Optimal. Leaf size=151 \[ \frac{3 (10 A+7 C) \sin (c+d x) (b \sec (c+d x))^{4/3} \text{Hypergeometric2F1}\left (-\frac{2}{3},\frac{1}{2},\frac{1}{3},\cos ^2(c+d x)\right )}{40 d \sqrt{\sin ^2(c+d x)}}+\frac{3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \text{Hypergeometric2F1}\left (-\frac{7}{6},\frac{1}{2},-\frac{1}{6},\cos ^2(c+d x)\right )}{7 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x) (b \sec (c+d x))^{7/3}}{10 b d} \]
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Rubi [A] time = 0.154035, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {16, 4047, 3772, 2643, 4046} \[ \frac{3 (10 A+7 C) \sin (c+d x) (b \sec (c+d x))^{4/3} \, _2F_1\left (-\frac{2}{3},\frac{1}{2};\frac{1}{3};\cos ^2(c+d x)\right )}{40 d \sqrt{\sin ^2(c+d x)}}+\frac{3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \, _2F_1\left (-\frac{7}{6},\frac{1}{2};-\frac{1}{6};\cos ^2(c+d x)\right )}{7 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x) (b \sec (c+d x))^{7/3}}{10 b d} \]
Antiderivative was successfully verified.
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Rule 16
Rule 4047
Rule 3772
Rule 2643
Rule 4046
Rubi steps
\begin{align*} \int \sec (c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{\int (b \sec (c+d x))^{7/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b}\\ &=\frac{\int (b \sec (c+d x))^{7/3} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b}+\frac{B \int (b \sec (c+d x))^{10/3} \, dx}{b^2}\\ &=\frac{3 C (b \sec (c+d x))^{7/3} \tan (c+d x)}{10 b d}+\frac{(10 A+7 C) \int (b \sec (c+d x))^{7/3} \, dx}{10 b}+\frac{\left (B \sqrt [3]{\frac{\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \frac{1}{\left (\frac{\cos (c+d x)}{b}\right )^{10/3}} \, dx}{b^2}\\ &=\frac{3 C (b \sec (c+d x))^{7/3} \tan (c+d x)}{10 b d}+\frac{3 b B \, _2F_1\left (-\frac{7}{6},\frac{1}{2};-\frac{1}{6};\cos ^2(c+d x)\right ) \sec (c+d x) \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{7 d \sqrt{\sin ^2(c+d x)}}+\frac{\left ((10 A+7 C) \sqrt [3]{\frac{\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \frac{1}{\left (\frac{\cos (c+d x)}{b}\right )^{7/3}} \, dx}{10 b}\\ &=\frac{3 (10 A+7 C) \, _2F_1\left (-\frac{2}{3},\frac{1}{2};\frac{1}{3};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{4/3} \sin (c+d x)}{40 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C (b \sec (c+d x))^{7/3} \tan (c+d x)}{10 b d}+\frac{3 b B \, _2F_1\left (-\frac{7}{6},\frac{1}{2};-\frac{1}{6};\cos ^2(c+d x)\right ) \sec (c+d x) \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{7 d \sqrt{\sin ^2(c+d x)}}\\ \end{align*}
Mathematica [C] time = 6.9811, size = 465, normalized size = 3.08 \[ \frac{\frac{\cos ^4(c+d x) (b \sec (c+d x))^{7/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{3 \sec (c) \sec (c+d x) (70 A \sin (d x)+40 B \sin (c)+49 C \sin (d x))}{140 d}+\frac{3 (10 A+7 C) \tan (c)}{20 d}+\frac{3 \sec (c) \sec ^2(c+d x) (10 B \sin (d x)+7 C \sin (c))}{35 d}+\frac{24 B \csc (c) \cos (d x)}{7 d}+\frac{3 C \sec (c) \sin (d x) \sec ^3(c+d x)}{5 d}\right )}{A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C}-\frac{3 i e^{-i (c+d x)} \sqrt [3]{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} (b \sec (c+d x))^{7/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (7 \left (-1+e^{2 i c}\right ) (10 A+7 C) e^{i (c+d x)} \sqrt [3]{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{3},\frac{7}{6},-e^{2 i (c+d x)}\right )+160 B \left (-1+e^{2 i c}\right ) \sqrt [3]{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{1}{3},\frac{2}{3},-e^{2 i (c+d x)}\right )+160 B \left (1+e^{2 i (c+d x)}\right )\right )}{70\ 2^{2/3} \left (-1+e^{2 i c}\right ) d \sec ^{\frac{13}{3}}(c+d x) (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}}{b} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.155, size = 0, normalized size = 0. \begin{align*} \int \sec \left ( dx+c \right ) \left ( b\sec \left ( dx+c \right ) \right ) ^{{\frac{4}{3}}} \left ( A+B\sec \left ( dx+c \right ) +C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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 ({\left (C b \sec \left (d x + c\right )^{4} + B b \sec \left (d x + c\right )^{3} + A b \sec \left (d x + c\right )^{2}\right )} \left (b \sec \left (d x + c\right )\right )^{\frac{1}{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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{\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{4}{3}} \sec \left (d x + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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