3.75.19 \(\int \frac {-240+264 x+(48-48 x) \log (x)+(-12+12 x) \log (\frac {2}{1-2 x+x^2})}{-256+256 x+(-16+16 x) \log ^2(x)+(-32+32 x) \log (\frac {2}{1-2 x+x^2})+(-1+x) \log ^2(\frac {2}{1-2 x+x^2})+\log (x) (128-128 x+(8-8 x) \log (\frac {2}{1-2 x+x^2}))} \, dx\)

Optimal. Leaf size=23 \[ \frac {3 x}{4+\frac {1}{4} \log \left (\frac {2}{(-1+x)^2}\right )-\log (x)} \]

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Rubi [F]  time = 0.51, 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 {-240+264 x+(48-48 x) \log (x)+(-12+12 x) \log \left (\frac {2}{1-2 x+x^2}\right )}{-256+256 x+(-16+16 x) \log ^2(x)+(-32+32 x) \log \left (\frac {2}{1-2 x+x^2}\right )+(-1+x) \log ^2\left (\frac {2}{1-2 x+x^2}\right )+\log (x) \left (128-128 x+(8-8 x) \log \left (\frac {2}{1-2 x+x^2}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-240 + 264*x + (48 - 48*x)*Log[x] + (-12 + 12*x)*Log[2/(1 - 2*x + x^2)])/(-256 + 256*x + (-16 + 16*x)*Log
[x]^2 + (-32 + 32*x)*Log[2/(1 - 2*x + x^2)] + (-1 + x)*Log[2/(1 - 2*x + x^2)]^2 + Log[x]*(128 - 128*x + (8 - 8
*x)*Log[2/(1 - 2*x + x^2)])),x]

[Out]

72*Defer[Int][(16 + Log[2/(-1 + x)^2] - 4*Log[x])^(-2), x] + 24*Defer[Int][1/((-1 + x)*(16 + Log[2/(-1 + x)^2]
 - 4*Log[x])^2), x] + 12*Defer[Int][(16 + Log[2/(-1 + x)^2] - 4*Log[x])^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {12 \left (20-22 x-(-1+x) \log \left (\frac {2}{(-1+x)^2}\right )+4 (-1+x) \log (x)\right )}{(1-x) \left (16+\log \left (\frac {2}{(-1+x)^2}\right )-4 \log (x)\right )^2} \, dx\\ &=12 \int \frac {20-22 x-(-1+x) \log \left (\frac {2}{(-1+x)^2}\right )+4 (-1+x) \log (x)}{(1-x) \left (16+\log \left (\frac {2}{(-1+x)^2}\right )-4 \log (x)\right )^2} \, dx\\ &=12 \int \left (\frac {2 (-2+3 x)}{(-1+x) \left (16+\log \left (\frac {2}{(-1+x)^2}\right )-4 \log (x)\right )^2}+\frac {1}{16+\log \left (\frac {2}{(-1+x)^2}\right )-4 \log (x)}\right ) \, dx\\ &=12 \int \frac {1}{16+\log \left (\frac {2}{(-1+x)^2}\right )-4 \log (x)} \, dx+24 \int \frac {-2+3 x}{(-1+x) \left (16+\log \left (\frac {2}{(-1+x)^2}\right )-4 \log (x)\right )^2} \, dx\\ &=12 \int \frac {1}{16+\log \left (\frac {2}{(-1+x)^2}\right )-4 \log (x)} \, dx+24 \int \left (\frac {3}{\left (16+\log \left (\frac {2}{(-1+x)^2}\right )-4 \log (x)\right )^2}+\frac {1}{(-1+x) \left (16+\log \left (\frac {2}{(-1+x)^2}\right )-4 \log (x)\right )^2}\right ) \, dx\\ &=12 \int \frac {1}{16+\log \left (\frac {2}{(-1+x)^2}\right )-4 \log (x)} \, dx+24 \int \frac {1}{(-1+x) \left (16+\log \left (\frac {2}{(-1+x)^2}\right )-4 \log (x)\right )^2} \, dx+72 \int \frac {1}{\left (16+\log \left (\frac {2}{(-1+x)^2}\right )-4 \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 19, normalized size = 0.83 \begin {gather*} \frac {12 x}{16+\log \left (\frac {2}{(-1+x)^2}\right )-4 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-240 + 264*x + (48 - 48*x)*Log[x] + (-12 + 12*x)*Log[2/(1 - 2*x + x^2)])/(-256 + 256*x + (-16 + 16*
x)*Log[x]^2 + (-32 + 32*x)*Log[2/(1 - 2*x + x^2)] + (-1 + x)*Log[2/(1 - 2*x + x^2)]^2 + Log[x]*(128 - 128*x +
(8 - 8*x)*Log[2/(1 - 2*x + x^2)])),x]

[Out]

(12*x)/(16 + Log[2/(-1 + x)^2] - 4*Log[x])

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fricas [A]  time = 0.65, size = 26, normalized size = 1.13 \begin {gather*} -\frac {12 \, x}{4 \, \log \relax (x) - \log \left (\frac {2}{x^{2} - 2 \, x + 1}\right ) - 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-48*x+48)*log(x)+(12*x-12)*log(2/(x^2-2*x+1))+264*x-240)/((16*x-16)*log(x)^2+((-8*x+8)*log(2/(x^2-
2*x+1))-128*x+128)*log(x)+(x-1)*log(2/(x^2-2*x+1))^2+(32*x-32)*log(2/(x^2-2*x+1))+256*x-256),x, algorithm="fri
cas")

