Integrand size = 23, antiderivative size = 172 \[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\frac {\left (B-\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) e (1+m)}+\frac {\left (B+\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) e (1+m)} \]
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Time = 0.19 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {844, 66} \[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\frac {(e x)^{m+1} \left (B-\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{e (m+1) \left (b-\sqrt {b^2-4 a c}\right )}+\frac {(e x)^{m+1} \left (\frac {b B-2 A c}{\sqrt {b^2-4 a c}}+B\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{e (m+1) \left (\sqrt {b^2-4 a c}+b\right )} \]
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Rule 66
Rule 844
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (B+\frac {-b B+2 A c}{\sqrt {b^2-4 a c}}\right ) (e x)^m}{b-\sqrt {b^2-4 a c}+2 c x}+\frac {\left (B-\frac {-b B+2 A c}{\sqrt {b^2-4 a c}}\right ) (e x)^m}{b+\sqrt {b^2-4 a c}+2 c x}\right ) \, dx \\ & = \left (B-\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) \int \frac {(e x)^m}{b-\sqrt {b^2-4 a c}+2 c x} \, dx+\left (B+\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) \int \frac {(e x)^m}{b+\sqrt {b^2-4 a c}+2 c x} \, dx \\ & = \frac {\left (B-\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) (e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) e (1+m)}+\frac {\left (B+\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) (e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) e (1+m)} \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.78 \[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\frac {x (e x)^m \left (\left (-2 a B+A \left (b+\sqrt {b^2-4 a c}\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+\left (2 a B+A \left (-b+\sqrt {b^2-4 a c}\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )\right )}{2 a \sqrt {b^2-4 a c} (1+m)} \]
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\[\int \frac {\left (e x \right )^{m} \left (B x +A \right )}{c \,x^{2}+b x +a}d x\]
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\[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\int { \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{c x^{2} + b x + a} \,d x } \]
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\[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\int \frac {\left (e x\right )^{m} \left (A + B x\right )}{a + b x + c x^{2}}\, dx \]
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\[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\int { \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{c x^{2} + b x + a} \,d x } \]
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\[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\int { \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{c x^{2} + b x + a} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x\right )}{c\,x^2+b\,x+a} \,d x \]
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