3.100.22 \(\int \frac {(-2 x+x^2+(2-2 x) \log (x)) \log (3 x)-x^2 \log (x) \log (3 x) \log (\log (x))+((2 x-x^2) \log (x)+(-2 x+x^2) \log (x) \log (3 x)+((2 x^2-x^3) \log (x)+(-2 x^2+x^3) \log (x) \log (3 x)) \log (\log (x))) \log (\frac {-1-x \log (\log (x))}{-2 x+x^2}) \log (\log (\frac {-1-x \log (\log (x))}{-2 x+x^2}))+((2-x) \log (x)+(2 x-x^2) \log (x) \log (\log (x))) \log (\frac {-1-x \log (\log (x))}{-2 x+x^2}) \log (\log (\frac {-1-x \log (\log (x))}{-2 x+x^2})) \log (\log (\log (\frac {-1-x \log (\log (x))}{-2 x+x^2})))}{((-2 x+x^2) \log (x) \log ^2(3 x)+(-2 x^2+x^3) \log (x) \log ^2(3 x) \log (\log (x))) \log (\frac {-1-x \log (\log (x))}{-2 x+x^2}) \log (\log (\frac {-1-x \log (\log (x))}{-2 x+x^2}))} \, dx\)

Optimal. Leaf size=27 \[ \frac {x+\log \left (\log \left (\log \left (\frac {\frac {1}{x}+\log (\log (x))}{2-x}\right )\right )\right )}{\log (3 x)} \]

