Integrand size = 61, antiderivative size = 29 \[ \int \frac {4+(4-x) \log (4-x)+\left (8 x+6 x^2-2 x^3\right ) \log (x)+\left (4 x^2-x^3\right ) \log ^2(x)}{-8 x^2+2 x^3} \, dx=\frac {1-x+\log (4-x)-\left (x+x^2\right ) \log ^2(x)}{2 x} \]
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Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31, number of steps used = 16, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.180, Rules used = {1607, 6874, 46, 2442, 36, 31, 29, 2388, 2338, 2332, 2333} \[ \int \frac {4+(4-x) \log (4-x)+\left (8 x+6 x^2-2 x^3\right ) \log (x)+\left (4 x^2-x^3\right ) \log ^2(x)}{-8 x^2+2 x^3} \, dx=\frac {1}{2 x}-\frac {1}{2} x \log ^2(x)-\frac {\log ^2(x)}{2}+\frac {\log (4-x)}{2 x} \]
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Rule 29
Rule 31
Rule 36
Rule 46
Rule 1607
Rule 2332
Rule 2333
Rule 2338
Rule 2388
Rule 2442
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {4+(4-x) \log (4-x)+\left (8 x+6 x^2-2 x^3\right ) \log (x)+\left (4 x^2-x^3\right ) \log ^2(x)}{x^2 (-8+2 x)} \, dx \\ & = \int \left (\frac {4+4 \log (4-x)-x \log (4-x)}{2 (-4+x) x^2}-\frac {(1+x) \log (x)}{x}-\frac {\log ^2(x)}{2}\right ) \, dx \\ & = \frac {1}{2} \int \frac {4+4 \log (4-x)-x \log (4-x)}{(-4+x) x^2} \, dx-\frac {1}{2} \int \log ^2(x) \, dx-\int \frac {(1+x) \log (x)}{x} \, dx \\ & = -\frac {1}{2} x \log ^2(x)+\frac {1}{2} \int \left (\frac {4}{(-4+x) x^2}-\frac {\log (4-x)}{x^2}\right ) \, dx-\int \frac {\log (x)}{x} \, dx \\ & = -\frac {1}{2} \log ^2(x)-\frac {1}{2} x \log ^2(x)-\frac {1}{2} \int \frac {\log (4-x)}{x^2} \, dx+2 \int \frac {1}{(-4+x) x^2} \, dx \\ & = \frac {\log (4-x)}{2 x}-\frac {\log ^2(x)}{2}-\frac {1}{2} x \log ^2(x)+\frac {1}{2} \int \frac {1}{(4-x) x} \, dx+2 \int \left (\frac {1}{16 (-4+x)}-\frac {1}{4 x^2}-\frac {1}{16 x}\right ) \, dx \\ & = \frac {1}{2 x}+\frac {1}{8} \log (4-x)+\frac {\log (4-x)}{2 x}-\frac {\log (x)}{8}-\frac {\log ^2(x)}{2}-\frac {1}{2} x \log ^2(x)+\frac {1}{8} \int \frac {1}{4-x} \, dx+\frac {1}{8} \int \frac {1}{x} \, dx \\ & = \frac {1}{2 x}+\frac {\log (4-x)}{2 x}-\frac {\log ^2(x)}{2}-\frac {1}{2} x \log ^2(x) \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {4+(4-x) \log (4-x)+\left (8 x+6 x^2-2 x^3\right ) \log (x)+\left (4 x^2-x^3\right ) \log ^2(x)}{-8 x^2+2 x^3} \, dx=\frac {1}{2} \left (-\frac {-1-\log (4-x)}{x}-\log ^2(x)-x \log ^2(x)\right ) \]
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Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17
method | result | size |
risch | \(\frac {\ln \left (-x +4\right )}{2 x}-\frac {x^{2} \ln \left (x \right )^{2}+x \ln \left (x \right )^{2}-1}{2 x}\) | \(34\) |
parallelrisch | \(\frac {-8 x^{2} \ln \left (x \right )^{2}+8-8 x \ln \left (x \right )^{2}-x \ln \left (x -4\right )+\ln \left (-x +4\right ) x +x +8 \ln \left (-x +4\right )}{16 x}\) | \(48\) |
default | \(-\frac {x \ln \left (x \right )^{2}}{2}+\frac {\ln \left (-x \right )}{8}+\frac {\ln \left (-x +4\right ) \left (-x +4\right )}{8 x}+\frac {1}{2 x}-\frac {\ln \left (x \right )}{8}+\frac {\ln \left (x -4\right )}{8}-\frac {\ln \left (x \right )^{2}}{2}\) | \(52\) |
parts | \(-\frac {x \ln \left (x \right )^{2}}{2}+\frac {\ln \left (-x \right )}{8}+\frac {\ln \left (-x +4\right ) \left (-x +4\right )}{8 x}+\frac {1}{2 x}-\frac {\ln \left (x \right )}{8}+\frac {\ln \left (x -4\right )}{8}-\frac {\ln \left (x \right )^{2}}{2}\) | \(52\) |
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Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {4+(4-x) \log (4-x)+\left (8 x+6 x^2-2 x^3\right ) \log (x)+\left (4 x^2-x^3\right ) \log ^2(x)}{-8 x^2+2 x^3} \, dx=-\frac {{\left (x^{2} + x\right )} \log \left (x\right )^{2} - \log \left (-x + 4\right ) - 1}{2 \, x} \]
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Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {4+(4-x) \log (4-x)+\left (8 x+6 x^2-2 x^3\right ) \log (x)+\left (4 x^2-x^3\right ) \log ^2(x)}{-8 x^2+2 x^3} \, dx=\left (- \frac {x}{2} - \frac {1}{2}\right ) \log {\left (x \right )}^{2} + \frac {\log {\left (4 - x \right )}}{2 x} + \frac {1}{2 x} \]
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Time = 0.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66 \[ \int \frac {4+(4-x) \log (4-x)+\left (8 x+6 x^2-2 x^3\right ) \log (x)+\left (4 x^2-x^3\right ) \log ^2(x)}{-8 x^2+2 x^3} \, dx=-\frac {4 \, {\left (x^{2} + x\right )} \log \left (x\right )^{2} - x \log \left (x\right ) + {\left (x - 4\right )} \log \left (-x + 4\right )}{8 \, x} + \frac {1}{2 \, x} + \frac {1}{8} \, \log \left (x - 4\right ) - \frac {1}{8} \, \log \left (x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {4+(4-x) \log (4-x)+\left (8 x+6 x^2-2 x^3\right ) \log (x)+\left (4 x^2-x^3\right ) \log ^2(x)}{-8 x^2+2 x^3} \, dx=-\frac {1}{2} \, {\left (x + 1\right )} \log \left (x\right )^{2} + \frac {\log \left (-x + 4\right )}{2 \, x} + \frac {1}{2 \, x} \]
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Time = 8.56 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {4+(4-x) \log (4-x)+\left (8 x+6 x^2-2 x^3\right ) \log (x)+\left (4 x^2-x^3\right ) \log ^2(x)}{-8 x^2+2 x^3} \, dx=\frac {1}{2\,x}+\frac {\ln \left (4-x\right )}{2\,x}-{\ln \left (x\right )}^2\,\left (\frac {x}{2}+\frac {1}{2}\right ) \]
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