Integrand size = 86, antiderivative size = 29 \[ \int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx=\frac {7}{10 \left (e^{3-e^2-e^{x+\log ^2(3)} x}+\log (3)\right )} \]
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Time = 0.73 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6820, 12, 6818} \[ \int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx=-\frac {7 e^3}{10 \log (3) \left (\log (3) e^{x e^{x+\log ^2(3)}+e^2}+e^3\right )} \]
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Rule 12
Rule 6818
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {7 \exp \left (x+e^{x+\log ^2(3)} x+3 \left (1+\frac {1}{3} \left (e^2+\log ^2(3)\right )\right )\right ) (1+x)}{10 \left (e^3+e^{e^2+e^{x+\log ^2(3)} x} \log (3)\right )^2} \, dx \\ & = \frac {7}{10} \int \frac {\exp \left (x+e^{x+\log ^2(3)} x+3 \left (1+\frac {1}{3} \left (e^2+\log ^2(3)\right )\right )\right ) (1+x)}{\left (e^3+e^{e^2+e^{x+\log ^2(3)} x} \log (3)\right )^2} \, dx \\ & = -\frac {7 e^3}{10 \log (3) \left (e^3+e^{e^2+e^{x+\log ^2(3)} x} \log (3)\right )} \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx=-\frac {7 e^3}{10 \log (3) \left (e^3+e^{e^2+e^{x+\log ^2(3)} x} \log (3)\right )} \]
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Time = 0.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86
method | result | size |
default | \(\frac {7}{10 \left (\ln \left (3\right )+{\mathrm e}^{-x \,{\mathrm e}^{\ln \left (3\right )^{2}+x}-{\mathrm e}^{2}+3}\right )}\) | \(25\) |
norman | \(\frac {7}{10 \left (\ln \left (3\right )+{\mathrm e}^{-x \,{\mathrm e}^{\ln \left (3\right )^{2}+x}-{\mathrm e}^{2}+3}\right )}\) | \(25\) |
risch | \(\frac {7}{10 \left (\ln \left (3\right )+{\mathrm e}^{-x \,{\mathrm e}^{\ln \left (3\right )^{2}+x}-{\mathrm e}^{2}+3}\right )}\) | \(25\) |
parallelrisch | \(\frac {7}{10 \left (\ln \left (3\right )+{\mathrm e}^{-x \,{\mathrm e}^{\ln \left (3\right )^{2}+x}-{\mathrm e}^{2}+3}\right )}\) | \(25\) |
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Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx=\frac {7 \, e^{\left (\log \left (3\right )^{2} + x\right )}}{10 \, {\left (e^{\left (\log \left (3\right )^{2} + x\right )} \log \left (3\right ) + e^{\left (-x e^{\left (\log \left (3\right )^{2} + x\right )} + \log \left (3\right )^{2} + x - e^{2} + 3\right )}\right )}} \]
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Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx=\frac {7}{10 e^{- x e^{x + \log {\left (3 \right )}^{2}} - e^{2} + 3} + 10 \log {\left (3 \right )}} \]
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Time = 0.39 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx=-\frac {7 \, e^{3}}{10 \, {\left (e^{\left (x e^{\left (\log \left (3\right )^{2} + x\right )} + e^{2}\right )} \log \left (3\right )^{2} + e^{3} \log \left (3\right )\right )}} \]
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Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx=-\frac {7 \, e^{3}}{10 \, {\left (e^{\left (x e^{\left (\log \left (3\right )^{2} + x\right )} + e^{2}\right )} \log \left (3\right ) + e^{3}\right )} \log \left (3\right )} \]
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Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx=\frac {7}{10\,\left (\ln \left (3\right )+{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x\,{\mathrm {e}}^{{\ln \left (3\right )}^2}\,{\mathrm {e}}^x}\,{\mathrm {e}}^3\right )} \]
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