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Rubi [F]  time = 31.29, 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 {\left (-2 x+x^2+(2-2 x) \log (x)\right ) \log (3 x)-x^2 \log (x) \log (3 x) \log (\log (x))+\left (\left (2 x-x^2\right ) \log (x)+\left (-2 x+x^2\right ) \log (x) \log (3 x)+\left (\left (2 x^2-x^3\right ) \log (x)+\left (-2 x^2+x^3\right ) \log (x) \log (3 x)\right ) \log (\log (x))\right ) \log \left (\frac {-1-x \log (\log (x))}{-2 x+x^2}\right ) \log \left (\log \left (\frac {-1-x \log (\log (x))}{-2 x+x^2}\right )\right )+\left ((2-x) \log (x)+\left (2 x-x^2\right ) \log (x) \log (\log (x))\right ) \log \left (\frac {-1-x \log (\log (x))}{-2 x+x^2}\right ) \log \left (\log \left (\frac {-1-x \log (\log (x))}{-2 x+x^2}\right )\right ) \log \left (\log \left (\log \left (\frac {-1-x \log (\log (x))}{-2 x+x^2}\right )\right )\right )}{\left (\left (-2 x+x^2\right ) \log (x) \log ^2(3 x)+\left (-2 x^2+x^3\right ) \log (x) \log ^2(3 x) \log (\log (x))\right ) \log \left (\frac {-1-x \log (\log (x))}{-2 x+x^2}\right ) \log \left (\log \left (\frac {-1-x \log (\log (x))}{-2 x+x^2}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1+\log (3 x)+\frac {((-2+x) x-2 (-1+x) \log (x)) \log (3 x)}{(-2+x) x \log (x) (1+x \log (\log (x))) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )}-\frac {x \log (3 x) \log (\log (x))}{(-2+x) (1+x \log (\log (x))) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )}-\frac {\log \left (\log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )\right )}{x}}{\log ^2(3 x)} \, dx\\ &=\int \left (\frac {-2 x \log (3 x)+x^2 \log (3 x)+2 \log (x) \log (3 x)-2 x \log (x) \log (3 x)-x^2 \log (x) \log (3 x) \log (\log (x))+2 x \log (x) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )-x^2 \log (x) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )-2 x \log (x) \log (3 x) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )+x^2 \log (x) \log (3 x) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )+2 x^2 \log (x) \log (\log (x)) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )-x^3 \log (x) \log (\log (x)) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )-2 x^2 \log (x) \log (3 x) \log (\log (x)) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )+x^3 \log (x) \log (3 x) \log (\log (x)) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )}{(-2+x) x \log (x) \log ^2(3 x) (1+x \log (\log (x))) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )}-\frac {\log \left (\log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )\right )}{x \log ^2(3 x)}\right ) \, dx\\ &=\int \frac {-2 x \log (3 x)+x^2 \log (3 x)+2 \log (x) \log (3 x)-2 x \log (x) \log (3 x)-x^2 \log (x) \log (3 x) \log (\log (x))+2 x \log (x) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )-x^2 \log (x) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )-2 x \log (x) \log (3 x) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )+x^2 \log (x) \log (3 x) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )+2 x^2 \log (x) \log (\log (x)) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )-x^3 \log (x) \log (\log (x)) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )-2 x^2 \log (x) \log (3 x) \log (\log (x)) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )+x^3 \log (x) \log (3 x) \log (\log (x)) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )}{(-2+x) x \log (x) \log ^2(3 x) (1+x \log (\log (x))) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )} \, dx-\int \frac {\log \left (\log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )\right )}{x \log ^2(3 x)} \, dx\\ &=\int \frac {-1+\frac {\log (3 x) \left ((-2+x) x+\log (x) \left (2-2 x+(-2+x) x \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )+x^2 \log (\log (x)) \left (-1+(-2+x) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )\right )\right )\right )}{(-2+x) x \log (x) (1+x \log (\log (x))) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )}}{\log ^2(3 x)} \, dx-\int \frac {\log \left (\log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )\right )}{x \log ^2(3 x)} \, dx\\ &=\int \left (\frac {-1+\log (3 x)}{\log ^2(3 x)}+\frac {-2 x+x^2+2 \log (x)-2 x \log (x)-x^2 \log (x) \log (\log (x))}{(-2+x) x \log (x) \log (3 x) (1+x \log (\log (x))) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )}\right ) \, dx-\int \frac {\log \left (\log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )\right )}{x \log ^2(3 x)} \, dx\\ &=\int \frac {-1+\log (3 x)}{\log ^2(3 x)} \, dx+\int \frac {-2 x+x^2+2 \log (x)-2 x \log (x)-x^2 \log (x) \log (\log (x))}{(-2+x) x \log (x) \log (3 x) (1+x \log (\log (x))) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )} \, dx-\int \frac {\log \left (\log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )\right )}{x \log ^2(3 x)} \, dx\\ &=\int \left (-\frac {1}{\log ^2(3 x)}+\frac {1}{\log (3 x)}\right ) \, dx+\int \left (\frac {-2 x+x^2+2 \log (x)-2 x \log (x)-x^2 \log (x) \log (\log (x))}{2 (-2+x) \log (x) \log (3 x) (1+x \log (\log (x))) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )}+\frac {2 x-x^2-2 \log (x)+2 x \log (x)+x^2 \log (x) \log (\log (x))}{2 x \log (x) \log (3 x) (1+x \log (\log (x))) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )}\right ) \, dx-\int \frac {\log \left (\log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )\right )}{x \log ^2(3 x)} \, dx\\ &=\frac {1}{2} \int \frac {-2 x+x^2+2 \log (x)-2 x \log (x)-x^2 \log (x) \log (\log (x))}{(-2+x) \log (x) \log (3 x) (1+x \log (\log (x))) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )} \, dx+\frac {1}{2} \int \frac {2 x-x^2-2 \log (x)+2 x \log (x)+x^2 \log (x) \log (\log (x))}{x \log (x) \log (3 x) (1+x \log (\log (x))) \log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right ) \log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )} \, dx-\int \frac {1}{\log ^2(3 x)} \, dx+\int \frac {1}{\log (3 x)} \, dx-\int \frac {\log \left (\log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )\right )}{x \log ^2(3 x)} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.47, size = 36, normalized size = 1.33 \begin {gather*} \frac {x}{\log (3 x)}+\frac {\log \left (\log \left (\log \left (-\frac {1+x \log (\log (x))}{(-2+x) x}\right )\right )\right )}{\log (3 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-2*x + x^2 + (2 - 2*x)*Log[x])*Log[3*x] - x^2*Log[x]*Log[3*x]*Log[Log[x]] + ((2*x - x^2)*Log[x] +
(-2*x + x^2)*Log[x]*Log[3*x] + ((2*x^2 - x^3)*Log[x] + (-2*x^2 + x^3)*Log[x]*Log[3*x])*Log[Log[x]])*Log[(-1 -
x*Log[Log[x]])/(-2*x + x^2)]*Log[Log[(-1 - x*Log[Log[x]])/(-2*x + x^2)]] + ((2 - x)*Log[x] + (2*x - x^2)*Log[x
]*Log[Log[x]])*Log[(-1 - x*Log[Log[x]])/(-2*x + x^2)]*Log[Log[(-1 - x*Log[Log[x]])/(-2*x + x^2)]]*Log[Log[Log[
(-1 - x*Log[Log[x]])/(-2*x + x^2)]]])/(((-2*x + x^2)*Log[x]*Log[3*x]^2 + (-2*x^2 + x^3)*Log[x]*Log[3*x]^2*Log[
Log[x]])*Log[(-1 - x*Log[Log[x]])/(-2*x + x^2)]*Log[Log[(-1 - x*Log[Log[x]])/(-2*x + x^2)]]),x]

