Optimal. Leaf size=455 \[ -\frac{c^5 \sin (e+f x) \left (a^3 A \left (m^2-6 m+8\right )+3 a^2 b B \left (m^2-5 m+4\right )+3 a A b^2 \left (m^2-5 m+4\right )+b^3 B \left (m^2-4 m+3\right )\right ) (c \sec (e+f x))^{m-5} \, _2F_1\left (\frac{1}{2},\frac{5-m}{2};\frac{7-m}{2};\cos ^2(e+f x)\right )}{f (1-m) (3-m) (5-m) \sqrt{\sin ^2(e+f x)}}-\frac{c^4 \sin (e+f x) \left (3 a^2 A b (3-m)+a^3 B (3-m)+3 a b^2 B (2-m)+A b^3 (2-m)\right ) (c \sec (e+f x))^{m-4} \, _2F_1\left (\frac{1}{2},\frac{4-m}{2};\frac{6-m}{2};\cos ^2(e+f x)\right )}{f (2-m) (4-m) \sqrt{\sin ^2(e+f x)}}-\frac{a c^4 \tan (e+f x) \left (a^2 A (2-m)+3 a b B (1-m)-2 A b^2 m\right ) (c \sec (e+f x))^{m-4}}{f (1-m) (3-m)}-\frac{a^2 c^4 \tan (e+f x) \sec (e+f x) (a B (1-m)-A b (m+1)) (c \sec (e+f x))^{m-4}}{f (1-m) (2-m)}-\frac{a A c^4 \tan (e+f x) (a \sec (e+f x)+b)^2 (c \sec (e+f x))^{m-4}}{f (1-m)} \]
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Rubi [A] time = 1.14548, antiderivative size = 455, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {2960, 4026, 4076, 4047, 3772, 2643, 4046} \[ -\frac{c^5 \sin (e+f x) \left (a^3 A \left (m^2-6 m+8\right )+3 a^2 b B \left (m^2-5 m+4\right )+3 a A b^2 \left (m^2-5 m+4\right )+b^3 B \left (m^2-4 m+3\right )\right ) (c \sec (e+f x))^{m-5} \, _2F_1\left (\frac{1}{2},\frac{5-m}{2};\frac{7-m}{2};\cos ^2(e+f x)\right )}{f (1-m) (3-m) (5-m) \sqrt{\sin ^2(e+f x)}}-\frac{c^4 \sin (e+f x) \left (3 a^2 A b (3-m)+a^3 B (3-m)+3 a b^2 B (2-m)+A b^3 (2-m)\right ) (c \sec (e+f x))^{m-4} \, _2F_1\left (\frac{1}{2},\frac{4-m}{2};\frac{6-m}{2};\cos ^2(e+f x)\right )}{f (2-m) (4-m) \sqrt{\sin ^2(e+f x)}}-\frac{a c^4 \tan (e+f x) \left (a^2 A (2-m)+3 a b B (1-m)-2 A b^2 m\right ) (c \sec (e+f x))^{m-4}}{f (1-m) (3-m)}-\frac{a^2 c^4 \tan (e+f x) \sec (e+f x) (a B (1-m)-A b (m+1)) (c \sec (e+f x))^{m-4}}{f (1-m) (2-m)}-\frac{a A c^4 \tan (e+f x) (a \sec (e+f x)+b)^2 (c \sec (e+f x))^{m-4}}{f (1-m)} \]
Antiderivative was successfully verified.
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Rule 2960
Rule 4026
Rule 4076
Rule 4047
Rule 3772
Rule 2643
Rule 4046
Rubi steps
\begin{align*} \int (a+b \cos (e+f x))^3 (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx &=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) \, _2F_1\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) \, _2F_1\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) \, _2F_1\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*}
Mathematica [A] time = 2.16282, size = 259, normalized size = 0.57 \[ \frac{\sqrt{-\tan ^2(e+f x)} \cot (e+f x) (c \sec (e+f x))^m \left (\frac{b^2 (3 a B+A b) \cos ^3(e+f x) \, _2F_1\left (\frac{1}{2},\frac{m-3}{2};\frac{m-1}{2};\sec ^2(e+f x)\right )}{m-3}+a \left (\frac{3 b (a B+A b) \cos ^2(e+f x) \, _2F_1\left (\frac{1}{2},\frac{m-2}{2};\frac{m}{2};\sec ^2(e+f x)\right )}{m-2}+a \left (\frac{(a B+3 A b) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{m-1}{2};\frac{m+1}{2};\sec ^2(e+f x)\right )}{m-1}+\frac{a A \, _2F_1\left (\frac{1}{2},\frac{m}{2};\frac{m+2}{2};\sec ^2(e+f x)\right )}{m}\right )\right )+\frac{b^3 B \cos ^4(e+f x) \, _2F_1\left (\frac{1}{2},\frac{m-4}{2};\frac{m-2}{2};\sec ^2(e+f x)\right )}{m-4}\right )}{f} \]
Antiderivative was successfully verified.
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Maple [F] time = 2.351, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\cos \left ( fx+e \right ) \right ) ^{3} \left ( A+B\cos \left ( fx+e \right ) \right ) \left ( c\sec \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B b^{3} \cos \left (f x + e\right )^{4} + A a^{3} +{\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (B a^{2} b + A a b^{2}\right )} \cos \left (f x + e\right )^{2} +{\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (f x + e\right )\right )} \left (c \sec \left (f x + e\right )\right )^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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