Integrand size = 22, antiderivative size = 47 \[ \int (b x)^{5/2} (\pi +d x)^n (e+f x)^p \, dx=\frac {2 e^p \pi ^n (b x)^{7/2} \operatorname {AppellF1}\left (\frac {7}{2},-n,-p,\frac {9}{2},-\frac {d x}{\pi },-\frac {f x}{e}\right )}{7 b} \]
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Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {138} \[ \int (b x)^{5/2} (\pi +d x)^n (e+f x)^p \, dx=\frac {2 \pi ^n e^p (b x)^{7/2} \operatorname {AppellF1}\left (\frac {7}{2},-n,-p,\frac {9}{2},-\frac {d x}{\pi },-\frac {f x}{e}\right )}{7 b} \]
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Rule 138
Rubi steps \begin{align*} \text {integral}& = \frac {2 e^p \pi ^n (b x)^{7/2} F_1\left (\frac {7}{2};-n,-p;\frac {9}{2};-\frac {d x}{\pi },-\frac {f x}{e}\right )}{7 b} \\ \end{align*}
Time = 0.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.96 \[ \int (b x)^{5/2} (\pi +d x)^n (e+f x)^p \, dx=\frac {2}{7} e^p \pi ^n x (b x)^{5/2} \operatorname {AppellF1}\left (\frac {7}{2},-n,-p,\frac {9}{2},-\frac {d x}{\pi },-\frac {f x}{e}\right ) \]
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\[\int \left (b x \right )^{\frac {5}{2}} \left (d x +\pi \right )^{n} \left ({\mathrm e}+f x \right )^{p}d x\]
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\[ \int (b x)^{5/2} (\pi +d x)^n (e+f x)^p \, dx=\int { \left (b x\right )^{\frac {5}{2}} {\left (\pi + d x\right )}^{n} {\left (f x + e\right )}^{p} \,d x } \]
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Timed out. \[ \int (b x)^{5/2} (\pi +d x)^n (e+f x)^p \, dx=\text {Timed out} \]
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\[ \int (b x)^{5/2} (\pi +d x)^n (e+f x)^p \, dx=\int { \left (b x\right )^{\frac {5}{2}} {\left (\pi + d x\right )}^{n} {\left (f x + e\right )}^{p} \,d x } \]
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\[ \int (b x)^{5/2} (\pi +d x)^n (e+f x)^p \, dx=\int { \left (b x\right )^{\frac {5}{2}} {\left (\pi + d x\right )}^{n} {\left (f x + e\right )}^{p} \,d x } \]
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Timed out. \[ \int (b x)^{5/2} (\pi +d x)^n (e+f x)^p \, dx=\int {\left (\mathrm {e}+f\,x\right )}^p\,{\left (b\,x\right )}^{5/2}\,{\left (\Pi +d\,x\right )}^n \,d x \]
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