Integrand size = 50, antiderivative size = 18 \[ \int \frac {256-127 x+16 x^2+\left (-128 x+32 x^2\right ) \log (2 x)}{x^2+\left (256 x-128 x^2+16 x^3\right ) \log (2 x)} \, dx=\log \left (4 x+64 (4-x)^2 \log (2 x)\right ) \]
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\[ \int \frac {256-127 x+16 x^2+\left (-128 x+32 x^2\right ) \log (2 x)}{x^2+\left (256 x-128 x^2+16 x^3\right ) \log (2 x)} \, dx=\int \frac {256-127 x+16 x^2+\left (-128 x+32 x^2\right ) \log (2 x)}{x^2+\left (256 x-128 x^2+16 x^3\right ) \log (2 x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2}{-4+x}+\frac {-1024+764 x-193 x^2+16 x^3}{(-4+x) x \left (x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)\right )}\right ) \, dx \\ & = 2 \log (4-x)+\int \frac {-1024+764 x-193 x^2+16 x^3}{(-4+x) x \left (x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)\right )} \, dx \\ & = 2 \log (4-x)+\int \left (-\frac {129}{x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)}-\frac {8}{(-4+x) \left (x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)\right )}+\frac {256}{x \left (x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)\right )}+\frac {16 x}{x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)}\right ) \, dx \\ & = 2 \log (4-x)-8 \int \frac {1}{(-4+x) \left (x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)\right )} \, dx+16 \int \frac {x}{x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)} \, dx-129 \int \frac {1}{x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)} \, dx+256 \int \frac {1}{x \left (x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)\right )} \, dx \\ & = 2 \log (4-x)-8 \int \frac {1}{(-4+x) \left (x+16 (-4+x)^2 \log (2 x)\right )} \, dx+16 \int \frac {x}{x+16 (-4+x)^2 \log (2 x)} \, dx-129 \int \frac {1}{x+16 (-4+x)^2 \log (2 x)} \, dx+256 \int \frac {1}{x \left (x+16 (-4+x)^2 \log (2 x)\right )} \, dx \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {256-127 x+16 x^2+\left (-128 x+32 x^2\right ) \log (2 x)}{x^2+\left (256 x-128 x^2+16 x^3\right ) \log (2 x)} \, dx=\log \left (x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)\right ) \]
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Time = 0.40 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44
method | result | size |
norman | \(\ln \left (16 x^{2} \ln \left (2 x \right )-128 x \ln \left (2 x \right )+256 \ln \left (2 x \right )+x \right )\) | \(26\) |
risch | \(2 \ln \left (x -4\right )+\ln \left (\ln \left (2 x \right )+\frac {x}{16 x^{2}-128 x +256}\right )\) | \(27\) |
parallelrisch | \(\ln \left (x^{2} \ln \left (2 x \right )-8 x \ln \left (2 x \right )+\frac {x}{16}+16 \ln \left (2 x \right )\right )\) | \(27\) |
derivativedivides | \(\ln \left (32 x^{2} \ln \left (2 x \right )-256 x \ln \left (2 x \right )+512 \ln \left (2 x \right )+2 x \right )\) | \(28\) |
default | \(\ln \left (32 x^{2} \ln \left (2 x \right )-256 x \ln \left (2 x \right )+512 \ln \left (2 x \right )+2 x \right )\) | \(28\) |
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Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (16) = 32\).
Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.94 \[ \int \frac {256-127 x+16 x^2+\left (-128 x+32 x^2\right ) \log (2 x)}{x^2+\left (256 x-128 x^2+16 x^3\right ) \log (2 x)} \, dx=2 \, \log \left (x - 4\right ) + \log \left (\frac {16 \, {\left (x^{2} - 8 \, x + 16\right )} \log \left (2 \, x\right ) + x}{x^{2} - 8 \, x + 16}\right ) \]
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Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {256-127 x+16 x^2+\left (-128 x+32 x^2\right ) \log (2 x)}{x^2+\left (256 x-128 x^2+16 x^3\right ) \log (2 x)} \, dx=2 \log {\left (x - 4 \right )} + \log {\left (\frac {x}{16 x^{2} - 128 x + 256} + \log {\left (2 x \right )} \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (16) = 32\).
Time = 0.32 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.94 \[ \int \frac {256-127 x+16 x^2+\left (-128 x+32 x^2\right ) \log (2 x)}{x^2+\left (256 x-128 x^2+16 x^3\right ) \log (2 x)} \, dx=2 \, \log \left (x - 4\right ) + \log \left (\frac {16 \, x^{2} \log \left (2\right ) - x {\left (128 \, \log \left (2\right ) - 1\right )} + 16 \, {\left (x^{2} - 8 \, x + 16\right )} \log \left (x\right ) + 256 \, \log \left (2\right )}{16 \, {\left (x^{2} - 8 \, x + 16\right )}}\right ) \]
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Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {256-127 x+16 x^2+\left (-128 x+32 x^2\right ) \log (2 x)}{x^2+\left (256 x-128 x^2+16 x^3\right ) \log (2 x)} \, dx=\log \left (16 \, x^{2} \log \left (2 \, x\right ) - 128 \, x \log \left (2 \, x\right ) + x + 256 \, \log \left (2 \, x\right )\right ) \]
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Time = 13.42 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {256-127 x+16 x^2+\left (-128 x+32 x^2\right ) \log (2 x)}{x^2+\left (256 x-128 x^2+16 x^3\right ) \log (2 x)} \, dx=\ln \left (x+256\,\ln \left (2\,x\right )-128\,x\,\ln \left (2\,x\right )+16\,x^2\,\ln \left (2\,x\right )\right ) \]
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