Integrand size = 31, antiderivative size = 452 \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\frac {d (2 A b c-3 a B c+a A d) (e x)^{1+m}}{4 a c (b c-a d)^2 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (A \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right )-a B c (b c (11-m)+a d (1+m))\right ) (e x)^{1+m}}{8 a c^2 (b c-a d)^3 e \left (c+d x^2\right )}+\frac {b^2 (A b (b c (1-m)-a d (7-m))+a B (a d (5-m)+b c (1+m))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^4 e (1+m)}-\frac {d \left (b^2 c^2 (B c (3-m)-A d (7-m)) (5-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (5+4 m-m^2\right )+A d \left (7-8 m+m^2\right )\right )\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{8 c^3 (b c-a d)^4 e (1+m)} \]
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Time = 0.86 (sec) , antiderivative size = 452, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {593, 598, 371} \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\frac {d (e x)^{m+1} \left (A \left (-a^2 d^2 (3-m)+a b c d (11-m)+4 b^2 c^2\right )-a B c (a d (m+1)+b c (11-m))\right )}{8 a c^2 e \left (c+d x^2\right ) (b c-a d)^3}-\frac {d (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right ) \left (-a^2 d^2 (1-m) (A d (3-m)+B c (m+1))+2 a b c d \left (A d \left (m^2-8 m+7\right )+B c \left (-m^2+4 m+5\right )\right )+b^2 c^2 (5-m) (B c (3-m)-A d (7-m))\right )}{8 c^3 e (m+1) (b c-a d)^4}+\frac {b^2 (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right ) (A b (b c (1-m)-a d (7-m))+a B (a d (5-m)+b c (m+1)))}{2 a^2 e (m+1) (b c-a d)^4}+\frac {d (e x)^{m+1} (a A d-3 a B c+2 A b c)}{4 a c e \left (c+d x^2\right )^2 (b c-a d)^2}+\frac {(e x)^{m+1} (A b-a B)}{2 a e \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)} \]
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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}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )^2}-\frac {\int \frac {(e x)^m \left (2 a A d-A b c (1-m)-a B c (1+m)-(A b-a B) d (5-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx}{2 a (b c-a d)} \\ & = \frac {d (2 A b c-3 a B c+a A d) (e x)^{1+m}}{4 a c (b c-a d)^2 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )^2}-\frac {\int \frac {(e x)^m \left (2 \left (A \left (8 a b c d-2 b^2 c^2 (1-m)-a^2 d^2 (3-m)\right )-a B c (2 b c+a d) (1+m)\right )-2 b d (2 A b c-3 a B c+a A d) (3-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx}{8 a c (b c-a d)^2} \\ & = \frac {d (2 A b c-3 a B c+a A d) (e x)^{1+m}}{4 a c (b c-a d)^2 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (A \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right )-a B c (b c (11-m)+a d (1+m))\right ) (e x)^{1+m}}{8 a c^2 (b c-a d)^3 e \left (c+d x^2\right )}-\frac {\int \frac {(e x)^m \left (-2 \left (a B c \left (4 b^2 c^2-a^2 d^2 (1-m)+a b c d (9-m)\right ) (1+m)-A \left (24 a b^2 c^2 d-4 b^3 c^3 (1-m)-a^2 b c d^2 \left (11-12 m+m^2\right )+a^3 d^3 \left (3-4 m+m^2\right )\right )\right )-2 b d (1-m) \left (A \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right )-a B c (b c (11-m)+a d (1+m))\right ) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{16 a c^2 (b c-a d)^3} \\ & = \frac {d (2 A b c-3 a B c+a A d) (e x)^{1+m}}{4 a c (b c-a d)^2 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (A \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right )-a B c (b c (11-m)+a d (1+m))\right ) (e x)^{1+m}}{8 a c^2 (b c-a d)^3 e \left (c+d x^2\right )}-\frac {\int \left (\frac {8 b^2 c^2 (-A b (b c (1-m)-a d (7-m))-a B (a d (5-m)+b c (1+m))) (e x)^m}{(b c-a d) \left (a+b x^2\right )}+\frac {2 a d \left (b^2 c^2 (B c (3-m)-A d (7-m)) (5-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (5+4 m-m^2\right )+A d \left (7-8 m+m^2\right )\right )\right ) (e x)^m}{(b c-a d) \left (c+d x^2\right )}\right ) \, dx}{16 a c^2 (b c-a d)^3} \\ & = \frac {d (2 A b c-3 a B c+a A d) (e x)^{1+m}}{4 a c (b c-a d)^2 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (A \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right )-a B c (b c (11-m)+a d (1+m))\right ) (e x)^{1+m}}{8 a c^2 (b c-a d)^3 e \left (c+d x^2\right )}+\frac {\left (b^2 (A b (b c (1-m)-a d (7-m))+a B (a d (5-m)+b c (1+m)))\right ) \int \frac {(e x)^m}{a+b x^2} \, dx}{2 a (b c-a d)^4}-\frac {\left (d \left (b^2 c^2 (B c (3-m)-A d (7-m)) (5-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (5+4 m-m^2\right )+A d \left (7-8 m+m^2\right )\right )\right )\right ) \int \frac {(e x)^m}{c+d x^2} \, dx}{8 c^2 (b c-a d)^4} \\ & = \frac {d (2 A b c-3 a B c+a A d) (e x)^{1+m}}{4 a c (b c-a d)^2 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (A \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right )-a B c (b c (11-m)+a d (1+m))\right ) (e x)^{1+m}}{8 a c^2 (b c-a d)^3 e \left (c+d x^2\right )}+\frac {b^2 (A b (b c (1-m)-a d (7-m))+a B (a d (5-m)+b c (1+m))) (e 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)^4 e (1+m)}-\frac {d \left (b^2 c^2 (B c (3-m)-A d (7-m)) (5-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (5+4 m-m^2\right )+A d \left (7-8 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)^4 e (1+m)} \\ \end{align*}
Time = 1.24 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.59 \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\frac {x (e x)^m \left (\frac {b^2 (b B c-3 A b d+2 a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a}-\frac {b d (b B c-3 A b d+2 a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c}+\frac {b^2 (-A b+a B) (-b c+a d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^2}-\frac {d (b c-a d) (b B c-2 A b d+a B d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c^2}+\frac {d (b c-a d)^2 (-B c+A d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c^3}\right )}{(b c-a d)^4 (1+m)} \]
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\[\int \frac {\left (e x \right )^{m} \left (x^{2} B +A \right )}{\left (b \,x^{2}+a \right )^{2} \left (d \,x^{2}+c \right )^{3}}d x\]
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\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\text {Timed out} \]
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\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{3}} \,d x } \]
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\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m}{{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^3} \,d x \]
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