Integrand size = 37, antiderivative size = 30 \[ \int \frac {32+4 x^2-29 x^3+x^5+8 x^6}{32 x-32 x^4+8 x^7} \, dx=4+\frac {2+x}{16 \left (\frac {2}{x^2}-x\right )}+\log (75)-\log \left (\frac {3}{x}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {1608, 28, 1843, 21, 29} \[ \int \frac {32+4 x^2-29 x^3+x^5+8 x^6}{32 x-32 x^4+8 x^7} \, dx=\frac {x \left (x^2+2 x\right )}{16 \left (2-x^3\right )}+\log (x) \]
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Rule 21
Rule 28
Rule 29
Rule 1608
Rule 1843
Rubi steps \begin{align*} \text {integral}& = \int \frac {32+4 x^2-29 x^3+x^5+8 x^6}{x \left (32-32 x^3+8 x^6\right )} \, dx \\ & = 8 \int \frac {32+4 x^2-29 x^3+x^5+8 x^6}{x \left (-16+8 x^3\right )^2} \, dx \\ & = \frac {x \left (2 x+x^2\right )}{16 \left (2-x^3\right )}+\frac {1}{384} \int \frac {-6144+3072 x^3}{x \left (-16+8 x^3\right )} \, dx \\ & = \frac {x \left (2 x+x^2\right )}{16 \left (2-x^3\right )}+\int \frac {1}{x} \, dx \\ & = \frac {x \left (2 x+x^2\right )}{16 \left (2-x^3\right )}+\log (x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {32+4 x^2-29 x^3+x^5+8 x^6}{32 x-32 x^4+8 x^7} \, dx=\frac {1}{8} \left (\frac {-1-x^2}{-2+x^3}+8 \log (x)\right ) \]
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Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.63
method | result | size |
norman | \(\frac {-\frac {x^{2}}{8}-\frac {1}{8}}{x^{3}-2}+\ln \left (x \right )\) | \(19\) |
risch | \(\frac {-\frac {x^{2}}{8}-\frac {1}{8}}{x^{3}-2}+\ln \left (x \right )\) | \(19\) |
default | \(\ln \left (x \right )+\frac {-x^{2}-1}{8 x^{3}-16}\) | \(20\) |
parallelrisch | \(\frac {8 x^{3} \ln \left (x \right )-1-x^{2}-16 \ln \left (x \right )}{8 x^{3}-16}\) | \(28\) |
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Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {32+4 x^2-29 x^3+x^5+8 x^6}{32 x-32 x^4+8 x^7} \, dx=-\frac {x^{2} - 8 \, {\left (x^{3} - 2\right )} \log \left (x\right ) + 1}{8 \, {\left (x^{3} - 2\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.50 \[ \int \frac {32+4 x^2-29 x^3+x^5+8 x^6}{32 x-32 x^4+8 x^7} \, dx=\frac {- x^{2} - 1}{8 x^{3} - 16} + \log {\left (x \right )} \]
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none
Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.57 \[ \int \frac {32+4 x^2-29 x^3+x^5+8 x^6}{32 x-32 x^4+8 x^7} \, dx=-\frac {x^{2} + 1}{8 \, {\left (x^{3} - 2\right )}} + \log \left (x\right ) \]
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Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.60 \[ \int \frac {32+4 x^2-29 x^3+x^5+8 x^6}{32 x-32 x^4+8 x^7} \, dx=-\frac {x^{2} + 1}{8 \, {\left (x^{3} - 2\right )}} + \log \left ({\left | x \right |}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.63 \[ \int \frac {32+4 x^2-29 x^3+x^5+8 x^6}{32 x-32 x^4+8 x^7} \, dx=\ln \left (x\right )-\frac {\frac {x^2}{8}+\frac {1}{8}}{x^3-2} \]
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