Integrand size = 23, antiderivative size = 29 \[ \int \frac {1+x^2+x^4}{\left (1+x^2\right ) \left (4+x^2\right )^2} \, dx=-\frac {13 x}{24 \left (4+x^2\right )}+\frac {25}{144} \arctan \left (\frac {x}{2}\right )+\frac {\arctan (x)}{9} \]
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Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6857, 209, 205} \[ \int \frac {1+x^2+x^4}{\left (1+x^2\right ) \left (4+x^2\right )^2} \, dx=\frac {25}{144} \arctan \left (\frac {x}{2}\right )+\frac {\arctan (x)}{9}-\frac {13 x}{24 \left (x^2+4\right )} \]
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Rule 205
Rule 209
Rule 6857
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{9 \left (1+x^2\right )}-\frac {13}{3 \left (4+x^2\right )^2}+\frac {8}{9 \left (4+x^2\right )}\right ) \, dx \\ & = \frac {1}{9} \int \frac {1}{1+x^2} \, dx+\frac {8}{9} \int \frac {1}{4+x^2} \, dx-\frac {13}{3} \int \frac {1}{\left (4+x^2\right )^2} \, dx \\ & = -\frac {13 x}{24 \left (4+x^2\right )}+\frac {4}{9} \arctan \left (\frac {x}{2}\right )+\frac {\arctan (x)}{9}-\frac {13}{24} \int \frac {1}{4+x^2} \, dx \\ & = -\frac {13 x}{24 \left (4+x^2\right )}+\frac {25}{144} \arctan \left (\frac {x}{2}\right )+\frac {\arctan (x)}{9} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {1+x^2+x^4}{\left (1+x^2\right ) \left (4+x^2\right )^2} \, dx=-\frac {13 x}{24 \left (4+x^2\right )}+\frac {25}{144} \arctan \left (\frac {x}{2}\right )+\frac {\arctan (x)}{9} \]
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Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76
method | result | size |
default | \(-\frac {13 x}{24 \left (x^{2}+4\right )}+\frac {25 \arctan \left (\frac {x}{2}\right )}{144}+\frac {\arctan \left (x \right )}{9}\) | \(22\) |
risch | \(-\frac {13 x}{24 \left (x^{2}+4\right )}+\frac {25 \arctan \left (\frac {x}{2}\right )}{144}+\frac {\arctan \left (x \right )}{9}\) | \(22\) |
parallelrisch | \(-\frac {25 i \ln \left (x -2 i\right ) x^{2}+16 i \ln \left (x -i\right ) x^{2}-16 i \ln \left (x +i\right ) x^{2}-25 i \ln \left (x +2 i\right ) x^{2}+100 i \ln \left (x -2 i\right )+64 i \ln \left (x -i\right )-64 i \ln \left (x +i\right )-100 i \ln \left (x +2 i\right )+156 x}{288 \left (x^{2}+4\right )}\) | \(90\) |
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Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {1+x^2+x^4}{\left (1+x^2\right ) \left (4+x^2\right )^2} \, dx=\frac {25 \, {\left (x^{2} + 4\right )} \arctan \left (\frac {1}{2} \, x\right ) + 16 \, {\left (x^{2} + 4\right )} \arctan \left (x\right ) - 78 \, x}{144 \, {\left (x^{2} + 4\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {1+x^2+x^4}{\left (1+x^2\right ) \left (4+x^2\right )^2} \, dx=- \frac {13 x}{24 x^{2} + 96} + \frac {25 \operatorname {atan}{\left (\frac {x}{2} \right )}}{144} + \frac {\operatorname {atan}{\left (x \right )}}{9} \]
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Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {1+x^2+x^4}{\left (1+x^2\right ) \left (4+x^2\right )^2} \, dx=-\frac {13 \, x}{24 \, {\left (x^{2} + 4\right )}} + \frac {25}{144} \, \arctan \left (\frac {1}{2} \, x\right ) + \frac {1}{9} \, \arctan \left (x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {1+x^2+x^4}{\left (1+x^2\right ) \left (4+x^2\right )^2} \, dx=-\frac {13 \, x}{24 \, {\left (x^{2} + 4\right )}} + \frac {25}{144} \, \arctan \left (\frac {1}{2} \, x\right ) + \frac {1}{9} \, \arctan \left (x\right ) \]
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Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {1+x^2+x^4}{\left (1+x^2\right ) \left (4+x^2\right )^2} \, dx=\frac {25\,\mathrm {atan}\left (\frac {x}{2}\right )}{144}+\frac {\mathrm {atan}\left (x\right )}{9}-\frac {13\,x}{24\,\left (x^2+4\right )} \]
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