[Out]

x/Log[3*x] + Log[Log[Log[-((1 + x*Log[Log[x]])/((-2 + x)*x))]]]/Log[3*x]

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fricas [A]  time = 1.31, size = 31, normalized size = 1.15 \begin {gather*} \frac {x + \log \left (\log \left (\log \left (-\frac {x \log \left (\log \relax (x)\right ) + 1}{x^{2} - 2 \, x}\right )\right )\right )}{\log \relax (3) + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2+2*x)*log(x)*log(log(x))+(2-x)*log(x))*log((-x*log(log(x))-1)/(x^2-2*x))*log(log((-x*log(log(
x))-1)/(x^2-2*x)))*log(log(log((-x*log(log(x))-1)/(x^2-2*x))))+(((x^3-2*x^2)*log(x)*log(3*x)+(-x^3+2*x^2)*log(
x))*log(log(x))+(x^2-2*x)*log(x)*log(3*x)+(-x^2+2*x)*log(x))*log((-x*log(log(x))-1)/(x^2-2*x))*log(log((-x*log
(log(x))-1)/(x^2-2*x)))-x^2*log(x)*log(3*x)*log(log(x))+((-2*x+2)*log(x)+x^2-2*x)*log(3*x))/((x^3-2*x^2)*log(x
)*log(3*x)^2*log(log(x))+(x^2-2*x)*log(x)*log(3*x)^2)/log((-x*log(log(x))-1)/(x^2-2*x))/log(log((-x*log(log(x)
)-1)/(x^2-2*x))),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 2.66, size = 40, normalized size = 1.48 \begin {gather*} \frac {x}{\log \relax (3) + \log \relax (x)} + \frac {\log \left (\log \left (\log \left (-x \log \left (\log \relax (x)\right ) - 1\right ) - \log \left (x - 2\right ) - \log \relax (x)\right )\right )}{\log \relax (3) + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2+2*x)*log(x)*log(log(x))+(2-x)*log(x))*log((-x*log(log(x))-1)/(x^2-2*x))*log(log((-x*log(log(
x))-1)/(x^2-2*x)))*log(log(log((-x*log(log(x))-1)/(x^2-2*x))))+(((x^3-2*x^2)*log(x)*log(3*x)+(-x^3+2*x^2)*log(
x))*log(log(x))+(x^2-2*x)*log(x)*log(3*x)+(-x^2+2*x)*log(x))*log((-x*log(log(x))-1)/(x^2-2*x))*log(log((-x*log
(log(x))-1)/(x^2-2*x)))-x^2*log(x)*log(3*x)*log(log(x))+((-2*x+2)*log(x)+x^2-2*x)*log(3*x))/((x^3-2*x^2)*log(x
)*log(3*x)^2*log(log(x))+(x^2-2*x)*log(x)*log(3*x)^2)/log((-x*log(log(x))-1)/(x^2-2*x))/log(log((-x*log(log(x)
)-1)/(x^2-2*x))),x, algorithm="giac")

