Optimal. Leaf size=182 \[ \frac{a^3 b^2 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}+\frac{\sec ^4(c+d x) (a-b \sin (c+d x))}{4 d \left (a^2-b^2\right )}-\frac{\sec ^2(c+d x) \left (4 a^3-b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac{b (3 a+b) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac{b (3 a-b) \log (\sin (c+d x)+1)}{16 d (a-b)^3} \]
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Rubi [A] time = 0.372706, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2837, 12, 1647, 823, 801} \[ \frac{a^3 b^2 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}+\frac{\sec ^4(c+d x) (a-b \sin (c+d x))}{4 d \left (a^2-b^2\right )}-\frac{\sec ^2(c+d x) \left (4 a^3-b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac{b (3 a+b) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac{b (3 a-b) \log (\sin (c+d x)+1)}{16 d (a-b)^3} \]
Antiderivative was successfully verified.
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Rule 2837
Rule 12
Rule 1647
Rule 823
Rule 801
Rubi steps
\begin{align*} \int \frac{\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{x^3}{b^3 (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^2 \operatorname{Subst}\left (\int \frac{x^3}{(a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}+\frac{\operatorname{Subst}\left (\int \frac{\frac{a b^4}{a^2-b^2}-\frac{b^2 \left (4 a^2-b^2\right ) x}{a^2-b^2}}{(a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=\frac{\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac{\sec ^2(c+d x) \left (\frac{4 a^3}{a^2-b^2}-\frac{b \left (5 a^2-b^2\right ) \sin (c+d x)}{a^2-b^2}\right )}{8 \left (a^2-b^2\right ) d}-\frac{\operatorname{Subst}\left (\int \frac{\frac{a b^4 \left (3 a^2+b^2\right )}{a^2-b^2}-\frac{b^4 \left (5 a^2-b^2\right ) x}{a^2-b^2}}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 b^2 \left (a^2-b^2\right ) d}\\ &=\frac{\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac{\sec ^2(c+d x) \left (\frac{4 a^3}{a^2-b^2}-\frac{b \left (5 a^2-b^2\right ) \sin (c+d x)}{a^2-b^2}\right )}{8 \left (a^2-b^2\right ) d}-\frac{\operatorname{Subst}\left (\int \left (-\frac{b^3 (-a+b) (3 a+b)}{2 (a+b)^2 (b-x)}-\frac{8 a^3 b^4}{(a-b)^2 (a+b)^2 (a+x)}+\frac{(3 a-b) b^3 (a+b)}{2 (a-b)^2 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^2 \left (a^2-b^2\right ) d}\\ &=\frac{b (3 a+b) \log (1-\sin (c+d x))}{16 (a+b)^3 d}-\frac{(3 a-b) b \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac{a^3 b^2 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}+\frac{\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac{\sec ^2(c+d x) \left (\frac{4 a^3}{a^2-b^2}-\frac{b \left (5 a^2-b^2\right ) \sin (c+d x)}{a^2-b^2}\right )}{8 \left (a^2-b^2\right ) d}\\ \end{align*}
Mathematica [A] time = 1.43084, size = 166, normalized size = 0.91 \[ \frac{\frac{16 a^3 b^2 \log (a+b \sin (c+d x))}{(a-b)^3 (a+b)^3}+\frac{3 a+b}{(a+b)^2 (\sin (c+d x)-1)}+\frac{b-3 a}{(a-b)^2 (\sin (c+d x)+1)}+\frac{1}{(a+b) (\sin (c+d x)-1)^2}+\frac{1}{(a-b) (\sin (c+d x)+1)^2}+\frac{b (3 a+b) \log (1-\sin (c+d x))}{(a+b)^3}-\frac{b (3 a-b) \log (\sin (c+d x)+1)}{(a-b)^3}}{16 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.082, size = 261, normalized size = 1.4 \begin{align*}{\frac{{a}^{3}{b}^{2}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}}}+{\frac{1}{2\,d \left ( 8\,a+8\,b \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}+{\frac{3\,a}{16\,d \left ( a+b \right ) ^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{b}{16\,d \left ( a+b \right ) ^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{3\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) ab}{16\,d \left ( a+b \right ) ^{3}}}+{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ){b}^{2}}{16\,d \left ( a+b \right ) ^{3}}}+{\frac{1}{2\,d \left ( 8\,a-8\,b \right ) \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,a}{16\,d \left ( a-b \right ) ^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{b}{16\,d \left ( a-b \right ) ^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{3\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) ab}{16\,d \left ( a-b \right ) ^{3}}}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ){b}^{2}}{16\,d \left ( a-b \right ) ^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.00847, size = 360, normalized size = 1.98 \begin{align*} \frac{\frac{16 \, a^{3} b^{2} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac{{\left (3 \, a b - b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{{\left (3 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac{2 \,{\left (4 \, a^{3} \sin \left (d x + c\right )^{2} -{\left (5 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{3} - 2 \, a^{3} - 2 \, a b^{2} +{\left (3 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.41788, size = 579, normalized size = 3.18 \begin{align*} \frac{16 \, a^{3} b^{2} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) -{\left (3 \, a^{4} b + 8 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (3 \, a^{4} b - 8 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{5} - 8 \, a^{3} b^{2} + 4 \, a b^{4} - 8 \,{\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} -{\left (5 \, a^{4} b - 6 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \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 = 1.25635, size = 440, normalized size = 2.42 \begin{align*} \frac{\frac{16 \, a^{3} b^{3} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac{{\left (3 \, a b - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{{\left (3 \, a b + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac{2 \,{\left (6 \, a^{3} b^{2} \sin \left (d x + c\right )^{4} - 5 \, a^{4} b \sin \left (d x + c\right )^{3} + 6 \, a^{2} b^{3} \sin \left (d x + c\right )^{3} - b^{5} \sin \left (d x + c\right )^{3} + 4 \, a^{5} \sin \left (d x + c\right )^{2} - 16 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} + 3 \, a^{4} b \sin \left (d x + c\right ) - 2 \, a^{2} b^{3} \sin \left (d x + c\right ) - b^{5} \sin \left (d x + c\right ) - 2 \, a^{5} + 6 \, a^{3} b^{2} + 2 \, a b^{4}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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