Integrand size = 26, antiderivative size = 51 \[ \int \frac {(d+e x)^m}{a^2+2 a b x+b^2 x^2} \, dx=\frac {e (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^2 (1+m)} \]
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Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {27, 70} \[ \int \frac {(d+e x)^m}{a^2+2 a b x+b^2 x^2} \, dx=\frac {e (d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (2,m+1,m+2,\frac {b (d+e x)}{b d-a e}\right )}{(m+1) (b d-a e)^2} \]
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Rule 27
Rule 70
Rubi steps \begin{align*} \text {integral}& = \int \frac {(d+e x)^m}{(a+b x)^2} \, dx \\ & = \frac {e (d+e x)^{1+m} \, _2F_1\left (2,1+m;2+m;\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^2 (1+m)} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.02 \[ \int \frac {(d+e x)^m}{a^2+2 a b x+b^2 x^2} \, dx=\frac {e (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {b (d+e x)}{-b d+a e}\right )}{(-b d+a e)^2 (1+m)} \]
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\[\int \frac {\left (e x +d \right )^{m}}{b^{2} x^{2}+2 a b x +a^{2}}d x\]
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\[ \int \frac {(d+e x)^m}{a^2+2 a b x+b^2 x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{b^{2} x^{2} + 2 \, a b x + a^{2}} \,d x } \]
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\[ \int \frac {(d+e x)^m}{a^2+2 a b x+b^2 x^2} \, dx=\int \frac {\left (d + e x\right )^{m}}{\left (a + b x\right )^{2}}\, dx \]
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\[ \int \frac {(d+e x)^m}{a^2+2 a b x+b^2 x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{b^{2} x^{2} + 2 \, a b x + a^{2}} \,d x } \]
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\[ \int \frac {(d+e x)^m}{a^2+2 a b x+b^2 x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{b^{2} x^{2} + 2 \, a b x + a^{2}} \,d x } \]
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Timed out. \[ \int \frac {(d+e x)^m}{a^2+2 a b x+b^2 x^2} \, dx=\int \frac {{\left (d+e\,x\right )}^m}{a^2+2\,a\,b\,x+b^2\,x^2} \,d x \]
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