Optimal. Leaf size=225 \[ -\frac{3 (A (3 m+4)+3 C m+C) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m-1}(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{6} (2-3 m),\frac{1}{6} (8-3 m),\cos ^2(c+d x)\right )}{d (2-3 m) (3 m+4) \sqrt{\sin ^2(c+d x)}}+\frac{3 B \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^m(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{6} (-3 m-1),\frac{1}{6} (5-3 m),\cos ^2(c+d x)\right )}{d (3 m+1) \sqrt{\sin ^2(c+d x)}}+\frac{3 C \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m+1}(c+d x)}{d (3 m+4)} \]
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Rubi [A] time = 0.188582, antiderivative size = 225, 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} \[ -\frac{3 (A (3 m+4)+3 C m+C) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m-1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (2-3 m);\frac{1}{6} (8-3 m);\cos ^2(c+d x)\right )}{d (2-3 m) (3 m+4) \sqrt{\sin ^2(c+d x)}}+\frac{3 B \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^m(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (-3 m-1);\frac{1}{6} (5-3 m);\cos ^2(c+d x)\right )}{d (3 m+1) \sqrt{\sin ^2(c+d x)}}+\frac{3 C \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m+1}(c+d x)}{d (3 m+4)} \]
Antiderivative was successfully verified.
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Rule 20
Rule 4047
Rule 3772
Rule 2643
Rule 4046
Rubi steps
\begin{align*} \int \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{\sqrt [3]{b \sec (c+d x)} \int \sec ^{\frac{1}{3}+m}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{\sqrt [3]{\sec (c+d x)}}\\ &=\frac{\sqrt [3]{b \sec (c+d x)} \int \sec ^{\frac{1}{3}+m}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx}{\sqrt [3]{\sec (c+d x)}}+\frac{\left (B \sqrt [3]{b \sec (c+d x)}\right ) \int \sec ^{\frac{4}{3}+m}(c+d x) \, dx}{\sqrt [3]{\sec (c+d x)}}\\ &=\frac{3 C \sec ^{1+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (4+3 m)}+\frac{\left (\left (C \left (\frac{1}{3}+m\right )+A \left (\frac{4}{3}+m\right )\right ) \sqrt [3]{b \sec (c+d x)}\right ) \int \sec ^{\frac{1}{3}+m}(c+d x) \, dx}{\left (\frac{4}{3}+m\right ) \sqrt [3]{\sec (c+d x)}}+\left (B \cos ^{\frac{1}{3}+m}(c+d x) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)}\right ) \int \cos ^{-\frac{4}{3}-m}(c+d x) \, dx\\ &=\frac{3 C \sec ^{1+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (4+3 m)}+\frac{3 B \, _2F_1\left (\frac{1}{2},\frac{1}{6} (-1-3 m);\frac{1}{6} (5-3 m);\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (1+3 m) \sqrt{\sin ^2(c+d x)}}+\frac{\left (\left (C \left (\frac{1}{3}+m\right )+A \left (\frac{4}{3}+m\right )\right ) \cos ^{\frac{1}{3}+m}(c+d x) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)}\right ) \int \cos ^{-\frac{1}{3}-m}(c+d x) \, dx}{\frac{4}{3}+m}\\ &=\frac{3 C \sec ^{1+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (4+3 m)}-\frac{3 (C+3 C m+A (4+3 m)) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (2-3 m);\frac{1}{6} (8-3 m);\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (2-3 m) (4+3 m) \sqrt{\sin ^2(c+d x)}}+\frac{3 B \, _2F_1\left (\frac{1}{2},\frac{1}{6} (-1-3 m);\frac{1}{6} (5-3 m);\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (1+3 m) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}
Mathematica [C] time = 7.23119, size = 494, normalized size = 2.2 \[ -\frac{3 i 2^{m+\frac{4}{3}} e^{-\frac{1}{3} i d (3 m+1) x} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m+\frac{1}{3}} \left (1+e^{2 i (c+d x)}\right )^{m+\frac{1}{3}} \sqrt [3]{b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{e^{i c} \left (\frac{e^{\frac{1}{3} i (3 c+d (3 m+7) x)} \left (2 (3 m+13) (A+2 C) \text{Hypergeometric2F1}\left (m+\frac{7}{3},\frac{1}{6} (3 m+7),\frac{1}{6} (3 m+13),-e^{2 i (c+d x)}\right )+A (3 m+7) e^{2 i (c+d x)} \text{Hypergeometric2F1}\left (m+\frac{7}{3},\frac{1}{6} (3 m+13),\frac{1}{6} (3 m+19),-e^{2 i (c+d x)}\right )\right )}{(3 m+7) (3 m+13)}+\frac{2 B e^{\frac{1}{3} i d (3 m+4) x} \text{Hypergeometric2F1}\left (m+\frac{7}{3},\frac{1}{6} (3 m+4),\frac{m}{2}+\frac{5}{3},-e^{2 i (c+d x)}\right )}{3 m+4}\right )}{d}+\frac{A e^{\frac{1}{3} i x (3 d m+d)} \text{Hypergeometric2F1}\left (m+\frac{7}{3},\frac{1}{6} (3 m+1),\frac{1}{6} (3 m+7),-e^{2 i (c+d x)}\right )}{3 d m+d}+\frac{2 B e^{\frac{1}{3} i (9 c+d (3 m+10) x)} \text{Hypergeometric2F1}\left (\frac{m}{2}+\frac{5}{3},m+\frac{7}{3},\frac{m}{2}+\frac{8}{3},-e^{2 i (c+d x)}\right )}{d (3 m+10)}\right )}{\sec ^{\frac{7}{3}}(c+d x) (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]
Warning: Unable to verify antiderivative.
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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}\sqrt [3]{b\sec \left ( dx+c \right ) } \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 \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{1}{3}} \sec \left (d x + c\right )^{m}, 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{1}{3}} \sec \left (d x + c\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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