Integrand size = 19, antiderivative size = 19 \[ \int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx=3 \log \left (\frac {5}{4} \left (-1-\frac {3 e^5}{2 x}\right )\right ) \]
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Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 629} \[ \int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx=3 \log \left (2 x+3 e^5\right )-3 \log (x) \]
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Rule 12
Rule 629
Rubi steps \begin{align*} \text {integral}& = -\left (\left (9 e^5\right ) \int \frac {1}{3 e^5 x+2 x^2} \, dx\right ) \\ & = -3 \log (x)+3 \log \left (3 e^5+2 x\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68 \[ \int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx=-9 e^5 \left (\frac {\log (x)}{3 e^5}-\frac {\log \left (3 e^5+2 x\right )}{3 e^5}\right ) \]
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Time = 1.70 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79
method | result | size |
parallelrisch | \(-3 \ln \left (x \right )+3 \ln \left (\frac {3 \,{\mathrm e}^{5}}{2}+x \right )\) | \(15\) |
norman | \(-3 \ln \left (x \right )+3 \ln \left (3 \,{\mathrm e}^{5}+2 x \right )\) | \(17\) |
risch | \(-3 \ln \left (x \right )+3 \ln \left (3 \,{\mathrm e}^{5}+2 x \right )\) | \(17\) |
meijerg | \(-3 \ln \left (x \right )+3 \ln \left (3\right )-3 \ln \left (2\right )+15+3 \ln \left (1+\frac {2 \,{\mathrm e}^{-5} x}{3}\right )\) | \(25\) |
default | \(-9 \,{\mathrm e}^{5} \left (\frac {\ln \left (x \right ) {\mathrm e}^{-5}}{3}-\frac {{\mathrm e}^{-5} \ln \left (3 \,{\mathrm e}^{5}+2 x \right )}{3}\right )\) | \(29\) |
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Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx=3 \, \log \left (2 \, x + 3 \, e^{5}\right ) - 3 \, \log \left (x\right ) \]
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Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53 \[ \int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx=- 9 \left (\frac {\log {\left (x \right )}}{3 e^{5}} - \frac {\log {\left (x + \frac {3 e^{5}}{2} \right )}}{3 e^{5}}\right ) e^{5} \]
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none
Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx=3 \, {\left (e^{\left (-5\right )} \log \left (2 \, x + 3 \, e^{5}\right ) - e^{\left (-5\right )} \log \left (x\right )\right )} e^{5} \]
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Leaf count of result is larger than twice the leaf count of optimal. 25 vs. \(2 (12) = 24\).
Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32 \[ \int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx=3 \, {\left (e^{\left (-5\right )} \log \left ({\left | 2 \, x + 3 \, e^{5} \right |}\right ) - e^{\left (-5\right )} \log \left ({\left | x \right |}\right )\right )} e^{5} \]
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Time = 0.27 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx=6\,\mathrm {atanh}\left (\frac {4\,x\,{\mathrm {e}}^{-5}}{3}+1\right ) \]
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