Integrand size = 31, antiderivative size = 169 \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \, dx=-\frac {3 A b \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7+3 m),\frac {1}{6} (13+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (7+3 m) \sqrt {\sin ^2(c+d x)}}-\frac {3 b B \cos ^{3+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (10+3 m),\frac {1}{6} (16+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (10+3 m) \sqrt {\sin ^2(c+d x)}} \]
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Time = 0.11 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {20, 2827, 2722} \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \, dx=-\frac {3 A b \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+7),\frac {1}{6} (3 m+13),\cos ^2(c+d x)\right )}{d (3 m+7) \sqrt {\sin ^2(c+d x)}}-\frac {3 b B \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+3}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+10),\frac {1}{6} (3 m+16),\cos ^2(c+d x)\right )}{d (3 m+10) \sqrt {\sin ^2(c+d x)}} \]
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Rule 20
Rule 2722
Rule 2827
Rubi steps \begin{align*} \text {integral}& = \frac {\left (b \sqrt [3]{b \cos (c+d x)}\right ) \int \cos ^{\frac {4}{3}+m}(c+d x) (A+B \cos (c+d x)) \, dx}{\sqrt [3]{\cos (c+d x)}} \\ & = \frac {\left (A b \sqrt [3]{b \cos (c+d x)}\right ) \int \cos ^{\frac {4}{3}+m}(c+d x) \, dx}{\sqrt [3]{\cos (c+d x)}}+\frac {\left (b B \sqrt [3]{b \cos (c+d x)}\right ) \int \cos ^{\frac {7}{3}+m}(c+d x) \, dx}{\sqrt [3]{\cos (c+d x)}} \\ & = -\frac {3 A b \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7+3 m),\frac {1}{6} (13+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (7+3 m) \sqrt {\sin ^2(c+d x)}}-\frac {3 b B \cos ^{3+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (10+3 m),\frac {1}{6} (16+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (10+3 m) \sqrt {\sin ^2(c+d x)}} \\ \end{align*}
Time = 0.33 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.83 \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \, dx=-\frac {3 \cos ^{1+m}(c+d x) (b \cos (c+d x))^{4/3} \csc (c+d x) \left (B (7+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3}+\frac {m}{2},\frac {8}{3}+\frac {m}{2},\cos ^2(c+d x)\right )+A (10+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7+3 m),\frac {1}{6} (13+3 m),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (7+3 m) (10+3 m)} \]
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\[\int \left (\cos ^{m}\left (d x +c \right )\right ) \left (\cos \left (d x +c \right ) b \right )^{\frac {4}{3}} \left (A +B \cos \left (d x +c \right )\right )d x\]
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\[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{m} \,d x } \]
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Timed out. \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \, dx=\text {Timed out} \]
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\[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{m} \,d x } \]
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\[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{m} \,d x } \]
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Timed out. \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \, dx=\int {\cos \left (c+d\,x\right )}^m\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}\,\left (A+B\,\cos \left (c+d\,x\right )\right ) \,d x \]
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