Optimal. Leaf size=140 \[ -\frac{3 a^2 b \cos ^5(c+d x)}{5 d}+\frac{a^3 \sin ^5(c+d x)}{5 d}-\frac{2 a^3 \sin ^3(c+d x)}{3 d}+\frac{a^3 \sin (c+d x)}{d}-\frac{3 a b^2 \sin ^5(c+d x)}{5 d}+\frac{a b^2 \sin ^3(c+d x)}{d}+\frac{b^3 \cos ^5(c+d x)}{5 d}-\frac{b^3 \cos ^3(c+d x)}{3 d} \]
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Rubi [A] time = 0.163265, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 12, 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} \[ -\frac{3 a^2 b \cos ^5(c+d x)}{5 d}+\frac{a^3 \sin ^5(c+d x)}{5 d}-\frac{2 a^3 \sin ^3(c+d x)}{3 d}+\frac{a^3 \sin (c+d x)}{d}-\frac{3 a b^2 \sin ^5(c+d x)}{5 d}+\frac{a b^2 \sin ^3(c+d x)}{d}+\frac{b^3 \cos ^5(c+d x)}{5 d}-\frac{b^3 \cos ^3(c+d x)}{3 d} \]
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 ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx &=\int \left (a^3 \cos ^5(c+d x)+3 a^2 b \cos ^4(c+d x) \sin (c+d x)+3 a b^2 \cos ^3(c+d x) \sin ^2(c+d x)+b^3 \cos ^2(c+d x) \sin ^3(c+d x)\right ) \, dx\\ &=a^3 \int \cos ^5(c+d x) \, dx+\left (3 a^2 b\right ) \int \cos ^4(c+d x) \sin (c+d x) \, dx+\left (3 a b^2\right ) \int \cos ^3(c+d x) \sin ^2(c+d x) \, dx+b^3 \int \cos ^2(c+d x) \sin ^3(c+d x) \, dx\\ &=-\frac{a^3 \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int x^4 \, dx,x,\cos (c+d x)\right )}{d}+\frac{\left (3 a b^2\right ) \operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac{b^3 \operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{3 a^2 b \cos ^5(c+d x)}{5 d}+\frac{a^3 \sin (c+d x)}{d}-\frac{2 a^3 \sin ^3(c+d x)}{3 d}+\frac{a^3 \sin ^5(c+d x)}{5 d}+\frac{\left (3 a b^2\right ) \operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac{b^3 \operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{b^3 \cos ^3(c+d x)}{3 d}-\frac{3 a^2 b \cos ^5(c+d x)}{5 d}+\frac{b^3 \cos ^5(c+d x)}{5 d}+\frac{a^3 \sin (c+d x)}{d}-\frac{2 a^3 \sin ^3(c+d x)}{3 d}+\frac{a b^2 \sin ^3(c+d x)}{d}+\frac{a^3 \sin ^5(c+d x)}{5 d}-\frac{3 a b^2 \sin ^5(c+d x)}{5 d}\\ \end{align*}
Mathematica [A] time = 0.28896, size = 150, normalized size = 1.07 \[ \frac{-30 b \left (3 a^2+b^2\right ) \cos (c+d x)-5 \left (9 a^2 b+b^3\right ) \cos (3 (c+d x))-9 a^2 b \cos (5 (c+d x))+150 a^3 \sin (c+d x)+25 a^3 \sin (3 (c+d x))+3 a^3 \sin (5 (c+d x))+90 a b^2 \sin (c+d x)-15 a b^2 \sin (3 (c+d x))-9 a b^2 \sin (5 (c+d x))+3 b^3 \cos (5 (c+d x))}{240 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.07, size = 125, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({b}^{3} \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 ) +3\,a{b}^{2} \left ( -1/5\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1/15\, \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) -{\frac{3\,{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}}+{\frac{{a}^{3}\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 = 1.23966, size = 144, normalized size = 1.03 \begin{align*} -\frac{9 \, a^{2} 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^{3} + 3 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a b^{2} -{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} b^{3}}{15 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.494217, size = 230, normalized size = 1.64 \begin{align*} -\frac{5 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{5} -{\left (3 \,{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{3} + 6 \, a b^{2} +{\left (4 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{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 = 2.78219, size = 182, normalized size = 1.3 \begin{align*} \begin{cases} \frac{8 a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{a^{3} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac{3 a^{2} b \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac{2 a b^{2} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac{a b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac{b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{2 b^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + b \sin{\left (c \right )}\right )^{3} \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 = 1.1482, size = 196, normalized size = 1.4 \begin{align*} -\frac{{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac{{\left (9 \, a^{2} b + b^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac{{\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )}{8 \, d} + \frac{{\left (a^{3} - 3 \, a b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{{\left (5 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{{\left (5 \, a^{3} + 3 \, a 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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