Integrand size = 43, antiderivative size = 18 \[ \int \frac {-4+x^2+e^x \left (-4-4 x+x^2-x^3\right )}{4 x+x^3+e^x \left (4 x+x^3\right )} \, dx=9+\log \left (\frac {\frac {4}{x}+x}{1+e^x}\right ) \]
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Time = 0.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {6873, 6857, 2320, 36, 29, 31, 1816, 266} \[ \int \frac {-4+x^2+e^x \left (-4-4 x+x^2-x^3\right )}{4 x+x^3+e^x \left (4 x+x^3\right )} \, dx=\log \left (x^2+4\right )-\log \left (e^x+1\right )-\log (x) \]
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Rule 29
Rule 31
Rule 36
Rule 266
Rule 1816
Rule 2320
Rule 6857
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {-4+x^2+e^x \left (-4-4 x+x^2-x^3\right )}{\left (1+e^x\right ) x \left (4+x^2\right )} \, dx \\ & = \int \left (\frac {1}{1+e^x}+\frac {-4-4 x+x^2-x^3}{x \left (4+x^2\right )}\right ) \, dx \\ & = \int \frac {1}{1+e^x} \, dx+\int \frac {-4-4 x+x^2-x^3}{x \left (4+x^2\right )} \, dx \\ & = \int \left (-1-\frac {1}{x}+\frac {2 x}{4+x^2}\right ) \, dx+\text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,e^x\right ) \\ & = -x-\log (x)+2 \int \frac {x}{4+x^2} \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )-\text {Subst}\left (\int \frac {1}{1+x} \, dx,x,e^x\right ) \\ & = -\log \left (1+e^x\right )-\log (x)+\log \left (4+x^2\right ) \\ \end{align*}
Time = 0.72 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {-4+x^2+e^x \left (-4-4 x+x^2-x^3\right )}{4 x+x^3+e^x \left (4 x+x^3\right )} \, dx=-x-2 \text {arctanh}\left (1+2 e^x\right )-\log (x)+\log \left (4+x^2\right ) \]
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Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06
method | result | size |
norman | \(-\ln \left (x \right )-\ln \left ({\mathrm e}^{x}+1\right )+\ln \left (x^{2}+4\right )\) | \(19\) |
risch | \(-\ln \left (x \right )-\ln \left ({\mathrm e}^{x}+1\right )+\ln \left (x^{2}+4\right )\) | \(19\) |
parallelrisch | \(-\ln \left (x \right )-\ln \left ({\mathrm e}^{x}+1\right )+\ln \left (x^{2}+4\right )\) | \(19\) |
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Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-4+x^2+e^x \left (-4-4 x+x^2-x^3\right )}{4 x+x^3+e^x \left (4 x+x^3\right )} \, dx=\log \left (x^{2} + 4\right ) - \log \left (x\right ) - \log \left (e^{x} + 1\right ) \]
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Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {-4+x^2+e^x \left (-4-4 x+x^2-x^3\right )}{4 x+x^3+e^x \left (4 x+x^3\right )} \, dx=- \log {\left (x \right )} + \log {\left (x^{2} + 4 \right )} - \log {\left (e^{x} + 1 \right )} \]
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Time = 0.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-4+x^2+e^x \left (-4-4 x+x^2-x^3\right )}{4 x+x^3+e^x \left (4 x+x^3\right )} \, dx=\log \left (x^{2} + 4\right ) - \log \left (x\right ) - \log \left (e^{x} + 1\right ) \]
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Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-4+x^2+e^x \left (-4-4 x+x^2-x^3\right )}{4 x+x^3+e^x \left (4 x+x^3\right )} \, dx=\log \left (x^{2} + 4\right ) - \log \left (x\right ) - \log \left (e^{x} + 1\right ) \]
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Time = 0.17 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-4+x^2+e^x \left (-4-4 x+x^2-x^3\right )}{4 x+x^3+e^x \left (4 x+x^3\right )} \, dx=\ln \left (x^2+4\right )-\ln \left ({\mathrm {e}}^x+1\right )-\ln \left (x\right ) \]
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