Optimal. Leaf size=261 \[ -\frac{\sin (c+d x) \left (a^2 A (m+1)+2 a b B m+A b^2 m\right ) \sec ^{m-1}(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1-m}{2},\frac{3-m}{2},\cos ^2(c+d x)\right )}{d \left (1-m^2\right ) \sqrt{\sin ^2(c+d x)}}+\frac{\sin (c+d x) \left (a (m+2) (a B+2 A b)+b^2 B (m+1)\right ) \sec ^m(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},-\frac{m}{2},\frac{2-m}{2},\cos ^2(c+d x)\right )}{d m (m+2) \sqrt{\sin ^2(c+d x)}}+\frac{b \sin (c+d x) (a B (m+3)+A b (m+2)) \sec ^{m+1}(c+d x)}{d (m+1) (m+2)}+\frac{b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))}{d (m+2)} \]
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Rubi [A] time = 0.406208, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {4026, 4047, 3772, 2643, 4046} \[ -\frac{\sin (c+d x) \left (a^2 A (m+1)+2 a b B m+A b^2 m\right ) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1-m}{2};\frac{3-m}{2};\cos ^2(c+d x)\right )}{d \left (1-m^2\right ) \sqrt{\sin ^2(c+d x)}}+\frac{\sin (c+d x) \left (a (m+2) (a B+2 A b)+b^2 B (m+1)\right ) \sec ^m(c+d x) \, _2F_1\left (\frac{1}{2},-\frac{m}{2};\frac{2-m}{2};\cos ^2(c+d x)\right )}{d m (m+2) \sqrt{\sin ^2(c+d x)}}+\frac{b \sin (c+d x) (a B (m+3)+A b (m+2)) \sec ^{m+1}(c+d x)}{d (m+1) (m+2)}+\frac{b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))}{d (m+2)} \]
Antiderivative was successfully verified.
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Rule 4026
Rule 4047
Rule 3772
Rule 2643
Rule 4046
Rubi steps
\begin{align*} \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\frac{b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{d (2+m)}+\frac{\int \sec ^m(c+d x) \left (a (b B m+a A (2+m))+\left (b^2 B (1+m)+a (2 A b+a B) (2+m)\right ) \sec (c+d x)+b (A b (2+m)+a B (3+m)) \sec ^2(c+d x)\right ) \, dx}{2+m}\\ &=\frac{b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{d (2+m)}+\frac{\int \sec ^m(c+d x) \left (a (b B m+a A (2+m))+b (A b (2+m)+a B (3+m)) \sec ^2(c+d x)\right ) \, dx}{2+m}+\left (2 a A b+a^2 B+\frac{b^2 B (1+m)}{2+m}\right ) \int \sec ^{1+m}(c+d x) \, dx\\ &=\frac{b (A b (2+m)+a B (3+m)) \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+m) (2+m)}+\frac{b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{d (2+m)}+\frac{\left (A b^2 m+2 a b B m+a^2 A (1+m)\right ) \int \sec ^m(c+d x) \, dx}{1+m}+\left (\left (2 a A b+a^2 B+\frac{b^2 B (1+m)}{2+m}\right ) \cos ^m(c+d x) \sec ^m(c+d x)\right ) \int \cos ^{-1-m}(c+d x) \, dx\\ &=\frac{b (A b (2+m)+a B (3+m)) \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+m) (2+m)}+\frac{b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{d (2+m)}+\frac{\left (2 a A b+a^2 B+\frac{b^2 B (1+m)}{2+m}\right ) \, _2F_1\left (\frac{1}{2},-\frac{m}{2};\frac{2-m}{2};\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d m \sqrt{\sin ^2(c+d x)}}+\frac{\left (\left (A b^2 m+2 a b B m+a^2 A (1+m)\right ) \cos ^m(c+d x) \sec ^m(c+d x)\right ) \int \cos ^{-m}(c+d x) \, dx}{1+m}\\ &=\frac{b (A b (2+m)+a B (3+m)) \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+m) (2+m)}+\frac{b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{d (2+m)}-\frac{\left (A b^2 m+2 a b B m+a^2 A (1+m)\right ) \, _2F_1\left (\frac{1}{2},\frac{1-m}{2};\frac{3-m}{2};\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d \left (1-m^2\right ) \sqrt{\sin ^2(c+d x)}}+\frac{\left (2 a A b+a^2 B+\frac{b^2 B (1+m)}{2+m}\right ) \, _2F_1\left (\frac{1}{2},-\frac{m}{2};\frac{2-m}{2};\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d m \sqrt{\sin ^2(c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.963626, size = 239, normalized size = 0.92 \[ \frac{\sqrt{-\tan ^2(c+d x)} \csc (c+d x) \sec ^{m+2}(c+d x) \left (a^2 A \left (m^3+6 m^2+11 m+6\right ) \cos ^3(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m}{2},\frac{m+2}{2},\sec ^2(c+d x)\right )+a m \left (m^2+5 m+6\right ) (a B+2 A b) \cos ^2(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},\sec ^2(c+d x)\right )+b m (m+1) \left ((m+3) (2 a B+A b) \cos (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},\sec ^2(c+d x)\right )+b B (m+2) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+3}{2},\frac{m+5}{2},\sec ^2(c+d x)\right )\right )\right )}{d m (m+1) (m+2) (m+3)} \]
Antiderivative was successfully verified.
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Maple [F] time = 1.058, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{m} \left ( a+b\sec \left ( dx+c \right ) \right ) ^{2} \left ( A+B\sec \left ( dx+c \right ) \right ) \, 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{\left (B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B b^{2} \sec \left (d x + c\right )^{3} + A a^{2} +{\left (2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )\right )} \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B \sec{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right )^{2} \sec ^{m}{\left (c + d x \right )}\, 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{\left (B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \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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