Integrand size = 24, antiderivative size = 70 \[ \int \frac {x^9}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {3 a^2 x^2}{2 b^4}-\frac {a x^4}{2 b^3}+\frac {x^6}{6 b^2}-\frac {a^4}{2 b^5 \left (a+b x^2\right )}-\frac {2 a^3 \log \left (a+b x^2\right )}{b^5} \]
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Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 272, 45} \[ \int \frac {x^9}{a^2+2 a b x^2+b^2 x^4} \, dx=-\frac {a^4}{2 b^5 \left (a+b x^2\right )}-\frac {2 a^3 \log \left (a+b x^2\right )}{b^5}+\frac {3 a^2 x^2}{2 b^4}-\frac {a x^4}{2 b^3}+\frac {x^6}{6 b^2} \]
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Rule 28
Rule 45
Rule 272
Rubi steps \begin{align*} \text {integral}& = b^2 \int \frac {x^9}{\left (a b+b^2 x^2\right )^2} \, dx \\ & = \frac {1}{2} b^2 \text {Subst}\left (\int \frac {x^4}{\left (a b+b^2 x\right )^2} \, dx,x,x^2\right ) \\ & = \frac {1}{2} b^2 \text {Subst}\left (\int \left (\frac {3 a^2}{b^6}-\frac {2 a x}{b^5}+\frac {x^2}{b^4}+\frac {a^4}{b^6 (a+b x)^2}-\frac {4 a^3}{b^6 (a+b x)}\right ) \, dx,x,x^2\right ) \\ & = \frac {3 a^2 x^2}{2 b^4}-\frac {a x^4}{2 b^3}+\frac {x^6}{6 b^2}-\frac {a^4}{2 b^5 \left (a+b x^2\right )}-\frac {2 a^3 \log \left (a+b x^2\right )}{b^5} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.86 \[ \int \frac {x^9}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {9 a^2 b x^2-3 a b^2 x^4+b^3 x^6-\frac {3 a^4}{a+b x^2}-12 a^3 \log \left (a+b x^2\right )}{6 b^5} \]
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Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90
method | result | size |
risch | \(\frac {3 a^{2} x^{2}}{2 b^{4}}-\frac {a \,x^{4}}{2 b^{3}}+\frac {x^{6}}{6 b^{2}}-\frac {a^{4}}{2 b^{5} \left (b \,x^{2}+a \right )}-\frac {2 a^{3} \ln \left (b \,x^{2}+a \right )}{b^{5}}\) | \(63\) |
default | \(\frac {\frac {1}{3} b^{2} x^{6}-a b \,x^{4}+3 a^{2} x^{2}}{2 b^{4}}-\frac {2 a^{3} \ln \left (b \,x^{2}+a \right )}{b^{5}}-\frac {a^{4}}{2 b^{5} \left (b \,x^{2}+a \right )}\) | \(64\) |
norman | \(\frac {\frac {a^{2} x^{4}}{b^{3}}+\frac {x^{8}}{6 b}-\frac {a \,x^{6}}{3 b^{2}}-\frac {2 a^{4}}{b^{5}}}{b \,x^{2}+a}-\frac {2 a^{3} \ln \left (b \,x^{2}+a \right )}{b^{5}}\) | \(64\) |
parallelrisch | \(-\frac {-b^{4} x^{8}+2 a \,b^{3} x^{6}-6 a^{2} b^{2} x^{4}+12 \ln \left (b \,x^{2}+a \right ) x^{2} a^{3} b +12 \ln \left (b \,x^{2}+a \right ) a^{4}+12 a^{4}}{6 b^{5} \left (b \,x^{2}+a \right )}\) | \(79\) |
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none
Time = 0.24 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.16 \[ \int \frac {x^9}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {b^{4} x^{8} - 2 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 9 \, a^{3} b x^{2} - 3 \, a^{4} - 12 \, {\left (a^{3} b x^{2} + a^{4}\right )} \log \left (b x^{2} + a\right )}{6 \, {\left (b^{6} x^{2} + a b^{5}\right )}} \]
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Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.94 \[ \int \frac {x^9}{a^2+2 a b x^2+b^2 x^4} \, dx=- \frac {a^{4}}{2 a b^{5} + 2 b^{6} x^{2}} - \frac {2 a^{3} \log {\left (a + b x^{2} \right )}}{b^{5}} + \frac {3 a^{2} x^{2}}{2 b^{4}} - \frac {a x^{4}}{2 b^{3}} + \frac {x^{6}}{6 b^{2}} \]
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Time = 0.19 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.93 \[ \int \frac {x^9}{a^2+2 a b x^2+b^2 x^4} \, dx=-\frac {a^{4}}{2 \, {\left (b^{6} x^{2} + a b^{5}\right )}} - \frac {2 \, a^{3} \log \left (b x^{2} + a\right )}{b^{5}} + \frac {b^{2} x^{6} - 3 \, a b x^{4} + 9 \, a^{2} x^{2}}{6 \, b^{4}} \]
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Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.14 \[ \int \frac {x^9}{a^2+2 a b x^2+b^2 x^4} \, dx=-\frac {2 \, a^{3} \log \left ({\left | b x^{2} + a \right |}\right )}{b^{5}} + \frac {b^{4} x^{6} - 3 \, a b^{3} x^{4} + 9 \, a^{2} b^{2} x^{2}}{6 \, b^{6}} + \frac {4 \, a^{3} b x^{2} + 3 \, a^{4}}{2 \, {\left (b x^{2} + a\right )} b^{5}} \]
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Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.97 \[ \int \frac {x^9}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {x^6}{6\,b^2}-\frac {a^4}{2\,b\,\left (b^5\,x^2+a\,b^4\right )}-\frac {a\,x^4}{2\,b^3}-\frac {2\,a^3\,\ln \left (b\,x^2+a\right )}{b^5}+\frac {3\,a^2\,x^2}{2\,b^4} \]
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