Optimal. Leaf size=172 \[ \frac{B \sin (c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (-m-n),\frac{1}{2} (-m-n+2),\cos ^2(c+d x)\right )}{d (m+n) \sqrt{\sin ^2(c+d x)}}-\frac{A \sin (c+d x) \sec ^{m-1}(c+d x) (b \sec (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (-m-n+1),\frac{1}{2} (-m-n+3),\cos ^2(c+d x)\right )}{d (-m-n+1) \sqrt{\sin ^2(c+d x)}} \]
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Rubi [A] time = 0.110761, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {20, 3787, 3772, 2643} \[ \frac{B \sin (c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-m-n);\frac{1}{2} (-m-n+2);\cos ^2(c+d x)\right )}{d (m+n) \sqrt{\sin ^2(c+d x)}}-\frac{A \sin (c+d x) \sec ^{m-1}(c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-m-n+1);\frac{1}{2} (-m-n+3);\cos ^2(c+d x)\right )}{d (-m-n+1) \sqrt{\sin ^2(c+d x)}} \]
Antiderivative was successfully verified.
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Rule 20
Rule 3787
Rule 3772
Rule 2643
Rubi steps
\begin{align*} \int \sec ^m(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx &=\left (\sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{m+n}(c+d x) (A+B \sec (c+d x)) \, dx\\ &=\left (A \sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{m+n}(c+d x) \, dx+\left (B \sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{1+m+n}(c+d x) \, dx\\ &=\left (A \cos ^{m+n}(c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n\right ) \int \cos ^{-m-n}(c+d x) \, dx+\left (B \cos ^{m+n}(c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n\right ) \int \cos ^{-1-m-n}(c+d x) \, dx\\ &=-\frac{A \, _2F_1\left (\frac{1}{2},\frac{1}{2} (1-m-n);\frac{1}{2} (3-m-n);\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (1-m-n) \sqrt{\sin ^2(c+d x)}}+\frac{B \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-m-n);\frac{1}{2} (2-m-n);\cos ^2(c+d x)\right ) \sec ^m(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (m+n) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.213094, size = 126, normalized size = 0.73 \[ \frac{\sqrt{-\tan ^2(c+d x)} \csc (c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n \left (A (m+n+1) \cos (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+n}{2},\frac{1}{2} (m+n+2),\sec ^2(c+d x)\right )+B (m+n) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (m+n+1),\frac{1}{2} (m+n+3),\sec ^2(c+d x)\right )\right )}{d (m+n) (m+n+1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.995, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{m} \left ( b\sec \left ( dx+c \right ) \right ) ^{n} \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 )\right )^{n} \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 \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \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 (b \sec{\left (c + d x \right )}\right )^{n} \left (A + B \sec{\left (c + d x \right )}\right ) \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 )\right )^{n} \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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