Optimal. Leaf size=190 \[ \frac{\left (7 a^2+4 b^2\right ) \cos ^3(c+d x)}{105 d}-\frac{\left (7 a^2+4 b^2\right ) \cos (c+d x)}{35 d}+\frac{\left (2 a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{35 d}+\frac{a b \sin ^5(c+d x) \cos (c+d x)}{21 d}+\frac{\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}-\frac{a b \sin ^3(c+d x) \cos (c+d x)}{12 d}-\frac{a b \sin (c+d x) \cos (c+d x)}{8 d}+\frac{a b x}{8} \]
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Rubi [A] time = 0.382177, antiderivative size = 190, normalized size of antiderivative = 1., number of steps used = 10, 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, 2633, 2635, 8} \[ \frac{\left (7 a^2+4 b^2\right ) \cos ^3(c+d x)}{105 d}-\frac{\left (7 a^2+4 b^2\right ) \cos (c+d x)}{35 d}+\frac{\left (2 a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{35 d}+\frac{a b \sin ^5(c+d x) \cos (c+d x)}{21 d}+\frac{\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}-\frac{a b \sin ^3(c+d x) \cos (c+d x)}{12 d}-\frac{a b \sin (c+d x) \cos (c+d x)}{8 d}+\frac{a b x}{8} \]
Antiderivative was successfully verified.
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Rule 2889
Rule 3050
Rule 3033
Rule 3023
Rule 2748
Rule 2633
Rule 2635
Rule 8
Rubi steps
\begin{align*} \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx &=\int \sin ^3(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 ^4(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac{1}{7} \int \sin ^3(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 ^5(c+d x)}{21 d}+\frac{\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac{1}{42} \int \sin ^3(c+d x) \left (18 a^2+14 a b \sin (c+d x)-6 \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 ^4(c+d x)}{35 d}+\frac{a b \cos (c+d x) \sin ^5(c+d x)}{21 d}+\frac{\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac{1}{210} \int \sin ^3(c+d x) \left (6 \left (7 a^2+4 b^2\right )+70 a b \sin (c+d x)\right ) \, dx\\ &=\frac{\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{35 d}+\frac{a b \cos (c+d x) \sin ^5(c+d x)}{21 d}+\frac{\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac{1}{3} (a b) \int \sin ^4(c+d x) \, dx+\frac{1}{35} \left (7 a^2+4 b^2\right ) \int \sin ^3(c+d x) \, dx\\ &=-\frac{a b \cos (c+d x) \sin ^3(c+d x)}{12 d}+\frac{\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{35 d}+\frac{a b \cos (c+d x) \sin ^5(c+d x)}{21 d}+\frac{\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac{1}{4} (a b) \int \sin ^2(c+d x) \, dx-\frac{\left (7 a^2+4 b^2\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{35 d}\\ &=-\frac{\left (7 a^2+4 b^2\right ) \cos (c+d x)}{35 d}+\frac{\left (7 a^2+4 b^2\right ) \cos ^3(c+d x)}{105 d}-\frac{a b \cos (c+d x) \sin (c+d x)}{8 d}-\frac{a b \cos (c+d x) \sin ^3(c+d x)}{12 d}+\frac{\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{35 d}+\frac{a b \cos (c+d x) \sin ^5(c+d x)}{21 d}+\frac{\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac{1}{8} (a b) \int 1 \, dx\\ &=\frac{a b x}{8}-\frac{\left (7 a^2+4 b^2\right ) \cos (c+d x)}{35 d}+\frac{\left (7 a^2+4 b^2\right ) \cos ^3(c+d x)}{105 d}-\frac{a b \cos (c+d x) \sin (c+d x)}{8 d}-\frac{a b \cos (c+d x) \sin ^3(c+d x)}{12 d}+\frac{\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{35 d}+\frac{a b \cos (c+d x) \sin ^5(c+d x)}{21 d}+\frac{\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{7 d}\\ \end{align*}
Mathematica [A] time = 0.551489, size = 132, normalized size = 0.69 \[ \frac{-105 \left (8 a^2+5 b^2\right ) \cos (c+d x)-35 \left (4 a^2+b^2\right ) \cos (3 (c+d x))+84 a^2 \cos (5 (c+d x))-210 a b \sin (2 (c+d x))-210 a b \sin (4 (c+d x))+70 a b \sin (6 (c+d x))+840 a b c+840 a b d x+63 b^2 \cos (5 (c+d x))-15 b^2 \cos (7 (c+d x))}{6720 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.043, size = 150, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{5}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15}} \right ) +2\,ab \left ( -1/6\, \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}-1/8\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +1/16\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/16\,dx+c/16 \right ) +{b}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{7}}-{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{35}}-{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{105}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.12241, size = 140, normalized size = 0.74 \begin{align*} \frac{224 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{2} - 35 \,{\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 )} a b - 32 \,{\left (15 \, \cos \left (d x + c\right )^{7} - 42 \, \cos \left (d x + c\right )^{5} + 35 \, \cos \left (d x + c\right )^{3}\right )} b^{2}}{3360 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.49346, size = 274, normalized size = 1.44 \begin{align*} -\frac{120 \, b^{2} \cos \left (d x + c\right )^{7} - 168 \,{\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 105 \, a b d x + 280 \,{\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{3} - 35 \,{\left (8 \, a b \cos \left (d x + c\right )^{5} - 14 \, 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 )}{840 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 7.76245, size = 275, normalized size = 1.45 \begin{align*} \begin{cases} - \frac{a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{2 a^{2} \cos ^{5}{\left (c + d x \right )}}{15 d} + \frac{a b x \sin ^{6}{\left (c + d x \right )}}{8} + \frac{3 a b x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac{3 a b x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac{a b x \cos ^{6}{\left (c + d x \right )}}{8} + \frac{a b \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{a b \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{a b \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} - \frac{b^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{4 b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac{8 b^{2} \cos ^{7}{\left (c + d x \right )}}{105 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right )^{2} \sin ^{3}{\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 = 1.19229, size = 190, normalized size = 1. \begin{align*} \frac{1}{8} \, a b x - \frac{b^{2} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac{a b \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac{a b \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac{a b \sin \left (2 \, d x + 2 \, c\right )}{32 \, d} + \frac{{\left (4 \, a^{2} + 3 \, b^{2}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac{{\left (4 \, a^{2} + b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac{{\left (8 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )}{64 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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