Integrand size = 17, antiderivative size = 22 \[ \int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx=\frac {5 x (-1+5 x)+\frac {e^3 \log (5)}{x}}{x^2} \]
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Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {14} \[ \int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx=\frac {e^3 \log (5)}{x^3}-\frac {5}{x} \]
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Rule 14
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {5}{x^2}-\frac {3 e^3 \log (5)}{x^4}\right ) \, dx \\ & = -\frac {5}{x}+\frac {e^3 \log (5)}{x^3} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx=-\frac {5}{x}+\frac {e^3 \log (5)}{x^3} \]
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Time = 0.39 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68
method | result | size |
default | \(-\frac {5}{x}+\frac {{\mathrm e}^{3} \ln \left (5\right )}{x^{3}}\) | \(15\) |
gosper | \(\frac {{\mathrm e}^{3} \ln \left (5\right )-5 x^{2}}{x^{3}}\) | \(16\) |
norman | \(\frac {{\mathrm e}^{3} \ln \left (5\right )-5 x^{2}}{x^{3}}\) | \(16\) |
risch | \(\frac {{\mathrm e}^{3} \ln \left (5\right )-5 x^{2}}{x^{3}}\) | \(16\) |
parallelrisch | \(\frac {{\mathrm e}^{3} \ln \left (5\right )-5 x^{2}}{x^{3}}\) | \(16\) |
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Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx=-\frac {5 \, x^{2} - e^{3} \log \left (5\right )}{x^{3}} \]
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Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx=\frac {- 5 x^{2} + e^{3} \log {\left (5 \right )}}{x^{3}} \]
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Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx=-\frac {5 \, x^{2} - e^{3} \log \left (5\right )}{x^{3}} \]
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none
Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx=-\frac {5 \, x^{2} - e^{3} \log \left (5\right )}{x^{3}} \]
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Time = 11.95 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx=\frac {{\mathrm {e}}^3\,\ln \left (5\right )-5\,x^2}{x^3} \]
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