Integrand size = 17, antiderivative size = 81 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx=-\frac {7}{10 b^2 x^5}+\frac {7 c}{6 b^3 x^3}-\frac {7 c^2}{2 b^4 x}+\frac {1}{2 b x^5 \left (b+c x^2\right )}-\frac {7 c^{5/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{9/2}} \]
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Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1598, 296, 331, 211} \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx=-\frac {7 c^{5/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{9/2}}-\frac {7 c^2}{2 b^4 x}+\frac {7 c}{6 b^3 x^3}-\frac {7}{10 b^2 x^5}+\frac {1}{2 b x^5 \left (b+c x^2\right )} \]
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Rule 211
Rule 296
Rule 331
Rule 1598
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^6 \left (b+c x^2\right )^2} \, dx \\ & = \frac {1}{2 b x^5 \left (b+c x^2\right )}+\frac {7 \int \frac {1}{x^6 \left (b+c x^2\right )} \, dx}{2 b} \\ & = -\frac {7}{10 b^2 x^5}+\frac {1}{2 b x^5 \left (b+c x^2\right )}-\frac {(7 c) \int \frac {1}{x^4 \left (b+c x^2\right )} \, dx}{2 b^2} \\ & = -\frac {7}{10 b^2 x^5}+\frac {7 c}{6 b^3 x^3}+\frac {1}{2 b x^5 \left (b+c x^2\right )}+\frac {\left (7 c^2\right ) \int \frac {1}{x^2 \left (b+c x^2\right )} \, dx}{2 b^3} \\ & = -\frac {7}{10 b^2 x^5}+\frac {7 c}{6 b^3 x^3}-\frac {7 c^2}{2 b^4 x}+\frac {1}{2 b x^5 \left (b+c x^2\right )}-\frac {\left (7 c^3\right ) \int \frac {1}{b+c x^2} \, dx}{2 b^4} \\ & = -\frac {7}{10 b^2 x^5}+\frac {7 c}{6 b^3 x^3}-\frac {7 c^2}{2 b^4 x}+\frac {1}{2 b x^5 \left (b+c x^2\right )}-\frac {7 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{9/2}} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx=-\frac {1}{5 b^2 x^5}+\frac {2 c}{3 b^3 x^3}-\frac {3 c^2}{b^4 x}-\frac {c^3 x}{2 b^4 \left (b+c x^2\right )}-\frac {7 c^{5/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{9/2}} \]
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Time = 0.10 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.83
method | result | size |
default | \(-\frac {1}{5 b^{2} x^{5}}-\frac {3 c^{2}}{b^{4} x}+\frac {2 c}{3 b^{3} x^{3}}-\frac {c^{3} \left (\frac {x}{2 c \,x^{2}+2 b}+\frac {7 \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}}\right )}{b^{4}}\) | \(67\) |
risch | \(\frac {-\frac {7 c^{3} x^{6}}{2 b^{4}}-\frac {7 c^{2} x^{4}}{3 b^{3}}+\frac {7 c \,x^{2}}{15 b^{2}}-\frac {1}{5 b}}{x^{5} \left (c \,x^{2}+b \right )}+\frac {7 \sqrt {-b c}\, c^{2} \ln \left (-c x +\sqrt {-b c}\right )}{4 b^{5}}-\frac {7 \sqrt {-b c}\, c^{2} \ln \left (-c x -\sqrt {-b c}\right )}{4 b^{5}}\) | \(106\) |
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Time = 0.27 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.44 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx=\left [-\frac {210 \, c^{3} x^{6} + 140 \, b c^{2} x^{4} - 28 \, b^{2} c x^{2} + 12 \, b^{3} - 105 \, {\left (c^{3} x^{7} + b c^{2} x^{5}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} - 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right )}{60 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}, -\frac {105 \, c^{3} x^{6} + 70 \, b c^{2} x^{4} - 14 \, b^{2} c x^{2} + 6 \, b^{3} + 105 \, {\left (c^{3} x^{7} + b c^{2} x^{5}\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right )}{30 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}\right ] \]
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Time = 0.19 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.56 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx=\frac {7 \sqrt {- \frac {c^{5}}{b^{9}}} \log {\left (- \frac {b^{5} \sqrt {- \frac {c^{5}}{b^{9}}}}{c^{3}} + x \right )}}{4} - \frac {7 \sqrt {- \frac {c^{5}}{b^{9}}} \log {\left (\frac {b^{5} \sqrt {- \frac {c^{5}}{b^{9}}}}{c^{3}} + x \right )}}{4} + \frac {- 6 b^{3} + 14 b^{2} c x^{2} - 70 b c^{2} x^{4} - 105 c^{3} x^{6}}{30 b^{5} x^{5} + 30 b^{4} c x^{7}} \]
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Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx=-\frac {105 \, c^{3} x^{6} + 70 \, b c^{2} x^{4} - 14 \, b^{2} c x^{2} + 6 \, b^{3}}{30 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}} - \frac {7 \, c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{4}} \]
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Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx=-\frac {7 \, c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{4}} - \frac {c^{3} x}{2 \, {\left (c x^{2} + b\right )} b^{4}} - \frac {45 \, c^{2} x^{4} - 10 \, b c x^{2} + 3 \, b^{2}}{15 \, b^{4} x^{5}} \]
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Time = 12.88 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx=-\frac {\frac {1}{5\,b}-\frac {7\,c\,x^2}{15\,b^2}+\frac {7\,c^2\,x^4}{3\,b^3}+\frac {7\,c^3\,x^6}{2\,b^4}}{c\,x^7+b\,x^5}-\frac {7\,c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{2\,b^{9/2}} \]
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