Integrand size = 24, antiderivative size = 95 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {35}{16 a^4 x}+\frac {1}{6 a x \left (a+b x^2\right )^3}+\frac {7}{24 a^2 x \left (a+b x^2\right )^2}+\frac {35}{48 a^3 x \left (a+b x^2\right )}-\frac {35 \sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{9/2}} \]
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Time = 0.04 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {28, 296, 331, 211} \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {35 \sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{9/2}}-\frac {35}{16 a^4 x}+\frac {35}{48 a^3 x \left (a+b x^2\right )}+\frac {7}{24 a^2 x \left (a+b x^2\right )^2}+\frac {1}{6 a x \left (a+b x^2\right )^3} \]
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Rule 28
Rule 211
Rule 296
Rule 331
Rubi steps \begin{align*} \text {integral}& = b^4 \int \frac {1}{x^2 \left (a b+b^2 x^2\right )^4} \, dx \\ & = \frac {1}{6 a x \left (a+b x^2\right )^3}+\frac {\left (7 b^3\right ) \int \frac {1}{x^2 \left (a b+b^2 x^2\right )^3} \, dx}{6 a} \\ & = \frac {1}{6 a x \left (a+b x^2\right )^3}+\frac {7}{24 a^2 x \left (a+b x^2\right )^2}+\frac {\left (35 b^2\right ) \int \frac {1}{x^2 \left (a b+b^2 x^2\right )^2} \, dx}{24 a^2} \\ & = \frac {1}{6 a x \left (a+b x^2\right )^3}+\frac {7}{24 a^2 x \left (a+b x^2\right )^2}+\frac {35}{48 a^3 x \left (a+b x^2\right )}+\frac {(35 b) \int \frac {1}{x^2 \left (a b+b^2 x^2\right )} \, dx}{16 a^3} \\ & = -\frac {35}{16 a^4 x}+\frac {1}{6 a x \left (a+b x^2\right )^3}+\frac {7}{24 a^2 x \left (a+b x^2\right )^2}+\frac {35}{48 a^3 x \left (a+b x^2\right )}-\frac {\left (35 b^2\right ) \int \frac {1}{a b+b^2 x^2} \, dx}{16 a^4} \\ & = -\frac {35}{16 a^4 x}+\frac {1}{6 a x \left (a+b x^2\right )^3}+\frac {7}{24 a^2 x \left (a+b x^2\right )^2}+\frac {35}{48 a^3 x \left (a+b x^2\right )}-\frac {35 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{9/2}} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {48 a^3+231 a^2 b x^2+280 a b^2 x^4+105 b^3 x^6}{48 a^4 x \left (a+b x^2\right )^3}-\frac {35 \sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{9/2}} \]
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Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.68
method | result | size |
default | \(-\frac {1}{a^{4} x}-\frac {b \left (\frac {\frac {19}{16} b^{2} x^{5}+\frac {17}{6} a b \,x^{3}+\frac {29}{16} a^{2} x}{\left (b \,x^{2}+a \right )^{3}}+\frac {35 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \sqrt {a b}}\right )}{a^{4}}\) | \(65\) |
risch | \(\frac {-\frac {35 b^{3} x^{6}}{16 a^{4}}-\frac {35 b^{2} x^{4}}{6 a^{3}}-\frac {77 b \,x^{2}}{16 a^{2}}-\frac {1}{a}}{x \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right ) \left (b \,x^{2}+a \right )}+\frac {35 \sqrt {-a b}\, \ln \left (-b x +\sqrt {-a b}\right )}{32 a^{5}}-\frac {35 \sqrt {-a b}\, \ln \left (-b x -\sqrt {-a b}\right )}{32 a^{5}}\) | \(120\) |
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Time = 0.28 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.82 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\left [-\frac {210 \, b^{3} x^{6} + 560 \, a b^{2} x^{4} + 462 \, a^{2} b x^{2} + 96 \, a^{3} - 105 \, {\left (b^{3} x^{7} + 3 \, a b^{2} x^{5} + 3 \, a^{2} b x^{3} + a^{3} x\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{96 \, {\left (a^{4} b^{3} x^{7} + 3 \, a^{5} b^{2} x^{5} + 3 \, a^{6} b x^{3} + a^{7} x\right )}}, -\frac {105 \, b^{3} x^{6} + 280 \, a b^{2} x^{4} + 231 \, a^{2} b x^{2} + 48 \, a^{3} + 105 \, {\left (b^{3} x^{7} + 3 \, a b^{2} x^{5} + 3 \, a^{2} b x^{3} + a^{3} x\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right )}{48 \, {\left (a^{4} b^{3} x^{7} + 3 \, a^{5} b^{2} x^{5} + 3 \, a^{6} b x^{3} + a^{7} x\right )}}\right ] \]
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Time = 0.26 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.46 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {35 \sqrt {- \frac {b}{a^{9}}} \log {\left (- \frac {a^{5} \sqrt {- \frac {b}{a^{9}}}}{b} + x \right )}}{32} - \frac {35 \sqrt {- \frac {b}{a^{9}}} \log {\left (\frac {a^{5} \sqrt {- \frac {b}{a^{9}}}}{b} + x \right )}}{32} + \frac {- 48 a^{3} - 231 a^{2} b x^{2} - 280 a b^{2} x^{4} - 105 b^{3} x^{6}}{48 a^{7} x + 144 a^{6} b x^{3} + 144 a^{5} b^{2} x^{5} + 48 a^{4} b^{3} x^{7}} \]
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Time = 0.27 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {105 \, b^{3} x^{6} + 280 \, a b^{2} x^{4} + 231 \, a^{2} b x^{2} + 48 \, a^{3}}{48 \, {\left (a^{4} b^{3} x^{7} + 3 \, a^{5} b^{2} x^{5} + 3 \, a^{6} b x^{3} + a^{7} x\right )}} - \frac {35 \, b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{4}} \]
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Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.72 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {35 \, b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{4}} - \frac {1}{a^{4} x} - \frac {57 \, b^{3} x^{5} + 136 \, a b^{2} x^{3} + 87 \, a^{2} b x}{48 \, {\left (b x^{2} + a\right )}^{3} a^{4}} \]
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Time = 13.23 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {\frac {1}{a}+\frac {77\,b\,x^2}{16\,a^2}+\frac {35\,b^2\,x^4}{6\,a^3}+\frac {35\,b^3\,x^6}{16\,a^4}}{a^3\,x+3\,a^2\,b\,x^3+3\,a\,b^2\,x^5+b^3\,x^7}-\frac {35\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{16\,a^{9/2}} \]
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