Optimal. Leaf size=154 \[ \frac{3 (8 A+5 C) \sin (c+d x) (b \sec (c+d x))^{2/3} \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{1}{2},\frac{2}{3},\cos ^2(c+d x)\right )}{16 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 B \sin (c+d x) (b \sec (c+d x))^{5/3} \text{Hypergeometric2F1}\left (-\frac{5}{6},\frac{1}{2},\frac{1}{6},\cos ^2(c+d x)\right )}{5 b^3 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x) (b \sec (c+d x))^{5/3}}{8 b^3 d} \]
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Rubi [A] time = 0.150722, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {16, 4047, 3772, 2643, 4046} \[ \frac{3 (8 A+5 C) \sin (c+d x) (b \sec (c+d x))^{2/3} \, _2F_1\left (-\frac{1}{3},\frac{1}{2};\frac{2}{3};\cos ^2(c+d x)\right )}{16 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 B \sin (c+d x) (b \sec (c+d x))^{5/3} \, _2F_1\left (-\frac{5}{6},\frac{1}{2};\frac{1}{6};\cos ^2(c+d x)\right )}{5 b^3 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x) (b \sec (c+d x))^{5/3}}{8 b^3 d} \]
Antiderivative was successfully verified.
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Rule 16
Rule 4047
Rule 3772
Rule 2643
Rule 4046
Rubi steps
\begin{align*} \int \frac{\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx &=\frac{\int (b \sec (c+d x))^{5/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b^3}\\ &=\frac{\int (b \sec (c+d x))^{5/3} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b^3}+\frac{B \int (b \sec (c+d x))^{8/3} \, dx}{b^4}\\ &=\frac{3 C (b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b^3 d}+\frac{(8 A+5 C) \int (b \sec (c+d x))^{5/3} \, dx}{8 b^3}+\frac{\left (B \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 )^{8/3}} \, dx}{b^4}\\ &=\frac{3 B \, _2F_1\left (-\frac{5}{6},\frac{1}{2};\frac{1}{6};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{5/3} \sin (c+d x)}{5 b^3 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C (b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b^3 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^3}\\ &=\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^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 B \, _2F_1\left (-\frac{5}{6},\frac{1}{2};\frac{1}{6};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{5/3} \sin (c+d x)}{5 b^3 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C (b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b^3 d}\\ \end{align*}
Mathematica [C] time = 6.34181, size = 699, normalized size = 4.54 \[ -\frac{6 i 2^{2/3} B \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \left (1+e^{2 i (c+d x)}\right )^{2/3} \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{2}{3},\frac{4}{3},-e^{2 i (c+d x)}\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{5 d \sec ^{\frac{2}{3}}(c+d x) (b \sec (c+d x))^{4/3} (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}+\frac{3 A \csc (c) e^{-i d x} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \left (1+e^{2 i (c+d x)}\right )^{2/3} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \text{Hypergeometric2F1}\left (\frac{2}{3},\frac{5}{6},\frac{11}{6},-e^{2 i (c+d x)}\right )-5 \sqrt [3]{1+e^{2 i (c+d x)}}\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{5 \sqrt [3]{2} d \sec ^{\frac{2}{3}}(c+d x) (b \sec (c+d x))^{4/3} (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}+\frac{3 C \csc (c) e^{-i d x} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \left (1+e^{2 i (c+d x)}\right )^{2/3} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \text{Hypergeometric2F1}\left (\frac{2}{3},\frac{5}{6},\frac{11}{6},-e^{2 i (c+d x)}\right )-5 \sqrt [3]{1+e^{2 i (c+d x)}}\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{8 \sqrt [3]{2} d \sec ^{\frac{2}{3}}(c+d x) (b \sec (c+d x))^{4/3} (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}+\frac{\left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{3 (8 A+5 C) \csc (c) \cos (d x)}{8 d}+\frac{3 \sec (c) \sec (c+d x) (8 B \sin (d x)+5 C \sin (c))}{20 d}+\frac{6 B \tan (c)}{5 d}+\frac{3 C \sec (c) \sin (d x) \sec ^2(c+d x)}{4 d}\right )}{(b \sec (c+d x))^{4/3} (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.164, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sec \left ( dx+c \right ) \right ) ^{3} \left ( A+B\sec \left ( dx+c \right ) +C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \right ) \left ( b\sec \left ( dx+c \right ) \right ) ^{-{\frac{4}{3}}}}\, 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} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{\left (b \sec \left (d x + c\right )\right )^{\frac{4}{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} + B \sec \left (d x + c\right )^{2} + A \sec \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{\frac{2}{3}}}{b^{2}}, 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 + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (b \sec{\left (c + d x \right )}\right )^{\frac{4}{3}}}\, 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} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{\left (b \sec \left (d x + c\right )\right )^{\frac{4}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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