Integrand size = 31, antiderivative size = 483 \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {(a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}+\frac {\left (a^2 d^2 (1-m) (A d (3-m)+B c (1+m))-2 a b c d (B c (1+m) (1-m-2 p)+A d (1-m) (3-m-2 p))+b^2 c^2 (1-m-2 p) (A d (3-m-2 p)+B c (1+m+2 p))\right ) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1+m}{2},-p,1,\frac {3+m}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{8 c^3 d (b c-a d)^2 e (1+m)}-\frac {b (a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (1+m+2 p) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},-p,\frac {3+m}{2},-\frac {b x^2}{a}\right )}{8 c^2 d (b c-a d)^2 e (1+m)} \]
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Time = 0.70 (sec) , antiderivative size = 483, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {593, 598, 372, 371, 525, 524} \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\frac {(e x)^{m+1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a^2 d^2 (1-m) (A d (3-m)+B c (m+1))-2 a b c d (A d (1-m) (-m-2 p+3)+B c (m+1) (-m-2 p+1))+b^2 c^2 (-m-2 p+1) (A d (-m-2 p+3)+B c (m+2 p+1))\right ) \operatorname {AppellF1}\left (\frac {m+1}{2},-p,1,\frac {m+3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{8 c^3 d e (m+1) (b c-a d)^2}-\frac {b (m+2 p+1) (e x)^{m+1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},-p,\frac {m+3}{2},-\frac {b x^2}{a}\right ) (a d (A d (3-m)+B c (m+1))+b c (B c (-m-2 p+1)-A d (-m-2 p+5)))}{8 c^2 d e (m+1) (b c-a d)^2}+\frac {(e x)^{m+1} \left (a+b x^2\right )^{p+1} (a d (A d (3-m)+B c (m+1))+b c (B c (-m-2 p+1)-A d (-m-2 p+5)))}{8 c^2 e \left (c+d x^2\right ) (b c-a d)^2}+\frac {(e x)^{m+1} \left (a+b x^2\right )^{p+1} (B c-A d)}{4 c e \left (c+d x^2\right )^2 (b c-a d)} \]
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Rule 371
Rule 372
Rule 524
Rule 525
Rule 593
Rule 598
Rubi steps \begin{align*} \text {integral}& = \frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {\int \frac {(e x)^m \left (a+b x^2\right )^p \left (4 A b c-a A d (3-m)-a B c (1+m)+b (B c-A d) (1-m-2 p) 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} \left (a+b x^2\right )^{1+p}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {(a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}+\frac {\int \frac {(e x)^m \left (a+b x^2\right )^p \left (a B c (1+m) (a d (1-m)-b c (3-m-2 p))+A \left (8 b^2 c^2+a^2 d^2 \left (3-4 m+m^2\right )-a b c d \left (9+m^2-2 m (3-p)+2 p\right )\right )+b (d (4 A b c-a A d (3-m)-a B c (1+m))-b c (B c-A d) (1-m-2 p)) (1+m+2 p) x^2\right )}{c+d x^2} \, dx}{8 c^2 (b c-a d)^2} \\ & = \frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {(a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}+\frac {\int \left (\frac {b (d (4 A b c-a A d (3-m)-a B c (1+m))-b c (B c-A d) (1-m-2 p)) (1+m+2 p) (e x)^m \left (a+b x^2\right )^p}{d}+\frac {\left (-b c (d (4 A b c-a A d (3-m)-a B c (1+m))-b c (B c-A d) (1-m-2 p)) (1+m+2 p)+d \left (a B c (1+m) (a d (1-m)-b c (3-m-2 p))+A \left (8 b^2 c^2+a^2 d^2 \left (3-4 m+m^2\right )-a b c d \left (9+m^2-2 m (3-p)+2 p\right )\right )\right )\right ) (e x)^m \left (a+b x^2\right )^p}{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} \left (a+b x^2\right )^{1+p}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {(a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}-\frac {(b (a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (1+m+2 p)) \int (e x)^m \left (a+b x^2\right )^p \, dx}{8 c^2 d (b c-a d)^2}+\frac {\left (a^2 d^2 (1-m) (A d (3-m)+B c (1+m))-2 a b c d (B c (1+m) (1-m-2 p)+A d (1-m) (3-m-2 p))+b^2 c^2 (1-m-2 p) (A d (3-m-2 p)+B c (1+m+2 p))\right ) \int \frac {(e x)^m \left (a+b x^2\right )^p}{c+d x^2} \, dx}{8 c^2 d (b c-a d)^2} \\ & = \frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {(a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}-\frac {\left (b (a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (1+m+2 p) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int (e x)^m \left (1+\frac {b x^2}{a}\right )^p \, dx}{8 c^2 d (b c-a d)^2}+\frac {\left (\left (a^2 d^2 (1-m) (A d (3-m)+B c (1+m))-2 a b c d (B c (1+m) (1-m-2 p)+A d (1-m) (3-m-2 p))+b^2 c^2 (1-m-2 p) (A d (3-m-2 p)+B c (1+m+2 p))\right ) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \frac {(e x)^m \left (1+\frac {b x^2}{a}\right )^p}{c+d x^2} \, dx}{8 c^2 d (b c-a d)^2} \\ & = \frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {(a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}+\frac {\left (a^2 d^2 (1-m) (A d (3-m)+B c (1+m))-2 a b c d (B c (1+m) (1-m-2 p)+A d (1-m) (3-m-2 p))+b^2 c^2 (1-m-2 p) (A d (3-m-2 p)+B c (1+m+2 p))\right ) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {1+m}{2};-p,1;\frac {3+m}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{8 c^3 d (b c-a d)^2 e (1+m)}-\frac {b (a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (1+m+2 p) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1+m}{2},-p;\frac {3+m}{2};-\frac {b x^2}{a}\right )}{8 c^2 d (b c-a d)^2 e (1+m)} \\ \end{align*}
Time = 0.55 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.27 \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\frac {x (e x)^m \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (B c \operatorname {AppellF1}\left (\frac {1+m}{2},-p,2,\frac {3+m}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+(-B c+A d) \operatorname {AppellF1}\left (\frac {1+m}{2},-p,3,\frac {3+m}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}{c^3 d (1+m)} \]
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\[\int \frac {\left (e x \right )^{m} \left (b \,x^{2}+a \right )^{p} \left (x^{2} B +A \right )}{\left (d \,x^{2}+c \right )^{3}}d x\]
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\[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{p} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\text {Timed out} \]
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\[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{p} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]
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\[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{p} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m\,{\left (b\,x^2+a\right )}^p}{{\left (d\,x^2+c\right )}^3} \,d x \]
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