Optimal. Leaf size=116 \[ \frac{\left (6 a^2+b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{\left (6 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} x \left (6 a^2+b^2\right )-\frac{7 a b \cos ^5(c+d x)}{30 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d} \]
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Rubi [A] time = 0.115815, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2692, 2669, 2635, 8} \[ \frac{\left (6 a^2+b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{\left (6 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} x \left (6 a^2+b^2\right )-\frac{7 a b \cos ^5(c+d x)}{30 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d} \]
Antiderivative was successfully verified.
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Rule 2692
Rule 2669
Rule 2635
Rule 8
Rubi steps
\begin{align*} \int \cos ^4(c+d x) (a+b \sin (c+d x))^2 \, dx &=-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}+\frac{1}{6} \int \cos ^4(c+d x) \left (6 a^2+b^2+7 a b \sin (c+d x)\right ) \, dx\\ &=-\frac{7 a b \cos ^5(c+d x)}{30 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}+\frac{1}{6} \left (6 a^2+b^2\right ) \int \cos ^4(c+d x) \, dx\\ &=-\frac{7 a b \cos ^5(c+d x)}{30 d}+\frac{\left (6 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}+\frac{1}{8} \left (6 a^2+b^2\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{7 a b \cos ^5(c+d x)}{30 d}+\frac{\left (6 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{\left (6 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}+\frac{1}{16} \left (6 a^2+b^2\right ) \int 1 \, dx\\ &=\frac{1}{16} \left (6 a^2+b^2\right ) x-\frac{7 a b \cos ^5(c+d x)}{30 d}+\frac{\left (6 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{\left (6 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\\ \end{align*}
Mathematica [A] time = 0.193004, size = 133, normalized size = 1.15 \[ \frac{240 a^2 \sin (2 (c+d x))+30 a^2 \sin (4 (c+d x))+360 a^2 c+360 a^2 d x-240 a b \cos (c+d x)-120 a b \cos (3 (c+d x))-24 a b \cos (5 (c+d x))+15 b^2 \sin (2 (c+d x))-15 b^2 \sin (4 (c+d x))-5 b^2 \sin (6 (c+d x))+60 b^2 c+60 b^2 d x}{960 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.043, size = 108, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{6}}+{\frac{\sin \left ( dx+c \right ) }{24} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{dx}{16}}+{\frac{c}{16}} \right ) -{\frac{2\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}}+{a}^{2} \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.962841, size = 119, normalized size = 1.03 \begin{align*} -\frac{384 \, a b \cos \left (d x + c\right )^{5} - 30 \,{\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^{2} - 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 = 2.30337, size = 217, normalized size = 1.87 \begin{align*} -\frac{96 \, a b \cos \left (d x + c\right )^{5} - 15 \,{\left (6 \, a^{2} + b^{2}\right )} d x + 5 \,{\left (8 \, b^{2} \cos \left (d x + c\right )^{5} - 2 \,{\left (6 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (6 \, 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 = 4.7273, size = 287, normalized size = 2.47 \begin{align*} \begin{cases} \frac{3 a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 a^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 a^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac{2 a b \cos ^{5}{\left (c + d x \right )}}{5 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} \cos ^{4}{\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 = 1.09804, size = 166, normalized size = 1.43 \begin{align*} \frac{1}{16} \,{\left (6 \, 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 )}{8 \, 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{{\left (2 \, a^{2} - b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{{\left (16 \, a^{2} + b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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