Integrand size = 41, antiderivative size = 226 \[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\frac {3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+3 m) (b \sec (c+d x))^{2/3}}-\frac {3 (A-C (2-3 m)+3 A m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5-3 m),\frac {1}{6} (11-3 m),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d (5-3 m) (1+3 m) (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2-3 m),\frac {1}{6} (8-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d (2-3 m) (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}} \]
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Time = 0.23 (sec) , antiderivative size = 226, 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 = {20, 4132, 3857, 2722, 4131} \[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=-\frac {3 (3 A m+A-C (2-3 m)) \sin (c+d x) \sec ^{m-1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5-3 m),\frac {1}{6} (11-3 m),\cos ^2(c+d x)\right )}{d (5-3 m) (3 m+1) \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}-\frac {3 B \sin (c+d x) \sec ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2-3 m),\frac {1}{6} (8-3 m),\cos ^2(c+d x)\right )}{d (2-3 m) \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac {3 C \sin (c+d x) \sec ^{m+1}(c+d x)}{d (3 m+1) (b \sec (c+d x))^{2/3}} \]
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Rule 20
Rule 2722
Rule 3857
Rule 4131
Rule 4132
Rubi steps \begin{align*} \text {integral}& = \frac {\sec ^{\frac {2}{3}}(c+d x) \int \sec ^{-\frac {2}{3}+m}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{(b \sec (c+d x))^{2/3}} \\ & = \frac {\sec ^{\frac {2}{3}}(c+d x) \int \sec ^{-\frac {2}{3}+m}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx}{(b \sec (c+d x))^{2/3}}+\frac {\left (B \sec ^{\frac {2}{3}}(c+d x)\right ) \int \sec ^{\frac {1}{3}+m}(c+d x) \, dx}{(b \sec (c+d x))^{2/3}} \\ & = \frac {3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+3 m) (b \sec (c+d x))^{2/3}}+\frac {\left (\left (C \left (-\frac {2}{3}+m\right )+A \left (\frac {1}{3}+m\right )\right ) \sec ^{\frac {2}{3}}(c+d x)\right ) \int \sec ^{-\frac {2}{3}+m}(c+d x) \, dx}{\left (\frac {1}{3}+m\right ) (b \sec (c+d x))^{2/3}}+\frac {\left (B \cos ^{\frac {1}{3}+m}(c+d x) \sec ^{1+m}(c+d x)\right ) \int \cos ^{-\frac {1}{3}-m}(c+d x) \, dx}{(b \sec (c+d x))^{2/3}} \\ & = \frac {3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+3 m) (b \sec (c+d x))^{2/3}}-\frac {3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2-3 m),\frac {1}{6} (8-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d (2-3 m) (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}+\frac {\left (\left (C \left (-\frac {2}{3}+m\right )+A \left (\frac {1}{3}+m\right )\right ) \cos ^{\frac {1}{3}+m}(c+d x) \sec ^{1+m}(c+d x)\right ) \int \cos ^{\frac {2}{3}-m}(c+d x) \, dx}{\left (\frac {1}{3}+m\right ) (b \sec (c+d x))^{2/3}} \\ & = \frac {3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+3 m) (b \sec (c+d x))^{2/3}}-\frac {3 (A-C (2-3 m)+3 A m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5-3 m),\frac {1}{6} (11-3 m),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d (5-3 m) (1+3 m) (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2-3 m),\frac {1}{6} (8-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d (2-3 m) (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}} \\ \end{align*}
Time = 2.16 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\frac {3 b \csc (c+d x) \sec ^m(c+d x) \left (A \left (4+15 m+9 m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-2+3 m),\frac {1}{6} (4+3 m),\sec ^2(c+d x)\right )+(-2+3 m) \left (B (4+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1+3 m),\frac {1}{6} (7+3 m),\sec ^2(c+d x)\right )+C (1+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4+3 m),\frac {5}{3}+\frac {m}{2},\sec ^2(c+d x)\right )\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (-2+3 m) (1+3 m) (4+3 m) (b \sec (c+d x))^{5/3}} \]
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\[\int \frac {\sec \left (d x +c \right )^{m} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {2}{3}}}d x\]
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\[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
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\[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{m}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]
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\[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
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\[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3}} \,d x \]
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