Integrand size = 33, antiderivative size = 455 \[ \int (a+b \cos (e+f x))^3 (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=-\frac {c^5 \left (a^3 A \left (8-6 m+m^2\right )+3 a A b^2 \left (4-5 m+m^2\right )+3 a^2 b B \left (4-5 m+m^2\right )+b^3 B \left (3-4 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5-m}{2},\frac {7-m}{2},\cos ^2(e+f x)\right ) (c \sec (e+f x))^{-5+m} \sin (e+f x)}{f (1-m) (3-m) (5-m) \sqrt {\sin ^2(e+f x)}}-\frac {c^4 \left (A b^3 (2-m)+3 a b^2 B (2-m)+3 a^2 A b (3-m)+a^3 B (3-m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4-m}{2},\frac {6-m}{2},\cos ^2(e+f x)\right ) (c \sec (e+f x))^{-4+m} \sin (e+f x)}{f (2-m) (4-m) \sqrt {\sin ^2(e+f x)}}-\frac {a c^4 \left (3 a b B (1-m)+a^2 A (2-m)-2 A b^2 m\right ) (c \sec (e+f x))^{-4+m} \tan (e+f x)}{f (1-m) (3-m)}-\frac {a^2 c^4 (a B (1-m)-A b (1+m)) \sec (e+f x) (c \sec (e+f x))^{-4+m} \tan (e+f x)}{f (1-m) (2-m)}-\frac {a A c^4 (c \sec (e+f x))^{-4+m} (b+a \sec (e+f x))^2 \tan (e+f x)}{f (1-m)} \]
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Time = 1.21 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3039, 4111, 4161, 4132, 3857, 2722, 4131} \[ \int (a+b \cos (e+f x))^3 (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=-\frac {a c^4 \tan (e+f x) \left (a^2 A (2-m)+3 a b B (1-m)-2 A b^2 m\right ) (c \sec (e+f x))^{m-4}}{f (1-m) (3-m)}-\frac {a^2 c^4 \tan (e+f x) \sec (e+f x) (a B (1-m)-A b (m+1)) (c \sec (e+f x))^{m-4}}{f (1-m) (2-m)}-\frac {c^5 \sin (e+f x) \left (a^3 A \left (m^2-6 m+8\right )+3 a^2 b B \left (m^2-5 m+4\right )+3 a A b^2 \left (m^2-5 m+4\right )+b^3 B \left (m^2-4 m+3\right )\right ) (c \sec (e+f x))^{m-5} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5-m}{2},\frac {7-m}{2},\cos ^2(e+f x)\right )}{f (1-m) (3-m) (5-m) \sqrt {\sin ^2(e+f x)}}-\frac {c^4 \sin (e+f x) \left (a^3 B (3-m)+3 a^2 A b (3-m)+3 a b^2 B (2-m)+A b^3 (2-m)\right ) (c \sec (e+f x))^{m-4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4-m}{2},\frac {6-m}{2},\cos ^2(e+f x)\right )}{f (2-m) (4-m) \sqrt {\sin ^2(e+f x)}}-\frac {a A c^4 \tan (e+f x) (a \sec (e+f x)+b)^2 (c \sec (e+f x))^{m-4}}{f (1-m)} \]
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Rule 2722
Rule 3039
Rule 3857
Rule 4111
Rule 4131
Rule 4132
Rule 4161
Rubi steps \begin{align*} \text {integral}& = c^4 \int (c \sec (e+f x))^{-4+m} (b+a \sec (e+f x))^3 (B+A \sec (e+f x)) \, dx \\ & = -\frac {a A c^4 (c \sec (e+f x))^{-4+m} (b+a \sec (e+f x))^2 \tan (e+f x)}{f (1-m)}-\frac {c^4 \int (c \sec (e+f x))^{-4+m} (b+a \sec (e+f x)) \left (-b (b B (1-m)+a A (4-m))-\left (b (A b+2 a B) (1-m)+a^2 A (2-m)\right ) \sec (e+f x)-a (a B (1-m)-A b (1+m)) \sec ^2(e+f x)\right ) \, dx}{1-m} \\ & = -\frac {a^2 c^4 (a B (1-m)-A b (1+m)) \sec (e+f x) (c \sec (e+f x))^{-4+m} \tan (e+f x)}{f (1-m) (2-m)}-\frac {a A c^4 (c \sec (e+f x))^{-4+m} (b+a \sec (e+f x))^2 \tan (e+f x)}{f (1-m)}+\frac {c^4 \int (c \sec (e+f x))^{-4+m} \left (b^2 (b B (1-m)+a A (4-m)) (2-m)+\left (A b^3 (2-m)+3 a b^2 B (2-m)+3 a^2 A b (3-m)+a^3 B (3-m)\right ) (1-m) \sec (e+f x)+a (2-m) \left (3 a b B (1-m)+a^2 A (2-m)-2 A b^2 m\right ) \sec ^2(e+f x)\right ) \, dx}{2-3 m+m^2} \\ & = -\frac {a^2 c^4 (a B (1-m)-A b (1+m)) \sec (e+f x) (c \sec (e+f x))^{-4+m} \tan (e+f x)}{f (1-m) (2-m)}-\frac {a A c^4 (c \sec (e+f x))^{-4+m} (b+a \sec (e+f x))^2 \tan (e+f x)}{f (1-m)}+\frac {\left (c^3 \left (A b^3 (2-m)+3 a b^2 B (2-m)+3 a^2 A b (3-m)+a^3 B (3-m)\right )\right ) \int (c \sec (e+f x))^{-3+m} \, dx}{2-m}+\frac {c^4 \int (c \sec (e+f x))^{-4+m} \left (b^2 (b B (1-m)+a A (4-m)) (2-m)+a (2-m) \left (3 a b B (1-m)+a^2 A (2-m)-2 A b^2 m\right ) \sec ^2(e+f x)\right ) \, dx}{2-3 m+m^2} \\ & = -\frac {a c^4 \left (3 a b B (1-m)+a^2 A (2-m)-2 A b^2 m\right ) (c \sec (e+f x))^{-4+m} \tan (e+f x)}{f (1-m) (3-m)}-\frac {a^2 c^4 (a B (1-m)-A b (1+m)) \sec (e+f x) (c \sec (e+f x))^{-4+m} \tan (e+f x)}{f (1-m) (2-m)}-\frac {a A c^4 (c \sec (e+f x))^{-4+m} (b+a \sec (e+f x))^2 \tan (e+f x)}{f (1-m)}+\frac {\left (c^4 \left (a^3 A \left (8-6 m+m^2\right )+3 a A b^2 \left (4-5 m+m^2\right )+3 a^2 b B \left (4-5 m+m^2\right )+b^3 B \left (3-4 m+m^2\right )\right )\right ) \int (c \sec (e+f x))^{-4+m} \, dx}{(1-m) (3-m)}+\frac {\left (c^3 \left (A b^3 (2-m)+3 a b^2 B (2-m)+3 a^2 A b (3-m)+a^3 B (3-m)\right ) \left (\frac {\cos (e+f x)}{c}\right )^m (c \sec (e+f x))^m\right ) \int \left (\frac {\cos (e+f x)}{c}\right )^{3-m} \, dx}{2-m} \\ & = -\frac {\left (A b^3 (2-m)+3 a b^2 B (2-m)+3 a^2 A b (3-m)+a^3 B (3-m)\right ) \cos ^4(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4-m}{2},\frac {6-m}{2},\cos ^2(e+f x)\right ) (c \sec (e+f x))^m \sin (e+f x)}{f (2-m) (4-m) \sqrt {\sin ^2(e+f x)}}-\frac {a c^4 \left (3 a b B (1-m)+a^2 A (2-m)-2 A b^2 m\right ) (c \sec (e+f x))^{-4+m} \tan (e+f x)}{f (1-m) (3-m)}-\frac {a^2 c^4 (a B (1-m)-A b (1+m)) \sec (e+f x) (c \sec (e+f x))^{-4+m} \tan (e+f x)}{f (1-m) (2-m)}-\frac {a A c^4 (c \sec (e+f x))^{-4+m} (b+a \sec (e+f x))^2 \tan (e+f x)}{f (1-m)}+\frac {\left (c^4 \left (a^3 A \left (8-6 m+m^2\right )+3 a A b^2 \left (4-5 m+m^2\right )+3 a^2 b B \left (4-5 m+m^2\right )+b^3 B \left (3-4 m+m^2\right )\right ) \left (\frac {\cos (e+f x)}{c}\right )^m (c \sec (e+f x))^m\right ) \int \left (\frac {\cos (e+f x)}{c}\right )^{4-m} \, dx}{(1-m) (3-m)} \\ & = -\frac {\left (A b^3 (2-m)+3 a b^2 B (2-m)+3 a^2 A b (3-m)+a^3 B (3-m)\right ) \cos ^4(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4-m}{2},\frac {6-m}{2},\cos ^2(e+f x)\right ) (c \sec (e+f x))^m \sin (e+f x)}{f (2-m) (4-m) \sqrt {\sin ^2(e+f x)}}-\frac {\left (a^3 A \left (8-6 m+m^2\right )+3 a A b^2 \left (4-5 m+m^2\right )+3 a^2 b B \left (4-5 m+m^2\right )+b^3 B \left (3-4 m+m^2\right )\right ) \cos ^5(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5-m}{2},\frac {7-m}{2},\cos ^2(e+f x)\right ) (c \sec (e+f x))^m \sin (e+f x)}{f (1-m) (3-m) (5-m) \sqrt {\sin ^2(e+f x)}}-\frac {a c^4 \left (3 a b B (1-m)+a^2 A (2-m)-2 A b^2 m\right ) (c \sec (e+f x))^{-4+m} \tan (e+f x)}{f (1-m) (3-m)}-\frac {a^2 c^4 (a B (1-m)-A b (1+m)) \sec (e+f x) (c \sec (e+f x))^{-4+m} \tan (e+f x)}{f (1-m) (2-m)}-\frac {a A c^4 (c \sec (e+f x))^{-4+m} (b+a \sec (e+f x))^2 \tan (e+f x)}{f (1-m)} \\ \end{align*}
