Optimal. Leaf size=178 \[ -\frac{\left (6 a^2-5 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}-\frac{\left (14 a^2-5 b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}-\frac{5}{16} x \left (6 a^2-b^2\right )-\frac{a^2 \cot (c+d x)}{d}+\frac{2 a b \cos ^5(c+d x)}{5 d}+\frac{2 a b \cos ^3(c+d x)}{3 d}+\frac{2 a b \cos (c+d x)}{d}-\frac{2 a b \tanh ^{-1}(\cos (c+d x))}{d}+\frac{b^2 \sin (c+d x) \cos ^5(c+d x)}{6 d} \]
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Rubi [A] time = 0.463483, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2911, 2592, 302, 206, 434, 456, 453, 203} \[ -\frac{\left (6 a^2-5 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}-\frac{\left (14 a^2-5 b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}-\frac{5}{16} x \left (6 a^2-b^2\right )-\frac{a^2 \cot (c+d x)}{d}+\frac{2 a b \cos ^5(c+d x)}{5 d}+\frac{2 a b \cos ^3(c+d x)}{3 d}+\frac{2 a b \cos (c+d x)}{d}-\frac{2 a b \tanh ^{-1}(\cos (c+d x))}{d}+\frac{b^2 \sin (c+d x) \cos ^5(c+d x)}{6 d} \]
Antiderivative was successfully verified.
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Rule 2911
Rule 2592
Rule 302
Rule 206
Rule 434
Rule 456
Rule 453
Rule 203
Rubi steps
\begin{align*} \int \cos ^4(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^2 \, dx &=(2 a b) \int \cos ^5(c+d x) \cot (c+d x) \, dx+\int \cos ^4(c+d x) \cot ^2(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{a^2+b^2+\frac{a^2}{x^2}}{\left (1+x^2\right )^4} \, dx,x,\tan (c+d x)\right )}{d}-\frac{(2 a b) \operatorname{Subst}\left (\int \frac{x^6}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{a^2+\left (a^2+b^2\right ) x^2}{x^2 \left (1+x^2\right )^4} \, dx,x,\tan (c+d x)\right )}{d}-\frac{(2 a b) \operatorname{Subst}\left (\int \left (-1-x^2-x^4+\frac{1}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{2 a b \cos (c+d x)}{d}+\frac{2 a b \cos ^3(c+d x)}{3 d}+\frac{2 a b \cos ^5(c+d x)}{5 d}+\frac{b^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{\operatorname{Subst}\left (\int \frac{-6 a^2-5 b^2 x^2}{x^2 \left (1+x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{6 d}-\frac{(2 a b) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{2 a b \tanh ^{-1}(\cos (c+d x))}{d}+\frac{2 a b \cos (c+d x)}{d}+\frac{2 a b \cos ^3(c+d x)}{3 d}+\frac{2 a b \cos ^5(c+d x)}{5 d}-\frac{\left (6 a^2-5 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{b^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac{\operatorname{Subst}\left (\int \frac{24 a^2-3 \left (6 a^2-5 b^2\right ) x^2}{x^2 \left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{24 d}\\ &=-\frac{2 a b \tanh ^{-1}(\cos (c+d x))}{d}+\frac{2 a b \cos (c+d x)}{d}+\frac{2 a b \cos ^3(c+d x)}{3 d}+\frac{2 a b \cos ^5(c+d x)}{5 d}-\frac{\left (14 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}-\frac{\left (6 a^2-5 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{b^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{\operatorname{Subst}\left (\int \frac{-48 a^2+3 \left (14 a^2-5 b^2\right ) x^2}{x^2 \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{48 d}\\ &=-\frac{2 a b \tanh ^{-1}(\cos (c+d x))}{d}+\frac{2 a b \cos (c+d x)}{d}+\frac{2 a b \cos ^3(c+d x)}{3 d}+\frac{2 a b \cos ^5(c+d x)}{5 d}-\frac{a^2 \cot (c+d x)}{d}-\frac{\left (14 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}-\frac{\left (6 a^2-5 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{b^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{\left (5 \left (6 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{16 d}\\ &=-\frac{5}{16} \left (6 a^2-b^2\right ) x-\frac{2 a b \tanh ^{-1}(\cos (c+d x))}{d}+\frac{2 a b \cos (c+d x)}{d}+\frac{2 a b \cos ^3(c+d x)}{3 d}+\frac{2 a b \cos ^5(c+d x)}{5 d}-\frac{a^2 \cot (c+d x)}{d}-\frac{\left (14 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}-\frac{\left (6 a^2-5 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{b^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}\\ \end{align*}
Mathematica [A] time = 0.400461, size = 220, normalized size = 1.24 \[ -\frac{15 a^2 (c+d x)}{8 d}-\frac{a^2 \sin (2 (c+d x))}{2 d}-\frac{a^2 \sin (4 (c+d x))}{32 d}-\frac{a^2 \cot (c+d x)}{d}+\frac{11 a b \cos (c+d x)}{4 d}+\frac{7 a b \cos (3 (c+d x))}{24 d}+\frac{a b \cos (5 (c+d x))}{40 d}+\frac{2 a b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}-\frac{2 a b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{5 b^2 (c+d x)}{16 d}+\frac{15 b^2 \sin (2 (c+d x))}{64 d}+\frac{3 b^2 \sin (4 (c+d x))}{64 d}+\frac{b^2 \sin (6 (c+d x))}{192 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.078, size = 250, normalized size = 1.4 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{d\sin \left ( dx+c \right ) }}-{\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 ) }{4\,d}}-{\frac{15\,{a}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}-{\frac{15\,{a}^{2}x}{8}}-{\frac{15\,{a}^{2}c}{8\,d}}+{\frac{2\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5\,d}}+{\frac{2\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+2\,{\frac{ab\cos \left ( dx+c \right ) }{d}}+2\,{\frac{ab\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) }{6\,d}}+{\frac{5\,{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{24\,d}}+{\frac{5\,{b}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{16\,d}}+{\frac{5\,{b}^{2}x}{16}}+{\frac{5\,{b}^{2}c}{16\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.48727, size = 232, normalized size = 1.3 \begin{align*} -\frac{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} - 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 b + 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 )} b^{2}}{960 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.88746, size = 489, normalized size = 2.75 \begin{align*} -\frac{40 \, b^{2} \cos \left (d x + c\right )^{7} - 10 \,{\left (6 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{5} - 25 \,{\left (6 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{3} + 240 \, a b \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 240 \, a b \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 75 \,{\left (6 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right ) -{\left (96 \, a b \cos \left (d x + c\right )^{5} + 160 \, a b \cos \left (d x + c\right )^{3} - 75 \,{\left (6 \, a^{2} - b^{2}\right )} d x + 480 \, a b \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 [B] time = 1.25425, size = 497, normalized size = 2.79 \begin{align*} \frac{480 \, a b \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 ) - 75 \,{\left (6 \, a^{2} - b^{2}\right )}{\left (d x + c\right )} - \frac{120 \,{\left (4 \, a b \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 (270 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 165 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 1440 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} + 570 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 25 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 4320 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 300 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 450 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 7360 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 300 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 450 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6720 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 570 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 25 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2976 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 270 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 165 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 736 \, a b\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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