3.18.10 \(\int \frac {-2+4 x+(8-4 x) \log (-2+x)}{(10-25 x+10 x^2) \log (-2+x) \log (\frac {\log ^2(-2+x)}{1-4 x+4 x^2})} \, dx\)

Optimal. Leaf size=20 \[ \frac {1}{5} \log \left (\log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )\right ) \]

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Rubi [F]  time = 1.51, 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 {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2 + 4*x + (8 - 4*x)*Log[-2 + x])/((10 - 25*x + 10*x^2)*Log[-2 + x]*Log[Log[-2 + x]^2/(1 - 4*x + 4*x^2)])
,x]

[Out]

(-4*Defer[Int][1/((-1 + 2*x)*Log[Log[-2 + x]^2/(-1 + 2*x)^2]), x])/5 + (2*Defer[Int][1/((-2 + x)*Log[-2 + x]*L
og[Log[-2 + x]^2/(-1 + 2*x)^2]), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {2 (1-2 x-4 \log (-2+x)+2 x \log (-2+x))}{15 (-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {4 (1-2 x-4 \log (-2+x)+2 x \log (-2+x))}{15 (-1+2 x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}\right ) \, dx\\ &=-\left (\frac {2}{15} \int \frac {1-2 x-4 \log (-2+x)+2 x \log (-2+x)}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx\right )+\frac {4}{15} \int \frac {1-2 x-4 \log (-2+x)+2 x \log (-2+x)}{(-1+2 x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx\\ &=-\left (\frac {2}{15} \int \left (-\frac {4}{(-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {2 x}{(-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {1}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}-\frac {2 x}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}\right ) \, dx\right )+\frac {4}{15} \int \left (-\frac {4}{(-1+2 x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {2 x}{(-1+2 x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {1}{(-1+2 x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}-\frac {2 x}{(-1+2 x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}\right ) \, dx\\ &=-\left (\frac {2}{15} \int \frac {1}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx\right )-\frac {4}{15} \int \frac {x}{(-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx+\frac {4}{15} \int \frac {x}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx+\frac {4}{15} \int \frac {1}{(-1+2 x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx+\frac {8}{15} \int \frac {1}{(-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx+\frac {8}{15} \int \frac {x}{(-1+2 x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx-\frac {8}{15} \int \frac {x}{(-1+2 x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx-\frac {16}{15} \int \frac {1}{(-1+2 x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx\\ &=-\left (\frac {2}{15} \int \frac {1}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx\right )-\frac {4}{15} \int \left (\frac {1}{\log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {2}{(-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}\right ) \, dx+\frac {4}{15} \int \left (\frac {1}{\log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {2}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}\right ) \, dx+\frac {4}{15} \int \frac {1}{(-1+2 x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx+\frac {8}{15} \int \left (\frac {1}{2 \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {1}{2 (-1+2 x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}\right ) \, dx-\frac {8}{15} \int \left (\frac {1}{2 \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {1}{2 (-1+2 x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}\right ) \, dx+\frac {8}{15} \int \frac {1}{(-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx-\frac {16}{15} \int \frac {1}{(-1+2 x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx\\ &=-\left (\frac {2}{15} \int \frac {1}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx\right )+\frac {4}{15} \int \frac {1}{(-1+2 x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx+\frac {8}{15} \int \frac {1}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx-\frac {16}{15} \int \frac {1}{(-1+2 x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 20, normalized size = 1.00 \begin {gather*} \frac {1}{5} \log \left (\log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + 4*x + (8 - 4*x)*Log[-2 + x])/((10 - 25*x + 10*x^2)*Log[-2 + x]*Log[Log[-2 + x]^2/(1 - 4*x + 4*
x^2)]),x]

[Out]

Log[Log[Log[-2 + x]^2/(-1 + 2*x)^2]]/5

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fricas [A]  time = 0.83, size = 23, normalized size = 1.15 \begin {gather*} \frac {1}{5} \, \log \left (\log \left (\frac {\log \left (x - 2\right )^{2}}{4 \, x^{2} - 4 \, x + 1}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x+8)*log(x-2)+4*x-2)/(10*x^2-25*x+10)/log(x-2)/log(log(x-2)^2/(4*x^2-4*x+1)),x, algorithm="fric
as")

