Integrand size = 58, antiderivative size = 31 \[ \int \frac {x \left (-3 x^2+48 x^3-144 x^4+\left (-1+144 x^2\right ) \log (2)\right )}{e^2 \left (-x^2-\log (2)\right ) \left (36 x^5+36 x^3 \log (2)\right )} \, dx=\frac {\left (4-\frac {1}{3 x}\right )^2}{4 e^2 \left (-x-\frac {\log (2)}{x}\right )} \]
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Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {12, 1598, 21, 1819, 30} \[ \int \frac {x \left (-3 x^2+48 x^3-144 x^4+\left (-1+144 x^2\right ) \log (2)\right )}{e^2 \left (-x^2-\log (2)\right ) \left (36 x^5+36 x^3 \log (2)\right )} \, dx=\frac {x (1-144 \log (2))+24 \log (2)}{36 e^2 \log (2) \left (x^2+\log (2)\right )}-\frac {1}{36 e^2 x \log (2)} \]
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Rule 12
Rule 21
Rule 30
Rule 1598
Rule 1819
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {x \left (-3 x^2+48 x^3-144 x^4+\left (-1+144 x^2\right ) \log (2)\right )}{\left (-x^2-\log (2)\right ) \left (36 x^5+36 x^3 \log (2)\right )} \, dx}{e^2} \\ & = \frac {\int \frac {-3 x^2+48 x^3-144 x^4+\left (-1+144 x^2\right ) \log (2)}{x^2 \left (-x^2-\log (2)\right ) \left (36 x^2+36 \log (2)\right )} \, dx}{e^2} \\ & = -\frac {\int \frac {-3 x^2+48 x^3-144 x^4+\left (-1+144 x^2\right ) \log (2)}{x^2 \left (-x^2-\log (2)\right )^2} \, dx}{36 e^2} \\ & = \frac {x (1-144 \log (2))+24 \log (2)}{36 e^2 \log (2) \left (x^2+\log (2)\right )}-\frac {\int \frac {2 x^2+\log (4)}{x^2 \left (-x^2-\log (2)\right )} \, dx}{72 e^2 \log (2)} \\ & = \frac {x (1-144 \log (2))+24 \log (2)}{36 e^2 \log (2) \left (x^2+\log (2)\right )}+\frac {\int \frac {1}{x^2} \, dx}{36 e^2 \log (2)} \\ & = -\frac {1}{36 e^2 x \log (2)}+\frac {x (1-144 \log (2))+24 \log (2)}{36 e^2 \log (2) \left (x^2+\log (2)\right )} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {x \left (-3 x^2+48 x^3-144 x^4+\left (-1+144 x^2\right ) \log (2)\right )}{e^2 \left (-x^2-\log (2)\right ) \left (36 x^5+36 x^3 \log (2)\right )} \, dx=-\frac {(1-12 x)^2}{36 e^2 x \left (x^2+\log (2)\right )} \]
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Time = 0.97 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {{\mathrm e}^{-2} \left (-4 x^{2}+\frac {2}{3} x -\frac {1}{36}\right )}{x \left (\ln \left (2\right )+x^{2}\right )}\) | \(25\) |
gosper | \(-\frac {\left (12 x -1\right )^{2} {\mathrm e}^{-2}}{36 x \left (\ln \left (2\right )+x^{2}\right )}\) | \(30\) |
norman | \(\frac {\frac {2 x^{2} {\mathrm e}^{-2}}{3}-\frac {x \,{\mathrm e}^{-2}}{36}-4 \,{\mathrm e}^{-2} x^{3}}{x^{2} \left (\ln \left (2\right )+x^{2}\right )}\) | \(39\) |
default | \(\frac {{\mathrm e}^{-2} \left (-\frac {\left (144 \ln \left (2\right )-1\right ) x -24 \ln \left (2\right )}{\ln \left (2\right ) \left (\ln \left (2\right )+x^{2}\right )}-\frac {1}{x \ln \left (2\right )}\right )}{36}\) | \(42\) |
parallelrisch | \(-\frac {\left (-24 x \ln \left (2\right )^{2}+288 x^{4} \ln \left (2\right )+144 x^{2} \ln \left (2\right )^{2}+2 x^{2} \ln \left (2\right )-48 x^{3} \ln \left (2\right )+\ln \left (2\right )^{2}+x^{4}+144 x^{6}-24 x^{5}\right ) {\mathrm e}^{-2}}{36 x \left (\ln \left (2\right )+x^{2}\right )^{3}}\) | \(86\) |
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Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {x \left (-3 x^2+48 x^3-144 x^4+\left (-1+144 x^2\right ) \log (2)\right )}{e^2 \left (-x^2-\log (2)\right ) \left (36 x^5+36 x^3 \log (2)\right )} \, dx=-\frac {144 \, x^{2} - 24 \, x + 1}{36 \, {\left (x^{3} e^{2} + x e^{2} \log \left (2\right )\right )}} \]
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Time = 0.53 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {x \left (-3 x^2+48 x^3-144 x^4+\left (-1+144 x^2\right ) \log (2)\right )}{e^2 \left (-x^2-\log (2)\right ) \left (36 x^5+36 x^3 \log (2)\right )} \, dx=\frac {- 144 x^{2} + 24 x - 1}{36 x^{3} e^{2} + 36 x e^{2} \log {\left (2 \right )}} \]
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Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {x \left (-3 x^2+48 x^3-144 x^4+\left (-1+144 x^2\right ) \log (2)\right )}{e^2 \left (-x^2-\log (2)\right ) \left (36 x^5+36 x^3 \log (2)\right )} \, dx=-\frac {{\left (144 \, x^{2} - 24 \, x + 1\right )} e^{\left (-2\right )}}{36 \, {\left (x^{3} + x \log \left (2\right )\right )}} \]
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {x \left (-3 x^2+48 x^3-144 x^4+\left (-1+144 x^2\right ) \log (2)\right )}{e^2 \left (-x^2-\log (2)\right ) \left (36 x^5+36 x^3 \log (2)\right )} \, dx=-\frac {{\left (144 \, x^{2} - 24 \, x + 1\right )} e^{\left (-2\right )}}{36 \, {\left (x^{3} + x \log \left (2\right )\right )}} \]
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Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {x \left (-3 x^2+48 x^3-144 x^4+\left (-1+144 x^2\right ) \log (2)\right )}{e^2 \left (-x^2-\log (2)\right ) \left (36 x^5+36 x^3 \log (2)\right )} \, dx=-\frac {{\mathrm {e}}^{-2}\,\left (24\,x^3+144\,\ln \left (2\right )\,x^2+\ln \left (2\right )\right )}{36\,x\,\ln \left (2\right )\,\left (x^2+\ln \left (2\right )\right )} \]
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