Integrand size = 31, antiderivative size = 165 \[ \int \frac {\sec ^m(c+d x) (A+B \sec (c+d x))}{\sqrt [3]{b \sec (c+d x)}} \, dx=-\frac {3 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4-3 m),\frac {1}{6} (10-3 m),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d (4-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d (1-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}} \]
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Time = 0.15 (sec) , antiderivative size = 165, 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} \[ \int \frac {\sec ^m(c+d x) (A+B \sec (c+d x))}{\sqrt [3]{b \sec (c+d x)}} \, dx=-\frac {3 A \sin (c+d x) \sec ^{m-1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4-3 m),\frac {1}{6} (10-3 m),\cos ^2(c+d x)\right )}{d (4-3 m) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac {3 B \sin (c+d x) \sec ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right )}{d (1-3 m) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}} \]
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Rule 20
Rule 2722
Rule 3857
Rule 3872
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [3]{\sec (c+d x)} \int \sec ^{-\frac {1}{3}+m}(c+d x) (A+B \sec (c+d x)) \, dx}{\sqrt [3]{b \sec (c+d x)}} \\ & = \frac {\left (A \sqrt [3]{\sec (c+d x)}\right ) \int \sec ^{-\frac {1}{3}+m}(c+d x) \, dx}{\sqrt [3]{b \sec (c+d x)}}+\frac {\left (B \sqrt [3]{\sec (c+d x)}\right ) \int \sec ^{\frac {2}{3}+m}(c+d x) \, dx}{\sqrt [3]{b \sec (c+d x)}} \\ & = \frac {\left (A \cos ^{\frac {2}{3}+m}(c+d x) \sec ^{1+m}(c+d x)\right ) \int \cos ^{\frac {1}{3}-m}(c+d x) \, dx}{\sqrt [3]{b \sec (c+d x)}}+\frac {\left (B \cos ^{\frac {2}{3}+m}(c+d x) \sec ^{1+m}(c+d x)\right ) \int \cos ^{-\frac {2}{3}-m}(c+d x) \, dx}{\sqrt [3]{b \sec (c+d x)}} \\ & = -\frac {3 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4-3 m),\frac {1}{6} (10-3 m),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d (4-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d (1-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^m(c+d x) (A+B \sec (c+d x))}{\sqrt [3]{b \sec (c+d x)}} \, dx=\frac {3 \csc (c+d x) \left (A (2+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1+3 m),\frac {1}{6} (5+3 m),\sec ^2(c+d x)\right )+B (-1+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\sec ^2(c+d x)\right )\right ) \sec ^m(c+d x) \sqrt {-\tan ^2(c+d x)}}{d (-1+3 m) (2+3 m) \sqrt [3]{b \sec (c+d x)}} \]
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\[\int \frac {\sec \left (d x +c \right )^{m} \left (A +B \sec \left (d x +c \right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]
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\[ \int \frac {\sec ^m(c+d x) (A+B \sec (c+d x))}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {\sec ^m(c+d x) (A+B \sec (c+d x))}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{m}{\left (c + d x \right )}}{\sqrt [3]{b \sec {\left (c + d x \right )}}}\, dx \]
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\[ \int \frac {\sec ^m(c+d x) (A+B \sec (c+d x))}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {\sec ^m(c+d x) (A+B \sec (c+d x))}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\sec ^m(c+d x) (A+B \sec (c+d x))}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \]
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