Integrand size = 41, antiderivative size = 154 \[ \int \cos ^3(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {3 b^3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 d (b \sec (c+d x))^{5/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 b^2 (2 A+5 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{10 d (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 A b^3 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}} \]
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Time = 0.21 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {16, 4132, 3857, 2722, 4130} \[ \int \cos ^3(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 A b^3 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}-\frac {3 b^2 (2 A+5 C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}-\frac {3 b^3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right )}{5 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{5/3}} \]
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Rule 16
Rule 2722
Rule 3857
Rule 4130
Rule 4132
Rubi steps \begin{align*} \text {integral}& = b^3 \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/3}} \, dx \\ & = b^3 \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/3}} \, dx+\left (b^2 B\right ) \int \frac {1}{(b \sec (c+d x))^{2/3}} \, dx \\ & = \frac {3 A b^3 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}+\frac {1}{5} (b (2 A+5 C)) \int \sqrt [3]{b \sec (c+d x)} \, dx+\left (b^2 B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \left (\frac {\cos (c+d x)}{b}\right )^{2/3} \, dx \\ & = -\frac {3 b B \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{5 d \sqrt {\sin ^2(c+d x)}}+\frac {3 A b^3 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}+\frac {1}{5} \left (b (2 A+5 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \frac {1}{\sqrt [3]{\frac {\cos (c+d x)}{b}}} \, dx \\ & = -\frac {3 b (2 A+5 C) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{10 d \sqrt {\sin ^2(c+d x)}}-\frac {3 b B \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{5 d \sqrt {\sin ^2(c+d x)}}+\frac {3 A b^3 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}} \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.77 \[ \int \cos ^3(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {3 b \cot (c+d x) \left (2 A \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\sec ^2(c+d x)\right )+5 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\sec ^2(c+d x)\right )-10 C \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sec ^2(c+d x)\right )\right ) \sqrt [3]{b \sec (c+d x)} \sqrt {-\tan ^2(c+d x)}}{10 d} \]
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\[\int \cos \left (d x +c \right )^{3} \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )d x\]
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\[ \int \cos ^3(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{3} \,d x } \]
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Timed out. \[ \int \cos ^3(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
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\[ \int \cos ^3(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{3} \,d x } \]
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\[ \int \cos ^3(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{3} \,d x } \]
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Timed out. \[ \int \cos ^3(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^3\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]
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