Optimal. Leaf size=158 \[ \frac{2 a^2 \cos ^5(c+d x)}{5 d}+\frac{2 a^2 \cos ^3(c+d x)}{3 d}+\frac{2 a^2 \cos (c+d x)}{d}-\frac{a^2 \cot (c+d x)}{d}+\frac{a^2 \sin ^5(c+d x) \cos (c+d x)}{6 d}-\frac{7 a^2 \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac{7 a^2 \sin (c+d x) \cos (c+d x)}{16 d}-\frac{2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{25 a^2 x}{16} \]
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Rubi [A] time = 0.226387, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2872, 3770, 3767, 8, 2638, 2633, 2635} \[ \frac{2 a^2 \cos ^5(c+d x)}{5 d}+\frac{2 a^2 \cos ^3(c+d x)}{3 d}+\frac{2 a^2 \cos (c+d x)}{d}-\frac{a^2 \cot (c+d x)}{d}+\frac{a^2 \sin ^5(c+d x) \cos (c+d x)}{6 d}-\frac{7 a^2 \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac{7 a^2 \sin (c+d x) \cos (c+d x)}{16 d}-\frac{2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{25 a^2 x}{16} \]
Antiderivative was successfully verified.
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Rule 2872
Rule 3770
Rule 3767
Rule 8
Rule 2638
Rule 2633
Rule 2635
Rubi steps
\begin{align*} \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac{\int \left (-2 a^8+2 a^8 \csc (c+d x)+a^8 \csc ^2(c+d x)-6 a^8 \sin (c+d x)+6 a^8 \sin ^3(c+d x)+2 a^8 \sin ^4(c+d x)-2 a^8 \sin ^5(c+d x)-a^8 \sin ^6(c+d x)\right ) \, dx}{a^6}\\ &=-2 a^2 x+a^2 \int \csc ^2(c+d x) \, dx-a^2 \int \sin ^6(c+d x) \, dx+\left (2 a^2\right ) \int \csc (c+d x) \, dx+\left (2 a^2\right ) \int \sin ^4(c+d x) \, dx-\left (2 a^2\right ) \int \sin ^5(c+d x) \, dx-\left (6 a^2\right ) \int \sin (c+d x) \, dx+\left (6 a^2\right ) \int \sin ^3(c+d x) \, dx\\ &=-2 a^2 x-\frac{2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{6 a^2 \cos (c+d x)}{d}-\frac{a^2 \cos (c+d x) \sin ^3(c+d x)}{2 d}+\frac{a^2 \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac{1}{6} \left (5 a^2\right ) \int \sin ^4(c+d x) \, dx+\frac{1}{2} \left (3 a^2\right ) \int \sin ^2(c+d x) \, dx-\frac{a^2 \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac{\left (6 a^2\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-2 a^2 x-\frac{2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{2 a^2 \cos (c+d x)}{d}+\frac{2 a^2 \cos ^3(c+d x)}{3 d}+\frac{2 a^2 \cos ^5(c+d x)}{5 d}-\frac{a^2 \cot (c+d x)}{d}-\frac{3 a^2 \cos (c+d x) \sin (c+d x)}{4 d}-\frac{7 a^2 \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac{a^2 \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac{1}{8} \left (5 a^2\right ) \int \sin ^2(c+d x) \, dx+\frac{1}{4} \left (3 a^2\right ) \int 1 \, dx\\ &=-\frac{5 a^2 x}{4}-\frac{2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{2 a^2 \cos (c+d x)}{d}+\frac{2 a^2 \cos ^3(c+d x)}{3 d}+\frac{2 a^2 \cos ^5(c+d x)}{5 d}-\frac{a^2 \cot (c+d x)}{d}-\frac{7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}-\frac{7 a^2 \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac{a^2 \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac{1}{16} \left (5 a^2\right ) \int 1 \, dx\\ &=-\frac{25 a^2 x}{16}-\frac{2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{2 a^2 \cos (c+d x)}{d}+\frac{2 a^2 \cos ^3(c+d x)}{3 d}+\frac{2 a^2 \cos ^5(c+d x)}{5 d}-\frac{a^2 \cot (c+d x)}{d}-\frac{7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}-\frac{7 a^2 \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac{a^2 \cos (c+d x) \sin ^5(c+d x)}{6 d}\\ \end{align*}
Mathematica [A] time = 0.307121, size = 110, normalized size = 0.7 \[ \frac{a^2 \left (-255 \sin (2 (c+d x))+15 \sin (4 (c+d x))+5 \sin (6 (c+d x))+2640 \cos (c+d x)+280 \cos (3 (c+d x))+24 \cos (5 (c+d x))-960 \cot (c+d x)+1920 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-1920 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-1500 c-1500 d x\right )}{960 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.072, size = 175, normalized size = 1.1 \begin{align*} -{\frac{5\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) }{6\,d}}-{\frac{25\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{24\,d}}-{\frac{25\,{a}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{16\,d}}-{\frac{25\,{a}^{2}x}{16}}-{\frac{25\,c{a}^{2}}{16\,d}}+{\frac{2\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5\,d}}+{\frac{2\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+2\,{\frac{{a}^{2}\cos \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{d\sin \left ( dx+c \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.76176, size = 234, normalized size = 1.48 \begin{align*} \frac{64 \,{\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 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 120 \,{\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}}{960 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.22021, size = 439, normalized size = 2.78 \begin{align*} -\frac{40 \, a^{2} \cos \left (d x + c\right )^{7} - 50 \, a^{2} \cos \left (d x + c\right )^{5} - 125 \, a^{2} \cos \left (d x + c\right )^{3} + 240 \, a^{2} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 240 \, a^{2} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 375 \, a^{2} \cos \left (d x + c\right ) -{\left (96 \, a^{2} \cos \left (d x + c\right )^{5} + 160 \, a^{2} \cos \left (d x + c\right )^{3} - 375 \, a^{2} d x + 480 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \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 = 1.25403, size = 370, normalized size = 2.34 \begin{align*} -\frac{375 \,{\left (d x + c\right )} a^{2} - 480 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 120 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{120 \,{\left (4 \, 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 )} - \frac{2 \,{\left (105 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 1440 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} + 595 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 4320 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 150 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 7360 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 150 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6720 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 595 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2976 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 105 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 736 \, a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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