Integrand size = 60, antiderivative size = 19 \[ \int \frac {-16+24 x^2-16 x^3+3 x^4+\left (-32+24 x^2-8 x^3\right ) \log (25)+\left (-24+6 x^2\right ) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx=\frac {\left (2 x-x^2+x \log (25)\right )^4}{x^5} \]
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Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {14} \[ \int \frac {-16+24 x^2-16 x^3+3 x^4+\left (-32+24 x^2-8 x^3\right ) \log (25)+\left (-24+6 x^2\right ) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx=x^3-4 x^2 (2+\log (25))+6 x (2+\log (25))^2+\frac {(2+\log (25))^4}{x} \]
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Rule 14
Rubi steps \begin{align*} \text {integral}& = \int \left (3 x^2-8 x (2+\log (25))+6 (2+\log (25))^2-\frac {(2+\log (25))^4}{x^2}\right ) \, dx \\ & = x^3-4 x^2 (2+\log (25))+6 x (2+\log (25))^2+\frac {(2+\log (25))^4}{x} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(40\) vs. \(2(19)=38\).
Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.11 \[ \int \frac {-16+24 x^2-16 x^3+3 x^4+\left (-32+24 x^2-8 x^3\right ) \log (25)+\left (-24+6 x^2\right ) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx=x^3-4 x^2 (2+\log (25))+6 x (2+\log (25))^2-3 (2+\log (25))^3+\frac {(2+\log (25))^4}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(57\) vs. \(2(20)=40\).
Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 3.05
method | result | size |
norman | \(\frac {x^{4}+\left (-8 \ln \left (5\right )-8\right ) x^{3}+\left (24 \ln \left (5\right )^{2}+48 \ln \left (5\right )+24\right ) x^{2}+16 \ln \left (5\right )^{4}+64 \ln \left (5\right )^{3}+96 \ln \left (5\right )^{2}+64 \ln \left (5\right )+16}{x}\) | \(58\) |
default | \(24 x \ln \left (5\right )^{2}-8 x^{2} \ln \left (5\right )+x^{3}+48 x \ln \left (5\right )-8 x^{2}+24 x -\frac {-16 \ln \left (5\right )^{4}-64 \ln \left (5\right )^{3}-96 \ln \left (5\right )^{2}-64 \ln \left (5\right )-16}{x}\) | \(61\) |
gosper | \(\frac {16 \ln \left (5\right )^{4}+24 x^{2} \ln \left (5\right )^{2}-8 x^{3} \ln \left (5\right )+x^{4}+64 \ln \left (5\right )^{3}+48 x^{2} \ln \left (5\right )-8 x^{3}+96 \ln \left (5\right )^{2}+24 x^{2}+64 \ln \left (5\right )+16}{x}\) | \(65\) |
parallelrisch | \(\frac {16 \ln \left (5\right )^{4}+24 x^{2} \ln \left (5\right )^{2}-8 x^{3} \ln \left (5\right )+x^{4}+64 \ln \left (5\right )^{3}+48 x^{2} \ln \left (5\right )-8 x^{3}+96 \ln \left (5\right )^{2}+24 x^{2}+64 \ln \left (5\right )+16}{x}\) | \(65\) |
risch | \(24 x \ln \left (5\right )^{2}-8 x^{2} \ln \left (5\right )+x^{3}+48 x \ln \left (5\right )-8 x^{2}+24 x +\frac {16 \ln \left (5\right )^{4}}{x}+\frac {64 \ln \left (5\right )^{3}}{x}+\frac {96 \ln \left (5\right )^{2}}{x}+\frac {64 \ln \left (5\right )}{x}+\frac {16}{x}\) | \(71\) |
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Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (18) = 36\).
Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.95 \[ \int \frac {-16+24 x^2-16 x^3+3 x^4+\left (-32+24 x^2-8 x^3\right ) \log (25)+\left (-24+6 x^2\right ) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx=\frac {x^{4} + 16 \, \log \left (5\right )^{4} - 8 \, x^{3} + 24 \, {\left (x^{2} + 4\right )} \log \left (5\right )^{2} + 64 \, \log \left (5\right )^{3} + 24 \, x^{2} - 8 \, {\left (x^{3} - 6 \, x^{2} - 8\right )} \log \left (5\right ) + 16}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (17) = 34\).
Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 3.05 \[ \int \frac {-16+24 x^2-16 x^3+3 x^4+\left (-32+24 x^2-8 x^3\right ) \log (25)+\left (-24+6 x^2\right ) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx=x^{3} + x^{2} \left (- 8 \log {\left (5 \right )} - 8\right ) + x \left (24 + 24 \log {\left (5 \right )}^{2} + 48 \log {\left (5 \right )}\right ) + \frac {16 + 64 \log {\left (5 \right )} + 16 \log {\left (5 \right )}^{4} + 96 \log {\left (5 \right )}^{2} + 64 \log {\left (5 \right )}^{3}}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (18) = 36\).
Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.79 \[ \int \frac {-16+24 x^2-16 x^3+3 x^4+\left (-32+24 x^2-8 x^3\right ) \log (25)+\left (-24+6 x^2\right ) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx=x^{3} - 8 \, x^{2} {\left (\log \left (5\right ) + 1\right )} + 24 \, {\left (\log \left (5\right )^{2} + 2 \, \log \left (5\right ) + 1\right )} x + \frac {16 \, {\left (\log \left (5\right )^{4} + 4 \, \log \left (5\right )^{3} + 6 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 1\right )}}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (18) = 36\).
Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 3.05 \[ \int \frac {-16+24 x^2-16 x^3+3 x^4+\left (-32+24 x^2-8 x^3\right ) \log (25)+\left (-24+6 x^2\right ) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx=x^{3} - 8 \, x^{2} \log \left (5\right ) + 24 \, x \log \left (5\right )^{2} - 8 \, x^{2} + 48 \, x \log \left (5\right ) + 24 \, x + \frac {16 \, {\left (\log \left (5\right )^{4} + 4 \, \log \left (5\right )^{3} + 6 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 1\right )}}{x} \]
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Time = 0.13 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.89 \[ \int \frac {-16+24 x^2-16 x^3+3 x^4+\left (-32+24 x^2-8 x^3\right ) \log (25)+\left (-24+6 x^2\right ) \log ^2(25)-8 \log ^3(25)-\log ^4(25)}{x^2} \, dx=x\,\left (48\,\ln \left (5\right )+24\,{\ln \left (5\right )}^2+24\right )-x^2\,\left (8\,\ln \left (5\right )+8\right )+\frac {32\,\ln \left (25\right )+24\,{\ln \left (25\right )}^2+8\,{\ln \left (25\right )}^3+{\ln \left (25\right )}^4+16}{x}+x^3 \]
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