3.81.96 \(\int \frac {16 \log (x)+(-2+2 \log (x)) \log (4 x^2) \log (\log ^2(4 x^2))}{x^2 \log (4 x^2) \log ^3(\log ^2(4 x^2))} \, dx\)

Optimal. Leaf size=18 \[ -\frac {2 \log (x)}{x \log ^2\left (\log ^2\left (4 x^2\right )\right )} \]

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Rubi [F]  time = 0.64, 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 {16 \log (x)+(-2+2 \log (x)) \log \left (4 x^2\right ) \log \left (\log ^2\left (4 x^2\right )\right )}{x^2 \log \left (4 x^2\right ) \log ^3\left (\log ^2\left (4 x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(16*Log[x] + (-2 + 2*Log[x])*Log[4*x^2]*Log[Log[4*x^2]^2])/(x^2*Log[4*x^2]*Log[Log[4*x^2]^2]^3),x]

[Out]

16*Defer[Int][Log[x]/(x^2*Log[4*x^2]*Log[Log[4*x^2]^2]^3), x] - 2*Defer[Int][1/(x^2*Log[Log[4*x^2]^2]^2), x] +
2*Defer[Int][Log[x]/(x^2*Log[Log[4*x^2]^2]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {16 \log (x)}{x^2 \log \left (4 x^2\right ) \log ^3\left (\log ^2\left (4 x^2\right )\right )}+\frac {2 (-1+\log (x))}{x^2 \log ^2\left (\log ^2\left (4 x^2\right )\right )}\right ) \, dx\\ &=2 \int \frac {-1+\log (x)}{x^2 \log ^2\left (\log ^2\left (4 x^2\right )\right )} \, dx+16 \int \frac {\log (x)}{x^2 \log \left (4 x^2\right ) \log ^3\left (\log ^2\left (4 x^2\right )\right )} \, dx\\ &=2 \int \left (-\frac {1}{x^2 \log ^2\left (\log ^2\left (4 x^2\right )\right )}+\frac {\log (x)}{x^2 \log ^2\left (\log ^2\left (4 x^2\right )\right )}\right ) \, dx+16 \int \frac {\log (x)}{x^2 \log \left (4 x^2\right ) \log ^3\left (\log ^2\left (4 x^2\right )\right )} \, dx\\ &=-\left (2 \int \frac {1}{x^2 \log ^2\left (\log ^2\left (4 x^2\right )\right )} \, dx\right )+2 \int \frac {\log (x)}{x^2 \log ^2\left (\log ^2\left (4 x^2\right )\right )} \, dx+16 \int \frac {\log (x)}{x^2 \log \left (4 x^2\right ) \log ^3\left (\log ^2\left (4 x^2\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 18, normalized size = 1.00 \begin {gather*} -\frac {2 \log (x)}{x \log ^2\left (\log ^2\left (4 x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16*Log[x] + (-2 + 2*Log[x])*Log[4*x^2]*Log[Log[4*x^2]^2])/(x^2*Log[4*x^2]*Log[Log[4*x^2]^2]^3),x]

[Out]

(-2*Log[x])/(x*Log[Log[4*x^2]^2]^2)

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fricas [A]  time = 0.63, size = 29, normalized size = 1.61 \begin {gather*} -\frac {2 \, \log \relax (x)}{x \log \left (4 \, \log \relax (2)^{2} + 8 \, \log \relax (2) \log \relax (x) + 4 \, \log \relax (x)^{2}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(x)-2)*log(4*x^2)*log(log(4*x^2)^2)+16*log(x))/x^2/log(4*x^2)/log(log(4*x^2)^2)^3,x, algorith
m="fricas")

[Out]

-2*log(x)/(x*log(4*log(2)^2 + 8*log(2)*log(x) + 4*log(x)^2)^2)

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giac [B]  time = 0.69, size = 257, normalized size = 14.28 \begin {gather*} -\frac {2 \, \log \relax (2)^{2} \log \left (4 \, x^{2}\right ) \log \relax (x) \log \left (\log \left (4 \, x^{2}\right )^{2}\right ) - \log \relax (2) \log \left (4 \, x^{2}\right )^{2} \log \relax (x) \log \left (\log \left (4 \, x^{2}\right )^{2}\right ) + 4 \, \log \relax (2) \log \left (4 \, x^{2}\right ) \log \relax (x)^{2} \log \left (\log \left (4 \, x^{2}\right )^{2}\right ) - \log \left (4 \, x^{2}\right )^{2} \log \relax (x)^{2} \log \left (\log \left (4 \, x^{2}\right )^{2}\right ) + 2 \, \log \left (4 \, x^{2}\right ) \log \relax (x)^{3} \log \left (\log \left (4 \, x^{2}\right )^{2}\right ) - 2 \, \log \relax (2)^{2} \log \left (4 \, x^{2}\right ) \log \left (\log \left (4 \, x^{2}\right )^{2}\right ) + \log \relax (2) \log \left (4 \, x^{2}\right )^{2} \log \left (\log \left (4 \, x^{2}\right )^{2}\right ) - 2 \, \log \relax (2) \log \left (4 \, x^{2}\right ) \log \relax (x) \log \left (\log \left (4 \, x^{2}\right )^{2}\right ) + 4 \, \log \relax (2) \log \left (4 \, x^{2}\right ) \log \relax (x) + 4 \, \log \left (4 \, x^{2}\right ) \log \relax (x)^{2}}{4 \, {\left (x \log \relax (2)^{2} \log \left (\log \left (4 \, x^{2}\right )^{2}\right )^{2} + 2 \, x \log \relax (2) \log \relax (x) \log \left (\log \left (4 \, x^{2}\right )^{2}\right )^{2} + x \log \relax (x)^{2} \log \left (\log \left (4 \, x^{2}\right )^{2}\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(x)-2)*log(4*x^2)*log(log(4*x^2)^2)+16*log(x))/x^2/log(4*x^2)/log(log(4*x^2)^2)^3,x, algorith
m="giac")

