3.40.2 \(\int \frac {-8 \log (x) \log (\log (x))+(-16+8 \log (2 x)) \log (2-\log (2 x))}{(-2 x \log (x)+x \log (x) \log (2 x)) \log ^2(2-\log (2 x))} \, dx\)

Optimal. Leaf size=18 \[ 1+\frac {8 \log (\log (x))}{\log (2-\log (2 x))} \]

________________________________________________________________________________________

Rubi [F]  time = 0.78, 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 {-8 \log (x) \log (\log (x))+(-16+8 \log (2 x)) \log (2-\log (2 x))}{(-2 x \log (x)+x \log (x) \log (2 x)) \log ^2(2-\log (2 x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8*Log[x]*Log[Log[x]] + (-16 + 8*Log[2*x])*Log[2 - Log[2*x]])/((-2*x*Log[x] + x*Log[x]*Log[2*x])*Log[2 -
Log[2*x]]^2),x]

[Out]

-8*Defer[Int][Log[Log[x]]/(x*(-2 + Log[2*x])*Log[2 - Log[2*x]]^2), x] + 8*Defer[Int][1/(x*Log[x]*Log[2 - Log[2
*x]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8 \log (x) \log (\log (x))-(-16+8 \log (2 x)) \log (2-\log (2 x))}{x \log (x) (2-\log (2 x)) \log ^2(2-\log (2 x))} \, dx\\ &=\int \frac {8 \left (-\frac {\log (\log (x))}{-2+\log (2 x)}+\frac {\log (2-\log (2 x))}{\log (x)}\right )}{x \log ^2(2-\log (2 x))} \, dx\\ &=8 \int \frac {-\frac {\log (\log (x))}{-2+\log (2 x)}+\frac {\log (2-\log (2 x))}{\log (x)}}{x \log ^2(2-\log (2 x))} \, dx\\ &=8 \int \left (-\frac {\log (\log (x))}{x (-2+\log (2 x)) \log ^2(2-\log (2 x))}+\frac {1}{x \log (x) \log (2-\log (2 x))}\right ) \, dx\\ &=-\left (8 \int \frac {\log (\log (x))}{x (-2+\log (2 x)) \log ^2(2-\log (2 x))} \, dx\right )+8 \int \frac {1}{x \log (x) \log (2-\log (2 x))} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 16, normalized size = 0.89 \begin {gather*} \frac {8 \log (\log (x))}{\log (2-\log (2 x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*Log[x]*Log[Log[x]] + (-16 + 8*Log[2*x])*Log[2 - Log[2*x]])/((-2*x*Log[x] + x*Log[x]*Log[2*x])*Lo
g[2 - Log[2*x]]^2),x]

[Out]

(8*Log[Log[x]])/Log[2 - Log[2*x]]

________________________________________________________________________________________

fricas [A]  time = 0.59, size = 18, normalized size = 1.00 \begin {gather*} \frac {8 \, \log \left (\log \relax (x)\right )}{\log \left (-\log \relax (2) - \log \relax (x) + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*log(x)*log(log(x))+(8*log(2*x)-16)*log(-log(2*x)+2))/(x*log(x)*log(2*x)-2*x*log(x))/log(-log(2*x
)+2)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.22, size = 18, normalized size = 1.00 \begin {gather*} \frac {8 \, \log \left (\log \relax (x)\right )}{\log \left (-\log \relax (2) - \log \relax (x) + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*log(x)*log(log(x))+(8*log(2*x)-16)*log(-log(2*x)+2))/(x*log(x)*log(2*x)-2*x*log(x))/log(-log(2*x
)+2)^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.06, size = 19, normalized size = 1.06

method	result	size
risch	\(\frac {8 \ln \left (\ln \relax (x )\right )}{\ln \left (-\ln \relax (2)-\ln \relax (x )+2\right )}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*ln(x)*ln(ln(x))+(8*ln(2*x)-16)*ln(-ln(2*x)+2))/(x*ln(x)*ln(2*x)-2*x*ln(x))/ln(-ln(2*x)+2)^2,x,method=_
RETURNVERBOSE)

[Out]

8*ln(ln(x))/ln(-ln(2)-ln(x)+2)

________________________________________________________________________________________

maxima [A]  time = 0.51, size = 18, normalized size = 1.00 \begin {gather*} \frac {8 \, \log \left (\log \relax (x)\right )}{\log \left (-\log \relax (2) - \log \relax (x) + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*log(x)*log(log(x))+(8*log(2*x)-16)*log(-log(2*x)+2))/(x*log(x)*log(2*x)-2*x*log(x))/log(-log(2*x
)+2)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 2.64, size = 16, normalized size = 0.89 \begin {gather*} \frac {8\,\ln \left (\ln \relax (x)\right )}{\ln \left (2-\ln \left (2\,x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*log(log(x))*log(x) - log(2 - log(2*x))*(8*log(2*x) - 16))/(log(2 - log(2*x))^2*(2*x*log(x) - x*log(2*x)
*log(x))),x)

[Out]

(8*log(log(x)))/log(2 - log(2*x))

________________________________________________________________________________________

sympy [A]  time = 0.34, size = 15, normalized size = 0.83 \begin {gather*} \frac {8 \log {\left (\log {\relax (x )} \right )}}{\log {\left (- \log {\relax (x )} - \log {\relax (2 )} + 2 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*ln(x)*ln(ln(x))+(8*ln(2*x)-16)*ln(-ln(2*x)+2))/(x*ln(x)*ln(2*x)-2*x*ln(x))/ln(-ln(2*x)+2)**2,x)

[Out]

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

________________________________________________________________________________________