Optimal. Leaf size=103 \[ \frac{a^2 \sin ^5(c+d x)}{5 d}-\frac{2 a^2 \sin ^3(c+d x)}{3 d}+\frac{a^2 \sin (c+d x)}{d}-\frac{2 a b \cos ^5(c+d x)}{5 d}-\frac{b^2 \sin ^5(c+d x)}{5 d}+\frac{b^2 \sin ^3(c+d x)}{3 d} \]
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Rubi [A] time = 0.121948, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3090, 2633, 2565, 30, 2564, 14} \[ \frac{a^2 \sin ^5(c+d x)}{5 d}-\frac{2 a^2 \sin ^3(c+d x)}{3 d}+\frac{a^2 \sin (c+d x)}{d}-\frac{2 a b \cos ^5(c+d x)}{5 d}-\frac{b^2 \sin ^5(c+d x)}{5 d}+\frac{b^2 \sin ^3(c+d x)}{3 d} \]
Antiderivative was successfully verified.
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Rule 3090
Rule 2633
Rule 2565
Rule 30
Rule 2564
Rule 14
Rubi steps
\begin{align*} \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx &=\int \left (a^2 \cos ^5(c+d x)+2 a b \cos ^4(c+d x) \sin (c+d x)+b^2 \cos ^3(c+d x) \sin ^2(c+d x)\right ) \, dx\\ &=a^2 \int \cos ^5(c+d x) \, dx+(2 a b) \int \cos ^4(c+d x) \sin (c+d x) \, dx+b^2 \int \cos ^3(c+d x) \sin ^2(c+d x) \, dx\\ &=-\frac{a^2 \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac{(2 a b) \operatorname{Subst}\left (\int x^4 \, dx,x,\cos (c+d x)\right )}{d}+\frac{b^2 \operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{2 a b \cos ^5(c+d x)}{5 d}+\frac{a^2 \sin (c+d x)}{d}-\frac{2 a^2 \sin ^3(c+d x)}{3 d}+\frac{a^2 \sin ^5(c+d x)}{5 d}+\frac{b^2 \operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{2 a b \cos ^5(c+d x)}{5 d}+\frac{a^2 \sin (c+d x)}{d}-\frac{2 a^2 \sin ^3(c+d x)}{3 d}+\frac{b^2 \sin ^3(c+d x)}{3 d}+\frac{a^2 \sin ^5(c+d x)}{5 d}-\frac{b^2 \sin ^5(c+d x)}{5 d}\\ \end{align*}
Mathematica [A] time = 0.171228, size = 116, normalized size = 1.13 \[ \frac{150 a^2 \sin (c+d x)+25 a^2 \sin (3 (c+d x))+3 a^2 \sin (5 (c+d x))-60 a b \cos (c+d x)-30 a b \cos (3 (c+d x))-6 a b \cos (5 (c+d x))+30 b^2 \sin (c+d x)-5 b^2 \sin (3 (c+d x))-3 b^2 \sin (5 (c+d x))}{240 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.059, size = 88, 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 ) ^{4}}{5}}+{\frac{ \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{15}} \right ) -{\frac{2\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}}+{\frac{{a}^{2}\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.09447, size = 104, normalized size = 1.01 \begin{align*} -\frac{6 \, a b \cos \left (d x + c\right )^{5} -{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{2} +{\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} b^{2}}{15 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.486424, size = 169, normalized size = 1.64 \begin{align*} -\frac{6 \, a b \cos \left (d x + c\right )^{5} -{\left (3 \,{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} +{\left (4 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, a^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )}{15 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 2.3534, size = 138, normalized size = 1.34 \begin{align*} \begin{cases} \frac{8 a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{a^{2} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac{2 a b \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac{2 b^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + b \sin{\left (c \right )}\right )^{2} \cos ^{3}{\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 = 1.16224, size = 154, normalized size = 1.5 \begin{align*} -\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{{\left (a^{2} - b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{{\left (5 \, a^{2} - b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{{\left (5 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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