3.83.54 \(\int \frac {-8+8 x-2 x^2+(-16 x+12 x^2-2 x^3-2 x \log (4)) \log (x)+4 x \log (x) \log (\log (x))}{(-16 x+24 x^2-12 x^3+2 x^4+(-4 x+4 x^2-x^3) \log (4)) \log (x)+(8 x-8 x^2+2 x^3) \log (x) \log (\log (x))} \, dx\)

Optimal. Leaf size=27 \[ 4+\frac {x}{2-x}-\log (-4-\log (4)+2 (x+\log (\log (x)))) \]

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Rubi [F]  time = 3.20, 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+8 x-2 x^2+\left (-16 x+12 x^2-2 x^3-2 x \log (4)\right ) \log (x)+4 x \log (x) \log (\log (x))}{\left (-16 x+24 x^2-12 x^3+2 x^4+\left (-4 x+4 x^2-x^3\right ) \log (4)\right ) \log (x)+\left (8 x-8 x^2+2 x^3\right ) \log (x) \log (\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8 + 8*x - 2*x^2 + (-16*x + 12*x^2 - 2*x^3 - 2*x*Log[4])*Log[x] + 4*x*Log[x]*Log[Log[x]])/((-16*x + 24*x^
2 - 12*x^3 + 2*x^4 + (-4*x + 4*x^2 - x^3)*Log[4])*Log[x] + (8*x - 8*x^2 + 2*x^3)*Log[x]*Log[Log[x]]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (-(-2+x)^2-x \log (x) \left (8-6 x+x^2+\log (4)-2 \log (\log (x))\right )\right )}{(2-x)^2 x \log (x) \left (2 x-4 \left (1+\frac {\log (2)}{2}\right )+2 \log (\log (x))\right )} \, dx\\ &=2 \int \frac {-(-2+x)^2-x \log (x) \left (8-6 x+x^2+\log (4)-2 \log (\log (x))\right )}{(2-x)^2 x \log (x) \left (2 x-4 \left (1+\frac {\log (2)}{2}\right )+2 \log (\log (x))\right )} \, dx\\ &=2 \int \frac {(-2+x)^2+x \log (x) \left (8-6 x+x^2+\log (4)-2 \log (\log (x))\right )}{2 (2-x)^2 x \log (x) \left (2-x-\log \left (\frac {\log (x)}{2}\right )\right )} \, dx\\ &=\int \frac {(-2+x)^2+x \log (x) \left (8-6 x+x^2+\log (4)-2 \log (\log (x))\right )}{(2-x)^2 x \log (x) \left (2-x-\log \left (\frac {\log (x)}{2}\right )\right )} \, dx\\ &=\int \left (\frac {6 x}{(-2+x)^2 \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )}-\frac {x^2}{(-2+x)^2 \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )}-\frac {8 \left (1+\frac {\log (2)}{4}\right )}{(-2+x)^2 \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )}-\frac {1}{x \log (x) \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )}+\frac {2 \log (\log (x))}{(-2+x)^2 \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )}\right ) \, dx\\ &=2 \int \frac {\log (\log (x))}{(-2+x)^2 \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )} \, dx+6 \int \frac {x}{(-2+x)^2 \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )} \, dx-\left (8 \left (1+\frac {\log (2)}{4}\right )\right ) \int \frac {1}{(-2+x)^2 \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )} \, dx-\int \frac {x^2}{(-2+x)^2 \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )} \, dx-\int \frac {1}{x \log (x) \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )} \, dx\\ &=2 \int \frac {\log (\log (x))}{(-2+x)^2 \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )} \, dx+6 \int \left (\frac {2}{(-2+x)^2 \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )}+\frac {1}{(-2+x) \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )}\right ) \, dx-\left (8 \left (1+\frac {\log (2)}{4}\right )\right ) \int \frac {1}{(-2+x)^2 \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )} \, dx-\int \frac {1}{x \log (x) \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )} \, dx-\int \left (\frac {1}{-2+x+\log \left (\frac {\log (x)}{2}\right )}+\frac {4}{(-2+x)^2 \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )}+\frac {4}{(-2+x) \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )}\right ) \, dx\\ &=2 \int \frac {\log (\log (x))}{(-2+x)^2 \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )} \, dx-4 \int \frac {1}{(-2+x)^2 \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )} \, dx-4 \int \frac {1}{(-2+x) \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )} \, dx+6 \int \frac {1}{(-2+x) \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )} \, dx+12 \int \frac {1}{(-2+x)^2 \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )} \, dx-\left (8 \left (1+\frac {\log (2)}{4}\right )\right ) \int \frac {1}{(-2+x)^2 \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )} \, dx-\int \frac {1}{-2+x+\log \left (\frac {\log (x)}{2}\right )} \, dx-\int \frac {1}{x \log (x) \left (-2+x+\log \left (\frac {\log (x)}{2}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.36, size = 25, normalized size = 0.93 \begin {gather*} -2 \left (\frac {1}{-2+x}+\frac {1}{2} \log (4-2 x+\log (4)-2 \log (\log (x)))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 + 8*x - 2*x^2 + (-16*x + 12*x^2 - 2*x^3 - 2*x*Log[4])*Log[x] + 4*x*Log[x]*Log[Log[x]])/((-16*x +
24*x^2 - 12*x^3 + 2*x^4 + (-4*x + 4*x^2 - x^3)*Log[4])*Log[x] + (8*x - 8*x^2 + 2*x^3)*Log[x]*Log[Log[x]]),x]