[Out]

-12*x/(4*log(x) - log(2/(x^2 - 2*x + 1)) - 16)

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giac [A]  time = 0.49, size = 24, normalized size = 1.04 \begin {gather*} \frac {12 \, x}{\log \relax (2) - \log \left (x^{2} - 2 \, x + 1\right ) - 4 \, \log \relax (x) + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-48*x+48)*log(x)+(12*x-12)*log(2/(x^2-2*x+1))+264*x-240)/((16*x-16)*log(x)^2+((-8*x+8)*log(2/(x^2-
2*x+1))-128*x+128)*log(x)+(x-1)*log(2/(x^2-2*x+1))^2+(32*x-32)*log(2/(x^2-2*x+1))+256*x-256),x, algorithm="gia
c")

[Out]

12*x/(log(2) - log(x^2 - 2*x + 1) - 4*log(x) + 16)

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maple [C]  time = 0.10, size = 83, normalized size = 3.61




method result size



risch \(\frac {24 i x}{-\pi \mathrm {csgn}\left (i \left (x -1\right )\right )^{2} \mathrm {csgn}\left (i \left (x -1\right )^{2}\right )+2 \pi \,\mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i \left (x -1\right )^{2}\right )^{2}-\pi \mathrm {csgn}\left (i \left (x -1\right )^{2}\right )^{3}+2 i \ln \relax (2)-8 i \ln \relax (x )-4 i \ln \left (x -1\right )+32 i}\) \(83\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-48*x+48)*ln(x)+(12*x-12)*ln(2/(x^2-2*x+1))+264*x-240)/((16*x-16)*ln(x)^2+((-8*x+8)*ln(2/(x^2-2*x+1))-12
8*x+128)*ln(x)+(x-1)*ln(2/(x^2-2*x+1))^2+(32*x-32)*ln(2/(x^2-2*x+1))+256*x-256),x,method=_RETURNVERBOSE)

[Out]

24*I*x/(-Pi*csgn(I*(x-1))^2*csgn(I*(x-1)^2)+2*Pi*csgn(I*(x-1))*csgn(I*(x-1)^2)^2-Pi*csgn(I*(x-1)^2)^3+2*I*ln(2
)-8*I*ln(x)-4*I*ln(x-1)+32*I)

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maxima [A]  time = 0.52, size = 19, normalized size = 0.83 \begin {gather*} \frac {12 \, x}{\log \relax (2) - 2 \, \log \left (x - 1\right ) - 4 \, \log \relax (x) + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-48*x+48)*log(x)+(12*x-12)*log(2/(x^2-2*x+1))+264*x-240)/((16*x-16)*log(x)^2+((-8*x+8)*log(2/(x^2-
2*x+1))-128*x+128)*log(x)+(x-1)*log(2/(x^2-2*x+1))^2+(32*x-32)*log(2/(x^2-2*x+1))+256*x-256),x, algorithm="max
ima")

[Out]

12*x/(log(2) - 2*log(x - 1) - 4*log(x) + 16)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {264\,x-\ln \relax (x)\,\left (48\,x-48\right )+\ln \left (\frac {2}{x^2-2\,x+1}\right )\,\left (12\,x-12\right )-240}{256\,x-\ln \relax (x)\,\left (128\,x+\ln \left (\frac {2}{x^2-2\,x+1}\right )\,\left (8\,x-8\right )-128\right )+\ln \left (\frac {2}{x^2-2\,x+1}\right )\,\left (32\,x-32\right )+{\ln \left (\frac {2}{x^2-2\,x+1}\right )}^2\,\left (x-1\right )+{\ln \relax (x)}^2\,\left (16\,x-16\right )-256} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((264*x - log(x)*(48*x - 48) + log(2/(x^2 - 2*x + 1))*(12*x - 12) - 240)/(256*x - log(x)*(128*x + log(2/(x^
2 - 2*x + 1))*(8*x - 8) - 128) + log(2/(x^2 - 2*x + 1))*(32*x - 32) + log(2/(x^2 - 2*x + 1))^2*(x - 1) + log(x
)^2*(16*x - 16) - 256),x)

[Out]

int((264*x - log(x)*(48*x - 48) + log(2/(x^2 - 2*x + 1))*(12*x - 12) - 240)/(256*x - log(x)*(128*x + log(2/(x^
2 - 2*x + 1))*(8*x - 8) - 128) + log(2/(x^2 - 2*x + 1))*(32*x - 32) + log(2/(x^2 - 2*x + 1))^2*(x - 1) + log(x
)^2*(16*x - 16) - 256), x)

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sympy [A]  time = 0.35, size = 20, normalized size = 0.87 \begin {gather*} \frac {12 x}{- 4 \log {\relax (x )} + \log {\left (\frac {2}{x^{2} - 2 x + 1} \right )} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-48*x+48)*ln(x)+(12*x-12)*ln(2/(x**2-2*x+1))+264*x-240)/((16*x-16)*ln(x)**2+((-8*x+8)*ln(2/(x**2-2
*x+1))-128*x+128)*ln(x)+(x-1)*ln(2/(x**2-2*x+1))**2+(32*x-32)*ln(2/(x**2-2*x+1))+256*x-256),x)

[Out]

12*x/(-4*log(x) + log(2/(x**2 - 2*x + 1)) + 16)

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