Integrand size = 31, antiderivative size = 665 \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^3} \, dx=-\frac {d \left (A \left (2 a^2 d^2-b^2 c^2 (3-m)+a b c d (13-m)\right )-a B c (a d (11-m)+b c (1+m))\right ) (e x)^{1+m}}{8 a^2 c (b c-a d)^3 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2 \left (c+d x^2\right )^2}+\frac {(A b (b c (3-m)-a d (11-m))+a B (a d (7-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (A (b c+a d) \left (b^2 c^2 (3-m)+a^2 d^2 (3-m)-2 a b c d (9-m)\right )+a B c \left (2 a b c d (11-m)+b^2 c^2 (1+m)+a^2 d^2 (1+m)\right )\right ) (e x)^{1+m}}{8 a^2 c^2 (b c-a d)^4 e \left (c+d x^2\right )}+\frac {b^2 \left (a B \left (b^2 c^2 \left (1-m^2\right )-2 a b c d \left (7+6 m-m^2\right )-a^2 d^2 \left (35-12 m+m^2\right )\right )+A b \left (a^2 d^2 \left (63-16 m+m^2\right )-2 a b c d \left (9-10 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{8 a^3 (b c-a d)^5 e (1+m)}+\frac {d^2 \left (b^2 c^2 (B c (5-m)-A d (9-m)) (7-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (7+6 m-m^2\right )+A d \left (9-10 m+m^2\right )\right )\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{8 c^3 (b c-a d)^5 e (1+m)} \]
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Time = 1.47 (sec) , antiderivative size = 665, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {593, 598, 371} \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^3} \, dx=\frac {d (e x)^{m+1} \left (A (a d+b c) \left (a^2 d^2 (3-m)-2 a b c d (9-m)+b^2 c^2 (3-m)\right )+a B c \left (a^2 d^2 (m+1)+2 a b c d (11-m)+b^2 c^2 (m+1)\right )\right )}{8 a^2 c^2 e \left (c+d x^2\right ) (b c-a d)^4}-\frac {d (e x)^{m+1} \left (A \left (2 a^2 d^2+a b c d (13-m)-b^2 c^2 (3-m)\right )-a B c (a d (11-m)+b c (m+1))\right )}{8 a^2 c e \left (c+d x^2\right )^2 (b c-a d)^3}+\frac {d^2 (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right ) \left (-a^2 d^2 (1-m) (A d (3-m)+B c (m+1))+2 a b c d \left (A d \left (m^2-10 m+9\right )+B c \left (-m^2+6 m+7\right )\right )+b^2 c^2 (7-m) (B c (5-m)-A d (9-m))\right )}{8 c^3 e (m+1) (b c-a d)^5}+\frac {(e x)^{m+1} (A b (b c (3-m)-a d (11-m))+a B (a d (7-m)+b c (m+1)))}{8 a^2 e \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)^2}+\frac {b^2 (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right ) \left (A b \left (a^2 d^2 \left (m^2-16 m+63\right )-2 a b c d \left (m^2-10 m+9\right )+b^2 c^2 \left (m^2-4 m+3\right )\right )+a B \left (-a^2 d^2 \left (m^2-12 m+35\right )-2 a b c d \left (-m^2+6 m+7\right )+b^2 c^2 \left (1-m^2\right )\right )\right )}{8 a^3 e (m+1) (b c-a d)^5}+\frac {(e x)^{m+1} (A b-a B)}{4 a e \left (a+b x^2\right )^2 \left (c+d x^2\right )^2 (b c-a d)} \]
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Rule 371
Rule 593
Rule 598
Rubi steps \begin{align*} \text {integral}& = \frac {(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2 \left (c+d x^2\right )^2}-\frac {\int \frac {(e x)^m \left (4 a A d-A b c (3-m)-a B c (1+m)-(A b-a B) d (7-m) x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx}{4 a (b c-a d)} \\ & = \frac {(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2 \left (c+d x^2\right )^2}+\frac {(A b (b c (3-m)-a d (11-m))+a B (a d (7-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {\int \frac {(e x)^m \left (-a B c (1+m) (a d (9-m)-b (c-c m))+A \left (8 a^2 d^2-a b c d \left (3-12 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )+d (5-m) (A b (b c (3-m)-a d (11-m))+a B (a d (7-m)+b c (1+m))) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx}{8 a^2 (b c-a d)^2} \\ & = -\frac {d \left (A \left (2 a^2 d^2-b^2 c^2 (3-m)+a b c d (13-m)\right )-a B c (a d (11-m)+b c (1+m))\right ) (e x)^{1+m}}{8 a^2 c (b c-a d)^3 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2 \left (c+d x^2\right )^2}+\frac {(A b (b c (3-m)-a d (11-m))+a B (a d (7-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {\int \frac {(e x)^m \left (-4 \left (a B c \left (2 a^2 d^2-b^2 c^2 (1-m)+a b c d (11-m)\right ) (1+m)-A \left (24 a^2 b c d^2-2 a^3 d^3 (3-m)-a b^2 c^2 d \left (9-14 m+m^2\right )+b^3 c^3 \left (3-4 m+m^2\right )\right )\right )-4 b d (3-m) \left (A \left (2 a^2 d^2-b^2 c^2 (3-m)+a b c d (13-m)\right )-a B c (a d (11-m)+b c (1+m))\right ) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx}{32 a^2 c (b c-a d)^3} \\ & = -\frac {d \left (A \left (2 a^2 d^2-b^2 c^2 (3-m)+a b c d (13-m)\right )-a B c (a d (11-m)+b c (1+m))\right ) (e x)^{1+m}}{8 a^2 c (b c-a d)^3 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2 \left (c+d x^2\right )^2}+\frac {(A b (b c (3-m)-a d (11-m))+a B (a d (7-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (A (b c+a d) \left (b^2 c^2 (3-m)+a^2 d^2 (3-m)-2 a b c d (9-m)\right )+a B c \left (2 a b c d (11-m)+b^2 c^2 (1+m)+a^2 d^2 (1+m)\right )\right ) (e x)^{1+m}}{8 a^2 c^2 (b c-a d)^4 e \left (c+d x^2\right )}+\frac {\int \frac {(e x)^m \left (8 \left (a B c (b c+a d) \left (b^2 c^2 (1-m)+a^2 d^2 (1-m)-2 a b c d (7-m)\right ) (1+m)+A \left (48 a^2 b^2 c^2 d^2-a b^3 c^3 d \left (15-16 m+m^2\right )-a^3 b c d^3 \left (15-16 m+m^2\right )+b^4 c^4 \left (3-4 m+m^2\right )+a^4 d^4 \left (3-4 m+m^2\right )\right )\right )+8 b d (1-m) \left (A (b c+a d) \left (b^2 c^2 (3-m)+a^2 d^2 (3-m)-2 a b c d (9-m)\right )+a B c \left (2 a b c d (11-m)+b^2 c^2 (1+m)+a^2 d^2 (1+m)\right )\right ) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{64 a^2 c^2 (b c-a d)^4} \\ & = -\frac {d \left (A \left (2 a^2 d^2-b^2 c^2 (3-m)+a b c d (13-m)\right )-a B c (a d (11-m)+b c (1+m))\right ) (e x)^{1+m}}{8 a^2 c (b c-a d)^3 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2 \left (c+d x^2\right )^2}+\frac {(A b (b c (3-m)-a d (11-m))+a B (a d (7-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (A (b c+a d) \left (b^2 c^2 (3-m)+a^2 d^2 (3-m)-2 a b c d (9-m)\right )+a B c \left (2 a b c d (11-m)+b^2 c^2 (1+m)+a^2 d^2 (1+m)\right )\right ) (e x)^{1+m}}{8 a^2 c^2 (b c-a d)^4 e \left (c+d x^2\right )}+\frac {\int \left (\frac {8 b^2 c^2 \left (a B \left (b^2 c^2 \left (1-m^2\right )-2 a b c d \left (7+6 m-m^2\right )-a^2 d^2 \left (35-12 m+m^2\right )\right )+A b \left (a^2 d^2 \left (63-16 m+m^2\right )-2 a b c d \left (9-10 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )\right ) (e x)^m}{(b c-a d) \left (a+b x^2\right )}+\frac {8 a^2 d^2 \left (b^2 c^2 (B c (5-m)-A d (9-m)) (7-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (7+6 m-m^2\right )+A d \left (9-10 m+m^2\right )\right )\right ) (e x)^m}{(b c-a d) \left (c+d x^2\right )}\right ) \, dx}{64 a^2 c^2 (b c-a d)^4} \\ & = -\frac {d \left (A \left (2 a^2 d^2-b^2 c^2 (3-m)+a b c d (13-m)\right )-a B c (a d (11-m)+b c (1+m))\right ) (e x)^{1+m}}{8 a^2 c (b c-a d)^3 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2 \left (c+d x^2\right )^2}+\frac {(A b (b c (3-m)-a d (11-m))+a B (a d (7-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (A (b c+a d) \left (b^2 c^2 (3-m)+a^2 d^2 (3-m)-2 a b c d (9-m)\right )+a B c \left (2 a b c d (11-m)+b^2 c^2 (1+m)+a^2 