Integrand size = 17, antiderivative size = 20 \[ \int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx=\log (3)+\frac {3+\log (3)}{x}+\frac {e^3 \log (16)}{x} \]
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Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 30} \[ \int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx=\frac {3+\log (3)+e^3 \log (16)}{x} \]
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Rule 12
Rule 30
Rubi steps \begin{align*} \text {integral}& = \left (-3-\log (3)-e^3 \log (16)\right ) \int \frac {1}{x^2} \, dx \\ & = \frac {3+\log (3)+e^3 \log (16)}{x} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx=\frac {3+\log (3)+e^3 \log (16)}{x} \]
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Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75
method | result | size |
gosper | \(\frac {4 \,{\mathrm e}^{3} \ln \left (2\right )+\ln \left (3\right )+3}{x}\) | \(15\) |
norman | \(\frac {4 \,{\mathrm e}^{3} \ln \left (2\right )+\ln \left (3\right )+3}{x}\) | \(15\) |
default | \(-\frac {-4 \,{\mathrm e}^{3} \ln \left (2\right )-\ln \left (3\right )-3}{x}\) | \(18\) |
parallelrisch | \(-\frac {-4 \,{\mathrm e}^{3} \ln \left (2\right )-\ln \left (3\right )-3}{x}\) | \(18\) |
risch | \(\frac {4 \,{\mathrm e}^{3} \ln \left (2\right )}{x}+\frac {\ln \left (3\right )}{x}+\frac {3}{x}\) | \(22\) |
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Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx=\frac {4 \, e^{3} \log \left (2\right ) + \log \left (3\right ) + 3}{x} \]
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Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx=- \frac {- 4 e^{3} \log {\left (2 \right )} - 3 - \log {\left (3 \right )}}{x} \]
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Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx=\frac {4 \, e^{3} \log \left (2\right ) + \log \left (3\right ) + 3}{x} \]
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Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx=\frac {4 \, e^{3} \log \left (2\right ) + \log \left (3\right ) + 3}{x} \]
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Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx=\frac {\ln \left (3\right )+4\,{\mathrm {e}}^3\,\ln \left (2\right )+3}{x} \]
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