Optimal. Leaf size=183 \[ \frac{a \left (2 a^2+15 b^2\right ) \cot (c+d x)}{15 d}+\frac{b \left (3 a^2+4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac{a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{30 d}+\frac{3 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{40 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}-\frac{3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d} \]
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Rubi [A] time = 0.568896, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {2889, 3048, 3047, 3031, 3021, 2748, 3767, 8, 3770} \[ \frac{a \left (2 a^2+15 b^2\right ) \cot (c+d x)}{15 d}+\frac{b \left (3 a^2+4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac{a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{30 d}+\frac{3 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{40 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}-\frac{3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d} \]
Antiderivative was successfully verified.
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Rule 2889
Rule 3048
Rule 3047
Rule 3031
Rule 3021
Rule 2748
Rule 3767
Rule 8
Rule 3770
Rubi steps
\begin{align*} \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3 \, dx &=\int \csc ^6(c+d x) (a+b \sin (c+d x))^3 \left (1-\sin ^2(c+d x)\right ) \, dx\\ &=-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}+\frac{1}{5} \int \csc ^5(c+d x) (a+b \sin (c+d x))^2 \left (3 b-a \sin (c+d x)-4 b \sin ^2(c+d x)\right ) \, dx\\ &=-\frac{3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}+\frac{1}{20} \int \csc ^4(c+d x) (a+b \sin (c+d x)) \left (-2 \left (2 a^2-3 b^2\right )-11 a b \sin (c+d x)-13 b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac{a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{30 d}-\frac{3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}-\frac{1}{60} \int \csc ^3(c+d x) \left (9 b \left (5 a^2-2 b^2\right )+4 a \left (2 a^2+15 b^2\right ) \sin (c+d x)+39 b^3 \sin ^2(c+d x)\right ) \, dx\\ &=\frac{3 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{40 d}+\frac{a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{30 d}-\frac{3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}-\frac{1}{120} \int \csc ^2(c+d x) \left (8 a \left (2 a^2+15 b^2\right )+15 b \left (3 a^2+4 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=\frac{3 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{40 d}+\frac{a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{30 d}-\frac{3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}-\frac{1}{8} \left (b \left (3 a^2+4 b^2\right )\right ) \int \csc (c+d x) \, dx-\frac{1}{15} \left (a \left (2 a^2+15 b^2\right )\right ) \int \csc ^2(c+d x) \, dx\\ &=\frac{b \left (3 a^2+4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac{3 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{40 d}+\frac{a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{30 d}-\frac{3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}+\frac{\left (a \left (2 a^2+15 b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{15 d}\\ &=\frac{b \left (3 a^2+4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac{a \left (2 a^2+15 b^2\right ) \cot (c+d x)}{15 d}+\frac{3 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{40 d}+\frac{a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{30 d}-\frac{3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}\\ \end{align*}
Mathematica [A] time = 1.24525, size = 344, normalized size = 1.88 \[ \frac{32 \left (2 a^3+15 a b^2\right ) \cot \left (\frac{1}{2} (c+d x)\right )+30 \left (3 a^2 b-4 b^3\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )+a \csc ^4\left (\frac{1}{2} (c+d x)\right ) \left (\left (a^2-60 b^2\right ) \sin (c+d x)-45 a b\right )+45 a^2 b \sec ^4\left (\frac{1}{2} (c+d x)\right )-90 a^2 b \sec ^2\left (\frac{1}{2} (c+d x)\right )-360 a^2 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+360 a^2 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-64 a^3 \tan \left (\frac{1}{2} (c+d x)\right )-16 a^3 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)-3 a^3 \sin (c+d x) \csc ^6\left (\frac{1}{2} (c+d x)\right )+6 a^3 \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right )-480 a b^2 \tan \left (\frac{1}{2} (c+d x)\right )+960 a b^2 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)+120 b^3 \sec ^2\left (\frac{1}{2} (c+d x)\right )-480 b^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+480 b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{960 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.105, size = 227, normalized size = 1.2 \begin{align*} -{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{2\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{3\,{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{3\,{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,{a}^{2}b\cos \left ( dx+c \right ) }{8\,d}}-{\frac{3\,{a}^{2}b\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{8\,d}}-{\frac{a{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{3}\cos \left ( dx+c \right ) }{2\,d}}-{\frac{{b}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.16253, size = 212, normalized size = 1.16 \begin{align*} -\frac{45 \, a^{2} b{\left (\frac{2 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 60 \, b^{3}{\left (\frac{2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac{240 \, a b^{2}}{\tan \left (d x + c\right )^{3}} + \frac{16 \,{\left (5 \, \tan \left (d x + c\right )^{2} + 3\right )} a^{3}}{\tan \left (d x + c\right )^{5}}}{240 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.53561, size = 675, normalized size = 3.69 \begin{align*} \frac{16 \,{\left (2 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} - 80 \,{\left (a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 15 \,{\left ({\left (3 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, a^{2} b + 4 \, b^{3} - 2 \,{\left (3 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 15 \,{\left ({\left (3 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, a^{2} b + 4 \, b^{3} - 2 \,{\left (3 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 30 \,{\left ({\left (3 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{3} +{\left (3 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\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 = 1.27483, size = 392, normalized size = 2.14 \begin{align*} \frac{6 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 45 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 10 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 120 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 120 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 60 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 360 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 120 \,{\left (3 \, a^{2} b + 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{822 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1096 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 60 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 360 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 120 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 10 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 120 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 45 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}}}{960 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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