Integrand size = 22, antiderivative size = 156 \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right )}+\frac {b (b c (1-m)-a d (3-m)) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^2 (1+m)}+\frac {d^2 x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c (b c-a d)^2 (1+m)} \]
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Time = 0.13 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {483, 598, 371} \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {b x^{m+1} (b c (1-m)-a d (3-m)) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{2 a^2 (m+1) (b c-a d)^2}+\frac {d^2 x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{c (m+1) (b c-a d)^2}+\frac {b x^{m+1}}{2 a \left (a+b x^2\right ) (b c-a d)} \]
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Rule 371
Rule 483
Rule 598
Rubi steps \begin{align*} \text {integral}& = \frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right )}-\frac {\int \frac {x^m \left (2 a d-b c (1-m)-b d (1-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{2 a (b c-a d)} \\ & = \frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right )}-\frac {\int \left (\frac {b (-b c (1-m)+a d (3-m)) x^m}{(b c-a d) \left (a+b x^2\right )}+\frac {2 a d^2 x^m}{(-b c+a d) \left (c+d x^2\right )}\right ) \, dx}{2 a (b c-a d)} \\ & = \frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right )}+\frac {d^2 \int \frac {x^m}{c+d x^2} \, dx}{(b c-a d)^2}+\frac {(b (b c (1-m)-a d (3-m))) \int \frac {x^m}{a+b x^2} \, dx}{2 a (b c-a d)^2} \\ & = \frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right )}-\frac {b (a d (3-m)-b (c-c m)) x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^2 (1+m)}+\frac {d^2 x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{c (b c-a d)^2 (1+m)} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.81 \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {x^{1+m} \left (-a b c d \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+a^2 d^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )+b c (b c-a d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )\right )}{a^2 c (b c-a d)^2 (1+m)} \]
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\[\int \frac {x^{m}}{\left (b \,x^{2}+a \right )^{2} \left (d \,x^{2}+c \right )}d x\]
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\[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\int { \frac {x^{m}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\int { \frac {x^{m}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}} \,d x } \]
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\[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\int { \frac {x^{m}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\int \frac {x^m}{{\left (b\,x^2+a\right )}^2\,\left (d\,x^2+c\right )} \,d x \]
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