Integrand size = 71, antiderivative size = 24 \[ \int \frac {-4-8 x-6 x^2-2 \log \left (\frac {16 x^2}{9}\right )}{4 x^4+4 x^5+x^6+\left (4 x^3+2 x^4\right ) \log \left (\frac {16 x^2}{9}\right )+x^2 \log ^2\left (\frac {16 x^2}{9}\right )} \, dx=5+\frac {2}{x^2 \left (2+x+\frac {\log \left (\frac {16 x^2}{9}\right )}{x}\right )} \]
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Time = 0.12 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {6820, 12, 6819} \[ \int \frac {-4-8 x-6 x^2-2 \log \left (\frac {16 x^2}{9}\right )}{4 x^4+4 x^5+x^6+\left (4 x^3+2 x^4\right ) \log \left (\frac {16 x^2}{9}\right )+x^2 \log ^2\left (\frac {16 x^2}{9}\right )} \, dx=\frac {2}{x \left (\log \left (\frac {16 x^2}{9}\right )+x (x+2)\right )} \]
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Rule 12
Rule 6819
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {2 \left (-2-4 x-3 x^2-\log \left (\frac {16 x^2}{9}\right )\right )}{x^2 \left (x (2+x)+\log \left (\frac {16 x^2}{9}\right )\right )^2} \, dx \\ & = 2 \int \frac {-2-4 x-3 x^2-\log \left (\frac {16 x^2}{9}\right )}{x^2 \left (x (2+x)+\log \left (\frac {16 x^2}{9}\right )\right )^2} \, dx \\ & = \frac {2}{x \left (x (2+x)+\log \left (\frac {16 x^2}{9}\right )\right )} \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-4-8 x-6 x^2-2 \log \left (\frac {16 x^2}{9}\right )}{4 x^4+4 x^5+x^6+\left (4 x^3+2 x^4\right ) \log \left (\frac {16 x^2}{9}\right )+x^2 \log ^2\left (\frac {16 x^2}{9}\right )} \, dx=\frac {2}{x \left (2 x+x^2+\log \left (\frac {16 x^2}{9}\right )\right )} \]
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Time = 0.13 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88
method | result | size |
norman | \(\frac {2}{x \left (x^{2}+\ln \left (\frac {16 x^{2}}{9}\right )+2 x \right )}\) | \(21\) |
risch | \(\frac {2}{x \left (x^{2}+\ln \left (\frac {16 x^{2}}{9}\right )+2 x \right )}\) | \(21\) |
parallelrisch | \(\frac {2}{x \left (x^{2}+\ln \left (\frac {16 x^{2}}{9}\right )+2 x \right )}\) | \(21\) |
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Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {-4-8 x-6 x^2-2 \log \left (\frac {16 x^2}{9}\right )}{4 x^4+4 x^5+x^6+\left (4 x^3+2 x^4\right ) \log \left (\frac {16 x^2}{9}\right )+x^2 \log ^2\left (\frac {16 x^2}{9}\right )} \, dx=\frac {2}{x^{3} + 2 \, x^{2} + x \log \left (\frac {16}{9} \, x^{2}\right )} \]
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Time = 0.33 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-4-8 x-6 x^2-2 \log \left (\frac {16 x^2}{9}\right )}{4 x^4+4 x^5+x^6+\left (4 x^3+2 x^4\right ) \log \left (\frac {16 x^2}{9}\right )+x^2 \log ^2\left (\frac {16 x^2}{9}\right )} \, dx=\frac {2}{x^{3} + 2 x^{2} + x \log {\left (\frac {16 x^{2}}{9} \right )}} \]
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Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {-4-8 x-6 x^2-2 \log \left (\frac {16 x^2}{9}\right )}{4 x^4+4 x^5+x^6+\left (4 x^3+2 x^4\right ) \log \left (\frac {16 x^2}{9}\right )+x^2 \log ^2\left (\frac {16 x^2}{9}\right )} \, dx=\frac {2}{x^{3} + 2 \, x^{2} - 2 \, x {\left (\log \left (3\right ) - 2 \, \log \left (2\right )\right )} + 2 \, x \log \left (x\right )} \]
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Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {-4-8 x-6 x^2-2 \log \left (\frac {16 x^2}{9}\right )}{4 x^4+4 x^5+x^6+\left (4 x^3+2 x^4\right ) \log \left (\frac {16 x^2}{9}\right )+x^2 \log ^2\left (\frac {16 x^2}{9}\right )} \, dx=\frac {2}{x^{3} + 2 \, x^{2} + x \log \left (\frac {16}{9} \, x^{2}\right )} \]
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Timed out. \[ \int \frac {-4-8 x-6 x^2-2 \log \left (\frac {16 x^2}{9}\right )}{4 x^4+4 x^5+x^6+\left (4 x^3+2 x^4\right ) \log \left (\frac {16 x^2}{9}\right )+x^2 \log ^2\left (\frac {16 x^2}{9}\right )} \, dx=\int -\frac {8\,x+2\,\ln \left (\frac {16\,x^2}{9}\right )+6\,x^2+4}{x^2\,{\ln \left (\frac {16\,x^2}{9}\right )}^2+\ln \left (\frac {16\,x^2}{9}\right )\,\left (2\,x^4+4\,x^3\right )+4\,x^4+4\,x^5+x^6} \,d x \]
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