Integrand size = 64, antiderivative size = 26 \[ \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{9 x-6 x^5+x^9} \, dx=\log \left (\frac {1}{9} x^{-2+\frac {2 e^x x}{\frac {3}{x^3}-x}}\right ) \]
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Time = 0.42 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1608, 28, 6874, 2326} \[ \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{9 x-6 x^5+x^9} \, dx=\frac {2 e^x x^3 \left (3 x \log (x)-x^5 \log (x)\right )}{\left (3-x^4\right )^2}-2 \log (x) \]
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Rule 28
Rule 1608
Rule 2326
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{x \left (9-6 x^4+x^8\right )} \, dx \\ & = \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{x \left (-3+x^4\right )^2} \, dx \\ & = \int \left (-\frac {2}{x}-\frac {2 e^x x^3 \left (-3+x^4-12 \log (x)-3 x \log (x)+x^5 \log (x)\right )}{\left (-3+x^4\right )^2}\right ) \, dx \\ & = -2 \log (x)-2 \int \frac {e^x x^3 \left (-3+x^4-12 \log (x)-3 x \log (x)+x^5 \log (x)\right )}{\left (-3+x^4\right )^2} \, dx \\ & = -2 \log (x)+\frac {2 e^x x^3 \left (3 x \log (x)-x^5 \log (x)\right )}{\left (3-x^4\right )^2} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{9 x-6 x^5+x^9} \, dx=-\frac {2 \left (-3+\left (1+e^x\right ) x^4\right ) \log (x)}{-3+x^4} \]
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Time = 0.60 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85
method | result | size |
risch | \(-\frac {2 x^{4} {\mathrm e}^{x} \ln \left (x \right )}{x^{4}-3}-2 \ln \left (x \right )\) | \(22\) |
parallelrisch | \(\frac {-6 x^{4} {\mathrm e}^{x} \ln \left (x \right )-6 x^{4} \ln \left (x \right )+18 \ln \left (x \right )}{3 x^{4}-9}\) | \(31\) |
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Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{9 x-6 x^5+x^9} \, dx=-\frac {2 \, {\left (x^{4} e^{x} + x^{4} - 3\right )} \log \left (x\right )}{x^{4} - 3} \]
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Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{9 x-6 x^5+x^9} \, dx=- \frac {2 x^{4} e^{x} \log {\left (x \right )}}{x^{4} - 3} - 2 \log {\left (x \right )} \]
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Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{9 x-6 x^5+x^9} \, dx=-\frac {2 \, x^{4} e^{x} \log \left (x\right )}{x^{4} - 3} - 2 \, \log \left (x\right ) \]
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Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{9 x-6 x^5+x^9} \, dx=-\frac {2 \, {\left (x^{4} e^{x} \log \left (x\right ) + x^{4} \log \left (x\right ) - 3 \, \log \left (x\right )\right )}}{x^{4} - 3} \]
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Time = 11.46 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{9 x-6 x^5+x^9} \, dx=-\frac {2\,\ln \left (x\right )\,\left (x^4\,{\mathrm {e}}^x+x^4-3\right )}{x^4-3} \]
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