Integrand size = 20, antiderivative size = 135 \[ \int (e x)^m (A+B x) \left (a+c x^2\right )^p \, dx=\frac {A (e x)^{1+m} \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},-p,\frac {3+m}{2},-\frac {c x^2}{a}\right )}{e (1+m)}+\frac {B (e x)^{2+m} \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {2+m}{2},-p,\frac {4+m}{2},-\frac {c x^2}{a}\right )}{e^2 (2+m)} \]
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Time = 0.04 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {822, 372, 371} \[ \int (e x)^m (A+B x) \left (a+c x^2\right )^p \, dx=\frac {A (e x)^{m+1} \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},-p,\frac {m+3}{2},-\frac {c x^2}{a}\right )}{e (m+1)}+\frac {B (e x)^{m+2} \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {m+2}{2},-p,\frac {m+4}{2},-\frac {c x^2}{a}\right )}{e^2 (m+2)} \]
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Rule 371
Rule 372
Rule 822
Rubi steps \begin{align*} \text {integral}& = A \int (e x)^m \left (a+c x^2\right )^p \, dx+\frac {B \int (e x)^{1+m} \left (a+c x^2\right )^p \, dx}{e} \\ & = \left (A \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p}\right ) \int (e x)^m \left (1+\frac {c x^2}{a}\right )^p \, dx+\frac {\left (B \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p}\right ) \int (e x)^{1+m} \left (1+\frac {c x^2}{a}\right )^p \, dx}{e} \\ & = \frac {A (e x)^{1+m} \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1+m}{2},-p;\frac {3+m}{2};-\frac {c x^2}{a}\right )}{e (1+m)}+\frac {B (e x)^{2+m} \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \, _2F_1\left (\frac {2+m}{2},-p;\frac {4+m}{2};-\frac {c x^2}{a}\right )}{e^2 (2+m)} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.79 \[ \int (e x)^m (A+B x) \left (a+c x^2\right )^p \, dx=\frac {x (e x)^m \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \left (B (1+m) x \operatorname {Hypergeometric2F1}\left (1+\frac {m}{2},-p,2+\frac {m}{2},-\frac {c x^2}{a}\right )+A (2+m) \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},-p,\frac {3+m}{2},-\frac {c x^2}{a}\right )\right )}{(1+m) (2+m)} \]
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\[\int \left (e x \right )^{m} \left (B x +A \right ) \left (c \,x^{2}+a \right )^{p}d x\]
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\[ \int (e x)^m (A+B x) \left (a+c x^2\right )^p \, dx=\int { {\left (B x + A\right )} {\left (c x^{2} + a\right )}^{p} \left (e x\right )^{m} \,d x } \]
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Result contains complex when optimal does not.
Time = 40.91 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.79 \[ \int (e x)^m (A+B x) \left (a+c x^2\right )^p \, dx=\frac {A a^{p} e^{m} x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {B a^{p} e^{m} x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + 2\right )} \]
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\[ \int (e x)^m (A+B x) \left (a+c x^2\right )^p \, dx=\int { {\left (B x + A\right )} {\left (c x^{2} + a\right )}^{p} \left (e x\right )^{m} \,d x } \]
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\[ \int (e x)^m (A+B x) \left (a+c x^2\right )^p \, dx=\int { {\left (B x + A\right )} {\left (c x^{2} + a\right )}^{p} \left (e x\right )^{m} \,d x } \]
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Timed out. \[ \int (e x)^m (A+B x) \left (a+c x^2\right )^p \, dx=\int {\left (e\,x\right )}^m\,{\left (c\,x^2+a\right )}^p\,\left (A+B\,x\right ) \,d x \]
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