Integrand size = 15, antiderivative size = 15 \[ \int \frac {2 e^4+6 x \log (9)}{e^4} \, dx=-5+2 x+\frac {3 x^2 \log (9)}{e^4} \]
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Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {9} \[ \int \frac {2 e^4+6 x \log (9)}{e^4} \, dx=\frac {\left (3 x \log (9)+e^4\right )^2}{3 e^4 \log (9)} \]
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Rule 9
Rubi steps \begin{align*} \text {integral}& = \frac {\left (e^4+3 x \log (9)\right )^2}{3 e^4 \log (9)} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {2 e^4+6 x \log (9)}{e^4} \, dx=2 x+\frac {3 x^2 \log (9)}{e^4} \]
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Time = 0.20 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93
method | result | size |
risch | \(2 x +6 \ln \left (3\right ) x^{2} {\mathrm e}^{-4}\) | \(14\) |
gosper | \(2 x \left ({\mathrm e}^{4}+3 x \ln \left (3\right )\right ) {\mathrm e}^{-4}\) | \(18\) |
default | \(2 \,{\mathrm e}^{-4} \left (3 x^{2} \ln \left (3\right )+x \,{\mathrm e}^{4}\right )\) | \(21\) |
parallelrisch | \({\mathrm e}^{-4} \left (2 x \,{\mathrm e}^{4}+6 x^{2} \ln \left (3\right )\right )\) | \(21\) |
norman | \(\left (2 \,{\mathrm e}^{2} x +6 \,{\mathrm e}^{-2} \ln \left (3\right ) x^{2}\right ) {\mathrm e}^{-2}\) | \(23\) |
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none
Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {2 e^4+6 x \log (9)}{e^4} \, dx=2 \, {\left (3 \, x^{2} \log \left (3\right ) + x e^{4}\right )} e^{\left (-4\right )} \]
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Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {2 e^4+6 x \log (9)}{e^4} \, dx=\frac {6 x^{2} \log {\left (3 \right )}}{e^{4}} + 2 x \]
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none
Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {2 e^4+6 x \log (9)}{e^4} \, dx=2 \, {\left (3 \, x^{2} \log \left (3\right ) + x e^{4}\right )} e^{\left (-4\right )} \]
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none
Time = 0.29 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {2 e^4+6 x \log (9)}{e^4} \, dx=2 \, {\left (3 \, x^{2} \log \left (3\right ) + x e^{4}\right )} e^{\left (-4\right )} \]
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Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.20 \[ \int \frac {2 e^4+6 x \log (9)}{e^4} \, dx=\frac {{\mathrm {e}}^{-4}\,{\left ({\mathrm {e}}^4+6\,x\,\ln \left (3\right )\right )}^2}{6\,\ln \left (3\right )} \]
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