Optimal. Leaf size=234 \[ -\frac{3 (-3 A m+A+C (4-3 m)) \sin (c+d x) \sec ^{m-2}(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{6} (7-3 m),\frac{1}{6} (13-3 m),\cos ^2(c+d x)\right )}{b d (1-3 m) (7-3 m) \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac{3 B \sin (c+d x) \sec ^{m-1}(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{6} (4-3 m),\frac{1}{6} (10-3 m),\cos ^2(c+d x)\right )}{b d (4-3 m) \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac{3 C \sin (c+d x) \sec ^m(c+d x)}{b d (1-3 m) \sqrt [3]{b \sec (c+d x)}} \]
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Rubi [A] time = 0.204798, antiderivative size = 234, 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 = {20, 4047, 3772, 2643, 4046} \[ -\frac{3 (-3 A m+A+C (4-3 m)) \sin (c+d x) \sec ^{m-2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (7-3 m);\frac{1}{6} (13-3 m);\cos ^2(c+d x)\right )}{b d (1-3 m) (7-3 m) \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac{3 B \sin (c+d x) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (4-3 m);\frac{1}{6} (10-3 m);\cos ^2(c+d x)\right )}{b d (4-3 m) \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac{3 C \sin (c+d x) \sec ^m(c+d x)}{b d (1-3 m) \sqrt [3]{b \sec (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 20
Rule 4047
Rule 3772
Rule 2643
Rule 4046
Rubi steps
\begin{align*} \int \frac{\sec ^m(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{\sqrt [3]{\sec (c+d x)} \int \sec ^{-\frac{4}{3}+m}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b \sqrt [3]{b \sec (c+d x)}}\\ &=\frac{\sqrt [3]{\sec (c+d x)} \int \sec ^{-\frac{4}{3}+m}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx}{b \sqrt [3]{b \sec (c+d x)}}+\frac{\left (B \sqrt [3]{\sec (c+d x)}\right ) \int \sec ^{-\frac{1}{3}+m}(c+d x) \, dx}{b \sqrt [3]{b \sec (c+d x)}}\\ &=-\frac{3 C \sec ^m(c+d x) \sin (c+d x)}{b d (1-3 m) \sqrt [3]{b \sec (c+d x)}}+\frac{\left (\left (C \left (-\frac{4}{3}+m\right )+A \left (-\frac{1}{3}+m\right )\right ) \sqrt [3]{\sec (c+d x)}\right ) \int \sec ^{-\frac{4}{3}+m}(c+d x) \, dx}{b \left (-\frac{1}{3}+m\right ) \sqrt [3]{b \sec (c+d x)}}+\frac{\left (B \cos ^{\frac{2}{3}+m}(c+d x) \sec ^{1+m}(c+d x)\right ) \int \cos ^{\frac{1}{3}-m}(c+d x) \, dx}{b \sqrt [3]{b \sec (c+d x)}}\\ &=-\frac{3 C \sec ^m(c+d x) \sin (c+d x)}{b d (1-3 m) \sqrt [3]{b \sec (c+d x)}}-\frac{3 B \, _2F_1\left (\frac{1}{2},\frac{1}{6} (4-3 m);\frac{1}{6} (10-3 m);\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{b d (4-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt{\sin ^2(c+d x)}}+\frac{\left (\left (C \left (-\frac{4}{3}+m\right )+A \left (-\frac{1}{3}+m\right )\right ) \cos ^{\frac{2}{3}+m}(c+d x) \sec ^{1+m}(c+d x)\right ) \int \cos ^{\frac{4}{3}-m}(c+d x) \, dx}{b \left (-\frac{1}{3}+m\right ) \sqrt [3]{b \sec (c+d x)}}\\ &=-\frac{3 C \sec ^m(c+d x) \sin (c+d x)}{b d (1-3 m) \sqrt [3]{b \sec (c+d x)}}-\frac{3 (A (1-3 m)+C (4-3 m)) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (7-3 m);\frac{1}{6} (13-3 m);\cos ^2(c+d x)\right ) \sec ^{-2+m}(c+d x) \sin (c+d x)}{b d (1-3 m) (7-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt{\sin ^2(c+d x)}}-\frac{3 B \, _2F_1\left (\frac{1}{2},\frac{1}{6} (4-3 m);\frac{1}{6} (10-3 m);\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{b d (4-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt{\sin ^2(c+d x)}}\\ \end{align*}
Mathematica [C] time = 8.79981, size = 492, normalized size = 2.1 \[ -\frac{3 i 2^{m-\frac{1}{3}} e^{-\frac{1}{3} i (6 c+d (3 m+2) x)} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m+\frac{2}{3}} \left (1+e^{2 i (c+d x)}\right )^{m+\frac{2}{3}} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (e^{2 i c} \left (\frac{2 (A+2 C) e^{\frac{1}{3} i d (3 m+2) x} \text{Hypergeometric2F1}\left (m+\frac{2}{3},\frac{1}{6} (3 m+2),\frac{1}{6} (3 m+8),-e^{2 i (c+d x)}\right )}{3 m+2}+\frac{A e^{\frac{1}{3} i (6 c+d (3 m+8) x)} \text{Hypergeometric2F1}\left (m+\frac{2}{3},\frac{1}{6} (3 m+8),\frac{m}{2}+\frac{7}{3},-e^{2 i (c+d x)}\right )}{3 m+8}+\frac{2 B e^{\frac{1}{3} i (3 c+d (3 m+5) x)} \text{Hypergeometric2F1}\left (m+\frac{2}{3},\frac{1}{6} (3 m+5),\frac{1}{6} (3 m+11),-e^{2 i (c+d x)}\right )}{3 m+5}\right )+\frac{A e^{\frac{1}{3} i d (3 m-4) x} \text{Hypergeometric2F1}\left (m+\frac{2}{3},\frac{1}{6} (3 m-4),\frac{1}{6} (3 m+2),-e^{2 i (c+d x)}\right )}{3 m-4}+\frac{2 B e^{\frac{1}{3} i (3 c+d (3 m-1) x)} \text{Hypergeometric2F1}\left (m+\frac{2}{3},\frac{1}{6} (3 m-1),\frac{1}{6} (3 m+5),-e^{2 i (c+d x)}\right )}{3 m-1}\right )}{d \sec ^{\frac{2}{3}}(c+d x) (b \sec (c+d x))^{4/3} (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.184, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sec \left ( dx+c \right ) \right ) ^{m} \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 )^{m}}{\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}} \sec \left (d x + c\right )^{m}}{b^{2} \sec \left (d x + c\right )^{2}}, 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 \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 )^{m}}{\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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