Integrand size = 31, antiderivative size = 167 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\frac {3 A b \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1-3 m),\frac {1}{6} (5-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (1+3 m) \sqrt {\sin ^2(c+d x)}}+\frac {3 b B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-4-3 m),\frac {1}{6} (2-3 m),\cos ^2(c+d x)\right ) \sec ^{1+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (4+3 m) \sqrt {\sin ^2(c+d x)}} \]
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Time = 0.16 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {20, 3872, 3857, 2722} \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\frac {3 A b \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-3 m-1),\frac {1}{6} (5-3 m),\cos ^2(c+d x)\right )}{d (3 m+1) \sqrt {\sin ^2(c+d x)}}+\frac {3 b B \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-3 m-4),\frac {1}{6} (2-3 m),\cos ^2(c+d x)\right )}{d (3 m+4) \sqrt {\sin ^2(c+d x)}} \]
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Rule 20
Rule 2722
Rule 3857
Rule 3872
Rubi steps \begin{align*} \text {integral}& = \frac {\left (b \sqrt [3]{b \sec (c+d x)}\right ) \int \sec ^{\frac {4}{3}+m}(c+d x) (A+B \sec (c+d x)) \, dx}{\sqrt [3]{\sec (c+d x)}} \\ & = \frac {\left (A b \sqrt [3]{b \sec (c+d x)}\right ) \int \sec ^{\frac {4}{3}+m}(c+d x) \, dx}{\sqrt [3]{\sec (c+d x)}}+\frac {\left (b B \sqrt [3]{b \sec (c+d x)}\right ) \int \sec ^{\frac {7}{3}+m}(c+d x) \, dx}{\sqrt [3]{\sec (c+d x)}} \\ & = \left (A b \cos ^{\frac {1}{3}+m}(c+d x) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)}\right ) \int \cos ^{-\frac {4}{3}-m}(c+d x) \, dx+\left (b B \cos ^{\frac {1}{3}+m}(c+d x) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)}\right ) \int \cos ^{-\frac {7}{3}-m}(c+d x) \, dx \\ & = \frac {3 A b \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1-3 m),\frac {1}{6} (5-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (1+3 m) \sqrt {\sin ^2(c+d x)}}+\frac {3 b B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-4-3 m),\frac {1}{6} (2-3 m),\cos ^2(c+d x)\right ) \sec ^{1+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (4+3 m) \sqrt {\sin ^2(c+d x)}} \\ \end{align*}
Time = 0.31 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.84 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\frac {3 \csc (c+d x) \left (A (7+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4+3 m),\frac {5}{3}+\frac {m}{2},\sec ^2(c+d x)\right )+B (4+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7+3 m),\frac {1}{6} (13+3 m),\sec ^2(c+d x)\right )\right ) \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \sqrt {-\tan ^2(c+d x)}}{d (4+3 m) (7+3 m)} \]
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\[\int \sec \left (d x +c \right )^{m} \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )\right )d x\]
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\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{m} \,d x } \]
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Timed out. \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
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\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{m} \,d x } \]
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\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{m} \,d x } \]
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Timed out. \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m \,d x \]
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