Integrand size = 113, antiderivative size = 20 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=-2+e^{1+x-\frac {4}{3} \log ^2\left (\log \left ((1+x)^4\right )\right )} \]
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Time = 0.42 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6820, 12, 6838} \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e^{x-\frac {4}{3} \log ^2\left (\log \left ((x+1)^4\right )\right )+1} \]
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Rule 12
Rule 6820
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{1+x-\frac {4}{3} \log ^2\left (\log \left ((1+x)^4\right )\right )} \left (3 (1+x) \log \left ((1+x)^4\right )-32 \log \left (\log \left ((1+x)^4\right )\right )\right )}{3 (1+x) \log \left ((1+x)^4\right )} \, dx \\ & = \frac {1}{3} \int \frac {e^{1+x-\frac {4}{3} \log ^2\left (\log \left ((1+x)^4\right )\right )} \left (3 (1+x) \log \left ((1+x)^4\right )-32 \log \left (\log \left ((1+x)^4\right )\right )\right )}{(1+x) \log \left ((1+x)^4\right )} \, dx \\ & = e^{1+x-\frac {4}{3} \log ^2\left (\log \left ((1+x)^4\right )\right )} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e^{1+x-\frac {4}{3} \log ^2\left (\log \left ((1+x)^4\right )\right )} \]
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Time = 0.54 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55
method | result | size |
parallelrisch | \({\mathrm e} \,{\mathrm e}^{-\frac {4 {\ln \left (\ln \left (x^{4}+4 x^{3}+6 x^{2}+4 x +1\right )\right )}^{2}}{3}+x}\) | \(31\) |
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Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e^{\left (-\frac {4}{3} \, \log \left (\log \left (x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1\right )\right )^{2} + x + 1\right )} \]
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Time = 0.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e e^{x - \frac {4 \log {\left (\log {\left (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1 \right )} \right )}^{2}}{3}} \]
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Time = 0.44 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e^{\left (-\frac {16}{3} \, \log \left (2\right )^{2} - \frac {16}{3} \, \log \left (2\right ) \log \left (\log \left (x + 1\right )\right ) - \frac {4}{3} \, \log \left (\log \left (x + 1\right )\right )^{2} + x + 1\right )} \]
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Time = 1.34 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e^{\left (-\frac {4}{3} \, \log \left (\log \left (x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1\right )\right )^{2} + x + 1\right )} \]
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Time = 8.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=\mathrm {e}\,{\mathrm {e}}^{-\frac {4\,{\ln \left (\ln \left (x^4+4\,x^3+6\,x^2+4\,x+1\right )\right )}^2}{3}}\,{\mathrm {e}}^x \]
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