Optimal. Leaf size=148 \[ \frac{\left (2 a^2+5 b^2\right ) \cot (c+d x)}{15 d}+\frac{\left (a^2-2 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 d}+\frac{a b \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac{a b \cot (c+d x) \csc ^3(c+d x)}{10 d}+\frac{a b \cot (c+d x) \csc (c+d x)}{4 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2}{5 d} \]
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Rubi [A] time = 0.386162, antiderivative size = 148, 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, 3031, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac{\left (2 a^2+5 b^2\right ) \cot (c+d x)}{15 d}+\frac{\left (a^2-2 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 d}+\frac{a b \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac{a b \cot (c+d x) \csc ^3(c+d x)}{10 d}+\frac{a b \cot (c+d x) \csc (c+d x)}{4 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2}{5 d} \]
Antiderivative was successfully verified.
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Rule 2889
Rule 3048
Rule 3031
Rule 3021
Rule 2748
Rule 3768
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2 \, dx &=\int \csc ^6(c+d x) (a+b \sin (c+d x))^2 \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))^2}{5 d}+\frac{1}{5} \int \csc ^5(c+d x) (a+b \sin (c+d x)) \left (2 b-a \sin (c+d x)-3 b \sin ^2(c+d x)\right ) \, dx\\ &=-\frac{a b \cot (c+d x) \csc ^3(c+d x)}{10 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2}{5 d}-\frac{1}{20} \int \csc ^4(c+d x) \left (4 \left (a^2-2 b^2\right )+10 a b \sin (c+d x)+12 b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac{\left (a^2-2 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 d}-\frac{a b \cot (c+d x) \csc ^3(c+d x)}{10 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2}{5 d}-\frac{1}{60} \int \csc ^3(c+d x) \left (30 a b+4 \left (2 a^2+5 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=\frac{\left (a^2-2 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 d}-\frac{a b \cot (c+d x) \csc ^3(c+d x)}{10 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2}{5 d}-\frac{1}{2} (a b) \int \csc ^3(c+d x) \, dx-\frac{1}{15} \left (2 a^2+5 b^2\right ) \int \csc ^2(c+d x) \, dx\\ &=\frac{a b \cot (c+d x) \csc (c+d x)}{4 d}+\frac{\left (a^2-2 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 d}-\frac{a b \cot (c+d x) \csc ^3(c+d x)}{10 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2}{5 d}-\frac{1}{4} (a b) \int \csc (c+d x) \, dx+\frac{\left (2 a^2+5 b^2\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{15 d}\\ &=\frac{a b \tanh ^{-1}(\cos (c+d x))}{4 d}+\frac{\left (2 a^2+5 b^2\right ) \cot (c+d x)}{15 d}+\frac{a b \cot (c+d x) \csc (c+d x)}{4 d}+\frac{\left (a^2-2 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 d}-\frac{a b \cot (c+d x) \csc ^3(c+d x)}{10 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2}{5 d}\\ \end{align*}
Mathematica [A] time = 0.819497, size = 236, normalized size = 1.59 \[ \frac{\csc ^5(c+d x) \left (-40 \left (4 a^2+b^2\right ) \cos (c+d x)+20 \left (b^2-2 a^2\right ) \cos (3 (c+d x))+8 a^2 \cos (5 (c+d x))-180 a b \sin (2 (c+d x))-30 a b \sin (4 (c+d x))-150 a b \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+75 a b \sin (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-15 a b \sin (5 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+150 a b \sin (c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-75 a b \sin (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+15 a b \sin (5 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+20 b^2 \cos (5 (c+d x))\right )}{960 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.082, size = 156, normalized size = 1.1 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{2\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{ab\cos \left ( dx+c \right ) }{4\,d}}-{\frac{ab\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{4\,d}}-{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.08261, size = 146, normalized size = 0.99 \begin{align*} -\frac{15 \, a 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 )} + \frac{40 \, b^{2}}{\tan \left (d x + c\right )^{3}} + \frac{8 \,{\left (5 \, \tan \left (d x + c\right )^{2} + 3\right )} a^{2}}{\tan \left (d x + c\right )^{5}}}{120 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.39364, size = 518, normalized size = 3.5 \begin{align*} \frac{8 \,{\left (2 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 40 \,{\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{3} + 15 \,{\left (a b \cos \left (d x + c\right )^{4} - 2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 15 \,{\left (a b \cos \left (d x + c\right )^{4} - 2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 30 \,{\left (a b \cos \left (d x + c\right )^{3} + a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \,{\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 = 1.28206, size = 300, normalized size = 2.03 \begin{align*} \frac{3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 5 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 20 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 120 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 30 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 60 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{274 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 30 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 60 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 5 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 20 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 15 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}}}{480 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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