Integrand size = 21, antiderivative size = 356 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/3} \, dx=\frac {3 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {21 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{10 d (1+\sec (c+d x))}-\frac {7\ 3^{3/4} a \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) (a+a \sec (c+d x))^{2/3} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{10 \sqrt [3]{2} d (1-\sec (c+d x)) (1+\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \]
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Time = 0.40 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3913, 3912, 52, 65, 231} \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/3} \, dx=-\frac {7\ 3^{3/4} a \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{10 \sqrt [3]{2} d (1-\sec (c+d x)) (\sec (c+d x)+1) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}+\frac {3 a \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}+\frac {21 a \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{10 d (\sec (c+d x)+1)} \]
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Rule 52
Rule 65
Rule 231
Rule 3912
Rule 3913
Rubi steps \begin{align*} \text {integral}& = \frac {\left (a (a+a \sec (c+d x))^{2/3}\right ) \int \sec (c+d x) (1+\sec (c+d x))^{5/3} \, dx}{(1+\sec (c+d x))^{2/3}} \\ & = -\frac {\left (a (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {(1+x)^{7/6}}{\sqrt {1-x}} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}} \\ & = \frac {3 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}-\frac {\left (7 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {\sqrt [6]{1+x}}{\sqrt {1-x}} \, dx,x,\sec (c+d x)\right )}{5 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}} \\ & = \frac {3 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {21 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{10 d (1+\sec (c+d x))}-\frac {\left (7 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} (1+x)^{5/6}} \, dx,x,\sec (c+d x)\right )}{10 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}} \\ & = \frac {3 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {21 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{10 d (1+\sec (c+d x))}-\frac {\left (21 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {2-x^6}} \, dx,x,\sqrt [6]{1+\sec (c+d x)}\right )}{5 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}} \\ & = \frac {3 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {21 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{10 d (1+\sec (c+d x))}-\frac {7\ 3^{3/4} a \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) (a+a \sec (c+d x))^{2/3} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{10 \sqrt [3]{2} d (1-\sec (c+d x)) (1+\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.19 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/3} \, dx=\frac {4 \sqrt [6]{2} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x))\right ) (a (1+\sec (c+d x)))^{5/3} \tan (c+d x)}{d (1+\sec (c+d x))^{13/6}} \]
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\[\int \sec \left (d x +c \right ) \left (a +a \sec \left (d x +c \right )\right )^{\frac {5}{3}}d x\]
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\[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/3} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \sec \left (d x + c\right ) \,d x } \]
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\[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/3} \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{3}} \sec {\left (c + d x \right )}\, dx \]
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\[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/3} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \sec \left (d x + c\right ) \,d x } \]
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\[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/3} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \sec \left (d x + c\right ) \,d x } \]
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Timed out. \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/3} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/3}}{\cos \left (c+d\,x\right )} \,d x \]
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