Integrand size = 18, antiderivative size = 36 \[ \int \frac {4+3 x^4}{x^2 \left (1+x^2\right )^3} \, dx=-\frac {4}{x}-\frac {7 x}{4 \left (1+x^2\right )^2}-\frac {25 x}{8 \left (1+x^2\right )}-\frac {57 \arctan (x)}{8} \]
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Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1274, 467, 464, 209} \[ \int \frac {4+3 x^4}{x^2 \left (1+x^2\right )^3} \, dx=-\frac {57 \arctan (x)}{8}-\frac {25 x}{8 \left (x^2+1\right )}-\frac {7 x}{4 \left (x^2+1\right )^2}-\frac {4}{x} \]
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Rule 209
Rule 464
Rule 467
Rule 1274
Rubi steps \begin{align*} \text {integral}& = -\frac {7 x}{4 \left (1+x^2\right )^2}-\frac {1}{4} \int \frac {-16+9 x^2}{x^2 \left (1+x^2\right )^2} \, dx \\ & = -\frac {7 x}{4 \left (1+x^2\right )^2}-\frac {25 x}{8 \left (1+x^2\right )}+\frac {1}{8} \int \frac {32-25 x^2}{x^2 \left (1+x^2\right )} \, dx \\ & = -\frac {4}{x}-\frac {7 x}{4 \left (1+x^2\right )^2}-\frac {25 x}{8 \left (1+x^2\right )}-\frac {57}{8} \int \frac {1}{1+x^2} \, dx \\ & = -\frac {4}{x}-\frac {7 x}{4 \left (1+x^2\right )^2}-\frac {25 x}{8 \left (1+x^2\right )}-\frac {57 \arctan (x)}{8} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.92 \[ \int \frac {4+3 x^4}{x^2 \left (1+x^2\right )^3} \, dx=-\frac {32+103 x^2+57 x^4}{8 x \left (1+x^2\right )^2}-\frac {57 \arctan (x)}{8} \]
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Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.81
method | result | size |
default | \(-\frac {\frac {25}{8} x^{3}+\frac {39}{8} x}{\left (x^{2}+1\right )^{2}}-\frac {57 \arctan \left (x \right )}{8}-\frac {4}{x}\) | \(29\) |
risch | \(\frac {-\frac {57}{8} x^{4}-\frac {103}{8} x^{2}-4}{\left (x^{2}+1\right )^{2} x}-\frac {57 \arctan \left (x \right )}{8}\) | \(29\) |
meijerg | \(-\frac {15 x^{4}+25 x^{2}+8}{2 x \left (x^{2}+1\right )^{2}}-\frac {57 \arctan \left (x \right )}{8}-\frac {x \left (-3 x^{2}+3\right )}{8 \left (x^{2}+1\right )^{2}}\) | \(47\) |
parallelrisch | \(\frac {57 i \ln \left (x -i\right ) x^{5}-57 i \ln \left (x +i\right ) x^{5}-64+114 i \ln \left (x -i\right ) x^{3}-114 i \ln \left (x +i\right ) x^{3}-114 x^{4}+57 i \ln \left (x -i\right ) x -57 i \ln \left (x +i\right ) x -206 x^{2}}{16 x \left (x^{2}+1\right )^{2}}\) | \(87\) |
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Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11 \[ \int \frac {4+3 x^4}{x^2 \left (1+x^2\right )^3} \, dx=-\frac {57 \, x^{4} + 103 \, x^{2} + 57 \, {\left (x^{5} + 2 \, x^{3} + x\right )} \arctan \left (x\right ) + 32}{8 \, {\left (x^{5} + 2 \, x^{3} + x\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.89 \[ \int \frac {4+3 x^4}{x^2 \left (1+x^2\right )^3} \, dx=\frac {- 57 x^{4} - 103 x^{2} - 32}{8 x^{5} + 16 x^{3} + 8 x} - \frac {57 \operatorname {atan}{\left (x \right )}}{8} \]
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Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.86 \[ \int \frac {4+3 x^4}{x^2 \left (1+x^2\right )^3} \, dx=-\frac {57 \, x^{4} + 103 \, x^{2} + 32}{8 \, {\left (x^{5} + 2 \, x^{3} + x\right )}} - \frac {57}{8} \, \arctan \left (x\right ) \]
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Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.78 \[ \int \frac {4+3 x^4}{x^2 \left (1+x^2\right )^3} \, dx=-\frac {25 \, x^{3} + 39 \, x}{8 \, {\left (x^{2} + 1\right )}^{2}} - \frac {4}{x} - \frac {57}{8} \, \arctan \left (x\right ) \]
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Time = 0.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.81 \[ \int \frac {4+3 x^4}{x^2 \left (1+x^2\right )^3} \, dx=-\frac {57\,\mathrm {atan}\left (x\right )}{8}-\frac {\frac {57\,x^4}{8}+\frac {103\,x^2}{8}+4}{x\,{\left (x^2+1\right )}^2} \]
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