Integrand size = 24, antiderivative size = 90 \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^p \, dx=-\frac {2 d^2 (b+2 c x)^3 \left (\frac {1}{4} \left (4 a-\frac {b^2}{c}\right )+\frac {(b+2 c x)^2}{4 c}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,\frac {5}{2}+p,\frac {5}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{3 \left (b^2-4 a c\right )} \]
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Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.94, 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} \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^p \, dx=\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^p \left (1-\frac {(b+2 c x)^2}{b^2-4 a c}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{6 c} \]
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Rule 371
Rule 372
Rule 708
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int x^2 \left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )^p \, 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 x^2 \left (1+\frac {x^2}{4 \left (a-\frac {b^2}{4 c}\right ) c d^2}\right )^p \, dx,x,b d+2 c d x\right )}{c d} \\ & = \frac {2^{-1+2 p} d^2 (b+2 c x)^3 \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 {3}{2},-p;\frac {5}{2};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{3 c} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^p \, dx=\frac {2^{-1-2 p} d^2 (b+2 c x)^3 (a+x (b+c x))^p \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{3 c} \]
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\[\int \left (2 c d x +b d \right )^{2} \left (c \,x^{2}+b x +a \right )^{p}d x\]
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\[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^p \, dx=\int { {\left (2 \, c d x + b d\right )}^{2} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \]
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\[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^p \, dx=d^{2} \left (\int b^{2} \left (a + b x + c x^{2}\right )^{p}\, dx + \int 4 c^{2} x^{2} \left (a + b x + c x^{2}\right )^{p}\, dx + \int 4 b c x \left (a + b x + c x^{2}\right )^{p}\, dx\right ) \]
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\[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^p \, dx=\int { {\left (2 \, c d x + b d\right )}^{2} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \]
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\[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^p \, dx=\int { {\left (2 \, c d x + b d\right )}^{2} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \]
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Timed out. \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^p \, dx=\int {\left (b\,d+2\,c\,d\,x\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \]
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