Integrand size = 108, antiderivative size = 21 \[ \int \frac {9+e^4+2 x+\log (3)}{-27 x-3 e^4 x-3 x^2+\left (-27-3 e^4-3 x\right ) \log (3)+\left (9 x+e^4 x+x^2+\left (9+e^4+x\right ) \log (3)\right ) \log \left (9+e^4+x\right )+\left (-9 x-e^4 x-x^2+\left (-9-e^4-x\right ) \log (3)\right ) \log \left (\frac {5}{x+\log (3)}\right )} \, dx=\log \left (3-\log \left (9+e^4+x\right )+\log \left (\frac {5}{x+\log (3)}\right )\right ) \]
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Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {6, 6820, 6816} \[ \int \frac {9+e^4+2 x+\log (3)}{-27 x-3 e^4 x-3 x^2+\left (-27-3 e^4-3 x\right ) \log (3)+\left (9 x+e^4 x+x^2+\left (9+e^4+x\right ) \log (3)\right ) \log \left (9+e^4+x\right )+\left (-9 x-e^4 x-x^2+\left (-9-e^4-x\right ) \log (3)\right ) \log \left (\frac {5}{x+\log (3)}\right )} \, dx=\log \left (-\log \left (x+e^4+9\right )+\log \left (\frac {5}{x+\log (3)}\right )+3\right ) \]
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Rule 6
Rule 6816
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {9+e^4+2 x+\log (3)}{\left (-27-3 e^4\right ) x-3 x^2+\left (-27-3 e^4-3 x\right ) \log (3)+\left (9 x+e^4 x+x^2+\left (9+e^4+x\right ) \log (3)\right ) \log \left (9+e^4+x\right )+\left (-9 x-e^4 x-x^2+\left (-9-e^4-x\right ) \log (3)\right ) \log \left (\frac {5}{x+\log (3)}\right )} \, dx \\ & = \int \frac {-9-e^4-2 x-\log (3)}{\left (9+e^4+x\right ) (x+\log (3)) \left (3-\log \left (9+e^4+x\right )+\log \left (\frac {5}{x+\log (3)}\right )\right )} \, dx \\ & = \log \left (3-\log \left (9+e^4+x\right )+\log \left (\frac {5}{x+\log (3)}\right )\right ) \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {9+e^4+2 x+\log (3)}{-27 x-3 e^4 x-3 x^2+\left (-27-3 e^4-3 x\right ) \log (3)+\left (9 x+e^4 x+x^2+\left (9+e^4+x\right ) \log (3)\right ) \log \left (9+e^4+x\right )+\left (-9 x-e^4 x-x^2+\left (-9-e^4-x\right ) \log (3)\right ) \log \left (\frac {5}{x+\log (3)}\right )} \, dx=\log \left (-3+\log \left (9+e^4+x\right )-\log \left (\frac {5}{x+\log (3)}\right )\right ) \]
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Time = 1.59 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90
method | result | size |
risch | \(\ln \left (-3-\ln \left (5\right )+\ln \left ({\mathrm e}^{4}+x +9\right )+\ln \left (\ln \left (3\right )+x \right )\right )\) | \(19\) |
norman | \(\ln \left (\ln \left ({\mathrm e}^{4}+x +9\right )-\ln \left (\frac {5}{\ln \left (3\right )+x}\right )-3\right )\) | \(21\) |
parallelrisch | \(\ln \left (\ln \left ({\mathrm e}^{4}+x +9\right )-\ln \left (\frac {5}{\ln \left (3\right )+x}\right )-3\right )\) | \(21\) |
default | \(\ln \left (\ln \left (5\right )-\ln \left (\left (\frac {{\mathrm e}^{4}}{\ln \left (3\right )+x}-\frac {\ln \left (3\right )}{\ln \left (3\right )+x}+\frac {9}{\ln \left (3\right )+x}+1\right ) \left (\ln \left (3\right )+x \right )\right )+\ln \left (\frac {1}{\ln \left (3\right )+x}\right )+3\right )\) | \(50\) |
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Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {9+e^4+2 x+\log (3)}{-27 x-3 e^4 x-3 x^2+\left (-27-3 e^4-3 x\right ) \log (3)+\left (9 x+e^4 x+x^2+\left (9+e^4+x\right ) \log (3)\right ) \log \left (9+e^4+x\right )+\left (-9 x-e^4 x-x^2+\left (-9-e^4-x\right ) \log (3)\right ) \log \left (\frac {5}{x+\log (3)}\right )} \, dx=\log \left (\log \left (x + e^{4} + 9\right ) - \log \left (\frac {5}{x + \log \left (3\right )}\right ) - 3\right ) \]
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Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {9+e^4+2 x+\log (3)}{-27 x-3 e^4 x-3 x^2+\left (-27-3 e^4-3 x\right ) \log (3)+\left (9 x+e^4 x+x^2+\left (9+e^4+x\right ) \log (3)\right ) \log \left (9+e^4+x\right )+\left (-9 x-e^4 x-x^2+\left (-9-e^4-x\right ) \log (3)\right ) \log \left (\frac {5}{x+\log (3)}\right )} \, dx=\log {\left (\log {\left (\frac {5}{x + \log {\left (3 \right )}} \right )} - \log {\left (x + 9 + e^{4} \right )} + 3 \right )} \]
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Time = 0.38 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {9+e^4+2 x+\log (3)}{-27 x-3 e^4 x-3 x^2+\left (-27-3 e^4-3 x\right ) \log (3)+\left (9 x+e^4 x+x^2+\left (9+e^4+x\right ) \log (3)\right ) \log \left (9+e^4+x\right )+\left (-9 x-e^4 x-x^2+\left (-9-e^4-x\right ) \log (3)\right ) \log \left (\frac {5}{x+\log (3)}\right )} \, dx=\log \left (-\log \left (5\right ) + \log \left (x + e^{4} + 9\right ) + \log \left (x + \log \left (3\right )\right ) - 3\right ) \]
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Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {9+e^4+2 x+\log (3)}{-27 x-3 e^4 x-3 x^2+\left (-27-3 e^4-3 x\right ) \log (3)+\left (9 x+e^4 x+x^2+\left (9+e^4+x\right ) \log (3)\right ) \log \left (9+e^4+x\right )+\left (-9 x-e^4 x-x^2+\left (-9-e^4-x\right ) \log (3)\right ) \log \left (\frac {5}{x+\log (3)}\right )} \, dx=\log \left (\log \left (5\right ) - \log \left (x + e^{4} + 9\right ) - \log \left (x + \log \left (3\right )\right ) + 3\right ) \]
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Time = 131.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {9+e^4+2 x+\log (3)}{-27 x-3 e^4 x-3 x^2+\left (-27-3 e^4-3 x\right ) \log (3)+\left (9 x+e^4 x+x^2+\left (9+e^4+x\right ) \log (3)\right ) \log \left (9+e^4+x\right )+\left (-9 x-e^4 x-x^2+\left (-9-e^4-x\right ) \log (3)\right ) \log \left (\frac {5}{x+\log (3)}\right )} \, dx=\ln \left (\ln \left (\frac {5}{x+\ln \left (3\right )}\right )-\ln \left (x+{\mathrm {e}}^4+9\right )+3\right ) \]
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