Integrand size = 30, antiderivative size = 29 \[ \int \frac {-12 x^2-8 x^3+e^4 (8+4 x)}{2 x+x^2} \, dx=4 \left (x-x^2+e^4 (-3+\log (x))-\log \left (\frac {1}{2} (2+x)^2\right )\right ) \]
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Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1607, 1634} \[ \int \frac {-12 x^2-8 x^3+e^4 (8+4 x)}{2 x+x^2} \, dx=-4 x^2+4 x+4 e^4 \log (x)-8 \log (x+2) \]
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Rule 1607
Rule 1634
Rubi steps \begin{align*} \text {integral}& = \int \frac {-12 x^2-8 x^3+e^4 (8+4 x)}{x (2+x)} \, dx \\ & = \int \left (4+\frac {4 e^4}{x}-8 x-\frac {8}{2+x}\right ) \, dx \\ & = 4 x-4 x^2+4 e^4 \log (x)-8 \log (2+x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {-12 x^2-8 x^3+e^4 (8+4 x)}{2 x+x^2} \, dx=20 (2+x)-4 (2+x)^2+4 e^4 \log (x)-8 \log (2+x) \]
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Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76
method | result | size |
default | \(4 x -4 x^{2}+4 \,{\mathrm e}^{4} \ln \left (x \right )-8 \ln \left (2+x \right )\) | \(22\) |
norman | \(4 x -4 x^{2}+4 \,{\mathrm e}^{4} \ln \left (x \right )-8 \ln \left (2+x \right )\) | \(22\) |
parallelrisch | \(4 x -4 x^{2}+4 \,{\mathrm e}^{4} \ln \left (x \right )-8 \ln \left (2+x \right )\) | \(22\) |
risch | \(-4 x^{2}+4 x -8 \ln \left (2+x \right )+4 \,{\mathrm e}^{4} \ln \left (-x \right )\) | \(24\) |
meijerg | \(4 \,{\mathrm e}^{4} \ln \left (1+\frac {x}{2}\right )+4 \,{\mathrm e}^{4} \left (\ln \left (x \right )-\ln \left (2\right )-\ln \left (1+\frac {x}{2}\right )\right )+\frac {8 x \left (-\frac {3 x}{2}+6\right )}{3}-8 \ln \left (1+\frac {x}{2}\right )-12 x\) | \(50\) |
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none
Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {-12 x^2-8 x^3+e^4 (8+4 x)}{2 x+x^2} \, dx=-4 \, x^{2} + 4 \, e^{4} \log \left (x\right ) + 4 \, x - 8 \, \log \left (x + 2\right ) \]
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Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {-12 x^2-8 x^3+e^4 (8+4 x)}{2 x+x^2} \, dx=- 4 x^{2} + 4 x + 4 e^{4} \log {\left (x \right )} - 8 \log {\left (x + \frac {8 + 4 e^{4}}{4 + 2 e^{4}} \right )} \]
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none
Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {-12 x^2-8 x^3+e^4 (8+4 x)}{2 x+x^2} \, dx=-4 \, x^{2} + 4 \, e^{4} \log \left (x\right ) + 4 \, x - 8 \, \log \left (x + 2\right ) \]
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none
Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {-12 x^2-8 x^3+e^4 (8+4 x)}{2 x+x^2} \, dx=-4 \, x^{2} + 4 \, e^{4} \log \left ({\left | x \right |}\right ) + 4 \, x - 8 \, \log \left ({\left | x + 2 \right |}\right ) \]
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Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {-12 x^2-8 x^3+e^4 (8+4 x)}{2 x+x^2} \, dx=4\,x-8\,\ln \left (x+2\right )+4\,{\mathrm {e}}^4\,\ln \left (x\right )-4\,x^2 \]
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