Optimal. Leaf size=116 \[ \frac{a^2 \cos ^3(c+d x)}{3 d}+\frac{a^2 \cos (c+d x)}{d}-\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{a b \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac{3 a b \sin (c+d x) \cos (c+d x)}{4 d}+\frac{3 a b x}{4}-\frac{b^2 \cos ^5(c+d x)}{5 d} \]
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Rubi [A] time = 0.447365, antiderivative size = 190, normalized size of antiderivative = 1.64, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2895, 3049, 3033, 3023, 2735, 3770} \[ -\frac{\left (-14 a^2 b^2+a^4+3 b^4\right ) \cos (c+d x)}{15 b^2 d}-\frac{\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}-\frac{a \left (2 a^2-27 b^2\right ) \sin (c+d x) \cos (c+d x)}{60 b d}-\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac{\sin (c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{5 b d}+\frac{3 a b x}{4} \]
Antiderivative was successfully verified.
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Rule 2895
Rule 3049
Rule 3033
Rule 3023
Rule 2735
Rule 3770
Rubi steps
\begin{align*} \int \cos ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac{a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac{\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}-\frac{\int \csc (c+d x) (a+b \sin (c+d x))^2 \left (-20 b^2+2 a b \sin (c+d x)-2 \left (a^2-12 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{20 b^2}\\ &=-\frac{\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}+\frac{a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac{\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}-\frac{\int \csc (c+d x) (a+b \sin (c+d x)) \left (-60 a b^2+2 b \left (a^2-6 b^2\right ) \sin (c+d x)-2 a \left (2 a^2-27 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{60 b^2}\\ &=-\frac{a \left (2 a^2-27 b^2\right ) \cos (c+d x) \sin (c+d x)}{60 b d}-\frac{\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}+\frac{a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac{\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}-\frac{\int \csc (c+d x) \left (-120 a^2 b^2-90 a b^3 \sin (c+d x)-8 \left (a^4-14 a^2 b^2+3 b^4\right ) \sin ^2(c+d x)\right ) \, dx}{120 b^2}\\ &=-\frac{\left (a^4-14 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^2 d}-\frac{a \left (2 a^2-27 b^2\right ) \cos (c+d x) \sin (c+d x)}{60 b d}-\frac{\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}+\frac{a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac{\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}-\frac{\int \csc (c+d x) \left (-120 a^2 b^2-90 a b^3 \sin (c+d x)\right ) \, dx}{120 b^2}\\ &=\frac{3 a b x}{4}-\frac{\left (a^4-14 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^2 d}-\frac{a \left (2 a^2-27 b^2\right ) \cos (c+d x) \sin (c+d x)}{60 b d}-\frac{\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}+\frac{a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac{\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}+a^2 \int \csc (c+d x) \, dx\\ &=\frac{3 a b x}{4}-\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{\left (a^4-14 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^2 d}-\frac{a \left (2 a^2-27 b^2\right ) \cos (c+d x) \sin (c+d x)}{60 b d}-\frac{\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}+\frac{a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac{\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}\\ \end{align*}
Mathematica [A] time = 0.516802, size = 125, normalized size = 1.08 \[ \frac{30 \left (10 a^2-b^2\right ) \cos (c+d x)+5 \left (4 a^2-3 b^2\right ) \cos (3 (c+d x))+15 a \left (4 \left (4 a \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-4 a \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+3 b (c+d x)\right )+8 b \sin (2 (c+d x))+b \sin (4 (c+d x))\right )-3 b^2 \cos (5 (c+d x))}{240 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.083, size = 123, normalized size = 1.1 \begin{align*}{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{{a}^{2}\cos \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}+{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{2\,d}}+{\frac{3\,ab\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{4\,d}}+{\frac{3\,abx}{4}}+{\frac{3\,abc}{4\,d}}-{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.00125, size = 131, normalized size = 1.13 \begin{align*} -\frac{48 \, b^{2} \cos \left (d x + c\right )^{5} - 40 \,{\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 15 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a b}{240 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.94262, size = 309, normalized size = 2.66 \begin{align*} -\frac{12 \, b^{2} \cos \left (d x + c\right )^{5} - 20 \, a^{2} \cos \left (d x + c\right )^{3} - 45 \, a b d x - 60 \, a^{2} \cos \left (d x + c\right ) + 30 \, a^{2} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 30 \, a^{2} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 15 \,{\left (2 \, a b \cos \left (d x + c\right )^{3} + 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, d} \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 = 1.3056, size = 288, normalized size = 2.48 \begin{align*} \frac{45 \,{\left (d x + c\right )} a b + 60 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{2 \,{\left (75 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 120 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 60 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 30 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 360 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 440 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 120 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 30 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 280 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 75 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 80 \, a^{2} + 12 \, b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{5}}}{60 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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