Integrand size = 33, antiderivative size = 142 \[ \int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 C \cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^n \sin (c+d x)}{d (9+2 n)}-\frac {2 (C (7+2 n)+A (9+2 n)) \cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (7+2 n),\frac {1}{4} (11+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (7+2 n) (9+2 n) \sqrt {\sin ^2(c+d x)}} \]
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Time = 0.13 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {20, 3093, 2722} \[ \int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 C \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^n}{d (2 n+9)}-\frac {2 \left (\frac {A}{2 n+7}+\frac {C}{2 n+9}\right ) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n+7),\frac {1}{4} (2 n+11),\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}} \]
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Rule 20
Rule 2722
Rule 3093
Rubi steps \begin{align*} \text {integral}& = \left (\cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{\frac {5}{2}+n}(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx \\ & = \frac {2 C \cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^n \sin (c+d x)}{d (9+2 n)}+\frac {\left (\left (C \left (\frac {7}{2}+n\right )+A \left (\frac {9}{2}+n\right )\right ) \cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{\frac {5}{2}+n}(c+d x) \, dx}{\frac {9}{2}+n} \\ & = \frac {2 C \cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^n \sin (c+d x)}{d (9+2 n)}-\frac {2 (C (7+2 n)+A (9+2 n)) \cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (7+2 n),\frac {1}{4} (11+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (7+2 n) (9+2 n) \sqrt {\sin ^2(c+d x)}} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.99 \[ \int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {2 \cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^n \csc (c+d x) \left (A (11+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (7+2 n),\frac {1}{4} (11+2 n),\cos ^2(c+d x)\right )+C (7+2 n) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (11+2 n),\frac {1}{4} (15+2 n),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (7+2 n) (11+2 n)} \]
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\[\int \left (\cos ^{\frac {5}{2}}\left (d x +c \right )\right ) \left (\cos \left (d x +c \right ) b \right )^{n} \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )d x\]
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\[ \int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
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Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]
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\[ \int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
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\[ \int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
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Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^n \,d x \]
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