Integrand size = 29, antiderivative size = 120 \[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{c+d x^2} \, dx=-\frac {(b B c-A b d-a B d) (e x)^{1+m}}{d^2 e (1+m)}+\frac {b B (e x)^{3+m}}{d e^3 (3+m)}+\frac {(b c-a d) (B c-A d) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c d^2 e (1+m)} \]
[Out]
Time = 0.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {584, 371} \[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{c+d x^2} \, dx=\frac {(e x)^{m+1} (b c-a d) (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{c d^2 e (m+1)}-\frac {(e x)^{m+1} (-a B d-A b d+b B c)}{d^2 e (m+1)}+\frac {b B (e x)^{m+3}}{d e^3 (m+3)} \]
[In]
[Out]
Rule 371
Rule 584
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {(b B c-A b d-a B d) (e x)^m}{d^2}+\frac {b B (e x)^{2+m}}{d e^2}+\frac {\left (b B c^2-A b c d-a B c d+a A d^2\right ) (e x)^m}{d^2 \left (c+d x^2\right )}\right ) \, dx \\ & = -\frac {(b B c-A b d-a B d) (e x)^{1+m}}{d^2 e (1+m)}+\frac {b B (e x)^{3+m}}{d e^3 (3+m)}+\frac {((b c-a d) (B c-A d)) \int \frac {(e x)^m}{c+d x^2} \, dx}{d^2} \\ & = -\frac {(b B c-A b d-a B d) (e x)^{1+m}}{d^2 e (1+m)}+\frac {b B (e x)^{3+m}}{d e^3 (3+m)}+\frac {(b c-a d) (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 d^2 e (1+m)} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.78 \[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{c+d x^2} \, dx=\frac {x (e x)^m \left (\frac {-b B c+A b d+a B d}{1+m}+\frac {b B d x^2}{3+m}+\frac {(b c-a d) (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c (1+m)}\right )}{d^2} \]
[In]
[Out]
\[\int \frac {\left (e x \right )^{m} \left (b \,x^{2}+a \right ) \left (x^{2} B +A \right )}{d \,x^{2}+c}d x\]
[In]
[Out]
\[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{c+d x^2} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )} \left (e x\right )^{m}}{d x^{2} + c} \,d x } \]
[In]
[Out]
Result contains complex when optimal does not.
Time = 3.50 (sec) , antiderivative size = 418, normalized size of antiderivative = 3.48 \[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{c+d x^2} \, dx=\frac {A a e^{m} m x^{m + 1} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {A a e^{m} x^{m + 1} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {A b e^{m} m x^{m + 3} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 A b e^{m} x^{m + 3} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {B a e^{m} m x^{m + 3} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 B a e^{m} x^{m + 3} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {B b e^{m} m x^{m + 5} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {5}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {5 B b e^{m} x^{m + 5} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {5}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} \]
[In]
[Out]
\[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{c+d x^2} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )} \left (e x\right )^{m}}{d x^{2} + c} \,d x } \]
[In]
[Out]
\[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{c+d x^2} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )} \left (e x\right )^{m}}{d x^{2} + c} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {(e x)^m \left (a+b x^2\right ) \left (A+B x^2\right )}{c+d x^2} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m\,\left (b\,x^2+a\right )}{d\,x^2+c} \,d x \]
[In]
[Out]