Integrand size = 35, antiderivative size = 446 \[ \int (a+a \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 C (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {3 \sqrt {2} A \operatorname {AppellF1}\left (\frac {7}{6},\frac {1}{2},1,\frac {13}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{7 d \sqrt {1-\sec (c+d x)}}+\frac {3 (5 B+2 C) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{10 d (1+\sec (c+d x))}-\frac {3^{3/4} (5 B+2 C) \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) (a+a \sec (c+d x))^{2/3} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{10 \sqrt [3]{2} d (1-\sec (c+d x)) (1+\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \]
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Time = 0.86 (sec) , antiderivative size = 446, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4139, 4009, 3864, 3863, 141, 3913, 3912, 52, 65, 231} \[ \int (a+a \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \sqrt {2} A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},\frac {1}{2},1,\frac {13}{6},\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{7 d \sqrt {1-\sec (c+d x)}}-\frac {3^{3/4} (5 B+2 C) \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{10 \sqrt [3]{2} d (1-\sec (c+d x)) (\sec (c+d x)+1) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}+\frac {3 (5 B+2 C) \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{10 d (\sec (c+d x)+1)}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d} \]
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Rule 52
Rule 65
Rule 141
Rule 231
Rule 3863
Rule 3864
Rule 3912
Rule 3913
Rule 4009
Rule 4139
Rubi steps \begin{align*} \text {integral}& = \frac {3 C (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {3 \int (a+a \sec (c+d x))^{2/3} \left (\frac {5 a A}{3}+\frac {1}{3} a (5 B+2 C) \sec (c+d x)\right ) \, dx}{5 a} \\ & = \frac {3 C (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+A \int (a+a \sec (c+d x))^{2/3} \, dx+\frac {1}{5} (5 B+2 C) \int \sec (c+d x) (a+a \sec (c+d x))^{2/3} \, dx \\ & = \frac {3 C (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {\left (A (a+a \sec (c+d x))^{2/3}\right ) \int (1+\sec (c+d x))^{2/3} \, dx}{(1+\sec (c+d x))^{2/3}}+\frac {\left ((5 B+2 C) (a+a \sec (c+d x))^{2/3}\right ) \int \sec (c+d x) (1+\sec (c+d x))^{2/3} \, dx}{5 (1+\sec (c+d x))^{2/3}} \\ & = \frac {3 C (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}-\frac {\left (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} x} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}-\frac {\left ((5 B+2 C) (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 C (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {3 \sqrt {2} A \operatorname {AppellF1}\left (\frac {7}{6},\frac {1}{2},1,\frac {13}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{7 d \sqrt {1-\sec (c+d x)}}+\frac {3 (5 B+2 C) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{10 d (1+\sec (c+d x))}-\frac {\left ((5 B+2 C) (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 C (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {3 \sqrt {2} A \operatorname {AppellF1}\left (\frac {7}{6},\frac {1}{2},1,\frac {13}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{7 d \sqrt {1-\sec (c+d x)}}+\frac {3 (5 B+2 C) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{10 d (1+\sec (c+d x))}-\frac {\left (3 (5 B+2 C) (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 C (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {3 \sqrt {2} A \operatorname {AppellF1}\left (\frac {7}{6},\frac {1}{2},1,\frac {13}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{7 d \sqrt {1-\sec (c+d x)}}+\frac {3 (5 B+2 C) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{10 d (1+\sec (c+d x))}-\frac {3^{3/4} (5 B+2 C) \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*}
Leaf count is larger than twice the leaf count of optimal. \(5449\) vs. \(2(446)=892\).
Time = 21.90 (sec) , antiderivative size = 5449, normalized size of antiderivative = 12.22 \[ \int (a+a \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]
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\[\int \left (a +a \sec \left (d x +c \right )\right )^{\frac {2}{3}} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )d x\]
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Timed out. \[ \int (a+a \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
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\[ \int (a+a \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {2}{3}} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \]
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\[ \int (a+a \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \]
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\[ \int (a+a \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \]
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Timed out. \[ \int (a+a \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{2/3}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]
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