Integrand size = 26, antiderivative size = 169 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {5}{8 a^2 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {15 \left (a+b x^2\right )}{8 a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {15 \sqrt {b} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]
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Time = 0.05 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1126, 296, 331, 211} \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {5}{8 a^2 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )}-\frac {15 \sqrt {b} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {15 \left (a+b x^2\right )}{8 a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}} \]
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Rule 211
Rule 296
Rule 331
Rule 1126
Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^2 \left (a b+b^2 x^2\right )\right ) \int \frac {1}{x^2 \left (a b+b^2 x^2\right )^3} \, dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {1}{4 a x \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (5 b \left (a b+b^2 x^2\right )\right ) \int \frac {1}{x^2 \left (a b+b^2 x^2\right )^2} \, dx}{4 a \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {5}{8 a^2 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (15 \left (a b+b^2 x^2\right )\right ) \int \frac {1}{x^2 \left (a b+b^2 x^2\right )} \, dx}{8 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {5}{8 a^2 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {15 \left (a+b x^2\right )}{8 a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (15 b \left (a b+b^2 x^2\right )\right ) \int \frac {1}{a b+b^2 x^2} \, dx}{8 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {5}{8 a^2 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {15 \left (a+b x^2\right )}{8 a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {15 \sqrt {b} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ \end{align*}
Time = 1.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.55 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {-\sqrt {a} \left (8 a^2+25 a b x^2+15 b^2 x^4\right )-15 \sqrt {b} x \left (a+b x^2\right )^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} x \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \]
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Time = 0.78 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.70
method | result | size |
default | \(-\frac {\left (15 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{3} x^{5}+15 \sqrt {a b}\, b^{2} x^{4}+30 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a \,b^{2} x^{3}+25 \sqrt {a b}\, a b \,x^{2}+15 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2} b x +8 \sqrt {a b}\, a^{2}\right ) \left (b \,x^{2}+a \right )}{8 \sqrt {a b}\, x \,a^{3} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(119\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {15 b^{2} x^{4}}{8 a^{3}}-\frac {25 b \,x^{2}}{8 a^{2}}-\frac {1}{a}\right )}{\left (b \,x^{2}+a \right )^{3} x}+\frac {15 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, \ln \left (-b x +\sqrt {-a b}\right )}{16 \left (b \,x^{2}+a \right ) a^{4}}-\frac {15 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, \ln \left (-b x -\sqrt {-a b}\right )}{16 \left (b \,x^{2}+a \right ) a^{4}}\) | \(140\) |
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Time = 0.27 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.20 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\left [-\frac {30 \, b^{2} x^{4} + 50 \, a b x^{2} - 15 \, {\left (b^{2} x^{5} + 2 \, a b x^{3} + a^{2} x\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) + 16 \, a^{2}}{16 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}}, -\frac {15 \, b^{2} x^{4} + 25 \, a b x^{2} + 15 \, {\left (b^{2} x^{5} + 2 \, a b x^{3} + a^{2} x\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 8 \, a^{2}}{8 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}}\right ] \]
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\[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {1}{x^{2} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
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Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.42 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {15 \, b^{2} x^{4} + 25 \, a b x^{2} + 8 \, a^{2}}{8 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}} - \frac {15 \, b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} \]
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Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.51 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {15 \, b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {7 \, b^{2} x^{3} + 9 \, a b x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {1}{a^{3} x \mathrm {sgn}\left (b x^{2} + a\right )} \]
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Timed out. \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {1}{x^2\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \]
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