Integrand size = 11, antiderivative size = 12 \[ \int \left (1+e^x (3+x) \log (5)\right ) \, dx=-3+x+e^x (2+x) \log (5) \]
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Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.50, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2207, 2225} \[ \int \left (1+e^x (3+x) \log (5)\right ) \, dx=x-e^x \log (5)+e^x (x+3) \log (5) \]
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Rule 2207
Rule 2225
Rubi steps \begin{align*} \text {integral}& = x+\log (5) \int e^x (3+x) \, dx \\ & = x+e^x (3+x) \log (5)-\log (5) \int e^x \, dx \\ & = x-e^x \log (5)+e^x (3+x) \log (5) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \left (1+e^x (3+x) \log (5)\right ) \, dx=x+e^x (2+x) \log (5) \]
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Time = 0.07 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92
method | result | size |
risch | \({\mathrm e}^{x} \left (2+x \right ) \ln \left (5\right )+x\) | \(11\) |
default | \(x +\ln \left (5\right ) \left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right )\) | \(15\) |
norman | \(x +x \,{\mathrm e}^{x} \ln \left (5\right )+2 \,{\mathrm e}^{x} \ln \left (5\right )\) | \(15\) |
parallelrisch | \(x +\ln \left (5\right ) \left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right )\) | \(15\) |
parts | \(x +\ln \left (5\right ) \left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right )\) | \(15\) |
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none
Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \left (1+e^x (3+x) \log (5)\right ) \, dx={\left (x + 2\right )} e^{x} \log \left (5\right ) + x \]
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Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \left (1+e^x (3+x) \log (5)\right ) \, dx=x + \left (x \log {\left (5 \right )} + 2 \log {\left (5 \right )}\right ) e^{x} \]
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none
Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.33 \[ \int \left (1+e^x (3+x) \log (5)\right ) \, dx={\left ({\left (x - 1\right )} e^{x} + 3 \, e^{x}\right )} \log \left (5\right ) + x \]
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none
Time = 0.27 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \left (1+e^x (3+x) \log (5)\right ) \, dx={\left (x + 2\right )} e^{x} \log \left (5\right ) + x \]
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Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \left (1+e^x (3+x) \log (5)\right ) \, dx=x+2\,{\mathrm {e}}^x\,\ln \left (5\right )+x\,{\mathrm {e}}^x\,\ln \left (5\right ) \]
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