 16, normalized size = 0.80 \begin {gather*} \log \left (-2 \, x + \log \left (\frac {4 \, \log \left (\log \left (\log \relax (5)\right )\right )}{5 \, x}\right ) + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x-1)/(x*log(4/5*log(log(log(5)))/x)-2*x^2+2*x),x, algorithm="fricas")

[Out]

log(-2*x + log(4/5*log(log(log(5)))/x) + 2)

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giac [B]  time = 0.19, size = 65, normalized size = 3.25 \begin {gather*} \frac {\log \left (-\frac {4 \, \log \left (\frac {4 \, \log \left (\log \left (\log \relax (5)\right )\right )}{5 \, x}\right ) \log \left (\log \left (\log \relax (5)\right )\right )}{x} - \frac {8 \, \log \left (\log \left (\log \relax (5)\right )\right )}{x} + 8 \, \log \left (\log \left (\log \relax (5)\right )\right )\right ) \log \left (\log \left (\log \relax (5)\right )\right ) - \log \left (\frac {4 \, \log \left (\log \left (\log \relax (5)\right )\right )}{5 \, x}\right ) \log \left (\log \left (\log \relax (5)\right )\right )}{\log \left (\log \left (\log \relax (5)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x-1)/(x*log(4/5*log(log(log(5)))/x)-2*x^2+2*x),x, algorithm="giac")

[Out]

(log(-4*log(4/5*log(log(log(5)))/x)*log(log(log(5)))/x - 8*log(log(log(5)))/x + 8*log(log(log(5))))*log(log(lo
g(5))) - log(4/5*log(log(log(5)))/x)*log(log(log(5))))/log(log(log(5)))

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maple [A]  time = 0.09, size = 17, normalized size = 0.85

method	result	size
risch	\(\ln \left (\ln \left (\frac {4 \ln \left (\ln \left (\ln \relax (5)\right )\right )}{5 x}\right )+2-2 x \right )\)	\(17\)
norman	\(\ln \left (2 x -\ln \left (\frac {4 \ln \left (\ln \left (\ln \relax (5)\right )\right )}{5 x}\right )-2\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x-1)/(x*ln(4/5*ln(ln(ln(5)))/x)-2*x^2+2*x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(4/5*ln(ln(ln(5)))/x)+2-2*x)

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maxima [A]  time = 0.47, size = 21, normalized size = 1.05 \begin {gather*} \log \left (2 \, x + \log \relax (5) - 2 \, \log \relax (2) + \log \relax (x) - \log \left (\log \left (\log \left (\log \relax (5)\right )\right )\right ) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x-1)/(x*log(4/5*log(log(log(5)))/x)-2*x^2+2*x),x, algorithm="maxima")

[Out]

log(2*x + log(5) - 2*log(2) + log(x) - log(log(log(log(5)))) - 2)

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mupad [B]  time = 6.60, size = 29, normalized size = 1.45 \begin {gather*} \ln \left (x-\frac {\ln \left (-\ln \left (\ln \left (\ln \relax (5)\right )\right )\right )}{2}-\ln \relax (2)+\frac {\ln \relax (5)}{2}-\frac {\ln \left (-\frac {1}{x}\right )}{2}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + 1)/(2*x + x*log((4*log(log(log(5))))/(5*x)) - 2*x^2),x)

[Out]

log(x - log(-log(log(log(5))))/2 - log(2) + log(5)/2 - log(-1/x)/2 - 1)

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sympy [A]  time = 0.14, size = 19, normalized size = 0.95 \begin {gather*} \log {\left (- 2 x + \log {\left (\frac {4 \log {\left (\log {\left (\log {\relax (5 )} \right )} \right )}}{5 x} \right )} + 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x-1)/(x*ln(4/5*ln(ln(ln(5)))/x)-2*x**2+2*x),x)

[Out]

log(-2*x + log(4*log(log(log(5)))/(5*x)) + 2)

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