Optimal. Leaf size=154 \[ \frac{3 (13 A+10 C) \sin (c+d x) (b \sec (c+d x))^{7/3} \text{Hypergeometric2F1}\left (-\frac{7}{6},\frac{1}{2},-\frac{1}{6},\cos ^2(c+d x)\right )}{91 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 B \sin (c+d x) (b \sec (c+d x))^{10/3} \text{Hypergeometric2F1}\left (-\frac{5}{3},\frac{1}{2},-\frac{2}{3},\cos ^2(c+d x)\right )}{10 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x) (b \sec (c+d x))^{10/3}}{13 b^2 d} \]
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Rubi [A] time = 0.153347, 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} \[ \frac{3 (13 A+10 C) \sin (c+d x) (b \sec (c+d x))^{7/3} \, _2F_1\left (-\frac{7}{6},\frac{1}{2};-\frac{1}{6};\cos ^2(c+d x)\right )}{91 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 B \sin (c+d x) (b \sec (c+d x))^{10/3} \, _2F_1\left (-\frac{5}{3},\frac{1}{2};-\frac{2}{3};\cos ^2(c+d x)\right )}{10 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x) (b \sec (c+d x))^{10/3}}{13 b^2 d} \]
Antiderivative was successfully verified.
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Rule 16
Rule 4047
Rule 3772
Rule 2643
Rule 4046
Rubi steps
\begin{align*} \int \sec ^2(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))^{10/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b^2}\\ &=\frac{\int (b \sec (c+d x))^{10/3} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b^2}+\frac{B \int (b \sec (c+d x))^{13/3} \, dx}{b^3}\\ &=\frac{3 C (b \sec (c+d x))^{10/3} \tan (c+d x)}{13 b^2 d}+\frac{(13 A+10 C) \int (b \sec (c+d x))^{10/3} \, dx}{13 b^2}+\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 )^{13/3}} \, dx}{b^3}\\ &=\frac{3 C (b \sec (c+d x))^{10/3} \tan (c+d x)}{13 b^2 d}+\frac{3 b B \, _2F_1\left (-\frac{5}{3},\frac{1}{2};-\frac{2}{3};\cos ^2(c+d x)\right ) \sec ^2(c+d x) \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{10 d \sqrt{\sin ^2(c+d x)}}+\frac{\left ((13 A+10 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 )^{10/3}} \, dx}{13 b^2}\\ &=\frac{3 C (b \sec (c+d x))^{10/3} \tan (c+d x)}{13 b^2 d}+\frac{3 b (13 A+10 C) \, _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)}{91 d \sqrt{\sin ^2(c+d x)}}+\frac{3 b B \, _2F_1\left (-\frac{5}{3},\frac{1}{2};-\frac{2}{3};\cos ^2(c+d x)\right ) \sec ^2(c+d x) \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{10 d \sqrt{\sin ^2(c+d x)}}\\ \end{align*}
Mathematica [C] time = 6.48562, size = 444, normalized size = 2.88 \[ \frac{3 b \csc (c) e^{-i d x} \sqrt [3]{b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (40 \sqrt [3]{2} \left (-1+e^{2 i c}\right ) (13 A+10 C) e^{2 i d x} \sqrt [3]{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt [3]{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{2}{3},\frac{5}{3},-e^{2 i (c+d x)}\right )-\frac{\left (-1+e^{2 i c}\right ) e^{-i (c-d x)} \sqrt [3]{\sec (c+d x)} \left (637 B \left (1+e^{2 i (c+d x)}\right )^{13/3} \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{3},\frac{7}{6},-e^{2 i (c+d x)}\right )+80 e^{i (c+d x)} \left (13 A \left (5 e^{2 i (c+d x)}+2 e^{4 i (c+d x)}+1\right ) \left (1+e^{2 i (c+d x)}\right )^2+2 C \left (21 e^{2 i (c+d x)}+79 e^{4 i (c+d x)}+45 e^{6 i (c+d x)}+10 e^{8 i (c+d x)}+5\right )\right )+91 B \left (-30 e^{2 i (c+d x)}+30 e^{6 i (c+d x)}+7 e^{8 i (c+d x)}-7\right )\right )}{2 \left (1+e^{2 i (c+d x)}\right )^4}\right )}{1820 d \sec ^{\frac{7}{3}}(c+d x) (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.177, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{2} \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 )^{5} + B b \sec \left (d x + c\right )^{4} + A b \sec \left (d x + c\right )^{3}\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 )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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