Integrand size = 29, antiderivative size = 116 \[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\frac {3 A \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{4/3} \sin (c+d x)}{4 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/3} \sin (c+d x)}{7 b d \sqrt {\sin ^2(c+d x)}} \]
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Time = 0.13 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {16, 3872, 3857, 2722} \[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\frac {3 A \sin (c+d x) (b \sec (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )}{7 b d \sqrt {\sin ^2(c+d x)}} \]
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Rule 16
Rule 2722
Rule 3857
Rule 3872
Rubi steps \begin{align*} \text {integral}& = \frac {\int (b \sec (c+d x))^{7/3} (A+B \sec (c+d x)) \, dx}{b} \\ & = \frac {A \int (b \sec (c+d x))^{7/3} \, dx}{b}+\frac {B \int (b \sec (c+d x))^{10/3} \, dx}{b^2} \\ & = \frac {\left (A \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{7/3}} \, dx}{b}+\frac {\left (B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{10/3}} \, dx}{b^2} \\ & = \frac {3 A \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{4/3} \sin (c+d x)}{4 d \sqrt {\sin ^2(c+d x)}}+\frac {3 b B \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right ) \sec (c+d x) \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{7 d \sqrt {\sin ^2(c+d x)}} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.78 \[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\frac {3 \csc (c+d x) \left (10 A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\sec ^2(c+d x)\right )+7 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^{7/3} \sqrt {-\tan ^2(c+d x)}}{70 b d} \]
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\[\int \sec \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )\right )d x\]
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\[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right ) \,d x } \]
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\[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{\frac {4}{3}} \left (A + B \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]
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\[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right ) \,d x } \]
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\[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right ) \,d x } \]
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Timed out. \[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}}{\cos \left (c+d\,x\right )} \,d x \]
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