Integrand size = 59, antiderivative size = 16 \[ \int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+\left (3 e^8 x+4 x^2\right ) \log \left (\frac {3 e^8+4 x}{x}\right )} \, dx=\log \left (12 \left (x+\log \left (4+\frac {3 e^8}{x}\right )\right )\right ) \]
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Time = 0.13 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6873, 6816} \[ \int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+\left (3 e^8 x+4 x^2\right ) \log \left (\frac {3 e^8+4 x}{x}\right )} \, dx=\log \left (x+\log \left (\frac {3 e^8}{x}+4\right )\right ) \]
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Rule 6816
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {-3 e^8+3 e^8 x+4 x^2}{x \left (3 e^8+4 x\right ) \left (x+\log \left (4+\frac {3 e^8}{x}\right )\right )} \, dx \\ & = \log \left (x+\log \left (4+\frac {3 e^8}{x}\right )\right ) \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+\left (3 e^8 x+4 x^2\right ) \log \left (\frac {3 e^8+4 x}{x}\right )} \, dx=\log \left (x+\log \left (4+\frac {3 e^8}{x}\right )\right ) \]
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Time = 1.97 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06
method | result | size |
risch | \(\ln \left (\ln \left (\frac {3 \,{\mathrm e}^{8}+4 x}{x}\right )+x \right )\) | \(17\) |
norman | \(\ln \left (\ln \left (\frac {3 \,{\mathrm e}^{8}+4 x}{x}\right )+x \right )\) | \(19\) |
parallelrisch | \(\ln \left (\ln \left (\frac {3 \,{\mathrm e}^{8}+4 x}{x}\right )+x \right )\) | \(19\) |
derivativedivides | \(-\ln \left (\frac {3 \,{\mathrm e}^{8}}{x}\right )+\ln \left (3 \,{\mathrm e}^{8}+\ln \left (4+\frac {3 \,{\mathrm e}^{8}}{x}\right ) \left (4+\frac {3 \,{\mathrm e}^{8}}{x}\right )-4 \ln \left (4+\frac {3 \,{\mathrm e}^{8}}{x}\right )\right )\) | \(60\) |
default | \(-\ln \left (\frac {3 \,{\mathrm e}^{8}}{x}\right )+\ln \left (3 \,{\mathrm e}^{8}+\ln \left (4+\frac {3 \,{\mathrm e}^{8}}{x}\right ) \left (4+\frac {3 \,{\mathrm e}^{8}}{x}\right )-4 \ln \left (4+\frac {3 \,{\mathrm e}^{8}}{x}\right )\right )\) | \(60\) |
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none
Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+\left (3 e^8 x+4 x^2\right ) \log \left (\frac {3 e^8+4 x}{x}\right )} \, dx=\log \left (x + \log \left (\frac {4 \, x + 3 \, e^{8}}{x}\right )\right ) \]
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Time = 0.12 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+\left (3 e^8 x+4 x^2\right ) \log \left (\frac {3 e^8+4 x}{x}\right )} \, dx=\log {\left (x + \log {\left (\frac {4 x + 3 e^{8}}{x} \right )} \right )} \]
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Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+\left (3 e^8 x+4 x^2\right ) \log \left (\frac {3 e^8+4 x}{x}\right )} \, dx=\log \left (x + \log \left (4 \, x + 3 \, e^{8}\right ) - \log \left (x\right )\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (17) = 34\).
Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 4.50 \[ \int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+\left (3 e^8 x+4 x^2\right ) \log \left (\frac {3 e^8+4 x}{x}\right )} \, dx={\left (e^{8} \log \left (\frac {{\left (4 \, x + 3 \, e^{8}\right )} \log \left (\frac {4 \, x + 3 \, e^{8}}{x}\right )}{x} + 3 \, e^{8} - 4 \, \log \left (\frac {4 \, x + 3 \, e^{8}}{x}\right )\right ) - e^{8} \log \left (\frac {4 \, x + 3 \, e^{8}}{x} - 4\right )\right )} e^{\left (-8\right )} \]
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Time = 16.44 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+\left (3 e^8 x+4 x^2\right ) \log \left (\frac {3 e^8+4 x}{x}\right )} \, dx=\ln \left (x+\ln \left (\frac {3\,{\mathrm {e}}^8}{x}+4\right )\right ) \]
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