Integrand size = 41, antiderivative size = 154 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\frac {3 (7 A+4 C) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{7 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{4/3} \sin (c+d x)}{4 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{4/3} \tan (c+d x)}{7 b^2 d} \]
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Time = 0.17 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {16, 4132, 3857, 2722, 4131} \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\frac {3 (7 A+4 C) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right )}{7 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )}{4 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{4/3}}{7 b^2 d} \]
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Rule 16
Rule 2722
Rule 3857
Rule 4131
Rule 4132
Rubi steps \begin{align*} \text {integral}& = \frac {\int (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b^2} \\ & = \frac {\int (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b^2}+\frac {B \int (b \sec (c+d x))^{7/3} \, dx}{b^3} \\ & = \frac {3 C (b \sec (c+d x))^{4/3} \tan (c+d x)}{7 b^2 d}+\frac {(7 A+4 C) \int (b \sec (c+d x))^{4/3} \, dx}{7 b^2}+\frac {\left (B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{7/3}} \, dx}{b^3} \\ & = \frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{4/3} \sin (c+d x)}{4 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{4/3} \tan (c+d x)}{7 b^2 d}+\frac {\left ((7 A+4 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{4/3}} \, dx}{7 b^2} \\ & = \frac {3 (7 A+4 C) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{7 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{4/3} \sin (c+d x)}{4 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{4/3} \tan (c+d x)}{7 b^2 d} \\ \end{align*}
Time = 0.99 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.79 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\frac {3 \csc (c+d x) \sqrt [3]{b \sec (c+d x)} \left (35 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\sec ^2(c+d x)\right )+2 \sec (c+d x) \left (10 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\sec ^2(c+d x)\right )+7 C \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\sec ^2(c+d x)\right ) \sec (c+d x)\right )\right ) \sqrt {-\tan ^2(c+d x)}}{140 b d} \]
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\[\int \frac {\sec \left (d x +c \right )^{2} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {2}{3}}}d x\]
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\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
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\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]
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\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
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\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^2\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3}} \,d x \]
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