[Out]

-1/4*(2*log(2)^2*log(4*x^2)*log(x)*log(log(4*x^2)^2) - log(2)*log(4*x^2)^2*log(x)*log(log(4*x^2)^2) + 4*log(2)
*log(4*x^2)*log(x)^2*log(log(4*x^2)^2) - log(4*x^2)^2*log(x)^2*log(log(4*x^2)^2) + 2*log(4*x^2)*log(x)^3*log(l
og(4*x^2)^2) - 2*log(2)^2*log(4*x^2)*log(log(4*x^2)^2) + log(2)*log(4*x^2)^2*log(log(4*x^2)^2) - 2*log(2)*log(
4*x^2)*log(x)*log(log(4*x^2)^2) + 4*log(2)*log(4*x^2)*log(x) + 4*log(4*x^2)*log(x)^2)/(x*log(2)^2*log(log(4*x^
2)^2)^2 + 2*x*log(2)*log(x)*log(log(4*x^2)^2)^2 + x*log(x)^2*log(log(4*x^2)^2)^2)

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maple [C]  time = 0.80, size = 459, normalized size = 25.50

method	result	size
risch	\(\frac {8 \ln \relax (x )}{x \left (\pi \mathrm {csgn}\left (i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (2)+4 i \ln \relax (x )\right )\right )^{2} \mathrm {csgn}\left (i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (2)+4 i \ln \relax (x )\right )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (2)+4 i \ln \relax (x )\right )\right ) \mathrm {csgn}\left (i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (2)+4 i \ln \relax (x )\right )^{2}\right )^{2}-\pi \mathrm {csgn}\left (i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (2)+4 i \ln \relax (x )\right )^{2}\right )^{3}+2 \pi \mathrm {csgn}\left (i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (2)+4 i \ln \relax (x )\right )^{2}\right )^{2}-2 \pi -4 i \ln \relax (2)+4 i \ln \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (2)+4 i \ln \relax (x )\right )\right )^{2}}\)	\(459\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*ln(x)-2)*ln(4*x^2)*ln(ln(4*x^2)^2)+16*ln(x))/x^2/ln(4*x^2)/ln(ln(4*x^2)^2)^3,x,method=_RETURNVERBOSE)

[Out]

8*ln(x)/x/(Pi*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(2)+4*I*l
n(x)))^2*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(2)+4*I*ln(x))
^2)-2*Pi*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(2)+4*I*ln(x))
)*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(2)+4*I*ln(x))^2)^2-P
i*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(2)+4*I*ln(x))^2)^3+2
*Pi*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(2)+4*I*ln(x))^2)^2
-2*Pi-4*I*ln(2)+4*I*ln(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(2)+4*I*
ln(x)))^2

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maxima [A]  time = 0.49, size = 34, normalized size = 1.89 \begin {gather*} -\frac {\log \relax (x)}{2 \, {\left (x \log \relax (2)^{2} + 2 \, x \log \relax (2) \log \left (\log \relax (2) + \log \relax (x)\right ) + x \log \left (\log \relax (2) + \log \relax (x)\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(x)-2)*log(4*x^2)*log(log(4*x^2)^2)+16*log(x))/x^2/log(4*x^2)/log(log(4*x^2)^2)^3,x, algorith
m="maxima")

[Out]

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

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mupad [B]  time = 6.06, size = 31, normalized size = 1.72 \begin {gather*} -\frac {2\,\ln \relax (x)}{x\,{\ln \left ({\ln \left (x^2\right )}^2+4\,\ln \relax (2)\,\ln \left (x^2\right )+4\,{\ln \relax (2)}^2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*log(x) + log(log(4*x^2)^2)*log(4*x^2)*(2*log(x) - 2))/(x^2*log(log(4*x^2)^2)^3*log(4*x^2)),x)

[Out]

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

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sympy [A]  time = 0.31, size = 20, normalized size = 1.11 \begin {gather*} - \frac {2 \log {\relax (x )}}{x \log {\left (\left (2 \log {\relax (x )} + \log {\relax (4 )}\right )^{2} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*ln(x)-2)*ln(4*x**2)*ln(ln(4*x**2)**2)+16*ln(x))/x**2/ln(4*x**2)/ln(ln(4*x**2)**2)**3,x)

[Out]

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

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