Integrand size = 24, antiderivative size = 93 \[ \int \frac {x^8}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {35 x}{16 b^4}-\frac {x^7}{6 b \left (a+b x^2\right )^3}-\frac {7 x^5}{24 b^2 \left (a+b x^2\right )^2}-\frac {35 x^3}{48 b^3 \left (a+b x^2\right )}-\frac {35 \sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 b^{9/2}} \]
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Time = 0.03 (sec) , antiderivative size = 93, 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, 294, 327, 211} \[ \int \frac {x^8}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {35 \sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 b^{9/2}}-\frac {35 x^3}{48 b^3 \left (a+b x^2\right )}-\frac {7 x^5}{24 b^2 \left (a+b x^2\right )^2}-\frac {x^7}{6 b \left (a+b x^2\right )^3}+\frac {35 x}{16 b^4} \]
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Rule 28
Rule 211
Rule 294
Rule 327
Rubi steps \begin{align*} \text {integral}& = b^4 \int \frac {x^8}{\left (a b+b^2 x^2\right )^4} \, dx \\ & = -\frac {x^7}{6 b \left (a+b x^2\right )^3}+\frac {1}{6} \left (7 b^2\right ) \int \frac {x^6}{\left (a b+b^2 x^2\right )^3} \, dx \\ & = -\frac {x^7}{6 b \left (a+b x^2\right )^3}-\frac {7 x^5}{24 b^2 \left (a+b x^2\right )^2}+\frac {35}{24} \int \frac {x^4}{\left (a b+b^2 x^2\right )^2} \, dx \\ & = -\frac {x^7}{6 b \left (a+b x^2\right )^3}-\frac {7 x^5}{24 b^2 \left (a+b x^2\right )^2}-\frac {35 x^3}{48 b^3 \left (a+b x^2\right )}+\frac {35 \int \frac {x^2}{a b+b^2 x^2} \, dx}{16 b^2} \\ & = \frac {35 x}{16 b^4}-\frac {x^7}{6 b \left (a+b x^2\right )^3}-\frac {7 x^5}{24 b^2 \left (a+b x^2\right )^2}-\frac {35 x^3}{48 b^3 \left (a+b x^2\right )}-\frac {(35 a) \int \frac {1}{a b+b^2 x^2} \, dx}{16 b^3} \\ & = \frac {35 x}{16 b^4}-\frac {x^7}{6 b \left (a+b x^2\right )^3}-\frac {7 x^5}{24 b^2 \left (a+b x^2\right )^2}-\frac {35 x^3}{48 b^3 \left (a+b x^2\right )}-\frac {35 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 b^{9/2}} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.83 \[ \int \frac {x^8}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {105 a^3 x+280 a^2 b x^3+231 a b^2 x^5+48 b^3 x^7}{48 b^4 \left (a+b x^2\right )^3}-\frac {35 \sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 b^{9/2}} \]
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Time = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.67
method | result | size |
default | \(\frac {x}{b^{4}}-\frac {a \left (\frac {-\frac {29}{16} b^{2} x^{5}-\frac {17}{6} a b \,x^{3}-\frac {19}{16} a^{2} x}{\left (b \,x^{2}+a \right )^{3}}+\frac {35 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \sqrt {a b}}\right )}{b^{4}}\) | \(62\) |
risch | \(\frac {x}{b^{4}}+\frac {\frac {29}{16} b^{2} x^{5} a +\frac {17}{6} a^{2} b \,x^{3}+\frac {19}{16} a^{3} x}{b^{4} \left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}+\frac {35 \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x -a \right )}{32 b^{5}}-\frac {35 \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x -a \right )}{32 b^{5}}\) | \(114\) |
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Time = 0.28 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.88 \[ \int \frac {x^8}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\left [\frac {96 \, b^{3} x^{7} + 462 \, a b^{2} x^{5} + 560 \, a^{2} b x^{3} + 210 \, a^{3} x + 105 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right )}{96 \, {\left (b^{7} x^{6} + 3 \, a b^{6} x^{4} + 3 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}, \frac {48 \, b^{3} x^{7} + 231 \, a b^{2} x^{5} + 280 \, a^{2} b x^{3} + 105 \, a^{3} x - 105 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right )}{48 \, {\left (b^{7} x^{6} + 3 \, a b^{6} x^{4} + 3 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}\right ] \]
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Time = 0.27 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.41 \[ \int \frac {x^8}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {35 \sqrt {- \frac {a}{b^{9}}} \log {\left (- b^{4} \sqrt {- \frac {a}{b^{9}}} + x \right )}}{32} - \frac {35 \sqrt {- \frac {a}{b^{9}}} \log {\left (b^{4} \sqrt {- \frac {a}{b^{9}}} + x \right )}}{32} + \frac {57 a^{3} x + 136 a^{2} b x^{3} + 87 a b^{2} x^{5}}{48 a^{3} b^{4} + 144 a^{2} b^{5} x^{2} + 144 a b^{6} x^{4} + 48 b^{7} x^{6}} + \frac {x}{b^{4}} \]
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Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.97 \[ \int \frac {x^8}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {87 \, a b^{2} x^{5} + 136 \, a^{2} b x^{3} + 57 \, a^{3} x}{48 \, {\left (b^{7} x^{6} + 3 \, a b^{6} x^{4} + 3 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}} - \frac {35 \, a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} b^{4}} + \frac {x}{b^{4}} \]
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Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.70 \[ \int \frac {x^8}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {35 \, a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} b^{4}} + \frac {x}{b^{4}} + \frac {87 \, a b^{2} x^{5} + 136 \, a^{2} b x^{3} + 57 \, a^{3} x}{48 \, {\left (b x^{2} + a\right )}^{3} b^{4}} \]
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Time = 0.10 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.92 \[ \int \frac {x^8}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {x}{b^4}+\frac {\frac {19\,a^3\,x}{16}+\frac {17\,a^2\,b\,x^3}{6}+\frac {29\,a\,b^2\,x^5}{16}}{a^3\,b^4+3\,a^2\,b^5\,x^2+3\,a\,b^6\,x^4+b^7\,x^6}-\frac {35\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{16\,b^{9/2}} \]
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