Optimal. Leaf size=434 \[ \frac {3 \sqrt {2} A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} F_1\left (\frac {7}{6};\frac {1}{2},1;\frac {13}{6};\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{7 d \sqrt {1-\sec (c+d x)}}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d (\sec (c+d x)+1)}-\frac {3^{3/4} C \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{5 \sqrt [3]{2} d (1-\sec (c+d x)) (\sec (c+d x)+1) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}} \]
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Rubi [A] time = 0.70, antiderivative size = 434, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {4055, 3924, 3779, 3778, 136, 3828, 3827, 50, 63, 225} \[ \frac {3 \sqrt {2} A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} F_1\left (\frac {7}{6};\frac {1}{2},1;\frac {13}{6};\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{7 d \sqrt {1-\sec (c+d x)}}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d (\sec (c+d x)+1)}-\frac {3^{3/4} C \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{5 \sqrt [3]{2} d (1-\sec (c+d x)) (\sec (c+d x)+1) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}} \]
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 136
Rule 225
Rule 3778
Rule 3779
Rule 3827
Rule 3828
Rule 3924
Rule 4055
Rubi steps
\begin {align*} \int (a+a \sec (c+d x))^{2/3} \left (A+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 \int (a+a \sec (c+d x))^{2/3} \left (\frac {5 a A}{3}+\frac {2}{3} a 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} (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 (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 ) \operatorname {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 (2 C (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname {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 F_1\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 C (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d (1+\sec (c+d x))}-\frac {\left (C (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} (1+x)^{5/6}} \, 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 F_1\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 C (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d (1+\sec (c+d x))}-\frac {\left (6 C (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname {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 F_1\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 C (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d (1+\sec (c+d x))}-\frac {3^{3/4} C F\left (\cos ^{-1}\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)}{5 \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*}
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Mathematica [F] time = 4.58, size = 0, normalized size = 0.00 \[ \int (a+a \sec (c+d x))^{2/3} \left (A+C \sec ^2(c+d x)\right ) \, dx \]
Verification is Not applicable to the result.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.38, size = 0, normalized size = 0.00 \[ \int \left (a +a \sec \left (d x +c \right )\right )^{\frac {2}{3}} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{2/3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {2}{3}} \left (A + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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