[Out]

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

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fricas [A]  time = 0.99, size = 24, normalized size = 0.89 \begin {gather*} -\frac {{\left (x - 2\right )} \log \left (x - \log \relax (2) + \log \left (\log \relax (x)\right ) - 2\right ) + 2}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x)*log(log(x))+(-4*x*log(2)-2*x^3+12*x^2-16*x)*log(x)-2*x^2+8*x-8)/((2*x^3-8*x^2+8*x)*log(x
)*log(log(x))+(2*(-x^3+4*x^2-4*x)*log(2)+2*x^4-12*x^3+24*x^2-16*x)*log(x)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 21, normalized size = 0.78 \begin {gather*} -\frac {2}{x - 2} - \log \left (x - \log \relax (2) + \log \left (\log \relax (x)\right ) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x)*log(log(x))+(-4*x*log(2)-2*x^3+12*x^2-16*x)*log(x)-2*x^2+8*x-8)/((2*x^3-8*x^2+8*x)*log(x
)*log(log(x))+(2*(-x^3+4*x^2-4*x)*log(2)+2*x^4-12*x^3+24*x^2-16*x)*log(x)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.44, size = 22, normalized size = 0.81

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*ln(x)*ln(ln(x))+(-4*x*ln(2)-2*x^3+12*x^2-16*x)*ln(x)-2*x^2+8*x-8)/((2*x^3-8*x^2+8*x)*ln(x)*ln(ln(x))+
(2*(-x^3+4*x^2-4*x)*ln(2)+2*x^4-12*x^3+24*x^2-16*x)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.50, size = 21, normalized size = 0.78 \begin {gather*} -\frac {2}{x - 2} - \log \left (x - \log \relax (2) + \log \left (\log \relax (x)\right ) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x)*log(log(x))+(-4*x*log(2)-2*x^3+12*x^2-16*x)*log(x)-2*x^2+8*x-8)/((2*x^3-8*x^2+8*x)*log(x
)*log(log(x))+(2*(-x^3+4*x^2-4*x)*log(2)+2*x^4-12*x^3+24*x^2-16*x)*log(x)),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\ln \relax (x)\,\left (16\,x+4\,x\,\ln \relax (2)-12\,x^2+2\,x^3\right )-8\,x+2\,x^2-4\,x\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x)+8}{\ln \relax (x)\,\left (16\,x+2\,\ln \relax (2)\,\left (x^3-4\,x^2+4\,x\right )-24\,x^2+12\,x^3-2\,x^4\right )-\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\,\left (2\,x^3-8\,x^2+8\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(16*x + 4*x*log(2) - 12*x^2 + 2*x^3) - 8*x + 2*x^2 - 4*x*log(log(x))*log(x) + 8)/(log(x)*(16*x + 2
*log(2)*(4*x - 4*x^2 + x^3) - 24*x^2 + 12*x^3 - 2*x^4) - log(log(x))*log(x)*(8*x - 8*x^2 + 2*x^3)),x)

[Out]

int((log(x)*(16*x + 4*x*log(2) - 12*x^2 + 2*x^3) - 8*x + 2*x^2 - 4*x*log(log(x))*log(x) + 8)/(log(x)*(16*x + 2
*log(2)*(4*x - 4*x^2 + x^3) - 24*x^2 + 12*x^3 - 2*x^4) - log(log(x))*log(x)*(8*x - 8*x^2 + 2*x^3)), x)

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sympy [A]  time = 0.38, size = 19, normalized size = 0.70 \begin {gather*} - \log {\left (x + \log {\left (\log {\relax (x )} \right )} - 2 - \log {\relax (2 )} \right )} - \frac {2}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*ln(x)*ln(ln(x))+(-4*x*ln(2)-2*x**3+12*x**2-16*x)*ln(x)-2*x**2+8*x-8)/((2*x**3-8*x**2+8*x)*ln(x)
*ln(ln(x))+(2*(-x**3+4*x**2-4*x)*ln(2)+2*x**4-12*x**3+24*x**2-16*x)*ln(x)),x)

[Out]

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

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