Integrand size = 50, antiderivative size = 23 \[ \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+10 x^2+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \log ^4(4)} \, dx=3+\frac {12 e^{1+e}}{5+x^2-x \log ^2(4)} \]
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Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6, 12, 1694, 267} \[ \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+10 x^2+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \log ^4(4)} \, dx=\frac {12 e^{1+e}}{x^2-x \log ^2(4)+5} \]
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Rule 6
Rule 12
Rule 267
Rule 1694
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \left (10+\log ^4(4)\right )} \, dx \\ & = e^{1+e} \int \frac {-24 x+12 \log ^2(4)}{25+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \left (10+\log ^4(4)\right )} \, dx \\ & = e^{1+e} \text {Subst}\left (\int -\frac {384 x}{\left (20+4 x^2-\log ^4(4)\right )^2} \, dx,x,x-\frac {\log ^2(4)}{2}\right ) \\ & = -\left (\left (384 e^{1+e}\right ) \text {Subst}\left (\int \frac {x}{\left (20+4 x^2-\log ^4(4)\right )^2} \, dx,x,x-\frac {\log ^2(4)}{2}\right )\right ) \\ & = \frac {12 e^{1+e}}{5+x^2-x \log ^2(4)} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+10 x^2+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \log ^4(4)} \, dx=\frac {12 e^{1+e}}{5+x^2-x \log ^2(4)} \]
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Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {3 \,{\mathrm e}^{1+{\mathrm e}}}{x \ln \left (2\right )^{2}-\frac {x^{2}}{4}-\frac {5}{4}}\) | \(23\) |
gosper | \(-\frac {12 \,{\mathrm e}^{1+{\mathrm e}}}{4 x \ln \left (2\right )^{2}-x^{2}-5}\) | \(24\) |
norman | \(-\frac {12 \,{\mathrm e} \,{\mathrm e}^{{\mathrm e}}}{4 x \ln \left (2\right )^{2}-x^{2}-5}\) | \(24\) |
parallelrisch | \(-\frac {12 \,{\mathrm e}^{1+{\mathrm e}}}{4 x \ln \left (2\right )^{2}-x^{2}-5}\) | \(24\) |
default | \(\frac {24 \,{\mathrm e}^{1+{\mathrm e}} \left (10-8 \ln \left (2\right )^{4}\right )}{\left (20-16 \ln \left (2\right )^{4}\right ) \left (x^{2}-4 x \ln \left (2\right )^{2}+5\right )}\) | \(40\) |
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Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+10 x^2+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \log ^4(4)} \, dx=-\frac {12 \, e^{\left (e + 1\right )}}{4 \, x \log \left (2\right )^{2} - x^{2} - 5} \]
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Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+10 x^2+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \log ^4(4)} \, dx=\frac {12 e e^{e}}{x^{2} - 4 x \log {\left (2 \right )}^{2} + 5} \]
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Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+10 x^2+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \log ^4(4)} \, dx=-\frac {12 \, e^{\left (e + 1\right )}}{4 \, x \log \left (2\right )^{2} - x^{2} - 5} \]
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Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+10 x^2+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \log ^4(4)} \, dx=-\frac {12 \, e^{\left (e + 1\right )}}{4 \, x \log \left (2\right )^{2} - x^{2} - 5} \]
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Time = 12.37 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+10 x^2+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \log ^4(4)} \, dx=\frac {12\,{\mathrm {e}}^{\mathrm {e}+1}}{x^2-4\,{\ln \left (2\right )}^2\,x+5} \]
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