Integrand size = 33, antiderivative size = 23 \[ \int \frac {1-e^x+2 x \log (3)+2 x \log \left (4 e^6\right )}{\log (3)+\log \left (4 e^6\right )} \, dx=x^2+\frac {-e^x+x}{\log (3)+\log \left (4 e^6\right )} \]
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Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6, 12, 2225} \[ \int \frac {1-e^x+2 x \log (3)+2 x \log \left (4 e^6\right )}{\log (3)+\log \left (4 e^6\right )} \, dx=x^2+\frac {x}{6+\log (12)}-\frac {e^x}{6+\log (12)} \]
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Rule 6
Rule 12
Rule 2225
Rubi steps \begin{align*} \text {integral}& = \int \frac {1-e^x+2 x \left (\log (3)+\log \left (4 e^6\right )\right )}{\log (3)+\log \left (4 e^6\right )} \, dx \\ & = \frac {\int \left (1-e^x+2 x \left (\log (3)+\log \left (4 e^6\right )\right )\right ) \, dx}{6+\log (12)} \\ & = x^2+\frac {x}{6+\log (12)}-\frac {\int e^x \, dx}{6+\log (12)} \\ & = x^2-\frac {e^x}{6+\log (12)}+\frac {x}{6+\log (12)} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {1-e^x+2 x \log (3)+2 x \log \left (4 e^6\right )}{\log (3)+\log \left (4 e^6\right )} \, dx=\frac {-e^x+x+6 x^2+\frac {1}{2} x^2 \log (144)}{6+\log (12)} \]
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Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35
method | result | size |
norman | \(x^{2}+\frac {x}{6+2 \ln \left (2\right )+\ln \left (3\right )}-\frac {{\mathrm e}^{x}}{6+2 \ln \left (2\right )+\ln \left (3\right )}\) | \(31\) |
default | \(\frac {x +x^{2} \ln \left (3\right )+\ln \left (4 \,{\mathrm e}^{6}\right ) x^{2}-{\mathrm e}^{x}}{\ln \left (4 \,{\mathrm e}^{6}\right )+\ln \left (3\right )}\) | \(33\) |
parallelrisch | \(\frac {x +x^{2} \ln \left (3\right )+\ln \left (4 \,{\mathrm e}^{6}\right ) x^{2}-{\mathrm e}^{x}}{\ln \left (4 \,{\mathrm e}^{6}\right )+\ln \left (3\right )}\) | \(33\) |
parts | \(\frac {x +x^{2} \ln \left (3\right )+\ln \left (4 \,{\mathrm e}^{6}\right ) x^{2}}{\ln \left (4 \,{\mathrm e}^{6}\right )+\ln \left (3\right )}-\frac {{\mathrm e}^{x}}{\ln \left (4 \,{\mathrm e}^{6}\right )+\ln \left (3\right )}\) | \(44\) |
risch | \(\frac {2 x^{2} \ln \left (2\right )}{6+2 \ln \left (2\right )+\ln \left (3\right )}+\frac {x^{2} \ln \left (3\right )}{6+2 \ln \left (2\right )+\ln \left (3\right )}+\frac {6 x^{2}}{6+2 \ln \left (2\right )+\ln \left (3\right )}+\frac {x}{6+2 \ln \left (2\right )+\ln \left (3\right )}-\frac {{\mathrm e}^{x}}{6+2 \ln \left (2\right )+\ln \left (3\right )}\) | \(76\) |
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Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {1-e^x+2 x \log (3)+2 x \log \left (4 e^6\right )}{\log (3)+\log \left (4 e^6\right )} \, dx=\frac {x^{2} \log \left (3\right ) + 2 \, x^{2} \log \left (2\right ) + 6 \, x^{2} + x - e^{x}}{\log \left (3\right ) + 2 \, \log \left (2\right ) + 6} \]
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Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {1-e^x+2 x \log (3)+2 x \log \left (4 e^6\right )}{\log (3)+\log \left (4 e^6\right )} \, dx=x^{2} + \frac {x}{\log {\left (3 \right )} + 2 \log {\left (2 \right )} + 6} - \frac {e^{x}}{\log {\left (3 \right )} + 2 \log {\left (2 \right )} + 6} \]
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Time = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {1-e^x+2 x \log (3)+2 x \log \left (4 e^6\right )}{\log (3)+\log \left (4 e^6\right )} \, dx=\frac {x^{2} \log \left (3\right ) + x^{2} \log \left (4 \, e^{6}\right ) + x - e^{x}}{\log \left (3\right ) + \log \left (4 \, e^{6}\right )} \]
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Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {1-e^x+2 x \log (3)+2 x \log \left (4 e^6\right )}{\log (3)+\log \left (4 e^6\right )} \, dx=\frac {x^{2} \log \left (3\right ) + x^{2} \log \left (4 \, e^{6}\right ) + x - e^{x}}{\log \left (3\right ) + \log \left (4 \, e^{6}\right )} \]
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Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1-e^x+2 x \log (3)+2 x \log \left (4 e^6\right )}{\log (3)+\log \left (4 e^6\right )} \, dx=\frac {x-{\mathrm {e}}^x+x^2\,\left (\ln \left (12\right )+6\right )}{\ln \left (12\,{\mathrm {e}}^6\right )} \]
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