Optimal. Leaf size=140 \[ \frac{a^2 \cos ^5(c+d x)}{5 d}-\frac{a^2 \cos (c+d x)}{d}-\frac{2 a^2 \cot (c+d x)}{d}+\frac{a^2 \sin ^3(c+d x) \cos (c+d x)}{2 d}-\frac{9 a^2 \sin (c+d x) \cos (c+d x)}{4 d}+\frac{3 a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{15 a^2 x}{4} \]
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Rubi [A] time = 0.209124, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2872, 3770, 3767, 8, 3768, 2635, 2633} \[ \frac{a^2 \cos ^5(c+d x)}{5 d}-\frac{a^2 \cos (c+d x)}{d}-\frac{2 a^2 \cot (c+d x)}{d}+\frac{a^2 \sin ^3(c+d x) \cos (c+d x)}{2 d}-\frac{9 a^2 \sin (c+d x) \cos (c+d x)}{4 d}+\frac{3 a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{15 a^2 x}{4} \]
Antiderivative was successfully verified.
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Rule 2872
Rule 3770
Rule 3767
Rule 8
Rule 3768
Rule 2635
Rule 2633
Rubi steps
\begin{align*} \int \cos ^3(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac{\int \left (-6 a^8-2 a^8 \csc (c+d x)+2 a^8 \csc ^2(c+d x)+a^8 \csc ^3(c+d x)+6 a^8 \sin ^2(c+d x)+2 a^8 \sin ^3(c+d x)-2 a^8 \sin ^4(c+d x)-a^8 \sin ^5(c+d x)\right ) \, dx}{a^6}\\ &=-6 a^2 x+a^2 \int \csc ^3(c+d x) \, dx-a^2 \int \sin ^5(c+d x) \, dx-\left (2 a^2\right ) \int \csc (c+d x) \, dx+\left (2 a^2\right ) \int \csc ^2(c+d x) \, dx+\left (2 a^2\right ) \int \sin ^3(c+d x) \, dx-\left (2 a^2\right ) \int \sin ^4(c+d x) \, dx+\left (6 a^2\right ) \int \sin ^2(c+d x) \, dx\\ &=-6 a^2 x+\frac{2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{3 a^2 \cos (c+d x) \sin (c+d x)}{d}+\frac{a^2 \cos (c+d x) \sin ^3(c+d x)}{2 d}+\frac{1}{2} a^2 \int \csc (c+d x) \, dx-\frac{1}{2} \left (3 a^2\right ) \int \sin ^2(c+d x) \, dx+\left (3 a^2\right ) \int 1 \, dx+\frac{a^2 \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac{\left (2 a^2\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}-\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-3 a^2 x+\frac{3 a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^2 \cos (c+d x)}{d}+\frac{a^2 \cos ^5(c+d x)}{5 d}-\frac{2 a^2 \cot (c+d x)}{d}-\frac{a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{9 a^2 \cos (c+d x) \sin (c+d x)}{4 d}+\frac{a^2 \cos (c+d x) \sin ^3(c+d x)}{2 d}-\frac{1}{4} \left (3 a^2\right ) \int 1 \, dx\\ &=-\frac{15 a^2 x}{4}+\frac{3 a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^2 \cos (c+d x)}{d}+\frac{a^2 \cos ^5(c+d x)}{5 d}-\frac{2 a^2 \cot (c+d x)}{d}-\frac{a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{9 a^2 \cos (c+d x) \sin (c+d x)}{4 d}+\frac{a^2 \cos (c+d x) \sin ^3(c+d x)}{2 d}\\ \end{align*}
Mathematica [A] time = 6.06911, size = 174, normalized size = 1.24 \[ \frac{(a \sin (c+d x)+a)^2 \left (-300 (c+d x)-80 \sin (2 (c+d x))-5 \sin (4 (c+d x))-70 \cos (c+d x)+5 \cos (3 (c+d x))+\cos (5 (c+d x))+80 \tan \left (\frac{1}{2} (c+d x)\right )-80 \cot \left (\frac{1}{2} (c+d x)\right )-10 \csc ^2\left (\frac{1}{2} (c+d x)\right )+10 \sec ^2\left (\frac{1}{2} (c+d x)\right )-120 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+120 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{80 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.081, size = 199, normalized size = 1.4 \begin{align*} -{\frac{3\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{10\,d}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{2\,d}}-{\frac{3\,{a}^{2}\cos \left ( dx+c \right ) }{2\,d}}-{\frac{3\,{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-2\,{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{d\sin \left ( dx+c \right ) }}-2\,{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) }{d}}-{\frac{5\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{15\,{a}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{4\,d}}-{\frac{15\,{a}^{2}x}{4}}-{\frac{15\,c{a}^{2}}{4\,d}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.70991, size = 258, normalized size = 1.84 \begin{align*} \frac{2 \,{\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 5 \,{\left (4 \, \cos \left (d x + c\right )^{3} - \frac{6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 15 \,{\left (15 \, d x + 15 \, c + \frac{15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{2}}{60 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.20187, size = 497, normalized size = 3.55 \begin{align*} \frac{4 \, a^{2} \cos \left (d x + c\right )^{7} - 4 \, a^{2} \cos \left (d x + c\right )^{5} - 75 \, a^{2} d x \cos \left (d x + c\right )^{2} - 20 \, a^{2} \cos \left (d x + c\right )^{3} + 75 \, a^{2} d x + 30 \, a^{2} \cos \left (d x + c\right ) + 15 \,{\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 15 \,{\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 5 \,{\left (2 \, a^{2} \cos \left (d x + c\right )^{5} + 5 \, a^{2} \cos \left (d x + c\right )^{3} - 15 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{20 \,{\left (d \cos \left (d x + c\right )^{2} - d\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 = 1.28472, size = 329, normalized size = 2.35 \begin{align*} \frac{5 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 150 \,{\left (d x + c\right )} a^{2} - 60 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 40 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{5 \,{\left (18 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{2}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}} + \frac{4 \,{\left (45 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 50 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 80 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 80 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 50 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 80 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 45 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 16 \, a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{5}}}{40 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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