Optimal. Leaf size=154 \[ -\frac{3 (5 A+2 C) \sin (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{2},\frac{7}{6},\cos ^2(c+d x)\right )}{5 b d \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}+\frac{3 B \sin (c+d x) (b \sec (c+d x))^{2/3} \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{1}{2},\frac{2}{3},\cos ^2(c+d x)\right )}{2 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x) (b \sec (c+d x))^{2/3}}{5 b^2 d} \]
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Rubi [A] time = 0.150623, 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} \[ -\frac{3 (5 A+2 C) \sin (c+d x) \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};\cos ^2(c+d x)\right )}{5 b d \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}+\frac{3 B \sin (c+d x) (b \sec (c+d x))^{2/3} \, _2F_1\left (-\frac{1}{3},\frac{1}{2};\frac{2}{3};\cos ^2(c+d x)\right )}{2 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x) (b \sec (c+d x))^{2/3}}{5 b^2 d} \]
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 ^2(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))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b^2}\\ &=\frac{\int (b \sec (c+d x))^{2/3} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b^2}+\frac{B \int (b \sec (c+d x))^{5/3} \, dx}{b^3}\\ &=\frac{3 C (b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b^2 d}+\frac{(5 A+2 C) \int (b \sec (c+d x))^{2/3} \, dx}{5 b^2}+\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 )^{5/3}} \, dx}{b^3}\\ &=\frac{3 B \, _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)}{2 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C (b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b^2 d}+\frac{\left ((5 A+2 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 )^{2/3}} \, dx}{5 b^2}\\ &=\frac{3 B \, _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)}{2 b^2 d \sqrt{\sin ^2(c+d x)}}-\frac{3 (5 A+2 C) \cos (c+d x) \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{5 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C (b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b^2 d}\\ \end{align*}
Mathematica [C] time = 1.66668, size = 299, normalized size = 1.94 \[ \frac{\left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{3 (5 B \csc (c) \cos (d x)+2 C \tan (c+d x))}{d}-\frac{3 i 2^{2/3} e^{-i (c+d x)} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \left (\left (-1+e^{2 i c}\right ) (5 A+2 C) e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{2/3} \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{2}{3},\frac{4}{3},-e^{2 i (c+d x)}\right )+5 B \left (-1+e^{2 i c}\right ) \left (1+e^{2 i (c+d x)}\right )^{2/3} \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{2}{3},\frac{5}{6},-e^{2 i (c+d x)}\right )+5 B \left (1+e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) d \sec ^{\frac{2}{3}}(c+d x)}\right )}{5 (b \sec (c+d x))^{4/3} (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.151, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sec \left ( dx+c \right ) \right ) ^{2} \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 )^{2}}{\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 )^{2} + B \sec \left (d x + c\right ) + A\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 ^{2}{\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 )^{2}}{\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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