Integrand size = 35, antiderivative size = 24 \[ \int \frac {20-23 x^2+2 x^4+e^{12} \left (-8 x+2 x^3\right )}{-4 x+x^3} \, dx=\left (e^{12}+x\right )^2+5 \left (4-\log \left (x \left (8-2 x^2\right )\right )\right ) \]
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Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {1607, 1816, 266} \[ \int \frac {20-23 x^2+2 x^4+e^{12} \left (-8 x+2 x^3\right )}{-4 x+x^3} \, dx=x^2-5 \log \left (4-x^2\right )+2 e^{12} x-5 \log (x) \]
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Rule 266
Rule 1607
Rule 1816
Rubi steps \begin{align*} \text {integral}& = \int \frac {20-23 x^2+2 x^4+e^{12} \left (-8 x+2 x^3\right )}{x \left (-4+x^2\right )} \, dx \\ & = \int \left (2 e^{12}-\frac {5}{x}+2 x-\frac {10 x}{-4+x^2}\right ) \, dx \\ & = 2 e^{12} x+x^2-5 \log (x)-10 \int \frac {x}{-4+x^2} \, dx \\ & = 2 e^{12} x+x^2-5 \log (x)-5 \log \left (4-x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {20-23 x^2+2 x^4+e^{12} \left (-8 x+2 x^3\right )}{-4 x+x^3} \, dx=2 e^{12} x+x^2-5 \log (x)-5 \log \left (4-x^2\right ) \]
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Time = 2.53 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83
method | result | size |
risch | \(2 x \,{\mathrm e}^{12}+x^{2}-5 \ln \left (x^{3}-4 x \right )\) | \(20\) |
default | \(x^{2}+2 x \,{\mathrm e}^{12}-5 \ln \left (2+x \right )-5 \ln \left (x \right )-5 \ln \left (-2+x \right )\) | \(26\) |
norman | \(x^{2}+2 x \,{\mathrm e}^{12}-5 \ln \left (2+x \right )-5 \ln \left (x \right )-5 \ln \left (-2+x \right )\) | \(26\) |
parallelrisch | \(x^{2}+2 x \,{\mathrm e}^{12}-5 \ln \left (2+x \right )-5 \ln \left (x \right )-5 \ln \left (-2+x \right )\) | \(26\) |
meijerg | \(-5 \ln \left (x \right )+5 \ln \left (2\right )-\frac {5 i \pi }{2}-5 \ln \left (1-\frac {x^{2}}{4}\right )-2 i {\mathrm e}^{12} \left (i x -2 i \operatorname {arctanh}\left (\frac {x}{2}\right )\right )+4 \,{\mathrm e}^{12} \operatorname {arctanh}\left (\frac {x}{2}\right )+x^{2}\) | \(52\) |
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Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {20-23 x^2+2 x^4+e^{12} \left (-8 x+2 x^3\right )}{-4 x+x^3} \, dx=x^{2} + 2 \, x e^{12} - 5 \, \log \left (x^{3} - 4 \, x\right ) \]
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Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {20-23 x^2+2 x^4+e^{12} \left (-8 x+2 x^3\right )}{-4 x+x^3} \, dx=x^{2} + 2 x e^{12} - 5 \log {\left (x^{3} - 4 x \right )} \]
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Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {20-23 x^2+2 x^4+e^{12} \left (-8 x+2 x^3\right )}{-4 x+x^3} \, dx=x^{2} + 2 \, x e^{12} - 5 \, \log \left (x + 2\right ) - 5 \, \log \left (x - 2\right ) - 5 \, \log \left (x\right ) \]
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Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {20-23 x^2+2 x^4+e^{12} \left (-8 x+2 x^3\right )}{-4 x+x^3} \, dx=x^{2} + 2 \, x e^{12} - 5 \, \log \left ({\left | x + 2 \right |}\right ) - 5 \, \log \left ({\left | x - 2 \right |}\right ) - 5 \, \log \left ({\left | x \right |}\right ) \]
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Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {20-23 x^2+2 x^4+e^{12} \left (-8 x+2 x^3\right )}{-4 x+x^3} \, dx=2\,x\,{\mathrm {e}}^{12}-5\,\ln \left (x^3-4\,x\right )+x^2 \]
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