Integrand size = 31, antiderivative size = 125 \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\frac {(A b-a B) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a (b c-a d) e (1+m)}+\frac {(B c-A d) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c (b c-a d) e (1+m)} \]
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Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {598, 371} \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\frac {(e x)^{m+1} (A b-a B) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{a e (m+1) (b c-a d)}+\frac {(e x)^{m+1} (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{c e (m+1) (b c-a d)} \]
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Rule 371
Rule 598
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(A b-a B) (e x)^m}{(b c-a d) \left (a+b x^2\right )}+\frac {(B c-A d) (e x)^m}{(b c-a d) \left (c+d x^2\right )}\right ) \, dx \\ & = \frac {(A b-a B) \int \frac {(e x)^m}{a+b x^2} \, dx}{b c-a d}+\frac {(B c-A d) \int \frac {(e x)^m}{c+d x^2} \, dx}{b c-a d} \\ & = \frac {(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) e (1+m)}+\frac {(B c-A d) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{c (b c-a d) e (1+m)} \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.80 \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\frac {x (e x)^m \left ((-A b c+a B c) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+a (-B c+A d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )\right )}{a c (-b c+a d) (1+m)} \]
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\[\int \frac {\left (e x \right )^{m} \left (x^{2} B +A \right )}{\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}d x\]
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\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}} \,d x } \]
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\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\int \frac {\left (e x\right )^{m} \left (A + B x^{2}\right )}{\left (a + b x^{2}\right ) \left (c + d x^{2}\right )}\, dx \]
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\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}} \,d x } \]
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\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m}{\left (b\,x^2+a\right )\,\left (d\,x^2+c\right )} \,d x \]
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