Integrand size = 23, antiderivative size = 318 \[ \int \frac {(e x)^m (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=\frac {(e x)^{1+m} \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{a \left (b^2-4 a c\right ) e \left (a+b x+c x^2\right )}-\frac {c \left (A b \left (b+\sqrt {b^2-4 a c}\right ) m-2 a \left (b B-2 A c (1-m)+B \sqrt {b^2-4 a c} m\right )\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) e (1+m)}-\frac {c \left ((A b-2 a B) m+\frac {2 a (b B-2 A c (1-m))-A b^2 m}{\sqrt {b^2-4 a c}}\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b+\sqrt {b^2-4 a c}\right ) e (1+m)} \]
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Time = 0.52 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {836, 844, 66} \[ \int \frac {(e x)^m (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {c (e x)^{m+1} \left (A b m \left (\sqrt {b^2-4 a c}+b\right )-2 a \left (B m \sqrt {b^2-4 a c}-2 A c (1-m)+b B\right )\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{a e (m+1) \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {c (e x)^{m+1} \left (\frac {-4 a A c (1-m)+2 a b B-A b^2 m}{\sqrt {b^2-4 a c}}+m (A b-2 a B)\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{a e (m+1) \left (b^2-4 a c\right ) \left (\sqrt {b^2-4 a c}+b\right )}+\frac {(e x)^{m+1} \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{a e \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \]
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Rule 66
Rule 836
Rule 844
Rubi steps \begin{align*} \text {integral}& = \frac {(e x)^{1+m} \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{a \left (b^2-4 a c\right ) e \left (a+b x+c x^2\right )}-\frac {\int \frac {(e x)^m \left (e^2 \left (2 a A c (1-m)+A b^2 m-a b B (1+m)\right )+(A b-2 a B) c e^2 m x\right )}{a+b x+c x^2} \, dx}{a \left (b^2-4 a c\right ) e^2} \\ & = \frac {(e x)^{1+m} \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{a \left (b^2-4 a c\right ) e \left (a+b x+c x^2\right )}-\frac {\int \left (\frac {\left ((A b-2 a B) c e^2 m+\frac {c e^2 \left (-2 a b B+4 a A c+A b^2 m-4 a A c m\right )}{\sqrt {b^2-4 a c}}\right ) (e x)^m}{b-\sqrt {b^2-4 a c}+2 c x}+\frac {\left ((A b-2 a B) c e^2 m-\frac {c e^2 \left (-2 a b B+4 a A c+A b^2 m-4 a A c m\right )}{\sqrt {b^2-4 a c}}\right ) (e x)^m}{b+\sqrt {b^2-4 a c}+2 c x}\right ) \, dx}{a \left (b^2-4 a c\right ) e^2} \\ & = \frac {(e x)^{1+m} \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{a \left (b^2-4 a c\right ) e \left (a+b x+c x^2\right )}-\frac {\left (c \left ((A b-2 a B) m-\frac {2 a (b B-2 A c (1-m))-A b^2 m}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {(e x)^m}{b-\sqrt {b^2-4 a c}+2 c x} \, dx}{a \left (b^2-4 a c\right )}-\frac {\left (c \left ((A b-2 a B) m+\frac {2 a b B-4 a A c (1-m)-A b^2 m}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {(e x)^m}{b+\sqrt {b^2-4 a c}+2 c x} \, dx}{a \left (b^2-4 a c\right )} \\ & = \frac {(e x)^{1+m} \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{a \left (b^2-4 a c\right ) e \left (a+b x+c x^2\right )}-\frac {c \left (A b \left (b+\sqrt {b^2-4 a c}\right ) m-2 a \left (b B-2 A c (1-m)+B \sqrt {b^2-4 a c} m\right )\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 )}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) e (1+m)}-\frac {c \left ((A b-2 a B) m+\frac {2 a b B-4 a A c (1-m)-A b^2 m}{\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 )}{a \left (b^2-4 a c\right ) \left (b+\sqrt {b^2-4 a c}\right ) e (1+m)} \\ \end{align*}
Time = 0.58 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.80 \[ \int \frac {(e x)^m (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=\frac {(e x)^m \left (\frac {x \left (-a B (b+2 c x)+A \left (b^2-2 a c+b c x\right )\right )}{a+x (b+c x)}+\frac {c x \left (-\frac {\left ((A b-2 a B) m+\frac {-2 a (b B+2 A c (-1+m))+A b^2 m}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )}{b-\sqrt {b^2-4 a c}}-\frac {\left ((A b-2 a B) m+\frac {2 a b B+4 a A c (-1+m)-A b^2 m}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{b+\sqrt {b^2-4 a c}}\right )}{1+m}\right )}{a \left (b^2-4 a c\right )} \]
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\[\int \frac {\left (e x \right )^{m} \left (B x +A \right )}{\left (c \,x^{2}+b x +a \right )^{2}}d x\]
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\[ \int \frac {(e x)^m (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=\int { \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{{\left (c x^{2} + b x + a\right )}^{2}} \,d x } \]
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\[ \int \frac {(e x)^m (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=\int \frac {\left (e x\right )^{m} \left (A + B x\right )}{\left (a + b x + c x^{2}\right )^{2}}\, dx \]
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\[ \int \frac {(e x)^m (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=\int { \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{{\left (c x^{2} + b x + a\right )}^{2}} \,d x } \]
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\[ \int \frac {(e x)^m (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=\int { \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{{\left (c x^{2} + b x + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx=\int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x+a\right )}^2} \,d x \]
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