Integrand size = 31, antiderivative size = 292 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\frac {b (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m)) (e x)^{1+m}}{8 c^2 d^3 e (1+m)}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{4 c d e \left (c+d x^2\right )^2}-\frac {(b c-a d) (e x)^{1+m} \left (a (A d (3-m)+B c (1+m))-b (A d (1+m)-B c (5+m)) x^2\right )}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac {(a d (a d (1-m)+b c (1+m)) (A d (3-m)+B c (1+m))-b c (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{8 c^3 d^3 e (1+m)} \]
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Time = 0.28 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {591, 470, 371} \[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\frac {(e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right ) (a d (a d (1-m)+b c (m+1)) (A d (3-m)+B c (m+1))-b c (a d (m+1)-b c (m+3)) (A d (m+1)-B c (m+5)))}{8 c^3 d^3 e (m+1)}+\frac {b (e x)^{m+1} (a d (m+1)-b c (m+3)) (A d (m+1)-B c (m+5))}{8 c^2 d^3 e (m+1)}-\frac {(e x)^{m+1} (b c-a d) \left (a (A d (3-m)+B c (m+1))-b x^2 (A d (m+1)-B c (m+5))\right )}{8 c^2 d^2 e \left (c+d x^2\right )}-\frac {\left (a+b x^2\right )^2 (e x)^{m+1} (B c-A d)}{4 c d e \left (c+d x^2\right )^2} \]
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Rule 371
Rule 470
Rule 591
Rubi steps \begin{align*} \text {integral}& = -\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{4 c d e \left (c+d x^2\right )^2}-\frac {\int \frac {(e x)^m \left (a+b x^2\right ) \left (-a (A d (3-m)+B c (1+m))+b (A d (1+m)-B c (5+m)) x^2\right )}{\left (c+d x^2\right )^2} \, dx}{4 c d} \\ & = -\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{4 c d e \left (c+d x^2\right )^2}-\frac {(b c-a d) (e x)^{1+m} \left (a (A d (3-m)+B c (1+m))-b (A d (1+m)-B c (5+m)) x^2\right )}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac {\int \frac {(e x)^m \left (a (a d (1-m)+b c (1+m)) (A d (3-m)+B c (1+m))+b (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m)) x^2\right )}{c+d x^2} \, dx}{8 c^2 d^2} \\ & = \frac {b (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m)) (e x)^{1+m}}{8 c^2 d^3 e (1+m)}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{4 c d e \left (c+d x^2\right )^2}-\frac {(b c-a d) (e x)^{1+m} \left (a (A d (3-m)+B c (1+m))-b (A d (1+m)-B c (5+m)) x^2\right )}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac {\left (a (a d (1-m)+b c (1+m)) (A d (3-m)+B c (1+m))-\frac {b c (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m))}{d}\right ) \int \frac {(e x)^m}{c+d x^2} \, dx}{8 c^2 d^2} \\ & = \frac {b (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m)) (e x)^{1+m}}{8 c^2 d^3 e (1+m)}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{4 c d e \left (c+d x^2\right )^2}-\frac {(b c-a d) (e x)^{1+m} \left (a (A d (3-m)+B c (1+m))-b (A d (1+m)-B c (5+m)) x^2\right )}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac {\left (a (a d (1-m)+b c (1+m)) (A d (3-m)+B c (1+m))-\frac {b c (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m))}{d}\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 d^2 e (1+m)} \\ \end{align*}
Time = 0.76 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.58 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\frac {x (e x)^m \left (b^2 B-\frac {b (3 b B c-A b d-2 a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c}+\frac {(b c-a d) (3 b B c-2 A b d-a B d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c^2}-\frac {(b c-a d)^2 (B c-A d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c^3}\right )}{d^3 (1+m)} \]
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\[\int \frac {\left (e x \right )^{m} \left (b \,x^{2}+a \right )^{2} \left (x^{2} B +A \right )}{\left (d \,x^{2}+c \right )^{3}}d x\]
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\[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]
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\[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int \frac {\left (e x\right )^{m} \left (A + B x^{2}\right ) \left (a + b x^{2}\right )^{2}}{\left (c + d x^{2}\right )^{3}}\, dx \]
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\[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]
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\[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m\,{\left (b\,x^2+a\right )}^2}{{\left (d\,x^2+c\right )}^3} \,d x \]
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