Integrand size = 41, antiderivative size = 230 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 b C \sec ^{2+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (7+3 m)}+\frac {3 b (C (4+3 m)+A (7+3 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1-3 m),\frac {1}{6} (5-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (1+3 m) (7+3 m) \sqrt {\sin ^2(c+d x)}}+\frac {3 b B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-4-3 m),\frac {1}{6} (2-3 m),\cos ^2(c+d x)\right ) \sec ^{1+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (4+3 m) \sqrt {\sin ^2(c+d x)}} \]
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Time = 0.24 (sec) , antiderivative size = 230, 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} \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 b (A (3 m+7)+C (3 m+4)) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-3 m-1),\frac {1}{6} (5-3 m),\cos ^2(c+d x)\right )}{d (3 m+1) (3 m+7) \sqrt {\sin ^2(c+d x)}}+\frac {3 b B \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-3 m-4),\frac {1}{6} (2-3 m),\cos ^2(c+d x)\right )}{d (3 m+4) \sqrt {\sin ^2(c+d x)}}+\frac {3 b C \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m+2}(c+d x)}{d (3 m+7)} \]
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Rule 20
Rule 2722
Rule 3857
Rule 4131
Rule 4132
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) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{\sqrt [3]{\sec (c+d x)}} \\ & = \frac {\left (b \sqrt [3]{b \sec (c+d x)}\right ) \int \sec ^{\frac {4}{3}+m}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, 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)}} \\ & = \frac {3 b C \sec ^{2+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (7+3 m)}+\frac {\left (b \left (C \left (\frac {4}{3}+m\right )+A \left (\frac {7}{3}+m\right )\right ) \sqrt [3]{b \sec (c+d x)}\right ) \int \sec ^{\frac {4}{3}+m}(c+d x) \, dx}{\left (\frac {7}{3}+m\right ) \sqrt [3]{\sec (c+d x)}}+\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 b C \sec ^{2+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (7+3 m)}+\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)}}+\frac {\left (b \left (C \left (\frac {4}{3}+m\right )+A \left (\frac {7}{3}+m\right )\right ) \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}{\frac {7}{3}+m} \\ & = \frac {3 b C \sec ^{2+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (7+3 m)}+\frac {3 b (C (4+3 m)+A (7+3 m)) \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) (7+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 = 1.48 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.89 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 b \csc (c+d x) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A \left (70+51 m+9 m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4+3 m),\frac {5}{3}+\frac {m}{2},\sec ^2(c+d x)\right )+(4+3 m) \left (C (7+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3}+\frac {m}{2},\frac {8}{3}+\frac {m}{2},\sec ^2(c+d x)\right )+B (10+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7+3 m),\frac {1}{6} (13+3 m),\sec ^2(c+d x)\right )\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (4+3 m) (7+3 m) (10+3 m)} \]
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\[\int \sec \left (d x +c \right )^{m} \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )d x\]
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\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{m} \,d x } \]
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Timed out. \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
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\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{m} \,d x } \]
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\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{m} \,d x } \]
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Timed out. \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]
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