Integrand size = 54, antiderivative size = 22 \[ \int \frac {39-9 \log \left (4+e^2\right )}{169-78 e^2 x+9 e^4 x^2+\left (-78+18 e^2 x\right ) \log \left (4+e^2\right )+9 \log ^2\left (4+e^2\right )} \, dx=\frac {x}{\frac {13}{3}-e^2 x-\log \left (4+e^2\right )} \]
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Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {12, 2006, 27, 32} \[ \int \frac {39-9 \log \left (4+e^2\right )}{169-78 e^2 x+9 e^4 x^2+\left (-78+18 e^2 x\right ) \log \left (4+e^2\right )+9 \log ^2\left (4+e^2\right )} \, dx=\frac {13-3 \log \left (4+e^2\right )}{e^2 \left (-3 e^2 x+13-3 \log \left (4+e^2\right )\right )} \]
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Rule 12
Rule 27
Rule 32
Rule 2006
Rubi steps \begin{align*} \text {integral}& = \left (3 \left (13-3 \log \left (4+e^2\right )\right )\right ) \int \frac {1}{169-78 e^2 x+9 e^4 x^2+\left (-78+18 e^2 x\right ) \log \left (4+e^2\right )+9 \log ^2\left (4+e^2\right )} \, dx \\ & = \left (3 \left (13-3 \log \left (4+e^2\right )\right )\right ) \int \frac {1}{9 e^4 x^2-6 e^2 x \left (13-3 \log \left (4+e^2\right )\right )+\left (13-3 \log \left (4+e^2\right )\right )^2} \, dx \\ & = \left (3 \left (13-3 \log \left (4+e^2\right )\right )\right ) \int \frac {1}{\left (-13+3 e^2 x+3 \log \left (4+e^2\right )\right )^2} \, dx \\ & = \frac {13-3 \log \left (4+e^2\right )}{e^2 \left (13-3 e^2 x-3 \log \left (4+e^2\right )\right )} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59 \[ \int \frac {39-9 \log \left (4+e^2\right )}{169-78 e^2 x+9 e^4 x^2+\left (-78+18 e^2 x\right ) \log \left (4+e^2\right )+9 \log ^2\left (4+e^2\right )} \, dx=-\frac {39-9 \log \left (4+e^2\right )}{3 e^2 \left (-13+3 e^2 x+3 \log \left (4+e^2\right )\right )} \]
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Time = 0.46 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41
method | result | size |
gosper | \(\frac {\left (3 \ln \left (4+{\mathrm e}^{2}\right )-13\right ) {\mathrm e}^{-2}}{3 \,{\mathrm e}^{2} x +3 \ln \left (4+{\mathrm e}^{2}\right )-13}\) | \(31\) |
norman | \(\frac {\left (3 \ln \left (4+{\mathrm e}^{2}\right )-13\right ) {\mathrm e}^{-2}}{3 \,{\mathrm e}^{2} x +3 \ln \left (4+{\mathrm e}^{2}\right )-13}\) | \(31\) |
parallelrisch | \(-\frac {\left (-9 \ln \left (4+{\mathrm e}^{2}\right )+39\right ) {\mathrm e}^{-2}}{3 \left (3 \,{\mathrm e}^{2} x +3 \ln \left (4+{\mathrm e}^{2}\right )-13\right )}\) | \(32\) |
risch | \(\frac {{\mathrm e}^{-2} \ln \left (4+{\mathrm e}^{2}\right )}{{\mathrm e}^{2} x +\ln \left (4+{\mathrm e}^{2}\right )-\frac {13}{3}}-\frac {13 \,{\mathrm e}^{-2}}{3 \left ({\mathrm e}^{2} x +\ln \left (4+{\mathrm e}^{2}\right )-\frac {13}{3}\right )}\) | \(40\) |
meijerg | \(\frac {9 \ln \left (4+{\mathrm e}^{2}\right ) \left (-3 \ln \left (4+{\mathrm e}^{2}\right )+13\right ) x}{{\left (3 \ln \left (4+{\mathrm e}^{2}\right )-13\right )}^{3} \left (1+\frac {3 x \,{\mathrm e}^{2}}{3 \ln \left (4+{\mathrm e}^{2}\right )-13}\right )}-\frac {39 \left (-3 \ln \left (4+{\mathrm e}^{2}\right )+13\right ) x}{{\left (3 \ln \left (4+{\mathrm e}^{2}\right )-13\right )}^{3} \left (1+\frac {3 x \,{\mathrm e}^{2}}{3 \ln \left (4+{\mathrm e}^{2}\right )-13}\right )}\) | \(93\) |
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Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {39-9 \log \left (4+e^2\right )}{169-78 e^2 x+9 e^4 x^2+\left (-78+18 e^2 x\right ) \log \left (4+e^2\right )+9 \log ^2\left (4+e^2\right )} \, dx=\frac {3 \, \log \left (e^{2} + 4\right ) - 13}{3 \, x e^{4} + 3 \, e^{2} \log \left (e^{2} + 4\right ) - 13 \, e^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (15) = 30\).
Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {39-9 \log \left (4+e^2\right )}{169-78 e^2 x+9 e^4 x^2+\left (-78+18 e^2 x\right ) \log \left (4+e^2\right )+9 \log ^2\left (4+e^2\right )} \, dx=- \frac {39 - 9 \log {\left (4 + e^{2} \right )}}{9 x e^{4} - 39 e^{2} + 9 e^{2} \log {\left (4 + e^{2} \right )}} \]
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Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {39-9 \log \left (4+e^2\right )}{169-78 e^2 x+9 e^4 x^2+\left (-78+18 e^2 x\right ) \log \left (4+e^2\right )+9 \log ^2\left (4+e^2\right )} \, dx=\frac {3 \, \log \left (e^{2} + 4\right ) - 13}{3 \, x e^{4} + 3 \, e^{2} \log \left (e^{2} + 4\right ) - 13 \, e^{2}} \]
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Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {39-9 \log \left (4+e^2\right )}{169-78 e^2 x+9 e^4 x^2+\left (-78+18 e^2 x\right ) \log \left (4+e^2\right )+9 \log ^2\left (4+e^2\right )} \, dx=\frac {{\left (3 \, \log \left (e^{2} + 4\right ) - 13\right )} e^{\left (-2\right )}}{3 \, x e^{2} + 3 \, \log \left (e^{2} + 4\right ) - 13} \]
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Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {39-9 \log \left (4+e^2\right )}{169-78 e^2 x+9 e^4 x^2+\left (-78+18 e^2 x\right ) \log \left (4+e^2\right )+9 \log ^2\left (4+e^2\right )} \, dx=\frac {{\mathrm {e}}^{-2}\,\left (3\,\ln \left ({\mathrm {e}}^2+4\right )-13\right )}{3\,\ln \left ({\mathrm {e}}^2+4\right )+3\,x\,{\mathrm {e}}^2-13} \]
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