Integrand size = 27, antiderivative size = 457 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx=-\frac {3 (A+C) \tan (c+d x)}{11 d (a+a \sec (c+d x))^{7/3}}-\frac {3 (4 A-7 C) \tan (c+d x)}{55 a^2 d (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 \sqrt {2} A \operatorname {AppellF1}\left (-\frac {5}{6},\frac {1}{2},1,\frac {1}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{5 a^2 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}+\frac {3^{3/4} (4 A-7 C) \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{55 \sqrt [3]{2} a^2 d (1-\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \]
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Time = 0.67 (sec) , antiderivative size = 457, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {4138, 4009, 3864, 3863, 141, 3913, 3912, 53, 65, 231} \[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx=-\frac {3 \sqrt {2} A \tan (c+d x) \operatorname {AppellF1}\left (-\frac {5}{6},\frac {1}{2},1,\frac {1}{6},\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{5 a^2 d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1) \sqrt [3]{a \sec (c+d x)+a}}+\frac {3^{3/4} (4 A-7 C) \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{55 \sqrt [3]{2} a^2 d (1-\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a}}-\frac {3 (4 A-7 C) \tan (c+d x)}{55 a^2 d (\sec (c+d x)+1) \sqrt [3]{a \sec (c+d x)+a}}-\frac {3 (A+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}} \]
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Rule 53
Rule 65
Rule 141
Rule 231
Rule 3863
Rule 3864
Rule 3912
Rule 3913
Rule 4009
Rule 4138
Rubi steps \begin{align*} \text {integral}& = -\frac {3 (A+C) \tan (c+d x)}{11 d (a+a \sec (c+d x))^{7/3}}-\frac {3 \int \frac {-\frac {11 a A}{3}+\frac {1}{3} a (4 A-7 C) \sec (c+d x)}{(a+a \sec (c+d x))^{4/3}} \, dx}{11 a^2} \\ & = -\frac {3 (A+C) \tan (c+d x)}{11 d (a+a \sec (c+d x))^{7/3}}+\frac {A \int \frac {1}{(a+a \sec (c+d x))^{4/3}} \, dx}{a}-\frac {(4 A-7 C) \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{4/3}} \, dx}{11 a} \\ & = -\frac {3 (A+C) \tan (c+d x)}{11 d (a+a \sec (c+d x))^{7/3}}+\frac {\left (A \sqrt [3]{1+\sec (c+d x)}\right ) \int \frac {1}{(1+\sec (c+d x))^{4/3}} \, dx}{a^2 \sqrt [3]{a+a \sec (c+d x)}}-\frac {\left ((4 A-7 C) \sqrt [3]{1+\sec (c+d x)}\right ) \int \frac {\sec (c+d x)}{(1+\sec (c+d x))^{4/3}} \, dx}{11 a^2 \sqrt [3]{a+a \sec (c+d x)}} \\ & = -\frac {3 (A+C) \tan (c+d x)}{11 d (a+a \sec (c+d x))^{7/3}}-\frac {(A \tan (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} x (1+x)^{11/6}} \, dx,x,\sec (c+d x)\right )}{a^2 d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}+\frac {((4 A-7 C) \tan (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} (1+x)^{11/6}} \, dx,x,\sec (c+d x)\right )}{11 a^2 d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}} \\ & = -\frac {3 (A+C) \tan (c+d x)}{11 d (a+a \sec (c+d x))^{7/3}}-\frac {3 (4 A-7 C) \tan (c+d x)}{55 a^2 d (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 \sqrt {2} A \operatorname {AppellF1}\left (-\frac {5}{6},\frac {1}{2},1,\frac {1}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{5 a^2 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}+\frac {((4 A-7 C) \tan (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} (1+x)^{5/6}} \, dx,x,\sec (c+d x)\right )}{55 a^2 d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}} \\ & = -\frac {3 (A+C) \tan (c+d x)}{11 d (a+a \sec (c+d x))^{7/3}}-\frac {3 (4 A-7 C) \tan (c+d x)}{55 a^2 d (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 \sqrt {2} A \operatorname {AppellF1}\left (-\frac {5}{6},\frac {1}{2},1,\frac {1}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{5 a^2 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}+\frac {(6 (4 A-7 C) \tan (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {2-x^6}} \, dx,x,\sqrt [6]{1+\sec (c+d x)}\right )}{55 a^2 d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}} \\ & = -\frac {3 (A+C) \tan (c+d x)}{11 d (a+a \sec (c+d x))^{7/3}}-\frac {3 (4 A-7 C) \tan (c+d x)}{55 a^2 d (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 \sqrt {2} A \operatorname {AppellF1}\left (-\frac {5}{6},\frac {1}{2},1,\frac {1}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{5 a^2 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}+\frac {3^{3/4} (4 A-7 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 ) \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)}{55 \sqrt [3]{2} a^2 d (1-\sec (c+d x)) \sqrt [3]{a+a \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*}
\[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx=\int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx \]
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\[\int \frac {A +C \sec \left (d x +c \right )^{2}}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {7}{3}}}d x\]
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Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx=\text {Timed out} \]
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\[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx=\int \frac {A + C \sec ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {7}{3}}}\, dx \]
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\[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {7}{3}}} \,d x } \]
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\[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {7}{3}}} \,d x } \]
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Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{7/3}} \,d x \]
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