d^2 (1+m)\right )\right ) (e x)^{1+m}}{8 a^2 c^2 (b c-a d)^4 e \left (c+d x^2\right )}+\frac {\left (d^2 \left (b^2 c^2 (B c (5-m)-A d (9-m)) (7-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (7+6 m-m^2\right )+A d \left (9-10 m+m^2\right )\right )\right )\right ) \int \frac {(e x)^m}{c+d x^2} \, dx}{8 c^2 (b c-a d)^5}+\frac {\left (b^2 \left (a B \left (b^2 c^2 \left (1-m^2\right )-2 a b c d \left (7+6 m-m^2\right )-a^2 d^2 \left (35-12 m+m^2\right )\right )+A b \left (a^2 d^2 \left (63-16 m+m^2\right )-2 a b c d \left (9-10 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )\right )\right ) \int \frac {(e x)^m}{a+b x^2} \, dx}{8 a^2 (b c-a d)^5} \\ & = -\frac {d \left (A \left (2 a^2 d^2-b^2 c^2 (3-m)+a b c d (13-m)\right )-a B c (a d (11-m)+b c (1+m))\right ) (e x)^{1+m}}{8 a^2 c (b c-a d)^3 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2 \left (c+d x^2\right )^2}+\frac {(A b (b c (3-m)-a d (11-m))+a B (a d (7-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (A (b c+a d) \left (b^2 c^2 (3-m)+a^2 d^2 (3-m)-2 a b c d (9-m)\right )+a B c \left (2 a b c d (11-m)+b^2 c^2 (1+m)+a^2 d^2 (1+m)\right )\right ) (e x)^{1+m}}{8 a^2 c^2 (b c-a d)^4 e \left (c+d x^2\right )}+\frac {b^2 \left (a B \left (b^2 c^2 \left (1-m^2\right )-2 a b c d \left (7+6 m-m^2\right )-a^2 d^2 \left (35-12 m+m^2\right )\right )+A b \left (a^2 d^2 \left (63-16 m+m^2\right )-2 a b c d \left (9-10 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )\right ) (e 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)^5 e (1+m)}+\frac {d^2 \left (b^2 c^2 (B c (5-m)-A d (9-m)) (7-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (7+6 m-m^2\right )+A d \left (9-10 m+m^2\right )\right )\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{8 c^3 (b c-a d)^5 e (1+m)} \\ \end{align*}
Time = 1.51 (sec) , antiderivative size = 329, normalized size of antiderivative = 0.49 \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^3} \, dx=\frac {x (e x)^m \left (-\frac {3 b^2 d (b B c-2 A b d+a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a}+\frac {3 b d^2 (b B c-2 A b d+a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c}+\frac {b^2 (b c-a d) (b B c-3 A b d+2 a B d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^2}+\frac {d^2 (b c-a d) (2 b B c-3 A b d+a B d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c^2}+\frac {b^2 (A b-a B) (b c-a d)^2 \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^3}+\frac {d^2 (b c-a d)^2 (B c-A d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c^3}\right )}{(b c-a d)^5 (1+m)} \]
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\[\int \frac {\left (e x \right )^{m} \left (x^{2} B +A \right )}{\left (b \,x^{2}+a \right )^{3} \left (d \,x^{2}+c \right )^{3}}d x\]
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\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{3} {\left (d x^{2} + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^3} \, dx=\text {Timed out} \]
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\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{3} {\left (d x^{2} + c\right )}^{3}} \,d x } \]
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\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{3} {\left (d x^{2} + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^3} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m}{{\left (b\,x^2+a\right )}^3\,{\left (d\,x^2+c\right )}^3} \,d x \]
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