Integrand size = 31, antiderivative size = 333 \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\frac {(B c-A d) (e x)^{1+m}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {(b c (B c (3-m)-A d (7-m))+a d (A d (3-m)+B c (1+m))) (e x)^{1+m}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}+\frac {b^2 (A b-a B) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a (b c-a d)^3 e (1+m)}+\frac {\left (b^2 c^2 (B c (1-m)-A d (5-m)) (3-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (3+2 m-m^2\right )+A d \left (5-6 m+m^2\right )\right )\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{8 c^3 (b c-a d)^3 e (1+m)} \]
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Time = 0.47 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {593, 598, 371} \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\frac {(e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right ) \left (-a^2 d^2 (1-m) (A d (3-m)+B c (m+1))+2 a b c d \left (A d \left (m^2-6 m+5\right )+B c \left (-m^2+2 m+3\right )\right )+b^2 c^2 (3-m) (B c (1-m)-A d (5-m))\right )}{8 c^3 e (m+1) (b c-a d)^3}+\frac {b^2 (e x)^{m+1} (A b-a B) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{a e (m+1) (b c-a d)^3}+\frac {(e x)^{m+1} (a d (A d (3-m)+B c (m+1))+b c (B c (3-m)-A d (7-m)))}{8 c^2 e \left (c+d x^2\right ) (b c-a d)^2}+\frac {(e x)^{m+1} (B c-A d)}{4 c e \left (c+d x^2\right )^2 (b c-a d)} \]
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Rule 371
Rule 593
Rule 598
Rubi steps \begin{align*} \text {integral}& = \frac {(B c-A d) (e x)^{1+m}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {\int \frac {(e x)^m \left (4 A b c-a A d (3-m)-a B c (1+m)+b (B c-A d) (3-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx}{4 c (b c-a d)} \\ & = \frac {(B c-A d) (e x)^{1+m}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {(b c (B c (3-m)-A d (7-m))+a d (A d (3-m)+B c (1+m))) (e x)^{1+m}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}+\frac {\int \frac {(e x)^m \left (a B c (a d (1-m)-b c (5-m)) (1+m)+A \left (8 b^2 c^2-a b c d \left (7-8 m+m^2\right )+a^2 d^2 \left (3-4 m+m^2\right )\right )+b (1-m) (b c (B c (3-m)-A d (7-m))+a d (A d (3-m)+B c (1+m))) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{8 c^2 (b c-a d)^2} \\ & = \frac {(B c-A d) (e x)^{1+m}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {(b c (B c (3-m)-A d (7-m))+a d (A d (3-m)+B c (1+m))) (e x)^{1+m}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}+\frac {\int \left (\frac {8 b^2 (A b-a B) c^2 (e x)^m}{(b c-a d) \left (a+b x^2\right )}+\frac {\left (b^2 c^2 (B c (1-m)-A d (5-m)) (3-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (3+2 m-m^2\right )+A d \left (5-6 m+m^2\right )\right )\right ) (e x)^m}{(b c-a d) \left (c+d x^2\right )}\right ) \, dx}{8 c^2 (b c-a d)^2} \\ & = \frac {(B c-A d) (e x)^{1+m}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {(b c (B c (3-m)-A d (7-m))+a d (A d (3-m)+B c (1+m))) (e x)^{1+m}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}+\frac {\left (b^2 (A b-a B)\right ) \int \frac {(e x)^m}{a+b x^2} \, dx}{(b c-a d)^3}+\frac {\left (b^2 c^2 (B c (1-m)-A d (5-m)) (3-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (3+2 m-m^2\right )+A d \left (5-6 m+m^2\right )\right )\right ) \int \frac {(e x)^m}{c+d x^2} \, dx}{8 c^2 (b c-a d)^3} \\ & = \frac {(B c-A d) (e x)^{1+m}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {(b c (B c (3-m)-A d (7-m))+a d (A d (3-m)+B c (1+m))) (e x)^{1+m}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}+\frac {b^2 (A b-a B) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{a (b c-a d)^3 e (1+m)}+\frac {\left (b^2 c^2 (B c (1-m)-A d (5-m)) (3-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (3+2 m-m^2\right )+A d \left (5-6 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)^3 e (1+m)} \\ \end{align*}
Time = 0.87 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.59 \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\frac {x (e x)^m \left (\frac {b^2 (A b-a B) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a}-\frac {b (A b-a B) d \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c}-\frac {(A b-a B) d (b c-a d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c^2}+\frac {(b c-a d)^2 (B c-A d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c^3}\right )}{(b c-a d)^3 (1+m)} \]
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\[\int \frac {\left (e x \right )^{m} \left (x^{2} B +A \right )}{\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )^{3}}d x\]
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\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\text {Timed out} \]
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\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}^{3}} \,d x } \]
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\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m}{\left (b\,x^2+a\right )\,{\left (d\,x^2+c\right )}^3} \,d x \]
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