Optimal. Leaf size=62 \[ -\frac{\cos (a+x (b-2 d)-2 c)}{4 (b-2 d)}-\frac{\cos (a+x (b+2 d)+2 c)}{4 (b+2 d)}-\frac{\cos (a+b x)}{2 b} \]
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Rubi [A] time = 0.0469182, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {4574, 2638} \[ -\frac{\cos (a+x (b-2 d)-2 c)}{4 (b-2 d)}-\frac{\cos (a+x (b+2 d)+2 c)}{4 (b+2 d)}-\frac{\cos (a+b x)}{2 b} \]
Antiderivative was successfully verified.
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Rule 4574
Rule 2638
Rubi steps
\begin{align*} \int \cos ^2(c+d x) \sin (a+b x) \, dx &=\int \left (\frac{1}{2} \sin (a+b x)+\frac{1}{4} \sin (a-2 c+(b-2 d) x)+\frac{1}{4} \sin (a+2 c+(b+2 d) x)\right ) \, dx\\ &=\frac{1}{4} \int \sin (a-2 c+(b-2 d) x) \, dx+\frac{1}{4} \int \sin (a+2 c+(b+2 d) x) \, dx+\frac{1}{2} \int \sin (a+b x) \, dx\\ &=-\frac{\cos (a+b x)}{2 b}-\frac{\cos (a-2 c+(b-2 d) x)}{4 (b-2 d)}-\frac{\cos (a+2 c+(b+2 d) x)}{4 (b+2 d)}\\ \end{align*}
Mathematica [A] time = 0.780621, size = 71, normalized size = 1.15 \[ \frac{1}{4} \left (-\frac{\cos (a+b x-2 c-2 d x)}{b-2 d}-\frac{\cos (a+b x+2 c+2 d x)}{b+2 d}+\frac{2 \sin (a) \sin (b x)}{b}-\frac{2 \cos (a) \cos (b x)}{b}\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.016, size = 57, normalized size = 0.9 \begin{align*} -{\frac{\cos \left ( bx+a \right ) }{2\,b}}-{\frac{\cos \left ( a-2\,c+ \left ( b-2\,d \right ) x \right ) }{4\,b-8\,d}}-{\frac{\cos \left ( a+2\,c+ \left ( b+2\,d \right ) x \right ) }{4\,b+8\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.21244, size = 559, normalized size = 9.02 \begin{align*} -\frac{{\left (b^{2} \cos \left (2 \, c\right ) - 2 \, b d \cos \left (2 \, c\right )\right )} \cos \left ({\left (b + 2 \, d\right )} x + a + 4 \, c\right ) +{\left (b^{2} \cos \left (2 \, c\right ) - 2 \, b d \cos \left (2 \, c\right )\right )} \cos \left ({\left (b + 2 \, d\right )} x + a\right ) +{\left (b^{2} \cos \left (2 \, c\right ) + 2 \, b d \cos \left (2 \, c\right )\right )} \cos \left (-{\left (b - 2 \, d\right )} x - a + 4 \, c\right ) +{\left (b^{2} \cos \left (2 \, c\right ) + 2 \, b d \cos \left (2 \, c\right )\right )} \cos \left (-{\left (b - 2 \, d\right )} x - a\right ) + 2 \,{\left (b^{2} \cos \left (2 \, c\right ) - 4 \, d^{2} \cos \left (2 \, c\right )\right )} \cos \left (b x + a + 2 \, c\right ) + 2 \,{\left (b^{2} \cos \left (2 \, c\right ) - 4 \, d^{2} \cos \left (2 \, c\right )\right )} \cos \left (b x + a - 2 \, c\right ) +{\left (b^{2} \sin \left (2 \, c\right ) - 2 \, b d \sin \left (2 \, c\right )\right )} \sin \left ({\left (b + 2 \, d\right )} x + a + 4 \, c\right ) -{\left (b^{2} \sin \left (2 \, c\right ) - 2 \, b d \sin \left (2 \, c\right )\right )} \sin \left ({\left (b + 2 \, d\right )} x + a\right ) +{\left (b^{2} \sin \left (2 \, c\right ) + 2 \, b d \sin \left (2 \, c\right )\right )} \sin \left (-{\left (b - 2 \, d\right )} x - a + 4 \, c\right ) -{\left (b^{2} \sin \left (2 \, c\right ) + 2 \, b d \sin \left (2 \, c\right )\right )} \sin \left (-{\left (b - 2 \, d\right )} x - a\right ) + 2 \,{\left (b^{2} \sin \left (2 \, c\right ) - 4 \, d^{2} \sin \left (2 \, c\right )\right )} \sin \left (b x + a + 2 \, c\right ) - 2 \,{\left (b^{2} \sin \left (2 \, c\right ) - 4 \, d^{2} \sin \left (2 \, c\right )\right )} \sin \left (b x + a - 2 \, c\right )}{8 \,{\left (b^{3} \cos \left (2 \, c\right )^{2} + b^{3} \sin \left (2 \, c\right )^{2} - 4 \,{\left (b \cos \left (2 \, c\right )^{2} + b \sin \left (2 \, c\right )^{2}\right )} d^{2}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.490526, size = 162, normalized size = 2.61 \begin{align*} -\frac{b^{2} \cos \left (b x + a\right ) \cos \left (d x + c\right )^{2} + 2 \, b d \cos \left (d x + c\right ) \sin \left (b x + a\right ) \sin \left (d x + c\right ) - 2 \, d^{2} \cos \left (b x + a\right )}{b^{3} - 4 \, b d^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 12.1257, size = 401, normalized size = 6.47 \begin{align*} \begin{cases} x \sin{\left (a \right )} \cos ^{2}{\left (c \right )} & \text{for}\: b = 0 \wedge d = 0 \\\left (\frac{x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{\sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d}\right ) \sin{\left (a \right )} & \text{for}\: b = 0 \\- \frac{x \sin{\left (a - 2 d x \right )} \sin ^{2}{\left (c + d x \right )}}{4} + \frac{x \sin{\left (a - 2 d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{x \sin{\left (c + d x \right )} \cos{\left (a - 2 d x \right )} \cos{\left (c + d x \right )}}{2} + \frac{\sin ^{2}{\left (c + d x \right )} \cos{\left (a - 2 d x \right )}}{8 d} + \frac{3 \cos{\left (a - 2 d x \right )} \cos ^{2}{\left (c + d x \right )}}{8 d} & \text{for}\: b = - 2 d \\- \frac{x \sin{\left (a + 2 d x \right )} \sin ^{2}{\left (c + d x \right )}}{4} + \frac{x \sin{\left (a + 2 d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} - \frac{x \sin{\left (c + d x \right )} \cos{\left (a + 2 d x \right )} \cos{\left (c + d x \right )}}{2} - \frac{\sin{\left (a + 2 d x \right )} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{4 d} - \frac{\cos{\left (a + 2 d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} & \text{for}\: b = 2 d \\- \frac{b^{2} \cos{\left (a + b x \right )} \cos ^{2}{\left (c + d x \right )}}{b^{3} - 4 b d^{2}} - \frac{2 b d \sin{\left (a + b x \right )} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{b^{3} - 4 b d^{2}} + \frac{2 d^{2} \sin ^{2}{\left (c + d x \right )} \cos{\left (a + b x \right )}}{b^{3} - 4 b d^{2}} + \frac{2 d^{2} \cos{\left (a + b x \right )} \cos ^{2}{\left (c + d x \right )}}{b^{3} - 4 b d^{2}} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.13856, size = 76, normalized size = 1.23 \begin{align*} -\frac{\cos \left (b x + 2 \, d x + a + 2 \, c\right )}{4 \,{\left (b + 2 \, d\right )}} - \frac{\cos \left (b x - 2 \, d x + a - 2 \, c\right )}{4 \,{\left (b - 2 \, d\right )}} - \frac{\cos \left (b x + a\right )}{2 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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