Integrand size = 31, antiderivative size = 162 \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{c+d x^2} \, dx=-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1+m}{2},-p,1,\frac {3+m}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c d e (1+m)}+\frac {B (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},-p,\frac {3+m}{2},-\frac {b x^2}{a}\right )}{d e (1+m)} \]
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Time = 0.11 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {598, 372, 371, 525, 524} \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{c+d x^2} \, dx=\frac {B (e x)^{m+1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},-p,\frac {m+3}{2},-\frac {b x^2}{a}\right )}{d e (m+1)}-\frac {(e x)^{m+1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} (B c-A d) \operatorname {AppellF1}\left (\frac {m+1}{2},-p,1,\frac {m+3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c d e (m+1)} \]
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Rule 371
Rule 372
Rule 524
Rule 525
Rule 598
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {B (e x)^m \left (a+b x^2\right )^p}{d}+\frac {(-B c+A d) (e x)^m \left (a+b x^2\right )^p}{d \left (c+d x^2\right )}\right ) \, dx \\ & = \frac {B \int (e x)^m \left (a+b x^2\right )^p \, dx}{d}+\frac {(-B c+A d) \int \frac {(e x)^m \left (a+b x^2\right )^p}{c+d x^2} \, dx}{d} \\ & = \frac {\left (B \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}{d}+\frac {\left ((-B c+A d) \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}{d} \\ & = -\frac {(B c-A d) (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 )}{c d e (1+m)}+\frac {B (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 )}{d e (1+m)} \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.73 \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{c+d x^2} \, dx=\frac {x (e x)^m \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left ((-B c+A d) \operatorname {AppellF1}\left (\frac {1+m}{2},-p,1,\frac {3+m}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+B c \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},-p,\frac {3+m}{2},-\frac {b x^2}{a}\right )\right )}{c d (1+m)} \]
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\[\int \frac {\left (e x \right )^{m} \left (b \,x^{2}+a \right )^{p} \left (x^{2} B +A \right )}{d \,x^{2}+c}d x\]
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\[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{c+d x^2} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{p} \left (e x\right )^{m}}{d x^{2} + c} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{c+d x^2} \, dx=\text {Timed out} \]
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\[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{c+d x^2} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{p} \left (e x\right )^{m}}{d x^{2} + c} \,d x } \]
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\[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{c+d x^2} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{p} \left (e x\right )^{m}}{d x^{2} + c} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{c+d x^2} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m\,{\left (b\,x^2+a\right )}^p}{d\,x^2+c} \,d x \]
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