Optimal. Leaf size=159 \[ \frac{b^2 (A b-a B) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac{\sec ^2(c+d x) (-(a A-b B) \sin (c+d x)-a B+A b)}{2 d \left (a^2-b^2\right )}-\frac{(a A+b (2 A+B)) \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac{(a A-b (2 A-B)) \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]
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Rubi [A] time = 0.289702, antiderivative size = 159, 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 = {2837, 823, 801} \[ \frac{b^2 (A b-a B) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac{\sec ^2(c+d x) (-(a A-b B) \sin (c+d x)-a B+A b)}{2 d \left (a^2-b^2\right )}-\frac{(a A+b (2 A+B)) \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac{(a A-b (2 A-B)) \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]
Antiderivative was successfully verified.
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Rule 2837
Rule 823
Rule 801
Rubi steps
\begin{align*} \int \frac{\sec ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx &=\frac{b^3 \operatorname{Subst}\left (\int \frac{A+\frac{B x}{b}}{(a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac{\sec ^2(c+d x) (A b-a B-(a A-b B) \sin (c+d x))}{2 \left (a^2-b^2\right ) d}-\frac{b \operatorname{Subst}\left (\int \frac{-a^2 A+2 A b^2-a b B-(a A-b B) x}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=-\frac{\sec ^2(c+d x) (A b-a B-(a A-b B) \sin (c+d x))}{2 \left (a^2-b^2\right ) d}-\frac{b \operatorname{Subst}\left (\int \left (\frac{(a-b) (-a A-b (2 A+B))}{2 b (a+b) (b-x)}+\frac{2 b (-A b+a B)}{(a-b) (a+b) (a+x)}+\frac{(a+b) (-a A+b (2 A-B))}{2 (a-b) b (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=-\frac{(a A+b (2 A+B)) \log (1-\sin (c+d x))}{4 (a+b)^2 d}+\frac{(a A-b (2 A-B)) \log (1+\sin (c+d x))}{4 (a-b)^2 d}+\frac{b^2 (A b-a B) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}-\frac{\sec ^2(c+d x) (A b-a B-(a A-b B) \sin (c+d x))}{2 \left (a^2-b^2\right ) d}\\ \end{align*}
Mathematica [A] time = 0.759357, size = 197, normalized size = 1.24 \[ \frac{\frac{4 b^2 (A b-a B) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}+\frac{A+B}{(a+b) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{B-A}{(a-b) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{2 (a A+b (2 A+B)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{(a+b)^2}+\frac{2 (a A+b (B-2 A)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{(a-b)^2}}{4 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.1, size = 297, normalized size = 1.9 \begin{align*}{\frac{{b}^{3}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) A}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}}}-{\frac{{b}^{2}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) aB}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}}}-{\frac{A}{d \left ( 4\,a+4\,b \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{B}{d \left ( 4\,a+4\,b \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) aA}{4\,d \left ( a+b \right ) ^{2}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) Ab}{2\,d \left ( a+b \right ) ^{2}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) Bb}{4\,d \left ( a+b \right ) ^{2}}}-{\frac{A}{d \left ( 4\,a-4\,b \right ) \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{B}{d \left ( 4\,a-4\,b \right ) \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) aA}{4\,d \left ( a-b \right ) ^{2}}}-{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) Ab}{2\,d \left ( a-b \right ) ^{2}}}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) Bb}{4\,d \left ( a-b \right ) ^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.00238, size = 236, normalized size = 1.48 \begin{align*} -\frac{\frac{4 \,{\left (B a b^{2} - A b^{3}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (A a -{\left (2 \, A - B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac{{\left (A a +{\left (2 \, A + B\right )} b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} + \frac{2 \,{\left (B a - A b +{\left (A a - B b\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 3.51595, size = 531, normalized size = 3.34 \begin{align*} \frac{2 \, B a^{3} - 2 \, A a^{2} b - 2 \, B a b^{2} + 2 \, A b^{3} - 4 \,{\left (B a b^{2} - A b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) +{\left (A a^{3} + B a^{2} b -{\left (3 \, A - 2 \, B\right )} a b^{2} -{\left (2 \, A - B\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A a^{3} + B a^{2} b -{\left (3 \, A + 2 \, B\right )} a b^{2} +{\left (2 \, A + B\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (A a^{3} - B a^{2} b - A a b^{2} + B b^{3}\right )} \sin \left (d x + c\right )}{4 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} 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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sin{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{a + b \sin{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.24874, size = 351, normalized size = 2.21 \begin{align*} -\frac{\frac{4 \,{\left (B a b^{3} - A b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} + \frac{{\left (A a + 2 \, A b + B b\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac{{\left (A a - 2 \, A b + B b\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} + \frac{2 \,{\left (B a b^{2} \sin \left (d x + c\right )^{2} - A b^{3} \sin \left (d x + c\right )^{2} + A a^{3} \sin \left (d x + c\right ) - B a^{2} b \sin \left (d x + c\right ) - A a b^{2} \sin \left (d x + c\right ) + B b^{3} \sin \left (d x + c\right ) + B a^{3} - A a^{2} b - 2 \, B a b^{2} + 2 \, A b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\sin \left (d x + c\right )^{2} - 1\right )}}}{4 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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