Optimal. Leaf size=23 \[ \log \left (\frac {-5+\frac {x}{\frac {1}{2}+\log (2+2 x)}}{-3+x}\right ) \]
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Rubi [F] time = 4.54, 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 {1-11 x+4 x^2+(-8-8 x) \log (2+2 x)+(-20-20 x) \log ^2(2+2 x)}{-15-4 x+9 x^2-2 x^3+\left (-60-28 x+28 x^2-4 x^3\right ) \log (2+2 x)+\left (-60-40 x+20 x^2\right ) \log ^2(2+2 x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1+11 x-4 x^2-(-8-8 x) \log (2+2 x)-(-20-20 x) \log ^2(2+2 x)}{\left (3+2 x-x^2\right ) (1+2 \log (2 (1+x))) (5-2 x+10 \log (2 (1+x)))} \, dx\\ &=\int \left (-\frac {1}{(-3+x) (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}+\frac {11 x}{(-3+x) (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}-\frac {4 x^2}{(-3+x) (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}+\frac {8 \log (2+2 x)}{(3-x) (1+\log (4)+2 \log (1+x)) (5-2 x+10 \log (2 (1+x)))}+\frac {20 \log ^2(2+2 x)}{(3-x) (1+\log (4)+2 \log (1+x)) (5-2 x+10 \log (2 (1+x)))}\right ) \, dx\\ &=-\left (4 \int \frac {x^2}{(-3+x) (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx\right )+8 \int \frac {\log (2+2 x)}{(3-x) (1+\log (4)+2 \log (1+x)) (5-2 x+10 \log (2 (1+x)))} \, dx+11 \int \frac {x}{(-3+x) (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx+20 \int \frac {\log ^2(2+2 x)}{(3-x) (1+\log (4)+2 \log (1+x)) (5-2 x+10 \log (2 (1+x)))} \, dx-\int \frac {1}{(-3+x) (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx\\ &=-\left (4 \int \left (\frac {1}{(1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}+\frac {9}{4 (-3+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}-\frac {1}{4 (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}\right ) \, dx\right )+8 \int \frac {\log (2+2 x)}{(3-x) (1+\log (4)+2 \log (1+x)) (5-2 x+10 \log (2 (1+x)))} \, dx+11 \int \left (\frac {3}{4 (-3+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}+\frac {1}{4 (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}\right ) \, dx+20 \int \frac {\log ^2(2+2 x)}{(3-x) (1+\log (4)+2 \log (1+x)) (5-2 x+10 \log (2 (1+x)))} \, dx-\int \left (\frac {1}{4 (-3+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}-\frac {1}{4 (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}\right ) \, dx\\ &=-\left (\frac {1}{4} \int \frac {1}{(-3+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx\right )+\frac {1}{4} \int \frac {1}{(1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx+\frac {11}{4} \int \frac {1}{(1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx-4 \int \frac {1}{(1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx+8 \int \frac {\log (2+2 x)}{(3-x) (1+\log (4)+2 \log (1+x)) (5-2 x+10 \log (2 (1+x)))} \, dx+\frac {33}{4} \int \frac {1}{(-3+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx-9 \int \frac {1}{(-3+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx+20 \int \frac {\log ^2(2+2 x)}{(3-x) (1+\log (4)+2 \log (1+x)) (5-2 x+10 \log (2 (1+x)))} \, dx+\int \frac {1}{(1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [F] time = 0.53, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1-11 x+4 x^2+(-8-8 x) \log (2+2 x)+(-20-20 x) \log ^2(2+2 x)}{-15-4 x+9 x^2-2 x^3+\left (-60-28 x+28 x^2-4 x^3\right ) \log (2+2 x)+\left (-60-40 x+20 x^2\right ) \log ^2(2+2 x)} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.65, size = 34, normalized size = 1.48 \begin {gather*} -\log \left (x - 3\right ) + \log \left (-2 \, x + 10 \, \log \left (2 \, x + 2\right ) + 5\right ) - \log \left (2 \, \log \left (2 \, x + 2\right ) + 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.43, size = 34, normalized size = 1.48 \begin {gather*} -\log \left (x - 3\right ) + \log \left (-2 \, x + 10 \, \log \left (2 \, x + 2\right ) + 5\right ) - \log \left (2 \, \log \left (2 \, x + 2\right ) + 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 31, normalized size = 1.35
method | result | size |
risch | \(-\ln \left (x -3\right )+\ln \left (\ln \left (2 x +2\right )-\frac {x}{5}+\frac {1}{2}\right )-\ln \left (\frac {1}{2}+\ln \left (2 x +2\right )\right )\) | \(31\) |
norman | \(-\ln \left (x -3\right )-\ln \left (2 \ln \left (2 x +2\right )+1\right )+\ln \left (2 x -10 \ln \left (2 x +2\right )-5\right )\) | \(35\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.72, size = 30, normalized size = 1.30 \begin {gather*} -\log \left (x - 3\right ) + \log \left (-\frac {1}{5} \, x + \log \relax (2) + \log \left (x + 1\right ) + \frac {1}{2}\right ) - \log \left (\log \relax (2) + \log \left (x + 1\right ) + \frac {1}{2}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.46, size = 51, normalized size = 2.22 \begin {gather*} \ln \left (\frac {10\,\ln \left (2\,x+2\right )-2\,x+5}{x+1}\right )-\ln \left (\frac {\left (2\,\ln \left (2\,x+2\right )+1\right )\,\left (x-4\right )}{x+1}\right )-2\,\mathrm {atanh}\left (2\,x-7\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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