3.29.52 \(\int \frac {-4+2 x+(-2+x) \log (4)+(-6 x^2+2 x^3) \log (x)+(4+2 \log (4)) \log (x) \log (\frac {1}{\log (x)})}{(8-8 x+2 x^2+(4-4 x+x^2) \log (4)) \log (x)} \, dx\)

Optimal. Leaf size=29 \[ 4+\frac {-\frac {x^2}{2+\log (4)}+\log \left (\frac {1}{\log (x)}\right )}{-1+\frac {2}{x}} \]

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Rubi [F]  time = 0.58, 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 {-4+2 x+(-2+x) \log (4)+\left (-6 x^2+2 x^3\right ) \log (x)+(4+2 \log (4)) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (8-8 x+2 x^2+\left (4-4 x+x^2\right ) \log (4)\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4 + 2*x + (-2 + x)*Log[4] + (-6*x^2 + 2*x^3)*Log[x] + (4 + 2*Log[4])*Log[x]*Log[Log[x]^(-1)])/((8 - 8*x
+ 2*x^2 + (4 - 4*x + x^2)*Log[4])*Log[x]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {-2+x}{\log (x)}+2 \left (\frac {(-3+x) x^2}{2+\log (4)}+\log \left (\frac {1}{\log (x)}\right )\right )}{(2-x)^2} \, dx\\ &=\int \left (\frac {-4 (1+\log (2))+2 x (1+\log (2))-6 x^2 \log (x)+2 x^3 \log (x)}{(2-x)^2 (2+\log (4)) \log (x)}+\frac {2 \log \left (\frac {1}{\log (x)}\right )}{(-2+x)^2}\right ) \, dx\\ &=2 \int \frac {\log \left (\frac {1}{\log (x)}\right )}{(-2+x)^2} \, dx+\frac {\int \frac {-4 (1+\log (2))+2 x (1+\log (2))-6 x^2 \log (x)+2 x^3 \log (x)}{(2-x)^2 \log (x)} \, dx}{2+\log (4)}\\ &=2 \int \frac {\log \left (\frac {1}{\log (x)}\right )}{(-2+x)^2} \, dx+\frac {\int \frac {2 \left ((-2+x) (1+\log (2))+(-3+x) x^2 \log (x)\right )}{(2-x)^2 \log (x)} \, dx}{2+\log (4)}\\ &=2 \int \frac {\log \left (\frac {1}{\log (x)}\right )}{(-2+x)^2} \, dx+\frac {2 \int \frac {(-2+x) (1+\log (2))+(-3+x) x^2 \log (x)}{(2-x)^2 \log (x)} \, dx}{2+\log (4)}\\ &=2 \int \frac {\log \left (\frac {1}{\log (x)}\right )}{(-2+x)^2} \, dx+\frac {2 \int \left (\frac {(-3+x) x^2}{(-2+x)^2}+\frac {1+\log (2)}{(-2+x) \log (x)}\right ) \, dx}{2+\log (4)}\\ &=2 \int \frac {\log \left (\frac {1}{\log (x)}\right )}{(-2+x)^2} \, dx+\frac {2 \int \frac {(-3+x) x^2}{(-2+x)^2} \, dx}{2+\log (4)}+\frac {(2 (1+\log (2))) \int \frac {1}{(-2+x) \log (x)} \, dx}{2+\log (4)}\\ &=-\frac {x^3}{(2-x) (2+\log (4))}+2 \int \frac {\log \left (\frac {1}{\log (x)}\right )}{(-2+x)^2} \, dx+\frac {(2 (1+\log (2))) \int \frac {1}{(-2+x) \log (x)} \, dx}{2+\log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 42, normalized size = 1.45 \begin {gather*} \frac {8-4 x+x^3-2 (2+\log (4)) \log \left (\frac {1}{\log (x)}\right )+(-2+x) (2+\log (4)) \log (\log (x))}{(-2+x) (2+\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + 2*x + (-2 + x)*Log[4] + (-6*x^2 + 2*x^3)*Log[x] + (4 + 2*Log[4])*Log[x]*Log[Log[x]^(-1)])/((8
- 8*x + 2*x^2 + (4 - 4*x + x^2)*Log[4])*Log[x]),x]

[Out]

(8 - 4*x + x^3 - 2*(2 + Log[4])*Log[Log[x]^(-1)] + (-2 + x)*(2 + Log[4])*Log[Log[x]])/((-2 + x)*(2 + Log[4]))

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fricas [A]  time = 0.51, size = 34, normalized size = 1.17 \begin {gather*} \frac {x^{3} - 2 \, {\left (x \log \relax (2) + x\right )} \log \left (\frac {1}{\log \relax (x)}\right ) - 4 \, x + 8}{2 \, {\left ({\left (x - 2\right )} \log \relax (2) + x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(2)+4)*log(x)*log(1/log(x))+(2*x^3-6*x^2)*log(x)+2*(x-2)*log(2)+2*x-4)/(2*(x^2-4*x+4)*log(2)+
2*x^2-8*x+8)/log(x),x, algorithm="fricas")

