Integrand size = 31, antiderivative size = 171 \[ \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 A \cos ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1+3 m),\frac {1}{6} (5+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (1-3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (2+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}} \]
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Time = 0.11 (sec) , antiderivative size = 171, 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} \[ \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 A \sin (c+d x) \cos ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m-1),\frac {1}{6} (3 m+5),\cos ^2(c+d x)\right )}{b d (1-3 m) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}}-\frac {3 B \sin (c+d x) \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+2),\frac {1}{6} (3 m+8),\cos ^2(c+d x)\right )}{b d (3 m+2) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}} \]
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Rule 20
Rule 2722
Rule 2827
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [3]{\cos (c+d x)} \int \cos ^{-\frac {4}{3}+m}(c+d x) (A+B \cos (c+d x)) \, dx}{b \sqrt [3]{b \cos (c+d x)}} \\ & = \frac {\left (A \sqrt [3]{\cos (c+d x)}\right ) \int \cos ^{-\frac {4}{3}+m}(c+d x) \, dx}{b \sqrt [3]{b \cos (c+d x)}}+\frac {\left (B \sqrt [3]{\cos (c+d x)}\right ) \int \cos ^{-\frac {1}{3}+m}(c+d x) \, dx}{b \sqrt [3]{b \cos (c+d x)}} \\ & = \frac {3 A \cos ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1+3 m),\frac {1}{6} (5+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (1-3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (2+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}} \\ \end{align*}
Time = 0.22 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{4/3}} \, dx=-\frac {3 \cos ^{1+m}(c+d x) \csc (c+d x) \left (A (2+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1+3 m),\frac {1}{6} (5+3 m),\cos ^2(c+d x)\right )+B (-1+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (-1+3 m) (2+3 m) (b \cos (c+d x))^{4/3}} \]
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\[\int \frac {\left (\cos ^{m}\left (d x +c \right )\right ) \left (A +B \cos \left (d x +c \right )\right )}{\left (\cos \left (d x +c \right ) b \right )^{\frac {4}{3}}}d x\]
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\[ \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{m}}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
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\[ \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{4/3}} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \cos ^{m}{\left (c + d x \right )}}{\left (b \cos {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]
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\[ \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{m}}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
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\[ \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{m}}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{4/3}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^m\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}} \,d x \]
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