[Out]

x/(log(3) + log(x)) + log(log(log(-x*log(log(x)) - 1) - log(x - 2) - log(x)))/(log(3) + log(x))

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maple [C]  time = 0.98, size = 274, normalized size = 10.15

method	result	size
risch	\(\frac {2 i \ln \left (\ln \left (i \pi -\ln \relax (x )-\ln \left (x -2\right )+\ln \left (x \ln \left (\ln \relax (x )\right )+1\right )-\frac {i \pi \,\mathrm {csgn}\left (\frac {i \left (x \ln \left (\ln \relax (x )\right )+1\right )}{x -2}\right ) \left (-\mathrm {csgn}\left (\frac {i \left (x \ln \left (\ln \relax (x )\right )+1\right )}{x -2}\right )+\mathrm {csgn}\left (\frac {i}{x -2}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \left (x \ln \left (\ln \relax (x )\right )+1\right )}{x -2}\right )+\mathrm {csgn}\left (i \left (x \ln \left (\ln \relax (x )\right )+1\right )\right )\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i \left (x \ln \left (\ln \relax (x )\right )+1\right )}{x \left (x -2\right )}\right ) \left (-\mathrm {csgn}\left (\frac {i \left (x \ln \left (\ln \relax (x )\right )+1\right )}{x \left (x -2\right )}\right )+\mathrm {csgn}\left (\frac {i}{x}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \left (x \ln \left (\ln \relax (x )\right )+1\right )}{x \left (x -2\right )}\right )+\mathrm {csgn}\left (\frac {i \left (x \ln \left (\ln \relax (x )\right )+1\right )}{x -2}\right )\right )}{2}+i \pi \mathrm {csgn}\left (\frac {i \left (x \ln \left (\ln \relax (x )\right )+1\right )}{x \left (x -2\right )}\right )^{2} \left (\mathrm {csgn}\left (\frac {i \left (x \ln \left (\ln \relax (x )\right )+1\right )}{x \left (x -2\right )}\right )-1\right )\right )\right )}{2 i \ln \relax (3)+2 i \ln \relax (x )}+\frac {2 i x}{2 i \ln \relax (3)+2 i \ln \relax (x )}\)	\(274\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^2+2*x)*ln(x)*ln(ln(x))+(2-x)*ln(x))*ln((-x*ln(ln(x))-1)/(x^2-2*x))*ln(ln((-x*ln(ln(x))-1)/(x^2-2*x))
)*ln(ln(ln((-x*ln(ln(x))-1)/(x^2-2*x))))+(((x^3-2*x^2)*ln(x)*ln(3*x)+(-x^3+2*x^2)*ln(x))*ln(ln(x))+(x^2-2*x)*l
n(x)*ln(3*x)+(-x^2+2*x)*ln(x))*ln((-x*ln(ln(x))-1)/(x^2-2*x))*ln(ln((-x*ln(ln(x))-1)/(x^2-2*x)))-x^2*ln(x)*ln(
3*x)*ln(ln(x))+((-2*x+2)*ln(x)+x^2-2*x)*ln(3*x))/((x^3-2*x^2)*ln(x)*ln(3*x)^2*ln(ln(x))+(x^2-2*x)*ln(x)*ln(3*x
)^2)/ln((-x*ln(ln(x))-1)/(x^2-2*x))/ln(ln((-x*ln(ln(x))-1)/(x^2-2*x))),x,method=_RETURNVERBOSE)

[Out]

