Integrand size = 31, antiderivative size = 132 \[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {b^4 (A (2-n)+C (3-n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4-n}{2},\frac {6-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-4+n} \sin (c+d x)}{d (2-n) (4-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^3 C (b \sec (c+d x))^{-3+n} \tan (c+d x)}{d (2-n)} \]
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Time = 0.17 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {16, 4131, 3857, 2722} \[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {b^4 (A (2-n)+C (3-n)) \sin (c+d x) (b \sec (c+d x))^{n-4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4-n}{2},\frac {6-n}{2},\cos ^2(c+d x)\right )}{d (2-n) (4-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^3 C \tan (c+d x) (b \sec (c+d x))^{n-3}}{d (2-n)} \]
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Rule 16
Rule 2722
Rule 3857
Rule 4131
Rubi steps \begin{align*} \text {integral}& = b^3 \int (b \sec (c+d x))^{-3+n} \left (A+C \sec ^2(c+d x)\right ) \, dx \\ & = -\frac {b^3 C (b \sec (c+d x))^{-3+n} \tan (c+d x)}{d (2-n)}+\left (b^3 \left (A+\frac {C (3-n)}{2-n}\right )\right ) \int (b \sec (c+d x))^{-3+n} \, dx \\ & = -\frac {b^3 C (b \sec (c+d x))^{-3+n} \tan (c+d x)}{d (2-n)}+\left (b^3 \left (A+\frac {C (3-n)}{2-n}\right ) \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n\right ) \int \left (\frac {\cos (c+d x)}{b}\right )^{3-n} \, dx \\ & = -\frac {\left (A+\frac {C (3-n)}{2-n}\right ) \cos ^4(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4-n}{2},\frac {6-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (4-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^3 C (b \sec (c+d x))^{-3+n} \tan (c+d x)}{d (2-n)} \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.89 \[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {b \cot (c+d x) \left (A (-1+n) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-3+n),\frac {1}{2} (-1+n),\sec ^2(c+d x)\right )+C (-3+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^{-1+n} \sqrt {-\tan ^2(c+d x)}}{d (-3+n) (-1+n)} \]
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\[\int \cos \left (d x +c \right )^{3} \left (b \sec \left (d x +c \right )\right )^{n} \left (A +C \sec \left (d x +c \right )^{2}\right )d x\]
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\[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{3} \,d x } \]
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\[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}\, dx \]
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\[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{3} \,d x } \]
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\[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{3} \,d x } \]
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Timed out. \[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^3\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]
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