Time = 1.42 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.57 \[ \int (a+b \cos (e+f x))^3 (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=\frac {\cot (e+f x) \left (\frac {b^3 B \cos ^4(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-4+m),\frac {1}{2} (-2+m),\sec ^2(e+f x)\right )}{-4+m}+\frac {b^2 (A b+3 a B) \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-3+m),\frac {1}{2} (-1+m),\sec ^2(e+f x)\right )}{-3+m}+a \left (\frac {3 b (A b+a B) \cos ^2(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2+m),\frac {m}{2},\sec ^2(e+f x)\right )}{-2+m}+a \left (\frac {(3 A b+a B) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+m),\frac {1+m}{2},\sec ^2(e+f x)\right )}{-1+m}+\frac {a A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(e+f x)\right )}{m}\right )\right )\right ) (c \sec (e+f x))^m \sqrt {-\tan ^2(e+f x)}}{f} \]
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\[\int \left (a +b \cos \left (f x +e \right )\right )^{3} \left (A +\cos \left (f x +e \right ) B \right ) \left (c \sec \left (f x +e \right )\right )^{m}d x\]
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\[ \int (a+b \cos (e+f x))^3 (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=\int { {\left (B \cos \left (f x + e\right ) + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{3} \left (c \sec \left (f x + e\right )\right )^{m} \,d x } \]
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\[ \int (a+b \cos (e+f x))^3 (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=\int \left (c \sec {\left (e + f x \right )}\right )^{m} \left (A + B \cos {\left (e + f x \right )}\right ) \left (a + b \cos {\left (e + f x \right )}\right )^{3}\, dx \]
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\[ \int (a+b \cos (e+f x))^3 (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=\int { {\left (B \cos \left (f x + e\right ) + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{3} \left (c \sec \left (f x + e\right )\right )^{m} \,d x } \]
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\[ \int (a+b \cos (e+f x))^3 (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=\int { {\left (B \cos \left (f x + e\right ) + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{3} \left (c \sec \left (f x + e\right )\right )^{m} \,d x } \]
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Timed out. \[ \int (a+b \cos (e+f x))^3 (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=\int {\left (\frac {c}{\cos \left (e+f\,x\right )}\right )}^m\,\left (A+B\,\cos \left (e+f\,x\right )\right )\,{\left (a+b\,\cos \left (e+f\,x\right )\right )}^3 \,d x \]
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