2*I/(2*I*ln(3)+2*I*ln(x))*ln(ln(I*Pi-ln(x)-ln(x-2)+ln(x*ln(ln(x))+1)-1/2*I*Pi*csgn(I/(x-2)*(x*ln(ln(x))+1))*(-
csgn(I/(x-2)*(x*ln(ln(x))+1))+csgn(I/(x-2)))*(-csgn(I/(x-2)*(x*ln(ln(x))+1))+csgn(I*(x*ln(ln(x))+1)))-1/2*I*Pi
*csgn(I/x/(x-2)*(x*ln(ln(x))+1))*(-csgn(I/x/(x-2)*(x*ln(ln(x))+1))+csgn(I/x))*(-csgn(I/x/(x-2)*(x*ln(ln(x))+1)
)+csgn(I/(x-2)*(x*ln(ln(x))+1)))+I*Pi*csgn(I/x/(x-2)*(x*ln(ln(x))+1))^2*(csgn(I/x/(x-2)*(x*ln(ln(x))+1))-1)))+
2*I*x/(2*I*ln(3)+2*I*ln(x))

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maxima [A]  time = 0.67, size = 33, normalized size = 1.22 \begin {gather*} \frac {x + \log \left (\log \left (\log \left (x \log \left (\log \relax (x)\right ) + 1\right ) - \log \relax (x) - \log \left (-x + 2\right )\right )\right )}{\log \relax (3) + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2+2*x)*log(x)*log(log(x))+(2-x)*log(x))*log((-x*log(log(x))-1)/(x^2-2*x))*log(log((-x*log(log(
x))-1)/(x^2-2*x)))*log(log(log((-x*log(log(x))-1)/(x^2-2*x))))+(((x^3-2*x^2)*log(x)*log(3*x)+(-x^3+2*x^2)*log(
x))*log(log(x))+(x^2-2*x)*log(x)*log(3*x)+(-x^2+2*x)*log(x))*log((-x*log(log(x))-1)/(x^2-2*x))*log(log((-x*log
(log(x))-1)/(x^2-2*x)))-x^2*log(x)*log(3*x)*log(log(x))+((-2*x+2)*log(x)+x^2-2*x)*log(3*x))/((x^3-2*x^2)*log(x
)*log(3*x)^2*log(log(x))+(x^2-2*x)*log(x)*log(3*x)^2)/log((-x*log(log(x))-1)/(x^2-2*x))/log(log((-x*log(log(x)
)-1)/(x^2-2*x))),x, algorithm="maxima")

[Out]

(x + log(log(log(x*log(log(x)) + 1) - log(x) - log(-x + 2))))/(log(3) + log(x))

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mupad [B]  time = 13.19, size = 31, normalized size = 1.15 \begin {gather*} \frac {x+\ln \left (\ln \left (\ln \left (\frac {x\,\ln \left (\ln \relax (x)\right )+1}{2\,x-x^2}\right )\right )\right )}{\ln \left (3\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**2+2*x)*ln(x)*ln(ln(x))+(2-x)*ln(x))*ln((-x*ln(ln(x))-1)/(x**2-2*x))*ln(ln((-x*ln(ln(x))-1)/(x
**2-2*x)))*ln(ln(ln((-x*ln(ln(x))-1)/(x**2-2*x))))+(((x**3-2*x**2)*ln(x)*ln(3*x)+(-x**3+2*x**2)*ln(x))*ln(ln(x
))+(x**2-2*x)*ln(x)*ln(3*x)+(-x**2+2*x)*ln(x))*ln((-x*ln(ln(x))-1)/(x**2-2*x))*ln(ln((-x*ln(ln(x))-1)/(x**2-2*
x)))-x**2*ln(x)*ln(3*x)*ln(ln(x))+((-2*x+2)*ln(x)+x**2-2*x)*ln(3*x))/((x**3-2*x**2)*ln(x)*ln(3*x)**2*ln(ln(x))
+(x**2-2*x)*ln(x)*ln(3*x)**2)/ln((-x*ln(ln(x))-1)/(x**2-2*x))/ln(ln((-x*ln(ln(x))-1)/(x**2-2*x))),x)

[Out]

Timed out

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