Integrand size = 59, antiderivative size = 30 \[ \int \frac {-32+16 x+30 x^2-16 x^3+2 x^4+(64-16 x) \log \left (x^2\right )+(-16+8 x) \log ^2\left (x^2\right )}{16 x^2-8 x^3+x^4} \, dx=-6+2 x+\frac {2+\frac {4 \log ^2(x+(-1+x) x)}{4-x}}{x} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.33 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.73, number of steps used = 23, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.220, Rules used = {1608, 27, 6874, 46, 45, 2380, 2341, 2379, 2438, 2404, 2355, 2354, 2342} \[ \int \frac {-32+16 x+30 x^2-16 x^3+2 x^4+(64-16 x) \log \left (x^2\right )+(-16+8 x) \log ^2\left (x^2\right )}{16 x^2-8 x^3+x^4} \, dx=2 \operatorname {PolyLog}\left (2,\frac {4}{x}\right )+2 \operatorname {PolyLog}\left (2,\frac {x}{4}\right )+\frac {x \log ^2\left (x^2\right )}{4 (4-x)}+\frac {\log ^2\left (x^2\right )}{x}-\log \left (1-\frac {4}{x}\right ) \log \left (x^2\right )+\log \left (1-\frac {x}{4}\right ) \log \left (x^2\right )+2 x+\frac {2}{x} \]
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Rule 27
Rule 45
Rule 46
Rule 1608
Rule 2341
Rule 2342
Rule 2354
Rule 2355
Rule 2379
Rule 2380
Rule 2404
Rule 2438
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-32+16 x+30 x^2-16 x^3+2 x^4+(64-16 x) \log \left (x^2\right )+(-16+8 x) \log ^2\left (x^2\right )}{x^2 \left (16-8 x+x^2\right )} \, dx \\ & = \int \frac {-32+16 x+30 x^2-16 x^3+2 x^4+(64-16 x) \log \left (x^2\right )+(-16+8 x) \log ^2\left (x^2\right )}{(-4+x)^2 x^2} \, dx \\ & = \int \left (\frac {30}{(-4+x)^2}-\frac {32}{(-4+x)^2 x^2}+\frac {16}{(-4+x)^2 x}-\frac {16 x}{(-4+x)^2}+\frac {2 x^2}{(-4+x)^2}-\frac {16 \log \left (x^2\right )}{(-4+x) x^2}+\frac {8 (-2+x) \log ^2\left (x^2\right )}{(-4+x)^2 x^2}\right ) \, dx \\ & = \frac {30}{4-x}+2 \int \frac {x^2}{(-4+x)^2} \, dx+8 \int \frac {(-2+x) \log ^2\left (x^2\right )}{(-4+x)^2 x^2} \, dx+16 \int \frac {1}{(-4+x)^2 x} \, dx-16 \int \frac {x}{(-4+x)^2} \, dx-16 \int \frac {\log \left (x^2\right )}{(-4+x) x^2} \, dx-32 \int \frac {1}{(-4+x)^2 x^2} \, dx \\ & = \frac {30}{4-x}+2 \int \left (1+\frac {16}{(-4+x)^2}+\frac {8}{-4+x}\right ) \, dx+4 \int \frac {\log \left (x^2\right )}{x^2} \, dx-4 \int \frac {\log \left (x^2\right )}{(-4+x) x} \, dx+8 \int \left (\frac {\log ^2\left (x^2\right )}{8 (-4+x)^2}-\frac {\log ^2\left (x^2\right )}{8 x^2}\right ) \, dx-16 \int \left (\frac {4}{(-4+x)^2}+\frac {1}{-4+x}\right ) \, dx+16 \int \left (\frac {1}{4 (-4+x)^2}-\frac {1}{16 (-4+x)}+\frac {1}{16 x}\right ) \, dx-32 \int \left (\frac {1}{16 (-4+x)^2}-\frac {1}{32 (-4+x)}+\frac {1}{16 x^2}+\frac {1}{32 x}\right ) \, dx \\ & = -\frac {6}{x}+2 x-\frac {4 \log \left (x^2\right )}{x}-\log \left (1-\frac {4}{x}\right ) \log \left (x^2\right )+2 \int \frac {\log \left (1-\frac {4}{x}\right )}{x} \, dx+\int \frac {\log ^2\left (x^2\right )}{(-4+x)^2} \, dx-\int \frac {\log ^2\left (x^2\right )}{x^2} \, dx \\ & = -\frac {6}{x}+2 x-\frac {4 \log \left (x^2\right )}{x}-\log \left (1-\frac {4}{x}\right ) \log \left (x^2\right )+\frac {\log ^2\left (x^2\right )}{x}+\frac {x \log ^2\left (x^2\right )}{4 (4-x)}+2 \text {Li}_2\left (\frac {4}{x}\right )-4 \int \frac {\log \left (x^2\right )}{x^2} \, dx+\int \frac {\log \left (x^2\right )}{-4+x} \, dx \\ & = \frac {2}{x}+2 x-\log \left (1-\frac {4}{x}\right ) \log \left (x^2\right )+\log \left (1-\frac {x}{4}\right ) \log \left (x^2\right )+\frac {\log ^2\left (x^2\right )}{x}+\frac {x \log ^2\left (x^2\right )}{4 (4-x)}+2 \text {Li}_2\left (\frac {4}{x}\right )-2 \int \frac {\log \left (1-\frac {x}{4}\right )}{x} \, dx \\ & = \frac {2}{x}+2 x-\log \left (1-\frac {4}{x}\right ) \log \left (x^2\right )+\log \left (1-\frac {x}{4}\right ) \log \left (x^2\right )+\frac {\log ^2\left (x^2\right )}{x}+\frac {x \log ^2\left (x^2\right )}{4 (4-x)}+2 \text {Li}_2\left (\frac {4}{x}\right )+2 \text {Li}_2\left (\frac {x}{4}\right ) \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-32+16 x+30 x^2-16 x^3+2 x^4+(64-16 x) \log \left (x^2\right )+(-16+8 x) \log ^2\left (x^2\right )}{16 x^2-8 x^3+x^4} \, dx=\frac {2 \left (-4+x-4 x^2+x^3-2 \log ^2\left (x^2\right )\right )}{(-4+x) x} \]
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Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93
method | result | size |
norman | \(\frac {-8-30 x +2 x^{3}-4 \ln \left (x^{2}\right )^{2}}{\left (x -4\right ) x}\) | \(28\) |
risch | \(-\frac {4 \ln \left (x^{2}\right )^{2}}{\left (x -4\right ) x}+\frac {2 x^{2}+2}{x}\) | \(28\) |
parallelrisch | \(\frac {-16+4 x^{3}-8 \ln \left (x^{2}\right )^{2}-60 x}{2 x \left (x -4\right )}\) | \(29\) |
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Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-32+16 x+30 x^2-16 x^3+2 x^4+(64-16 x) \log \left (x^2\right )+(-16+8 x) \log ^2\left (x^2\right )}{16 x^2-8 x^3+x^4} \, dx=\frac {2 \, {\left (x^{3} - 4 \, x^{2} - 2 \, \log \left (x^{2}\right )^{2} + x - 4\right )}}{x^{2} - 4 \, x} \]
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Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {-32+16 x+30 x^2-16 x^3+2 x^4+(64-16 x) \log \left (x^2\right )+(-16+8 x) \log ^2\left (x^2\right )}{16 x^2-8 x^3+x^4} \, dx=2 x - \frac {4 \log {\left (x^{2} \right )}^{2}}{x^{2} - 4 x} + \frac {2}{x} \]
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Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {-32+16 x+30 x^2-16 x^3+2 x^4+(64-16 x) \log \left (x^2\right )+(-16+8 x) \log ^2\left (x^2\right )}{16 x^2-8 x^3+x^4} \, dx=2 \, x - \frac {16 \, \log \left (x\right )^{2}}{x^{2} - 4 \, x} + \frac {4 \, {\left (x - 2\right )}}{x^{2} - 4 \, x} - \frac {2}{x - 4} \]
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Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {-32+16 x+30 x^2-16 x^3+2 x^4+(64-16 x) \log \left (x^2\right )+(-16+8 x) \log ^2\left (x^2\right )}{16 x^2-8 x^3+x^4} \, dx=-{\left (\frac {1}{x - 4} - \frac {1}{x}\right )} \log \left (x^{2}\right )^{2} + 2 \, x + \frac {2}{x} \]
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Time = 10.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-32+16 x+30 x^2-16 x^3+2 x^4+(64-16 x) \log \left (x^2\right )+(-16+8 x) \log ^2\left (x^2\right )}{16 x^2-8 x^3+x^4} \, dx=-\frac {-4\,x^3+15\,x^2+8\,{\ln \left (x^2\right )}^2+16}{2\,x\,\left (x-4\right )} \]
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