Optimal. Leaf size=163 \[ \frac{\left (2 a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac{\left (2 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} x \left (2 a^2+b^2\right )+\frac{2 a b \cos ^3(c+d x)}{15 d}-\frac{2 a b \cos (c+d x)}{5 d}+\frac{a b \sin ^4(c+d x) \cos (c+d x)}{15 d}+\frac{\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{6 d} \]
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Rubi [A] time = 0.381583, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2889, 3050, 3033, 3023, 2748, 2635, 8, 2633} \[ \frac{\left (2 a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac{\left (2 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} x \left (2 a^2+b^2\right )+\frac{2 a b \cos ^3(c+d x)}{15 d}-\frac{2 a b \cos (c+d x)}{5 d}+\frac{a b \sin ^4(c+d x) \cos (c+d x)}{15 d}+\frac{\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{6 d} \]
Antiderivative was successfully verified.
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Rule 2889
Rule 3050
Rule 3033
Rule 3023
Rule 2748
Rule 2635
Rule 8
Rule 2633
Rubi steps
\begin{align*} \int \cos ^2(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx &=\int \sin ^2(c+d x) (a+b \sin (c+d x))^2 \left (1-\sin ^2(c+d x)\right ) \, dx\\ &=\frac{\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{6 d}+\frac{1}{6} \int \sin ^2(c+d x) (a+b \sin (c+d x)) \left (3 a+b \sin (c+d x)-2 a \sin ^2(c+d x)\right ) \, dx\\ &=\frac{a b \cos (c+d x) \sin ^4(c+d x)}{15 d}+\frac{\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{6 d}+\frac{1}{30} \int \sin ^2(c+d x) \left (15 a^2+12 a b \sin (c+d x)-5 \left (2 a^2-b^2\right ) \sin ^2(c+d x)\right ) \, dx\\ &=\frac{\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac{a b \cos (c+d x) \sin ^4(c+d x)}{15 d}+\frac{\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{6 d}+\frac{1}{120} \int \sin ^2(c+d x) \left (15 \left (2 a^2+b^2\right )+48 a b \sin (c+d x)\right ) \, dx\\ &=\frac{\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac{a b \cos (c+d x) \sin ^4(c+d x)}{15 d}+\frac{\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{6 d}+\frac{1}{5} (2 a b) \int \sin ^3(c+d x) \, dx+\frac{1}{8} \left (2 a^2+b^2\right ) \int \sin ^2(c+d x) \, dx\\ &=-\frac{\left (2 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac{a b \cos (c+d x) \sin ^4(c+d x)}{15 d}+\frac{\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{6 d}+\frac{1}{16} \left (2 a^2+b^2\right ) \int 1 \, dx-\frac{(2 a b) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{5 d}\\ &=\frac{1}{16} \left (2 a^2+b^2\right ) x-\frac{2 a b \cos (c+d x)}{5 d}+\frac{2 a b \cos ^3(c+d x)}{15 d}-\frac{\left (2 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac{a b \cos (c+d x) \sin ^4(c+d x)}{15 d}+\frac{\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{6 d}\\ \end{align*}
Mathematica [A] time = 0.213002, size = 120, normalized size = 0.74 \[ \frac{-30 a^2 \sin (4 (c+d x))+120 a^2 c+120 a^2 d x-240 a b \cos (c+d x)-40 a b \cos (3 (c+d x))+24 a b \cos (5 (c+d x))-15 b^2 \sin (2 (c+d x))-15 b^2 \sin (4 (c+d x))+5 b^2 \sin (6 (c+d x))+60 b^2 c+60 b^2 d x}{960 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.042, size = 141, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{4}}+{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8}}+{\frac{dx}{8}}+{\frac{c}{8}} \right ) +2\,ab \left ( -1/5\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}-2/15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3} \right ) +{b}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{6}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{8}}+{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{16}}+{\frac{dx}{16}}+{\frac{c}{16}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.16233, size = 124, normalized size = 0.76 \begin{align*} \frac{30 \,{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} + 128 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a b - 5 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, 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 = 1.48997, size = 252, normalized size = 1.55 \begin{align*} \frac{96 \, a b \cos \left (d x + c\right )^{5} - 160 \, a b \cos \left (d x + c\right )^{3} + 15 \,{\left (2 \, a^{2} + b^{2}\right )} d x + 5 \,{\left (8 \, b^{2} \cos \left (d x + c\right )^{5} - 2 \,{\left (6 \, a^{2} + 7 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (2 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 4.69886, size = 309, normalized size = 1.9 \begin{align*} \begin{cases} \frac{a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{a^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{a^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac{2 a b \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{4 a b \cos ^{5}{\left (c + d x \right )}}{15 d} + \frac{b^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{3 b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{3 b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{b^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{b^{2} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} - \frac{b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac{b^{2} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right )^{2} \sin ^{2}{\left (c \right )} \cos ^{2}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.21461, size = 155, normalized size = 0.95 \begin{align*} \frac{1}{16} \,{\left (2 \, a^{2} + b^{2}\right )} x + \frac{a b \cos \left (5 \, d x + 5 \, c\right )}{40 \, d} - \frac{a b \cos \left (3 \, d x + 3 \, c\right )}{24 \, d} - \frac{a b \cos \left (d x + c\right )}{4 \, d} + \frac{b^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac{b^{2} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} - \frac{{\left (2 \, a^{2} + b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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