Integrand size = 24, antiderivative size = 88 \[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^2} \, dx=\frac {2 \left (\frac {1}{4} \left (4 a-\frac {b^2}{c}\right )+\frac {(b+2 c x)^2}{4 c}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+p,\frac {1}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right ) d^2 (b+2 c x)} \]
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Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {708, 372, 371} \[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^2} \, dx=-\frac {\left (a+b x+c x^2\right )^p \left (1-\frac {(b+2 c x)^2}{b^2-4 a c}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{2 c d^2 (b+2 c x)} \]
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Rule 371
Rule 372
Rule 708
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )^p}{x^2} \, dx,x,b d+2 c d x\right )}{2 c d} \\ & = \frac {\left (2^{-1+2 p} \left (a+b x+c x^2\right )^p \left (4+\frac {(b d+2 c d x)^2}{\left (a-\frac {b^2}{4 c}\right ) c d^2}\right )^{-p}\right ) \text {Subst}\left (\int \frac {\left (1+\frac {x^2}{4 \left (a-\frac {b^2}{4 c}\right ) c d^2}\right )^p}{x^2} \, dx,x,b d+2 c d x\right )}{c d} \\ & = -\frac {2^{-1+2 p} \left (a+b x+c x^2\right )^p \left (4-\frac {4 (b+2 c x)^2}{b^2-4 a c}\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{c d^2 (b+2 c x)} \\ \end{align*}
Time = 0.43 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^2} \, dx=-\frac {2^{-1-2 p} (a+x (b+c x))^p \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{c d^2 (b+2 c x)} \]
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\[\int \frac {\left (c \,x^{2}+b x +a \right )^{p}}{\left (2 c d x +b d \right )^{2}}d x\]
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\[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^2} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{{\left (2 \, c d x + b d\right )}^{2}} \,d x } \]
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\[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^2} \, dx=\frac {\int \frac {\left (a + b x + c x^{2}\right )^{p}}{b^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx}{d^{2}} \]
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\[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^2} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{{\left (2 \, c d x + b d\right )}^{2}} \,d x } \]
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\[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^2} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{{\left (2 \, c d x + b d\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^2} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^p}{{\left (b\,d+2\,c\,d\,x\right )}^2} \,d x \]
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