Integrand size = 35, antiderivative size = 22 \[ \int -\frac {1568 e^{\frac {1}{16} e^{4+\log ^2(2)}} x}{\left (49+16 x^2\right ) \left (196+64 x^2\right )} \, dx=\frac {e^{e^{(-2+\log (2))^2}}}{4+\frac {64 x^2}{49}} \]
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Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {12, 21, 267} \[ \int -\frac {1568 e^{\frac {1}{16} e^{4+\log ^2(2)}} x}{\left (49+16 x^2\right ) \left (196+64 x^2\right )} \, dx=\frac {49 e^{\frac {1}{16} e^{4+\log ^2(2)}}}{4 \left (16 x^2+49\right )} \]
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Rule 12
Rule 21
Rule 267
Rubi steps \begin{align*} \text {integral}& = -\left (\left (1568 e^{\frac {1}{16} e^{4+\log ^2(2)}}\right ) \int \frac {x}{\left (49+16 x^2\right ) \left (196+64 x^2\right )} \, dx\right ) \\ & = -\left (\left (392 e^{\frac {1}{16} e^{4+\log ^2(2)}}\right ) \int \frac {x}{\left (49+16 x^2\right )^2} \, dx\right ) \\ & = \frac {49 e^{\frac {1}{16} e^{4+\log ^2(2)}}}{4 \left (49+16 x^2\right )} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int -\frac {1568 e^{\frac {1}{16} e^{4+\log ^2(2)}} x}{\left (49+16 x^2\right ) \left (196+64 x^2\right )} \, dx=\frac {49 e^{\frac {1}{16} e^{4+\log ^2(2)}}}{4 \left (49+16 x^2\right )} \]
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Time = 1.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\frac {49 \,{\mathrm e}^{\frac {{\mathrm e}^{\ln \left (2\right )^{2}+4}}{16}}}{64 \left (x^{2}+\frac {49}{16}\right )}\) | \(20\) |
norman | \(\frac {49 \,{\mathrm e}^{\frac {{\mathrm e}^{\ln \left (2\right )^{2}+4}}{16}}}{4 \left (16 x^{2}+49\right )}\) | \(23\) |
gosper | \({\mathrm e}^{-\ln \left (\frac {64 x^{2}}{49}+4\right )+{\mathrm e}^{\ln \left (2\right )^{2}-4 \ln \left (2\right )+4}}\) | \(24\) |
derivativedivides | \({\mathrm e}^{-\ln \left (\frac {64 x^{2}}{49}+4\right )+{\mathrm e}^{\ln \left (2\right )^{2}-4 \ln \left (2\right )+4}}\) | \(24\) |
default | \({\mathrm e}^{-\ln \left (\frac {64 x^{2}}{49}+4\right )+{\mathrm e}^{\ln \left (2\right )^{2}-4 \ln \left (2\right )+4}}\) | \(24\) |
parallelrisch | \({\mathrm e}^{-\ln \left (\frac {64 x^{2}}{49}+4\right )+{\mathrm e}^{\ln \left (2\right )^{2}-4 \ln \left (2\right )+4}}\) | \(24\) |
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Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int -\frac {1568 e^{\frac {1}{16} e^{4+\log ^2(2)}} x}{\left (49+16 x^2\right ) \left (196+64 x^2\right )} \, dx=\frac {49 \, e^{\left (e^{\left (\log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 4\right )}\right )}}{4 \, {\left (16 \, x^{2} + 49\right )}} \]
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Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int -\frac {1568 e^{\frac {1}{16} e^{4+\log ^2(2)}} x}{\left (49+16 x^2\right ) \left (196+64 x^2\right )} \, dx=\frac {392 e^{\frac {e^{4} e^{\log {\left (2 \right )}^{2}}}{16}}}{512 x^{2} + 1568} \]
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Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int -\frac {1568 e^{\frac {1}{16} e^{4+\log ^2(2)}} x}{\left (49+16 x^2\right ) \left (196+64 x^2\right )} \, dx=\frac {49 \, e^{\left (\frac {1}{16} \, e^{\left (\log \left (2\right )^{2} + 4\right )}\right )}}{4 \, {\left (16 \, x^{2} + 49\right )}} \]
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Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int -\frac {1568 e^{\frac {1}{16} e^{4+\log ^2(2)}} x}{\left (49+16 x^2\right ) \left (196+64 x^2\right )} \, dx=\frac {49 \, e^{\left (\frac {1}{16} \, e^{\left (\log \left (2\right )^{2} + 4\right )}\right )}}{4 \, {\left (16 \, x^{2} + 49\right )}} \]
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Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int -\frac {1568 e^{\frac {1}{16} e^{4+\log ^2(2)}} x}{\left (49+16 x^2\right ) \left (196+64 x^2\right )} \, dx=\frac {49\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\ln \left (2\right )}^2}\,{\mathrm {e}}^4}{16}}}{4\,\left (16\,x^2+49\right )} \]
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