Optimal. Leaf size=182 \[ \frac{(A (n+2)+C (n+1)) \sin (c+d x) (b \sec (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},-\frac{n}{2},\frac{2-n}{2},\cos ^2(c+d x)\right )}{d n (n+2) \sqrt{\sin ^2(c+d x)}}+\frac{B \sin (c+d x) (b \sec (c+d x))^{n+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (-n-1),\frac{1-n}{2},\cos ^2(c+d x)\right )}{b d (n+1) \sqrt{\sin ^2(c+d x)}}+\frac{C \tan (c+d x) (b \sec (c+d x))^{n+1}}{b d (n+2)} \]
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Rubi [A] time = 0.185408, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {16, 4047, 3772, 2643, 4046} \[ \frac{(A (n+2)+C (n+1)) \sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac{1}{2},-\frac{n}{2};\frac{2-n}{2};\cos ^2(c+d x)\right )}{d n (n+2) \sqrt{\sin ^2(c+d x)}}+\frac{B \sin (c+d x) (b \sec (c+d x))^{n+1} \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-n-1);\frac{1-n}{2};\cos ^2(c+d x)\right )}{b d (n+1) \sqrt{\sin ^2(c+d x)}}+\frac{C \tan (c+d x) (b \sec (c+d x))^{n+1}}{b d (n+2)} \]
Antiderivative was successfully verified.
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Rule 16
Rule 4047
Rule 3772
Rule 2643
Rule 4046
Rubi steps
\begin{align*} \int \sec (c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{\int (b \sec (c+d x))^{1+n} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b}\\ &=\frac{\int (b \sec (c+d x))^{1+n} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b}+\frac{B \int (b \sec (c+d x))^{2+n} \, dx}{b^2}\\ &=\frac{C (b \sec (c+d x))^{1+n} \tan (c+d x)}{b d (2+n)}+\frac{\left (A+\frac{C (1+n)}{2+n}\right ) \int (b \sec (c+d x))^{1+n} \, dx}{b}+\frac{\left (B \left (\frac{\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n\right ) \int \left (\frac{\cos (c+d x)}{b}\right )^{-2-n} \, dx}{b^2}\\ &=\frac{B \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-1-n);\frac{1-n}{2};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{1+n} \sin (c+d x)}{b d (1+n) \sqrt{\sin ^2(c+d x)}}+\frac{C (b \sec (c+d x))^{1+n} \tan (c+d x)}{b d (2+n)}+\frac{\left (\left (A+\frac{C (1+n)}{2+n}\right ) \left (\frac{\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n\right ) \int \left (\frac{\cos (c+d x)}{b}\right )^{-1-n} \, dx}{b}\\ &=\frac{\left (A+\frac{C (1+n)}{2+n}\right ) \, _2F_1\left (\frac{1}{2},-\frac{n}{2};\frac{2-n}{2};\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d n \sqrt{\sin ^2(c+d x)}}+\frac{B \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-1-n);\frac{1-n}{2};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{1+n} \sin (c+d x)}{b d (1+n) \sqrt{\sin ^2(c+d x)}}+\frac{C (b \sec (c+d x))^{1+n} \tan (c+d x)}{b d (2+n)}\\ \end{align*}
Mathematica [C] time = 5.17132, size = 460, normalized size = 2.53 \[ -\frac{i 2^{n+2} e^{i (c-d n x)} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \left (1+e^{2 i (c+d x)}\right )^n \sec ^{-n-2}(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{A e^{i d (n+1) x} \text{Hypergeometric2F1}\left (\frac{n+1}{2},n+3,\frac{n+3}{2},-e^{2 i (c+d x)}\right )}{n+1}+\frac{2 A e^{i (2 c+d (n+3) x)} \text{Hypergeometric2F1}\left (\frac{n+3}{2},n+3,\frac{n+5}{2},-e^{2 i (c+d x)}\right )}{n+3}+\frac{A e^{i (4 c+d (n+5) x)} \text{Hypergeometric2F1}\left (n+3,\frac{n+5}{2},\frac{n+7}{2},-e^{2 i (c+d x)}\right )}{n+5}+\frac{2 B e^{i (c+d (n+2) x)} \text{Hypergeometric2F1}\left (\frac{n+2}{2},n+3,\frac{n+4}{2},-e^{2 i (c+d x)}\right )}{n+2}+\frac{2 B e^{i (3 c+d (n+4) x)} \text{Hypergeometric2F1}\left (n+3,\frac{n+4}{2},\frac{n+6}{2},-e^{2 i (c+d x)}\right )}{n+4}+\frac{4 C e^{i (2 c+d (n+3) x)} \text{Hypergeometric2F1}\left (\frac{n+3}{2},n+3,\frac{n+5}{2},-e^{2 i (c+d x)}\right )}{n+3}\right )}{d (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.871, size = 0, normalized size = 0. \begin{align*} \int \sec \left ( dx+c \right ) \left ( b\sec \left ( dx+c \right ) \right ) ^{n} \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] 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 )^{n} \sec \left (d x + c\right )\,{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 (C \sec \left (d x + c\right )^{3} + B \sec \left (d x + c\right )^{2} + A \sec \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{n}, 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 )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec{\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 (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 )^{n} \sec \left (d x + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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