Optimal. Leaf size=112 \[ -\frac{(a+b) (a A-b (A+2 B)) \log (1-\sin (c+d x))}{4 d}+\frac{(a-b) (a A+b (A-2 B)) \log (\sin (c+d x)+1)}{4 d}+\frac{\sec ^2(c+d x) (a+b \sin (c+d x)) ((a A+b B) \sin (c+d x)+a B+A b)}{2 d} \]
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Rubi [A] time = 0.179982, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {2837, 819, 633, 31} \[ -\frac{(a+b) (a A-b (A+2 B)) \log (1-\sin (c+d x))}{4 d}+\frac{(a-b) (a A+b (A-2 B)) \log (\sin (c+d x)+1)}{4 d}+\frac{\sec ^2(c+d x) (a+b \sin (c+d x)) ((a A+b B) \sin (c+d x)+a B+A b)}{2 d} \]
Antiderivative was successfully verified.
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Rule 2837
Rule 819
Rule 633
Rule 31
Rubi steps
\begin{align*} \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx &=\frac{b^3 \operatorname{Subst}\left (\int \frac{(a+x)^2 \left (A+\frac{B x}{b}\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\sec ^2(c+d x) (a+b \sin (c+d x)) (A b+a B+(a A+b B) \sin (c+d x))}{2 d}-\frac{b \operatorname{Subst}\left (\int \frac{-a^2 A+A b^2+2 a b B+2 b B x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{2 d}\\ &=\frac{\sec ^2(c+d x) (a+b \sin (c+d x)) (A b+a B+(a A+b B) \sin (c+d x))}{2 d}-\frac{((a-b) (a A+b (A-2 B))) \operatorname{Subst}\left (\int \frac{1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{4 d}+\frac{((a+b) (a A-b (A+2 B))) \operatorname{Subst}\left (\int \frac{1}{b-x} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=-\frac{(a+b) (a A-b (A+2 B)) \log (1-\sin (c+d x))}{4 d}+\frac{(a-b) (a A+b (A-2 B)) \log (1+\sin (c+d x))}{4 d}+\frac{\sec ^2(c+d x) (a+b \sin (c+d x)) (A b+a B+(a A+b B) \sin (c+d x))}{2 d}\\ \end{align*}
Mathematica [A] time = 1.46034, size = 174, normalized size = 1.55 \[ \frac{\left (-6 a^3 A b+4 a A b^3+2 b^4 B\right ) \tan ^2(c+d x)+\left (a^2-b^2\right ) ((a+b) (a A-b (A+2 B)) \log (1-\sin (c+d x))-(a-b) (a A+b (A-2 B)) \log (\sin (c+d x)+1))-2 \left (a^2-b^2\right ) \left (a^2 A+2 a b B+A b^2\right ) \tan (c+d x) \sec (c+d x)-2 a^3 (a B-A b) \sec ^2(c+d x)}{4 d \left (b^2-a^2\right )} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.095, size = 231, normalized size = 2.1 \begin{align*}{\frac{{a}^{2}A\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{B{a}^{2}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{Aab}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{Bab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{Bab\sin \left ( dx+c \right ) }{d}}-{\frac{Bab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{A{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{A{b}^{2}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{A{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{B{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{B{b}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.98654, size = 165, normalized size = 1.47 \begin{align*} \frac{{\left (A a^{2} - 2 \, B a b -{\left (A - 2 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A a^{2} - 2 \, B a b -{\left (A + 2 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (B a^{2} + 2 \, A a b + B b^{2} +{\left (A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.55613, size = 329, normalized size = 2.94 \begin{align*} \frac{{\left (A a^{2} - 2 \, B a b -{\left (A - 2 \, B\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A a^{2} - 2 \, B a b -{\left (A + 2 \, B\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, B a^{2} + 4 \, A a b + 2 \, B b^{2} + 2 \,{\left (A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \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 = 1.31049, size = 197, normalized size = 1.76 \begin{align*} \frac{{\left (A a^{2} - 2 \, B a b - A b^{2} + 2 \, B b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) -{\left (A a^{2} - 2 \, B a b - A b^{2} - 2 \, B b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (B b^{2} \sin \left (d x + c\right )^{2} + A a^{2} \sin \left (d x + c\right ) + 2 \, B a b \sin \left (d x + c\right ) + A b^{2} \sin \left (d x + c\right ) + B a^{2} + 2 \, A a b\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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