Integrand size = 22, antiderivative size = 18 \[ \int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx=-6+x-\frac {e^{2 (4+x)}}{3 \log (5)} \]
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Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 2225} \[ \int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx=x-\frac {e^{2 x+8}}{3 \log (5)} \]
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Rule 12
Rule 2225
Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (-2 e^{8+2 x}+3 \log (5)\right ) \, dx}{3 \log (5)} \\ & = x-\frac {2 \int e^{8+2 x} \, dx}{3 \log (5)} \\ & = x-\frac {e^{8+2 x}}{3 \log (5)} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx=x-\frac {e^{8+2 x}}{\log (125)} \]
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Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83
method | result | size |
norman | \(x -\frac {{\mathrm e}^{2 x +8}}{3 \ln \left (5\right )}\) | \(15\) |
risch | \(x -\frac {{\mathrm e}^{2 x +8}}{3 \ln \left (5\right )}\) | \(15\) |
parts | \(x -\frac {{\mathrm e}^{2 x +8}}{3 \ln \left (5\right )}\) | \(15\) |
default | \(\frac {-{\mathrm e}^{2 x +8}+3 x \ln \left (5\right )}{3 \ln \left (5\right )}\) | \(21\) |
parallelrisch | \(\frac {-{\mathrm e}^{2 x +8}+3 x \ln \left (5\right )}{3 \ln \left (5\right )}\) | \(21\) |
derivativedivides | \(\frac {-2 \,{\mathrm e}^{2 x +8}+3 \ln \left (5\right ) \ln \left ({\mathrm e}^{2 x +8}\right )}{6 \ln \left (5\right )}\) | \(27\) |
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Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx=\frac {3 \, x \log \left (5\right ) - e^{\left (2 \, x + 8\right )}}{3 \, \log \left (5\right )} \]
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Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx=x - \frac {e^{2 x + 8}}{3 \log {\left (5 \right )}} \]
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Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx=\frac {3 \, x \log \left (5\right ) - e^{\left (2 \, x + 8\right )}}{3 \, \log \left (5\right )} \]
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Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx=\frac {3 \, x \log \left (5\right ) - e^{\left (2 \, x + 8\right )}}{3 \, \log \left (5\right )} \]
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Time = 0.10 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx=x-\frac {{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^8}{3\,\ln \left (5\right )} \]
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