Integrand size = 42, antiderivative size = 15 \[ \int \frac {-1-18 e^4-81 e^8+\left (6 e^4+54 e^8\right ) \log (20)-9 e^8 \log ^2(20)}{e^8 x^2} \, dx=\frac {\left (9+\frac {1}{e^4}-3 \log (20)\right )^2}{x} \]
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Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {12, 30} \[ \int \frac {-1-18 e^4-81 e^8+\left (6 e^4+54 e^8\right ) \log (20)-9 e^8 \log ^2(20)}{e^8 x^2} \, dx=\frac {\left (1+3 e^4 (3-\log (20))\right )^2}{e^8 x} \]
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Rule 12
Rule 30
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (1+3 e^4 (3-\log (20))\right )^2 \int \frac {1}{x^2} \, dx}{e^8} \\ & = \frac {\left (1+3 e^4 (3-\log (20))\right )^2}{e^8 x} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33 \[ \int \frac {-1-18 e^4-81 e^8+\left (6 e^4+54 e^8\right ) \log (20)-9 e^8 \log ^2(20)}{e^8 x^2} \, dx=\frac {\left (1-3 e^4 (-3+\log (20))\right )^2}{e^8 x} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(48\) vs. \(2(16)=32\).
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 3.27
method | result | size |
gosper | \(\frac {\left (9 \,{\mathrm e}^{8} \ln \left (20\right )^{2}-54 \ln \left (20\right ) {\mathrm e}^{8}+81 \,{\mathrm e}^{8}-6 \ln \left (20\right ) {\mathrm e}^{4}+18 \,{\mathrm e}^{4}+1\right ) {\mathrm e}^{-8}}{x}\) | \(49\) |
norman | \(\frac {\left (9 \,{\mathrm e}^{8} \ln \left (20\right )^{2}-54 \ln \left (20\right ) {\mathrm e}^{8}+81 \,{\mathrm e}^{8}-6 \ln \left (20\right ) {\mathrm e}^{4}+18 \,{\mathrm e}^{4}+1\right ) {\mathrm e}^{-8}}{x}\) | \(49\) |
default | \(-\frac {\left (-9 \,{\mathrm e}^{8} \ln \left (20\right )^{2}+\left (54 \,{\mathrm e}^{8}+6 \,{\mathrm e}^{4}\right ) \ln \left (20\right )-81 \,{\mathrm e}^{8}-18 \,{\mathrm e}^{4}-1\right ) {\mathrm e}^{-8}}{x}\) | \(50\) |
parallelrisch | \(-\frac {\left (-9 \,{\mathrm e}^{8} \ln \left (20\right )^{2}+\left (54 \,{\mathrm e}^{8}+6 \,{\mathrm e}^{4}\right ) \ln \left (20\right )-81 \,{\mathrm e}^{8}-18 \,{\mathrm e}^{4}-1\right ) {\mathrm e}^{-8}}{x}\) | \(50\) |
risch | \(\frac {36 \,{\mathrm e}^{-8} \ln \left (2\right )^{2} {\mathrm e}^{8}}{x}+\frac {36 \,{\mathrm e}^{-8} \ln \left (2\right ) {\mathrm e}^{8} \ln \left (5\right )}{x}+\frac {9 \,{\mathrm e}^{-8} {\mathrm e}^{8} \ln \left (5\right )^{2}}{x}-\frac {108 \,{\mathrm e}^{-8} {\mathrm e}^{8} \ln \left (2\right )}{x}-\frac {12 \,{\mathrm e}^{-8} {\mathrm e}^{4} \ln \left (2\right )}{x}-\frac {54 \,{\mathrm e}^{-8} {\mathrm e}^{8} \ln \left (5\right )}{x}-\frac {6 \,{\mathrm e}^{-8} {\mathrm e}^{4} \ln \left (5\right )}{x}+\frac {81 \,{\mathrm e}^{-8} {\mathrm e}^{8}}{x}+\frac {18 \,{\mathrm e}^{-8} {\mathrm e}^{4}}{x}+\frac {{\mathrm e}^{-8}}{x}\) | \(109\) |
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Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (14) = 28\).
Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.33 \[ \int \frac {-1-18 e^4-81 e^8+\left (6 e^4+54 e^8\right ) \log (20)-9 e^8 \log ^2(20)}{e^8 x^2} \, dx=\frac {{\left (9 \, e^{8} \log \left (20\right )^{2} - 6 \, {\left (9 \, e^{8} + e^{4}\right )} \log \left (20\right ) + 81 \, e^{8} + 18 \, e^{4} + 1\right )} e^{\left (-8\right )}}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (14) = 28\).
Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.93 \[ \int \frac {-1-18 e^4-81 e^8+\left (6 e^4+54 e^8\right ) \log (20)-9 e^8 \log ^2(20)}{e^8 x^2} \, dx=- \frac {- 81 e^{8} - 9 e^{8} \log {\left (20 \right )}^{2} - 18 e^{4} - 1 + 6 e^{4} \log {\left (20 \right )} + 54 e^{8} \log {\left (20 \right )}}{x e^{8}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (14) = 28\).
Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.33 \[ \int \frac {-1-18 e^4-81 e^8+\left (6 e^4+54 e^8\right ) \log (20)-9 e^8 \log ^2(20)}{e^8 x^2} \, dx=\frac {{\left (9 \, e^{8} \log \left (20\right )^{2} - 6 \, {\left (9 \, e^{8} + e^{4}\right )} \log \left (20\right ) + 81 \, e^{8} + 18 \, e^{4} + 1\right )} e^{\left (-8\right )}}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (14) = 28\).
Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.33 \[ \int \frac {-1-18 e^4-81 e^8+\left (6 e^4+54 e^8\right ) \log (20)-9 e^8 \log ^2(20)}{e^8 x^2} \, dx=\frac {{\left (9 \, e^{8} \log \left (20\right )^{2} - 6 \, {\left (9 \, e^{8} + e^{4}\right )} \log \left (20\right ) + 81 \, e^{8} + 18 \, e^{4} + 1\right )} e^{\left (-8\right )}}{x} \]
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Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33 \[ \int \frac {-1-18 e^4-81 e^8+\left (6 e^4+54 e^8\right ) \log (20)-9 e^8 \log ^2(20)}{e^8 x^2} \, dx=\frac {{\mathrm {e}}^{-8}\,{\left (9\,{\mathrm {e}}^4-3\,{\mathrm {e}}^4\,\ln \left (20\right )+1\right )}^2}{x} \]
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