Optimal. Leaf size=95 \[ \frac{3 (8 A+5 C) \sin (c+d x) (b \sec (c+d x))^{2/3} \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{1}{2},\frac{2}{3},\cos ^2(c+d x)\right )}{16 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x) (b \sec (c+d x))^{5/3}}{8 b^2 d} \]
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Rubi [A] time = 0.0868797, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {16, 4046, 3772, 2643} \[ \frac{3 (8 A+5 C) \sin (c+d x) (b \sec (c+d x))^{2/3} \, _2F_1\left (-\frac{1}{3},\frac{1}{2};\frac{2}{3};\cos ^2(c+d x)\right )}{16 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x) (b \sec (c+d x))^{5/3}}{8 b^2 d} \]
Antiderivative was successfully verified.
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Rule 16
Rule 4046
Rule 3772
Rule 2643
Rubi steps
\begin{align*} \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{\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) \, _2F_1\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*}
Mathematica [C] time = 2.72516, size = 207, normalized size = 2.18 \[ \frac{3 i e^{-i (c+d x)} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{8/3} \left (A+C \sec ^2(c+d x)\right ) \left ((8 A+5 C) \left (1+e^{2 i (c+d x)}\right )^{8/3} \text{Hypergeometric2F1}\left (\frac{2}{3},\frac{5}{6},\frac{11}{6},-e^{2 i (c+d x)}\right )-5 \left (8 A \left (1+e^{2 i (c+d x)}\right )^2+C \left (14 e^{2 i (c+d x)}+5 e^{4 i (c+d x)}+1\right )\right )\right )}{20 \sqrt [3]{2} d \sec ^{\frac{5}{3}}(c+d x) \sqrt [3]{b \sec (c+d x)} (A \cos (2 (c+d x))+A+2 C)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.141, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sec \left ( dx+c \right ) \right ) ^{2} \left ( A+C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \right ){\frac{1}{\sqrt [3]{b\sec \left ( dx+c \right ) }}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \sec \left (d x + c\right )^{3} + A \sec \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{\frac{2}{3}}}{b}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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