[Out]

1/2*(x^3 - 2*(x*log(2) + x)*log(1/log(x)) - 4*x + 8)/((x - 2)*log(2) + x - 2)

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giac [A]  time = 0.71, size = 48, normalized size = 1.66 \begin {gather*} \frac {x^{2}}{2 \, {\left (\log \relax (2) + 1\right )}} + \frac {x}{\log \relax (2) + 1} + \frac {2 \, \log \left (\log \relax (x)\right )}{x - 2} + \frac {4}{x \log \relax (2) + x - 2 \, \log \relax (2) - 2} + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(2)+4)*log(x)*log(1/log(x))+(2*x^3-6*x^2)*log(x)+2*(x-2)*log(2)+2*x-4)/(2*(x^2-4*x+4)*log(2)+
2*x^2-8*x+8)/log(x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.54, size = 61, normalized size = 2.10

method	result	size
risch	\(\frac {2 \ln \left (\ln \relax (x )\right )}{x -2}+\frac {2 \ln \relax (2) \ln \left (\ln \relax (x )\right ) x +x^{3}-4 \ln \relax (2) \ln \left (\ln \relax (x )\right )+2 x \ln \left (\ln \relax (x )\right )-4 \ln \left (\ln \relax (x )\right )-4 x +8}{2 x \ln \relax (2)-4 \ln \relax (2)+2 x -4}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*ln(2)+4)*ln(x)*ln(1/ln(x))+(2*x^3-6*x^2)*ln(x)+2*(x-2)*ln(2)+2*x-4)/(2*(x^2-4*x+4)*ln(2)+2*x^2-8*x+8)/
ln(x),x,method=_RETURNVERBOSE)

[Out]

2/(x-2)*ln(ln(x))+1/2*(2*ln(2)*ln(ln(x))*x+x^3-4*ln(2)*ln(ln(x))+2*x*ln(ln(x))-4*ln(ln(x))-4*x+8)/(x*ln(2)-2*l
n(2)+x-2)

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maxima [A]  time = 0.67, size = 34, normalized size = 1.17 \begin {gather*} \frac {x^{3} + 2 \, x {\left (\log \relax (2) + 1\right )} \log \left (\log \relax (x)\right ) - 4 \, x + 8}{2 \, {\left (x {\left (\log \relax (2) + 1\right )} - 2 \, \log \relax (2) - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(2)+4)*log(x)*log(1/log(x))+(2*x^3-6*x^2)*log(x)+2*(x-2)*log(2)+2*x-4)/(2*(x^2-4*x+4)*log(2)+
2*x^2-8*x+8)/log(x),x, algorithm="maxima")

[Out]

1/2*(x^3 + 2*x*(log(2) + 1)*log(log(x)) - 4*x + 8)/(x*(log(2) + 1) - 2*log(2) - 2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {2\,x-\ln \relax (x)\,\left (6\,x^2-2\,x^3\right )+2\,\ln \relax (2)\,\left (x-2\right )+\ln \left (\frac {1}{\ln \relax (x)}\right )\,\ln \relax (x)\,\left (4\,\ln \relax (2)+4\right )-4}{\ln \relax (x)\,\left (2\,x^2-8\,x+2\,\ln \relax (2)\,\left (x^2-4\,x+4\right )+8\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - log(x)*(6*x^2 - 2*x^3) + 2*log(2)*(x - 2) + log(1/log(x))*log(x)*(4*log(2) + 4) - 4)/(log(x)*(2*x^2
- 8*x + 2*log(2)*(x^2 - 4*x + 4) + 8)),x)

[Out]

int((2*x - log(x)*(6*x^2 - 2*x^3) + 2*log(2)*(x - 2) + log(1/log(x))*log(x)*(4*log(2) + 4) - 4)/(log(x)*(2*x^2
- 8*x + 2*log(2)*(x^2 - 4*x + 4) + 8)), x)

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sympy [B]  time = 0.52, size = 48, normalized size = 1.66 \begin {gather*} \frac {x^{2}}{2 \log {\relax (2 )} + 2} + \frac {x}{\log {\relax (2 )} + 1} + \log {\left (\log {\relax (x )} \right )} + \frac {4}{x \left (\log {\relax (2 )} + 1\right ) - 2 - 2 \log {\relax (2 )}} - \frac {2 \log {\left (\frac {1}{\log {\relax (x )}} \right )}}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*ln(2)+4)*ln(x)*ln(1/ln(x))+(2*x**3-6*x**2)*ln(x)+2*(x-2)*ln(2)+2*x-4)/(2*(x**2-4*x+4)*ln(2)+2*x*
*2-8*x+8)/ln(x),x)

[Out]

x**2/(2*log(2) + 2) + x/(log(2) + 1) + log(log(x)) + 4/(x*(log(2) + 1) - 2 - 2*log(2)) - 2*log(1/log(x))/(x -
2)

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