Integrand size = 121, antiderivative size = 24 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log (4+x)+\log \left (\log \left (\log \left (25-2 \left (-5+\frac {2 e^3 x}{\log (3)}\right )\right )\right )\right ) \]
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Time = 0.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6820, 6816} \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log (x+4)+\log \left (\log \left (\log \left (35-\frac {4 e^3 x}{\log (3)}\right )\right )\right ) \]
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Rule 6816
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{4+x}+\frac {4 e^3}{\left (4 e^3 x-35 \log (3)\right ) \log \left (35-\frac {4 e^3 x}{\log (3)}\right ) \log \left (\log \left (35-\frac {4 e^3 x}{\log (3)}\right )\right )}\right ) \, dx \\ & = \log (4+x)+\left (4 e^3\right ) \int \frac {1}{\left (4 e^3 x-35 \log (3)\right ) \log \left (35-\frac {4 e^3 x}{\log (3)}\right ) \log \left (\log \left (35-\frac {4 e^3 x}{\log (3)}\right )\right )} \, dx \\ & = \log (4+x)+\log \left (\log \left (\log \left (35-\frac {4 e^3 x}{\log (3)}\right )\right )\right ) \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log (4+x)+\log \left (\log \left (\log \left (35-\frac {4 e^3 x}{\log (3)}\right )\right )\right ) \]
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Time = 0.45 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\ln \left (\ln \left (\ln \left (\frac {35 \ln \left (3\right )-4 x \,{\mathrm e}^{3}}{\ln \left (3\right )}\right )\right )\right )+\ln \left (4+x \right )\) | \(24\) |
risch | \(\ln \left (\ln \left (\ln \left (\frac {35 \ln \left (3\right )-4 x \,{\mathrm e}^{3}}{\ln \left (3\right )}\right )\right )\right )+\ln \left (4+x \right )\) | \(24\) |
parts | \(\ln \left (\ln \left (\ln \left (\frac {35 \ln \left (3\right )-4 x \,{\mathrm e}^{3}}{\ln \left (3\right )}\right )\right )\right )+\ln \left (4+x \right )\) | \(24\) |
default | \(\ln \left (4+x \right )+\ln \left (\ln \left (-\ln \left (\ln \left (3\right )\right )+\ln \left (35 \ln \left (3\right )-4 x \,{\mathrm e}^{3}\right )\right )\right )\) | \(25\) |
parallelrisch | \(\frac {\left (16 \ln \left (\ln \left (\ln \left (\frac {35 \ln \left (3\right )-4 x \,{\mathrm e}^{3}}{\ln \left (3\right )}\right )\right )\right ) {\mathrm e}^{6}+16 \,{\mathrm e}^{6} \ln \left (4+x \right )\right ) {\mathrm e}^{-6}}{16}\) | \(42\) |
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Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log \left (x + 4\right ) + \log \left (\log \left (\log \left (-\frac {4 \, x e^{3} - 35 \, \log \left (3\right )}{\log \left (3\right )}\right )\right )\right ) \]
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Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log {\left (x + 4 \right )} + \log {\left (\log {\left (\log {\left (\frac {- 4 x e^{3} + 35 \log {\left (3 \right )}}{\log {\left (3 \right )}} \right )} \right )} \right )} \]
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Time = 0.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log \left (x + 4\right ) + \log \left (\log \left (\log \left (-4 \, x e^{3} + 35 \, \log \left (3\right )\right ) - \log \left (\log \left (3\right )\right )\right )\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (19) = 38\).
Time = 0.32 (sec) , antiderivative size = 105, normalized size of antiderivative = 4.38 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\frac {35 \, \log \left (3\right ) \log \left (4 \, x e^{3} - 35 \, \log \left (3\right )\right ) - 35 \, \log \left (3\right ) \log \left (-4 \, x e^{3} + 35 \, \log \left (3\right )\right ) + 16 \, e^{3} \log \left (x + 4\right ) + 35 \, \log \left (3\right ) \log \left (x + 4\right ) + 16 \, e^{3} \log \left (\log \left (\log \left (-4 \, x e^{3} + 35 \, \log \left (3\right )\right ) - \log \left (\log \left (3\right )\right )\right )\right ) + 35 \, \log \left (3\right ) \log \left (\log \left (\log \left (-4 \, x e^{3} + 35 \, \log \left (3\right )\right ) - \log \left (\log \left (3\right )\right )\right )\right )}{16 \, e^{3} + 35 \, \log \left (3\right )} \]
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Time = 10.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\ln \left (x+4\right )+\ln \left (\ln \left (\ln \left (35\,\ln \left (3\right )-4\,x\,{\mathrm {e}}^3\right )-\ln \left (\ln \left (3\right )\right )\right )\right ) \]
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