Integrand size = 22, antiderivative size = 189 \[ \int (d+e x)^m \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {(d+e x)^{1+m} \left (a+b x+c x^2\right )^{5/2} \operatorname {AppellF1}\left (1+m,-\frac {5}{2},-\frac {5}{2},2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1+m) \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{5/2} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{5/2}} \]
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Time = 0.16 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {773, 138} \[ \int (d+e x)^m \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {\left (a+b x+c x^2\right )^{5/2} (d+e x)^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {5}{2},-\frac {5}{2},m+2,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (m+1) \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{5/2} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{5/2}} \]
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Rule 138
Rule 773
Rubi steps \begin{align*} \text {integral}& = \frac {\left (a+b x+c x^2\right )^{5/2} \text {Subst}\left (\int x^m \left (1-\frac {2 c x}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{5/2} \left (1-\frac {2 c x}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{5/2} \, dx,x,d+e x\right )}{e \left (1-\frac {d+e x}{d-\frac {\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c}}\right )^{5/2} \left (1-\frac {d+e x}{d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c}}\right )^{5/2}} \\ & = \frac {(d+e x)^{1+m} \left (a+b x+c x^2\right )^{5/2} F_1\left (1+m;-\frac {5}{2},-\frac {5}{2};2+m;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1+m) \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{5/2} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{5/2}} \\ \end{align*}
\[ \int (d+e x)^m \left (a+b x+c x^2\right )^{5/2} \, dx=\int (d+e x)^m \left (a+b x+c x^2\right )^{5/2} \, dx \]
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\[\int \left (e x +d \right )^{m} \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}d x\]
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\[ \int (d+e x)^m \left (a+b x+c x^2\right )^{5/2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} {\left (e x + d\right )}^{m} \,d x } \]
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Exception generated. \[ \int (d+e x)^m \left (a+b x+c x^2\right )^{5/2} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int (d+e x)^m \left (a+b x+c x^2\right )^{5/2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} {\left (e x + d\right )}^{m} \,d x } \]
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\[ \int (d+e x)^m \left (a+b x+c x^2\right )^{5/2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} {\left (e x + d\right )}^{m} \,d x } \]
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Timed out. \[ \int (d+e x)^m \left (a+b x+c x^2\right )^{5/2} \, dx=\int {\left (d+e\,x\right )}^m\,{\left (c\,x^2+b\,x+a\right )}^{5/2} \,d x \]
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