[Out]

1/5*log(log(log(x - 2)^2/(4*x^2 - 4*x + 1)))

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giac [A]  time = 0.34, size = 24, normalized size = 1.20 \begin {gather*} \frac {1}{5} \, \log \left (\log \left (4 \, x^{2} - 4 \, x + 1\right ) - \log \left (\log \left (x - 2\right )^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x+8)*log(x-2)+4*x-2)/(10*x^2-25*x+10)/log(x-2)/log(log(x-2)^2/(4*x^2-4*x+1)),x, algorithm="giac
")

[Out]

1/5*log(log(4*x^2 - 4*x + 1) - log(log(x - 2)^2))

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maple [A]  time = 0.49, size = 24, normalized size = 1.20

method	result	size
norman	\(\frac {\ln \left (\ln \left (\frac {\ln \left (x -2\right )^{2}}{4 x^{2}-4 x +1}\right )\right )}{5}\)	\(24\)
risch	\(\frac {\ln \left (\ln \left (\ln \left (x -2\right )\right )-\frac {i \left (\pi \,\mathrm {csgn}\left (\frac {i}{\left (x -\frac {1}{2}\right )^{2}}\right ) \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{\left (x -\frac {1}{2}\right )^{2}}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{\left (x -\frac {1}{2}\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{\left (x -\frac {1}{2}\right )^{2}}\right )^{2}-\pi \mathrm {csgn}\left (i \left (x -\frac {1}{2}\right )\right )^{2} \mathrm {csgn}\left (i \left (x -\frac {1}{2}\right )^{2}\right )+2 \pi \,\mathrm {csgn}\left (i \left (x -\frac {1}{2}\right )\right ) \mathrm {csgn}\left (i \left (x -\frac {1}{2}\right )^{2}\right )^{2}-\pi \mathrm {csgn}\left (i \left (x -\frac {1}{2}\right )^{2}\right )^{3}+\pi \mathrm {csgn}\left (i \ln \left (x -2\right )\right )^{2} \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \ln \left (x -2\right )\right ) \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )^{3}-\pi \,\mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{\left (x -\frac {1}{2}\right )^{2}}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{\left (x -\frac {1}{2}\right )^{2}}\right )^{3}-4 i \ln \left (x -\frac {1}{2}\right )-4 i \ln \relax (2)\right )}{4}\right )}{5}\)	\(255\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x+8)*ln(x-2)+4*x-2)/(10*x^2-25*x+10)/ln(x-2)/ln(ln(x-2)^2/(4*x^2-4*x+1)),x,method=_RETURNVERBOSE)

[Out]

1/5*ln(ln(ln(x-2)^2/(4*x^2-4*x+1)))

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maxima [A]  time = 0.55, size = 17, normalized size = 0.85 \begin {gather*} \frac {1}{5} \, \log \left (\log \left (2 \, x - 1\right ) - \log \left (\log \left (x - 2\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x+8)*log(x-2)+4*x-2)/(10*x^2-25*x+10)/log(x-2)/log(log(x-2)^2/(4*x^2-4*x+1)),x, algorithm="maxi
ma")

[Out]

1/5*log(log(2*x - 1) - log(log(x - 2)))

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mupad [B]  time = 1.65, size = 23, normalized size = 1.15 \begin {gather*} \frac {\ln \left (\ln \left (\frac {{\ln \left (x-2\right )}^2}{4\,x^2-4\,x+1}\right )\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x - 2)*(4*x - 8) - 4*x + 2)/(log(log(x - 2)^2/(4*x^2 - 4*x + 1))*log(x - 2)*(10*x^2 - 25*x + 10)),x)

[Out]

log(log(log(x - 2)^2/(4*x^2 - 4*x + 1)))/5

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sympy [A]  time = 0.74, size = 20, normalized size = 1.00 \begin {gather*} \frac {\log {\left (\log {\left (\frac {\log {\left (x - 2 \right )}^{2}}{4 x^{2} - 4 x + 1} \right )} \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x+8)*ln(x-2)+4*x-2)/(10*x**2-25*x+10)/ln(x-2)/ln(ln(x-2)**2/(4*x**2-4*x+1)),x)

[Out]

log(log(log(x - 2)**2/(4*x**2 - 4*x + 1)))/5

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