Integrand size = 20, antiderivative size = 19 \[ \int \frac {-5+96 x \log ^3\left (\frac {16}{e x^3}\right )}{x^2} \, dx=\frac {5}{x}-8 \log ^4\left (\frac {16}{e x^3}\right ) \]
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Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {14, 2339, 30} \[ \int \frac {-5+96 x \log ^3\left (\frac {16}{e x^3}\right )}{x^2} \, dx=\frac {5}{x}-8 \left (-\log \left (\frac {1}{x^3}\right )+1-\log (16)\right )^4 \]
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Rule 14
Rule 30
Rule 2339
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {5}{x^2}+\frac {96 \left (-1+\log (16)+\log \left (\frac {1}{x^3}\right )\right )^3}{x}\right ) \, dx \\ & = \frac {5}{x}+96 \int \frac {\left (-1+\log (16)+\log \left (\frac {1}{x^3}\right )\right )^3}{x} \, dx \\ & = \frac {5}{x}-32 \text {Subst}\left (\int x^3 \, dx,x,-1+\log (16)+\log \left (\frac {1}{x^3}\right )\right ) \\ & = \frac {5}{x}-8 \left (1-\log (16)-\log \left (\frac {1}{x^3}\right )\right )^4 \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-5+96 x \log ^3\left (\frac {16}{e x^3}\right )}{x^2} \, dx=\frac {5}{x}-8 \left (-1+\log (16)+\log \left (\frac {1}{x^3}\right )\right )^4 \]
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Time = 1.48 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {5}{x}-8 \ln \left (\frac {16 \,{\mathrm e}^{-1}}{x^{3}}\right )^{4}\) | \(19\) |
derivativedivides | \(\frac {5}{x}-8 \ln \left (\frac {16 \,{\mathrm e}^{-1}}{x^{3}}\right )^{4}\) | \(21\) |
default | \(\frac {5}{x}-8 \ln \left (\frac {16 \,{\mathrm e}^{-1}}{x^{3}}\right )^{4}\) | \(21\) |
parts | \(\frac {5}{x}-8 \ln \left (\frac {16 \,{\mathrm e}^{-1}}{x^{3}}\right )^{4}\) | \(21\) |
norman | \(\frac {5-8 \ln \left (\frac {16 \,{\mathrm e}^{-1}}{x^{3}}\right )^{4} x}{x}\) | \(22\) |
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Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-5+96 x \log ^3\left (\frac {16}{e x^3}\right )}{x^2} \, dx=-\frac {8 \, x \log \left (\frac {16 \, e^{\left (-1\right )}}{x^{3}}\right )^{4} - 5}{x} \]
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Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-5+96 x \log ^3\left (\frac {16}{e x^3}\right )}{x^2} \, dx=- 8 \log {\left (\frac {16}{e x^{3}} \right )}^{4} + \frac {5}{x} \]
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Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-5+96 x \log ^3\left (\frac {16}{e x^3}\right )}{x^2} \, dx=-8 \, \log \left (\frac {16 \, e^{\left (-1\right )}}{x^{3}}\right )^{4} + \frac {5}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (18) = 36\).
Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.11 \[ \int \frac {-5+96 x \log ^3\left (\frac {16}{e x^3}\right )}{x^2} \, dx=-8 \, \log \left (\frac {16}{x^{3}}\right )^{4} + 32 \, \log \left (\frac {16}{x^{3}}\right )^{3} - 48 \, \log \left (\frac {16}{x^{3}}\right )^{2} + \frac {5}{x} - 96 \, \log \left (x\right ) \]
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Time = 9.78 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-5+96 x \log ^3\left (\frac {16}{e x^3}\right )}{x^2} \, dx=\frac {5}{x}-8\,{\ln \left (\frac {16\,{\mathrm {e}}^{-1}}{x^3}\right )}^4 \]
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