Integrand size = 38, antiderivative size = 30 \[ \int \frac {3+3 e^3+8 x^2+\left (1+e^3+6 x^2\right ) \log (2)}{x^2+x^2 \log (2)} \, dx=5 x-\left (4+\frac {1+e^3}{x}-x\right ) \left (1+\frac {2}{1+\log (2)}\right ) \]
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Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {6, 12, 14} \[ \int \frac {3+3 e^3+8 x^2+\left (1+e^3+6 x^2\right ) \log (2)}{x^2+x^2 \log (2)} \, dx=\frac {2 x (4+\log (8))}{1+\log (2)}-\frac {\left (1+e^3\right ) (3+\log (2))}{x (1+\log (2))} \]
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Rule 6
Rule 12
Rule 14
Rubi steps \begin{align*} \text {integral}& = \int \frac {3+3 e^3+8 x^2+\left (1+e^3+6 x^2\right ) \log (2)}{x^2 (1+\log (2))} \, dx \\ & = \frac {\int \frac {3+3 e^3+8 x^2+\left (1+e^3+6 x^2\right ) \log (2)}{x^2} \, dx}{1+\log (2)} \\ & = \frac {\int \left (8 \left (1+\frac {3 \log (2)}{4}\right )+\frac {\left (1+e^3\right ) (3+\log (2))}{x^2}\right ) \, dx}{1+\log (2)} \\ & = -\frac {\left (1+e^3\right ) (3+\log (2))}{x (1+\log (2))}+\frac {2 x (4+\log (8))}{1+\log (2)} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {3+3 e^3+8 x^2+\left (1+e^3+6 x^2\right ) \log (2)}{x^2+x^2 \log (2)} \, dx=-\frac {3+\log (2)+e^3 (3+\log (2))-x^2 (8+\log (64))}{x (1+\log (2))} \]
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Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17
method | result | size |
default | \(\frac {6 x \ln \left (2\right )+8 x -\frac {{\mathrm e}^{3} \ln \left (2\right )+3 \,{\mathrm e}^{3}+\ln \left (2\right )+3}{x}}{1+\ln \left (2\right )}\) | \(35\) |
gosper | \(-\frac {-6 x^{2} \ln \left (2\right )+{\mathrm e}^{3} \ln \left (2\right )-8 x^{2}+3 \,{\mathrm e}^{3}+\ln \left (2\right )+3}{x \left (1+\ln \left (2\right )\right )}\) | \(37\) |
parallelrisch | \(-\frac {-6 x^{2} \ln \left (2\right )+{\mathrm e}^{3} \ln \left (2\right )-8 x^{2}+3 \,{\mathrm e}^{3}+\ln \left (2\right )+3}{x \left (1+\ln \left (2\right )\right )}\) | \(37\) |
norman | \(\frac {-\frac {{\mathrm e}^{3} \ln \left (2\right )+3 \,{\mathrm e}^{3}+\ln \left (2\right )+3}{1+\ln \left (2\right )}+\frac {2 \left (3 \ln \left (2\right )+4\right ) x^{2}}{1+\ln \left (2\right )}}{x}\) | \(44\) |
risch | \(\frac {6 x \ln \left (2\right )}{1+\ln \left (2\right )}+\frac {8 x}{1+\ln \left (2\right )}-\frac {{\mathrm e}^{3} \ln \left (2\right )^{2}}{\left (1+\ln \left (2\right )\right )^{2} x}-\frac {4 \,{\mathrm e}^{3} \ln \left (2\right )}{\left (1+\ln \left (2\right )\right )^{2} x}-\frac {\ln \left (2\right )^{2}}{\left (1+\ln \left (2\right )\right )^{2} x}-\frac {3 \,{\mathrm e}^{3}}{\left (1+\ln \left (2\right )\right )^{2} x}-\frac {4 \ln \left (2\right )}{\left (1+\ln \left (2\right )\right )^{2} x}-\frac {3}{\left (1+\ln \left (2\right )\right )^{2} x}\) | \(106\) |
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Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {3+3 e^3+8 x^2+\left (1+e^3+6 x^2\right ) \log (2)}{x^2+x^2 \log (2)} \, dx=\frac {8 \, x^{2} + {\left (6 \, x^{2} - e^{3} - 1\right )} \log \left (2\right ) - 3 \, e^{3} - 3}{x \log \left (2\right ) + x} \]
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Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {3+3 e^3+8 x^2+\left (1+e^3+6 x^2\right ) \log (2)}{x^2+x^2 \log (2)} \, dx=\frac {x \left (6 \log {\left (2 \right )} + 8\right ) + \frac {- 3 e^{3} - e^{3} \log {\left (2 \right )} - 3 - \log {\left (2 \right )}}{x}}{\log {\left (2 \right )} + 1} \]
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Time = 0.18 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {3+3 e^3+8 x^2+\left (1+e^3+6 x^2\right ) \log (2)}{x^2+x^2 \log (2)} \, dx=\frac {2 \, x {\left (3 \, \log \left (2\right ) + 4\right )}}{\log \left (2\right ) + 1} - \frac {{\left (e^{3} + 1\right )} \log \left (2\right ) + 3 \, e^{3} + 3}{x {\left (\log \left (2\right ) + 1\right )}} \]
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Time = 0.31 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {3+3 e^3+8 x^2+\left (1+e^3+6 x^2\right ) \log (2)}{x^2+x^2 \log (2)} \, dx=\frac {2 \, {\left (3 \, x \log \left (2\right ) + 4 \, x\right )}}{\log \left (2\right ) + 1} - \frac {e^{3} \log \left (2\right ) + 3 \, e^{3} + \log \left (2\right ) + 3}{x {\left (\log \left (2\right ) + 1\right )}} \]
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Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {3+3 e^3+8 x^2+\left (1+e^3+6 x^2\right ) \log (2)}{x^2+x^2 \log (2)} \, dx=\frac {x\,\left (\ln \left (64\right )+8\right )}{\ln \left (2\right )+1}-\frac {3\,{\mathrm {e}}^3+\ln \left (2\right )+{\mathrm {e}}^3\,\ln \left (2\right )+3}{x\,\left (\ln \left (2\right )+1\right )} \]
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