Optimal. Leaf size=123 \[ \frac{A+B}{16 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{3 A+B}{16 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{(2 A+B) \tanh ^{-1}(\sin (c+d x))}{8 a^2 d}-\frac{a (A-B)}{12 d (a \sin (c+d x)+a)^3}-\frac{A}{8 d (a \sin (c+d x)+a)^2} \]
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Rubi [A] time = 0.154616, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {2836, 77, 206} \[ \frac{A+B}{16 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{3 A+B}{16 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{(2 A+B) \tanh ^{-1}(\sin (c+d x))}{8 a^2 d}-\frac{a (A-B)}{12 d (a \sin (c+d x)+a)^3}-\frac{A}{8 d (a \sin (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 2836
Rule 77
Rule 206
Rubi steps
\begin{align*} \int \frac{\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx &=\frac{a^3 \operatorname{Subst}\left (\int \frac{A+\frac{B x}{a}}{(a-x)^2 (a+x)^4} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^3 \operatorname{Subst}\left (\int \left (\frac{A+B}{16 a^4 (a-x)^2}+\frac{A-B}{4 a^2 (a+x)^4}+\frac{A}{4 a^3 (a+x)^3}+\frac{3 A+B}{16 a^4 (a+x)^2}+\frac{2 A+B}{8 a^4 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{a (A-B)}{12 d (a+a \sin (c+d x))^3}-\frac{A}{8 d (a+a \sin (c+d x))^2}+\frac{A+B}{16 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{3 A+B}{16 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{(2 A+B) \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{8 a d}\\ &=\frac{(2 A+B) \tanh ^{-1}(\sin (c+d x))}{8 a^2 d}-\frac{a (A-B)}{12 d (a+a \sin (c+d x))^3}-\frac{A}{8 d (a+a \sin (c+d x))^2}+\frac{A+B}{16 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{3 A+B}{16 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 0.678867, size = 87, normalized size = 0.71 \[ -\frac{\frac{3 (A+B)}{\sin (c+d x)-1}+\frac{3 (3 A+B)}{\sin (c+d x)+1}+\frac{4 (A-B)}{(\sin (c+d x)+1)^3}-6 (2 A+B) \tanh ^{-1}(\sin (c+d x))+\frac{6 A}{(\sin (c+d x)+1)^2}}{48 a^2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.138, size = 207, normalized size = 1.7 \begin{align*} -{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) A}{8\,d{a}^{2}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) B}{16\,d{a}^{2}}}-{\frac{A}{16\,d{a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{B}{16\,d{a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{A}{8\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{A}{12\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{B}{12\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) A}{8\,d{a}^{2}}}+{\frac{B\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{16\,d{a}^{2}}}-{\frac{3\,A}{16\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{B}{16\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.03721, size = 188, normalized size = 1.53 \begin{align*} -\frac{\frac{2 \,{\left (3 \,{\left (2 \, A + B\right )} \sin \left (d x + c\right )^{3} + 6 \,{\left (2 \, A + B\right )} \sin \left (d x + c\right )^{2} +{\left (2 \, A + B\right )} \sin \left (d x + c\right ) - 8 \, A + 2 \, B\right )}}{a^{2} \sin \left (d x + c\right )^{4} + 2 \, a^{2} \sin \left (d x + c\right )^{3} - 2 \, a^{2} \sin \left (d x + c\right ) - a^{2}} - \frac{3 \,{\left (2 \, A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac{3 \,{\left (2 \, A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{48 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.52442, size = 597, normalized size = 4.85 \begin{align*} \frac{12 \,{\left (2 \, A + B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left ({\left (2 \, A + B\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (2 \, A + B\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \,{\left (2 \, A + B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (2 \, A + B\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (2 \, A + B\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \,{\left (2 \, A + B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (3 \,{\left (2 \, A + B\right )} \cos \left (d x + c\right )^{2} - 8 \, A - 4 \, B\right )} \sin \left (d x + c\right ) - 8 \, A - 16 \, B}{48 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{2}\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 [A] time = 1.36114, size = 228, normalized size = 1.85 \begin{align*} \frac{\frac{6 \,{\left (2 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac{6 \,{\left (2 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac{6 \,{\left (2 \, A \sin \left (d x + c\right ) + B \sin \left (d x + c\right ) - 3 \, A - 2 \, B\right )}}{a^{2}{\left (\sin \left (d x + c\right ) - 1\right )}} - \frac{22 \, A \sin \left (d x + c\right )^{3} + 11 \, B \sin \left (d x + c\right )^{3} + 84 \, A \sin \left (d x + c\right )^{2} + 39 \, B \sin \left (d x + c\right )^{2} + 114 \, A \sin \left (d x + c\right ) + 45 \, B \sin \left (d x + c\right ) + 60 \, A + 9 \, B}{a^{2}{\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{96 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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