Integrand size = 104, antiderivative size = 29 \[ \int \frac {8 x^5+e^8 \left (-48+8 x+16 x^2+2 x^3\right )+e^4 \left (192 x-80 x^3-8 x^4\right )}{-8 x^4+4 x^6+e^8 \left (-8+8 x+2 x^2-4 x^3+x^4\right )+e^4 \left (-16 x^2+8 x^3+8 x^4-4 x^5\right )} \, dx=\frac {12}{-1+\frac {2}{x-\frac {2 x^2}{e^4}}}+\log \left (2-x^2\right ) \]
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Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {2099, 643, 266} \[ \int \frac {8 x^5+e^8 \left (-48+8 x+16 x^2+2 x^3\right )+e^4 \left (192 x-80 x^3-8 x^4\right )}{-8 x^4+4 x^6+e^8 \left (-8+8 x+2 x^2-4 x^3+x^4\right )+e^4 \left (-16 x^2+8 x^3+8 x^4-4 x^5\right )} \, dx=\frac {24 e^4}{2 x^2-e^4 x+2 e^4}+\log \left (2-x^2\right ) \]
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Rule 266
Rule 643
Rule 2099
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {24 \left (e^8-4 e^4 x\right )}{\left (-2 e^4+e^4 x-2 x^2\right )^2}+\frac {2 x}{-2+x^2}\right ) \, dx \\ & = 2 \int \frac {x}{-2+x^2} \, dx+24 \int \frac {e^8-4 e^4 x}{\left (-2 e^4+e^4 x-2 x^2\right )^2} \, dx \\ & = \frac {24 e^4}{2 e^4-e^4 x+2 x^2}+\log \left (2-x^2\right ) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {8 x^5+e^8 \left (-48+8 x+16 x^2+2 x^3\right )+e^4 \left (192 x-80 x^3-8 x^4\right )}{-8 x^4+4 x^6+e^8 \left (-8+8 x+2 x^2-4 x^3+x^4\right )+e^4 \left (-16 x^2+8 x^3+8 x^4-4 x^5\right )} \, dx=-\frac {24 e^4}{e^4 (-2+x)-2 x^2}+\log \left (2-x^2\right ) \]
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Time = 0.44 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97
method | result | size |
norman | \(-\frac {24 \,{\mathrm e}^{4}}{x \,{\mathrm e}^{4}-2 x^{2}-2 \,{\mathrm e}^{4}}+\ln \left (x^{2}-2\right )\) | \(28\) |
risch | \(-\frac {24 \,{\mathrm e}^{4}}{x \,{\mathrm e}^{4}-2 x^{2}-2 \,{\mathrm e}^{4}}+\ln \left (x^{2}-2\right )\) | \(28\) |
parallelrisch | \(\frac {2 \,{\mathrm e}^{4} \ln \left (x^{2}-2\right ) x -4 \ln \left (x^{2}-2\right ) x^{2}-4 \,{\mathrm e}^{4} \ln \left (x^{2}-2\right )-48 \,{\mathrm e}^{4}}{2 x \,{\mathrm e}^{4}-4 x^{2}-4 \,{\mathrm e}^{4}}\) | \(56\) |
default | \(-12 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (4 \textit {\_Z}^{4}-4 \textit {\_Z}^{3} {\mathrm e}^{4}+\left (8 \,{\mathrm e}^{4}+{\mathrm e}^{8}\right ) \textit {\_Z}^{2}-4 \,{\mathrm e}^{8} \textit {\_Z} +4 \,{\mathrm e}^{8}\right )}{\sum }\frac {\left (-4 \textit {\_R} \,{\mathrm e}^{4}+{\mathrm e}^{8}\right ) \ln \left (x -\textit {\_R} \right )}{6 \textit {\_R}^{2} {\mathrm e}^{4}-8 \textit {\_R}^{3}-8 \textit {\_R} \,{\mathrm e}^{4}-\textit {\_R} \,{\mathrm e}^{8}+2 \,{\mathrm e}^{8}}\right )+\ln \left (x^{2}-2\right )\) | \(91\) |
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Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {8 x^5+e^8 \left (-48+8 x+16 x^2+2 x^3\right )+e^4 \left (192 x-80 x^3-8 x^4\right )}{-8 x^4+4 x^6+e^8 \left (-8+8 x+2 x^2-4 x^3+x^4\right )+e^4 \left (-16 x^2+8 x^3+8 x^4-4 x^5\right )} \, dx=\frac {{\left (2 \, x^{2} - {\left (x - 2\right )} e^{4}\right )} \log \left (x^{2} - 2\right ) + 24 \, e^{4}}{2 \, x^{2} - {\left (x - 2\right )} e^{4}} \]
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Time = 1.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {8 x^5+e^8 \left (-48+8 x+16 x^2+2 x^3\right )+e^4 \left (192 x-80 x^3-8 x^4\right )}{-8 x^4+4 x^6+e^8 \left (-8+8 x+2 x^2-4 x^3+x^4\right )+e^4 \left (-16 x^2+8 x^3+8 x^4-4 x^5\right )} \, dx=\log {\left (x^{2} - 2 \right )} + \frac {24 e^{4}}{2 x^{2} - x e^{4} + 2 e^{4}} \]
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Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {8 x^5+e^8 \left (-48+8 x+16 x^2+2 x^3\right )+e^4 \left (192 x-80 x^3-8 x^4\right )}{-8 x^4+4 x^6+e^8 \left (-8+8 x+2 x^2-4 x^3+x^4\right )+e^4 \left (-16 x^2+8 x^3+8 x^4-4 x^5\right )} \, dx=\frac {24 \, e^{4}}{2 \, x^{2} - x e^{4} + 2 \, e^{4}} + \log \left (x^{2} - 2\right ) \]
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Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {8 x^5+e^8 \left (-48+8 x+16 x^2+2 x^3\right )+e^4 \left (192 x-80 x^3-8 x^4\right )}{-8 x^4+4 x^6+e^8 \left (-8+8 x+2 x^2-4 x^3+x^4\right )+e^4 \left (-16 x^2+8 x^3+8 x^4-4 x^5\right )} \, dx=\frac {24 \, e^{4}}{2 \, x^{2} - x e^{4} + 2 \, e^{4}} + \log \left ({\left | x^{2} - 2 \right |}\right ) \]
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Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {8 x^5+e^8 \left (-48+8 x+16 x^2+2 x^3\right )+e^4 \left (192 x-80 x^3-8 x^4\right )}{-8 x^4+4 x^6+e^8 \left (-8+8 x+2 x^2-4 x^3+x^4\right )+e^4 \left (-16 x^2+8 x^3+8 x^4-4 x^5\right )} \, dx=\ln \left (x^2-2\right )+\frac {24\,{\mathrm {e}}^4}{2\,x^2-{\mathrm {e}}^4\,x+2\,{\mathrm {e}}^4} \]
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