Integrand size = 33, antiderivative size = 95 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\frac {3 (8 A+5 C) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{16 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b^2 d} \]
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Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {16, 4131, 3857, 2722} \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\frac {3 (8 A+5 C) \sin (c+d x) (b \sec (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )}{16 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{5/3}}{8 b^2 d} \]
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Rule 16
Rule 2722
Rule 3857
Rule 4131
Rubi steps \begin{align*} \text {integral}& = \frac {\int (b \sec (c+d x))^{5/3} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b^2} \\ & = \frac {3 C (b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b^2 d}+\frac {(8 A+5 C) \int (b \sec (c+d x))^{5/3} \, dx}{8 b^2} \\ & = \frac {3 C (b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b^2 d}+\frac {\left ((8 A+5 C) \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{5/3}} \, dx}{8 b^2} \\ & = \frac {3 (8 A+5 C) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{16 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b^2 d} \\ \end{align*}
Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.01 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\frac {3 \csc (c+d x) \left (11 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\sec ^2(c+d x)\right ) \sec (c+d x)+5 C \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {11}{6},\frac {17}{6},\sec ^2(c+d x)\right ) \sec ^3(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{55 d \sqrt [3]{b \sec (c+d x)}} \]
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\[\int \frac {\sec \left (d x +c \right )^{2} \left (A +C \sec \left (d x +c \right )^{2}\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]
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\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\sqrt [3]{b \sec {\left (c + d x \right )}}}\, dx \]
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\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^2\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \]
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