Integrand size = 31, antiderivative size = 90 \[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 (4 A+7 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{28 d (b \sec (c+d x))^{4/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 A b \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}} \]
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Time = 0.11 (sec) , antiderivative size = 90, 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, 4130, 3857, 2722} \[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\frac {3 A b \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}}-\frac {3 (4 A+7 C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{28 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}} \]
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Rule 16
Rule 2722
Rule 3857
Rule 4130
Rubi steps \begin{align*} \text {integral}& = b \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/3}} \, dx \\ & = \frac {3 A b \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}}+\frac {(4 A+7 C) \int \frac {1}{\sqrt [3]{b \sec (c+d x)}} \, dx}{7 b} \\ & = \frac {3 A b \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}}+\frac {\left ((4 A+7 C) \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}{7 b} \\ & = -\frac {3 (4 A+7 C) \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)}{28 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 A b \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 \cot (c+d x) \left (A \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\sec ^2(c+d x)\right )+7 C \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(c+d x)\right )\right ) \sqrt {-\tan ^2(c+d x)}}{7 b d \sqrt [3]{b \sec (c+d x)}} \]
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\[\int \frac {\cos \left (d x +c \right ) \left (A +C \sec \left (d x +c \right )^{2}\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}}d x\]
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\[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
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\[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]
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\[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
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\[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {\cos \left (c+d\,x\right )\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}} \,d x \]
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