Integrand size = 26, antiderivative size = 98 \[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{3/2} \, dx=-\frac {(b d+2 c d x)^{1+m} \left (4 a-\frac {b^2}{c}+\frac {(b+2 c x)^2}{c}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (1,\frac {6+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{16 \left (b^2-4 a c\right ) d (1+m)} \]
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Time = 0.08 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.14, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {708, 372, 371} \[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{3/2} \, dx=-\frac {\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (d (b+2 c x))^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m+1}{2},\frac {m+3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{8 c^2 d (m+1) \sqrt {1-\frac {(b+2 c x)^2}{b^2-4 a c}}} \]
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Rule 371
Rule 372
Rule 708
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int x^m \left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )^{3/2} \, dx,x,b d+2 c d x\right )}{2 c d} \\ & = \frac {\left (\left (a-\frac {b^2}{4 c}\right ) \sqrt {a+b x+c x^2}\right ) \text {Subst}\left (\int x^m \left (1+\frac {x^2}{4 \left (a-\frac {b^2}{4 c}\right ) c d^2}\right )^{3/2} \, dx,x,b d+2 c d x\right )}{c d \sqrt {4+\frac {(b d+2 c d x)^2}{\left (a-\frac {b^2}{4 c}\right ) c d^2}}} \\ & = -\frac {\left (b^2-4 a c\right ) (d (b+2 c x))^{1+m} \sqrt {a+b x+c x^2} \, _2F_1\left (-\frac {3}{2},\frac {1+m}{2};\frac {3+m}{2};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{8 c^2 d (1+m) \sqrt {1-\frac {(b+2 c x)^2}{b^2-4 a c}}} \\ \end{align*}
Time = 0.49 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.15 \[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{3/2} \, dx=-\frac {\left (b^2-4 a c\right ) (b+2 c x) (d (b+2 c x))^m \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{16 c^2 (1+m) \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \]
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\[\int \left (2 c d x +b d \right )^{m} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}d x\]
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\[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{3/2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} {\left (2 \, c d x + b d\right )}^{m} \,d x } \]
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\[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{3/2} \, dx=\int \left (d \left (b + 2 c x\right )\right )^{m} \left (a + b x + c x^{2}\right )^{\frac {3}{2}}\, dx \]
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\[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{3/2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} {\left (2 \, c d x + b d\right )}^{m} \,d x } \]
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\[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{3/2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} {\left (2 \, c d x + b d\right )}^{m} \,d x } \]
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Timed out. \[ \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^{3/2} \, dx=\int {\left (b\,d+2\,c\,d\,x\right )}^m\,{\left (c\,x^2+b\,x+a\right )}^{3/2} \,d x \]
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