3.84.72 \(\int \frac {(-8+2 x) \log (x)+x \log (x) \log (x^2) \log (\log (x^2))+((-2+2 x \log (x)) \log (x^2)+(4-x+(-4 x+x^2) \log (x)) \log (x^2) \log (\log (x^2))) \log (2+(-4+x) \log (\log (x^2)))}{(2 x \log (x) \log (x^2)+(-4 x+x^2) \log (x) \log (x^2) \log (\log (x^2))) \log (2+(-4+x) \log (\log (x^2)))} \, dx\) [8372]

Optimal. Leaf size=23 \[ x-\log \left (\frac {4 \log (x)}{\log \left (2+(-4+x) \log \left (\log \left (x^2\right )\right )\right )}\right ) \]

[Out]

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

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Rubi [F]
time = 0.82, 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+2 x) \log (x)+x \log (x) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+\left ((-2+2 x \log (x)) \log \left (x^2\right )+\left (4-x+\left (-4 x+x^2\right ) \log (x)\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right ) \log \left (2+(-4+x) \log \left (\log \left (x^2\right )\right )\right )}{\left (2 x \log (x) \log \left (x^2\right )+\left (-4 x+x^2\right ) \log (x) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right ) \log \left (2+(-4+x) \log \left (\log \left (x^2\right )\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]
time = 0.22, size = 20, normalized size = 0.87 \begin {gather*} x-\log (\log (x))+\log \left (\log \left (2+(-4+x) \log \left (\log \left (x^2\right )\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-8 + 2*x)*Log[x] + x*Log[x]*Log[x^2]*Log[Log[x^2]] + ((-2 + 2*x*Log[x])*Log[x^2] + (4 - x + (-4*x
+ x^2)*Log[x])*Log[x^2]*Log[Log[x^2]])*Log[2 + (-4 + x)*Log[Log[x^2]]])/((2*x*Log[x]*Log[x^2] + (-4*x + x^2)*L
og[x]*Log[x^2]*Log[Log[x^2]])*Log[2 + (-4 + x)*Log[Log[x^2]]]),x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 2.12, size = 50, normalized size = 2.17

method	result	size
risch	\(x -\ln \left (\ln \left (x \right )\right )+\ln \left (\ln \left (\left (x -4\right ) \ln \left (2 \ln \left (x \right )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right )+2\right )\right )\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x-ln(ln(x))+ln(ln((x-4)*ln(2*ln(x)-1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2)+2))

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Maxima [A]
time = 0.57, size = 26, normalized size = 1.13 \begin {gather*} x + \log \left (\log \left (x \log \left (2\right ) + {\left (x - 4\right )} \log \left (\log \left (x\right )\right ) - 4 \, \log \left (2\right ) + 2\right )\right ) - \log \left (\log \left (x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.42, size = 22, normalized size = 0.96 \begin {gather*} x - \log \left (2 \, \log \left (x\right )\right ) + \log \left (\log \left ({\left (x - 4\right )} \log \left (2 \, \log \left (x\right )\right ) + 2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.52, size = 20, normalized size = 0.87 \begin {gather*} x - \log {\left (\log {\left (x \right )} \right )} + \log {\left (\log {\left (\left (x - 4\right ) \log {\left (2 \log {\left (x \right )} \right )} + 2 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x**2-4*x)*ln(x)-x+4)*ln(x**2)*ln(ln(x**2))+(2*x*ln(x)-2)*ln(x**2))*ln((x-4)*ln(ln(x**2))+2)+x*ln
(x)*ln(x**2)*ln(ln(x**2))+(2*x-8)*ln(x))/((x**2-4*x)*ln(x)*ln(x**2)*ln(ln(x**2))+2*x*ln(x)*ln(x**2))/ln((x-4)*
ln(ln(x**2))+2),x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Simplification assuming t_nostep near 0Simplification assuming t_nostep near 0Simplification assuming t_nos
tep near 0S

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Mupad [B]
time = 5.91, size = 20, normalized size = 0.87 \begin {gather*} x+\ln \left (\ln \left (\ln \left (\ln \left (x^2\right )\right )\,\left (x-4\right )+2\right )\right )-\ln \left (\ln \left (x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(log(x^2))*(x - 4) + 2)*(log(x^2)*(2*x*log(x) - 2) - log(x^2)*log(log(x^2))*(x + log(x)*(4*x - x^2
) - 4)) + log(x)*(2*x - 8) + x*log(x^2)*log(log(x^2))*log(x))/(log(log(log(x^2))*(x - 4) + 2)*(2*x*log(x^2)*lo
g(x) - log(x^2)*log(log(x^2))*log(x)*(4*x - x^2))),x)

[Out]

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

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