Integrand size = 31, antiderivative size = 119 \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=-\frac {3 A b^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{7 d (b \sec (c+d x))^{7/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 b^2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 d (b \sec (c+d x))^{4/3} \sqrt {\sin ^2(c+d x)}} \]
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Time = 0.16 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {16, 3872, 3857, 2722} \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=-\frac {3 A b^3 \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}}-\frac {3 b^2 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}} \]
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Rule 16
Rule 2722
Rule 3857
Rule 3872
Rubi steps \begin{align*} \text {integral}& = b^2 \int \frac {A+B \sec (c+d x)}{(b \sec (c+d x))^{4/3}} \, dx \\ & = \left (A b^2\right ) \int \frac {1}{(b \sec (c+d x))^{4/3}} \, dx+(b B) \int \frac {1}{\sqrt [3]{b \sec (c+d x)}} \, dx \\ & = \left (A b^2 \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \left (\frac {\cos (c+d x)}{b}\right )^{4/3} \, dx+\left (b B \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \sqrt [3]{\frac {\cos (c+d x)}{b}} \, dx \\ & = -\frac {3 B \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{4 d \sqrt {\sin ^2(c+d x)}}-\frac {3 A \cos ^3(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{7 d \sqrt {\sin ^2(c+d x)}} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.74 \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=-\frac {3 b \cot (c+d x) \left (A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\sec ^2(c+d x)\right )+4 B \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(c+d x)\right )\right ) \sqrt {-\tan ^2(c+d x)}}{4 d \sqrt [3]{b \sec (c+d x)}} \]
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\[\int \cos \left (d x +c \right )^{2} \left (b \sec \left (d x +c \right )\right )^{\frac {2}{3}} \left (A +B \sec \left (d x +c \right )\right )d x\]
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\[ \int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \cos \left (d x + c\right )^{2} \,d x } \]
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Timed out. \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
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\[ \int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \cos \left (d x + c\right )^{2} \,d x } \]
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\[ \int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \cos \left (d x + c\right )^{2} \,d x } \]
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Timed out. \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\int {\cos \left (c+d\,x\right )}^2\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3} \,d x \]
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