Integrand size = 56, antiderivative size = 19 \[ \int \frac {-92 e-56 e \log \left (\frac {x}{15}\right )-8 e \log ^2\left (\frac {x}{15}\right )}{16 x^3+8 x^3 \log \left (\frac {x}{15}\right )+x^3 \log ^2\left (\frac {x}{15}\right )} \, dx=\frac {e \left (4-\frac {4}{4+\log \left (\frac {x}{15}\right )}\right )}{x^2} \]
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Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {6820, 12, 6874, 2343, 2346, 2209} \[ \int \frac {-92 e-56 e \log \left (\frac {x}{15}\right )-8 e \log ^2\left (\frac {x}{15}\right )}{16 x^3+8 x^3 \log \left (\frac {x}{15}\right )+x^3 \log ^2\left (\frac {x}{15}\right )} \, dx=\frac {4 e}{x^2}-\frac {4 e}{x^2 \left (\log \left (\frac {x}{15}\right )+4\right )} \]
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Rule 12
Rule 2209
Rule 2343
Rule 2346
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {4 e \left (-23-14 \log \left (\frac {x}{15}\right )-2 \log ^2\left (\frac {x}{15}\right )\right )}{x^3 \left (4+\log \left (\frac {x}{15}\right )\right )^2} \, dx \\ & = (4 e) \int \frac {-23-14 \log \left (\frac {x}{15}\right )-2 \log ^2\left (\frac {x}{15}\right )}{x^3 \left (4+\log \left (\frac {x}{15}\right )\right )^2} \, dx \\ & = (4 e) \int \left (-\frac {2}{x^3}+\frac {1}{x^3 \left (4+\log \left (\frac {x}{15}\right )\right )^2}+\frac {2}{x^3 \left (4+\log \left (\frac {x}{15}\right )\right )}\right ) \, dx \\ & = \frac {4 e}{x^2}+(4 e) \int \frac {1}{x^3 \left (4+\log \left (\frac {x}{15}\right )\right )^2} \, dx+(8 e) \int \frac {1}{x^3 \left (4+\log \left (\frac {x}{15}\right )\right )} \, dx \\ & = \frac {4 e}{x^2}-\frac {4 e}{x^2 \left (4+\log \left (\frac {x}{15}\right )\right )}+\frac {1}{225} (8 e) \text {Subst}\left (\int \frac {e^{-2 x}}{4+x} \, dx,x,\log \left (\frac {x}{15}\right )\right )-(8 e) \int \frac {1}{x^3 \left (4+\log \left (\frac {x}{15}\right )\right )} \, dx \\ & = \frac {4 e}{x^2}+\frac {8}{225} e^9 \text {Ei}\left (-2 \left (4+\log \left (\frac {x}{15}\right )\right )\right )-\frac {4 e}{x^2 \left (4+\log \left (\frac {x}{15}\right )\right )}-\frac {1}{225} (8 e) \text {Subst}\left (\int \frac {e^{-2 x}}{4+x} \, dx,x,\log \left (\frac {x}{15}\right )\right ) \\ & = \frac {4 e}{x^2}-\frac {4 e}{x^2 \left (4+\log \left (\frac {x}{15}\right )\right )} \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-92 e-56 e \log \left (\frac {x}{15}\right )-8 e \log ^2\left (\frac {x}{15}\right )}{16 x^3+8 x^3 \log \left (\frac {x}{15}\right )+x^3 \log ^2\left (\frac {x}{15}\right )} \, dx=-\frac {4 e \left (-1+\frac {1}{4+\log \left (\frac {x}{15}\right )}\right )}{x^2} \]
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Time = 2.48 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16
method | result | size |
derivativedivides | \(\frac {4 \,{\mathrm e} \left (\ln \left (\frac {x}{15}\right )+3\right )}{\left (4+\ln \left (\frac {x}{15}\right )\right ) x^{2}}\) | \(22\) |
default | \(\frac {4 \,{\mathrm e} \left (\ln \left (\frac {x}{15}\right )+3\right )}{\left (4+\ln \left (\frac {x}{15}\right )\right ) x^{2}}\) | \(22\) |
risch | \(\frac {4 \,{\mathrm e}}{x^{2}}-\frac {4 \,{\mathrm e}}{x^{2} \left (4+\ln \left (\frac {x}{15}\right )\right )}\) | \(24\) |
norman | \(\frac {4 \,{\mathrm e} \ln \left (\frac {x}{15}\right )+12 \,{\mathrm e}}{x^{2} \left (4+\ln \left (\frac {x}{15}\right )\right )}\) | \(26\) |
parallelrisch | \(\frac {4 \,{\mathrm e} \ln \left (\frac {x}{15}\right )+12 \,{\mathrm e}}{x^{2} \left (4+\ln \left (\frac {x}{15}\right )\right )}\) | \(26\) |
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Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {-92 e-56 e \log \left (\frac {x}{15}\right )-8 e \log ^2\left (\frac {x}{15}\right )}{16 x^3+8 x^3 \log \left (\frac {x}{15}\right )+x^3 \log ^2\left (\frac {x}{15}\right )} \, dx=\frac {4 \, {\left (e \log \left (\frac {1}{15} \, x\right ) + 3 \, e\right )}}{x^{2} \log \left (\frac {1}{15} \, x\right ) + 4 \, x^{2}} \]
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Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {-92 e-56 e \log \left (\frac {x}{15}\right )-8 e \log ^2\left (\frac {x}{15}\right )}{16 x^3+8 x^3 \log \left (\frac {x}{15}\right )+x^3 \log ^2\left (\frac {x}{15}\right )} \, dx=- \frac {4 e}{x^{2} \log {\left (\frac {x}{15} \right )} + 4 x^{2}} + \frac {4 e}{x^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (17) = 34\).
Time = 0.30 (sec) , antiderivative size = 38, normalized size of antiderivative = 2.00 \[ \int \frac {-92 e-56 e \log \left (\frac {x}{15}\right )-8 e \log ^2\left (\frac {x}{15}\right )}{16 x^3+8 x^3 \log \left (\frac {x}{15}\right )+x^3 \log ^2\left (\frac {x}{15}\right )} \, dx=\frac {4 \, {\left ({\left (\log \left (5\right ) + \log \left (3\right ) - 3\right )} e - e \log \left (x\right )\right )}}{x^{2} {\left (\log \left (5\right ) + \log \left (3\right ) - 4\right )} - x^{2} \log \left (x\right )} \]
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Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {-92 e-56 e \log \left (\frac {x}{15}\right )-8 e \log ^2\left (\frac {x}{15}\right )}{16 x^3+8 x^3 \log \left (\frac {x}{15}\right )+x^3 \log ^2\left (\frac {x}{15}\right )} \, dx=\frac {4 \, {\left (e \log \left (\frac {1}{15} \, x\right ) + 3 \, e\right )}}{x^{2} \log \left (\frac {1}{15} \, x\right ) + 4 \, x^{2}} \]
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Time = 9.37 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {-92 e-56 e \log \left (\frac {x}{15}\right )-8 e \log ^2\left (\frac {x}{15}\right )}{16 x^3+8 x^3 \log \left (\frac {x}{15}\right )+x^3 \log ^2\left (\frac {x}{15}\right )} \, dx=\frac {4\,\mathrm {e}\,\left (\ln \left (\frac {x}{15}\right )+3\right )}{x^2\,\left (\ln \left (\frac {x}{15}\right )+4\right )} \]
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