Integrand size = 22, antiderivative size = 234 \[ \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\frac {b x^{1+m}}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac {b (b c (3-m)-a d (7-m)) x^{1+m}}{8 a^2 (b c-a d)^2 \left (a+b x^2\right )}+\frac {b \left (a^2 d^2 \left (15-8 m+m^2\right )-2 a b c d \left (5-6 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right ) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{8 a^3 (b c-a d)^3 (1+m)}-\frac {d^3 x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c (b c-a d)^3 (1+m)} \]
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Time = 0.27 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {483, 593, 598, 371} \[ \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\frac {b x^{m+1} (b c (3-m)-a d (7-m))}{8 a^2 \left (a+b x^2\right ) (b c-a d)^2}+\frac {b x^{m+1} \left (a^2 d^2 \left (m^2-8 m+15\right )-2 a b c d \left (m^2-6 m+5\right )+b^2 c^2 \left (m^2-4 m+3\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{8 a^3 (m+1) (b c-a d)^3}-\frac {d^3 x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{c (m+1) (b c-a d)^3}+\frac {b x^{m+1}}{4 a \left (a+b x^2\right )^2 (b c-a d)} \]
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Rule 371
Rule 483
Rule 593
Rule 598
Rubi steps \begin{align*} \text {integral}& = \frac {b x^{1+m}}{4 a (b c-a d) \left (a+b x^2\right )^2}-\frac {\int \frac {x^m \left (4 a d-b c (3-m)-b d (3-m) x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx}{4 a (b c-a d)} \\ & = \frac {b x^{1+m}}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac {b (b c (3-m)-a d (7-m)) x^{1+m}}{8 a^2 (b c-a d)^2 \left (a+b x^2\right )}+\frac {\int \frac {x^m \left (8 a^2 d^2-a b c d \left (7-8 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )+b d (b c (3-m)-a d (7-m)) (1-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{8 a^2 (b c-a d)^2} \\ & = \frac {b x^{1+m}}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac {b (b c (3-m)-a d (7-m)) x^{1+m}}{8 a^2 (b c-a d)^2 \left (a+b x^2\right )}+\frac {\int \left (\frac {b \left (a^2 d^2 \left (15-8 m+m^2\right )-2 a b c d \left (5-6 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right ) x^m}{(b c-a d) \left (a+b x^2\right )}+\frac {8 a^2 d^3 x^m}{(-b c+a d) \left (c+d x^2\right )}\right ) \, dx}{8 a^2 (b c-a d)^2} \\ & = \frac {b x^{1+m}}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac {b (b c (3-m)-a d (7-m)) x^{1+m}}{8 a^2 (b c-a d)^2 \left (a+b x^2\right )}-\frac {d^3 \int \frac {x^m}{c+d x^2} \, dx}{(b c-a d)^3}+\frac {\left (b \left (a^2 d^2 \left (15-8 m+m^2\right )-2 a b c d \left (5-6 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )\right ) \int \frac {x^m}{a+b x^2} \, dx}{8 a^2 (b c-a d)^3} \\ & = \frac {b x^{1+m}}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac {b (b c (3-m)-a d (7-m)) x^{1+m}}{8 a^2 (b c-a d)^2 \left (a+b x^2\right )}+\frac {b \left (a^2 d^2 \left (15-8 m+m^2\right )-2 a b c d \left (5-6 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right ) x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{8 a^3 (b c-a d)^3 (1+m)}-\frac {d^3 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)^3 (1+m)} \\ \end{align*}
Time = 0.50 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.73 \[ \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\frac {x^{1+m} \left (-a^2 b c d^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+a^3 d^3 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )-b c (-b c+a d) \left (a d \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+(-b c+a d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )\right )\right )}{a^3 c (-b c+a d)^3 (1+m)} \]
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\[\int \frac {x^{m}}{\left (b \,x^{2}+a \right )^{3} \left (d \,x^{2}+c \right )}d x\]
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\[ \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\int { \frac {x^{m}}{{\left (b x^{2} + a\right )}^{3} {\left (d x^{2} + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\int { \frac {x^{m}}{{\left (b x^{2} + a\right )}^{3} {\left (d x^{2} + c\right )}} \,d x } \]
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\[ \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\int { \frac {x^{m}}{{\left (b x^{2} + a\right )}^{3} {\left (d x^{2} + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\int \frac {x^m}{{\left (b\,x^2+a\right )}^3\,\left (d\,x^2+c\right )} \,d x \]
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