Optimal. Leaf size=68 \[ -\frac {\cos (2 a+x (2 b-d)-c)}{4 (2 b-d)}+\frac {\cos (2 a+x (2 b+d)+c)}{4 (2 b+d)}-\frac {\cos (c+d x)}{2 d} \]
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Rubi [A] time = 0.05, antiderivative size = 68, normalized size of antiderivative = 1.00, 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 = {4569, 2638} \[ -\frac {\cos (2 a+x (2 b-d)-c)}{4 (2 b-d)}+\frac {\cos (2 a+x (2 b+d)+c)}{4 (2 b+d)}-\frac {\cos (c+d x)}{2 d} \]
Antiderivative was successfully verified.
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Rule 2638
Rule 4569
Rubi steps
\begin {align*} \int \sin ^2(a+b x) \sin (c+d x) \, dx &=\int \left (\frac {1}{4} \sin (2 a-c+(2 b-d) x)+\frac {1}{2} \sin (c+d x)-\frac {1}{4} \sin (2 a+c+(2 b+d) x)\right ) \, dx\\ &=\frac {1}{4} \int \sin (2 a-c+(2 b-d) x) \, dx-\frac {1}{4} \int \sin (2 a+c+(2 b+d) x) \, dx+\frac {1}{2} \int \sin (c+d x) \, dx\\ &=-\frac {\cos (2 a-c+(2 b-d) x)}{4 (2 b-d)}-\frac {\cos (c+d x)}{2 d}+\frac {\cos (2 a+c+(2 b+d) x)}{4 (2 b+d)}\\ \end {align*}
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Mathematica [A] time = 0.34, size = 80, normalized size = 1.18 \[ -\frac {\cos (2 a+2 b x-c-d x)}{4 (2 b-d)}+\frac {\cos (2 a+x (2 b+d)+c)}{4 (2 b+d)}+\frac {1}{2} \left (\frac {\sin (c) \sin (d x)}{d}-\frac {\cos (c) \cos (d x)}{d}\right ) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 69, normalized size = 1.01 \[ -\frac {2 \, b d \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (d x + c\right ) + {\left (d^{2} \cos \left (b x + a\right )^{2} + 2 \, b^{2} - d^{2}\right )} \cos \left (d x + c\right )}{4 \, b^{2} d - d^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.20, size = 61, normalized size = 0.90 \[ \frac {\cos \left (2 \, b x + d x + 2 \, a + c\right )}{4 \, {\left (2 \, b + d\right )}} - \frac {\cos \left (2 \, b x - d x + 2 \, a - c\right )}{4 \, {\left (2 \, b - d\right )}} - \frac {\cos \left (d x + c\right )}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.15, size = 63, normalized size = 0.93 \[ -\frac {\cos \left (2 a -c +\left (2 b -d \right ) x \right )}{4 \left (2 b -d \right )}-\frac {\cos \left (d x +c \right )}{2 d}+\frac {\cos \left (2 a +c +\left (2 b +d \right ) x \right )}{8 b +4 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.35, size = 371, normalized size = 5.46 \[ -\frac {{\left (2 \, b d \cos \relax (c) - d^{2} \cos \relax (c)\right )} \cos \left ({\left (2 \, b + d\right )} x + 2 \, a + 2 \, c\right ) + {\left (2 \, b d \cos \relax (c) - d^{2} \cos \relax (c)\right )} \cos \left ({\left (2 \, b + d\right )} x + 2 \, a\right ) - {\left (2 \, b d \cos \relax (c) + d^{2} \cos \relax (c)\right )} \cos \left (-{\left (2 \, b - d\right )} x - 2 \, a + 2 \, c\right ) - {\left (2 \, b d \cos \relax (c) + d^{2} \cos \relax (c)\right )} \cos \left (-{\left (2 \, b - d\right )} x - 2 \, a\right ) - 2 \, {\left (4 \, b^{2} \cos \relax (c) - d^{2} \cos \relax (c)\right )} \cos \left (d x + 2 \, c\right ) - 2 \, {\left (4 \, b^{2} \cos \relax (c) - d^{2} \cos \relax (c)\right )} \cos \left (d x\right ) + {\left (2 \, b d \sin \relax (c) - d^{2} \sin \relax (c)\right )} \sin \left ({\left (2 \, b + d\right )} x + 2 \, a + 2 \, c\right ) - {\left (2 \, b d \sin \relax (c) - d^{2} \sin \relax (c)\right )} \sin \left ({\left (2 \, b + d\right )} x + 2 \, a\right ) - {\left (2 \, b d \sin \relax (c) + d^{2} \sin \relax (c)\right )} \sin \left (-{\left (2 \, b - d\right )} x - 2 \, a + 2 \, c\right ) + {\left (2 \, b d \sin \relax (c) + d^{2} \sin \relax (c)\right )} \sin \left (-{\left (2 \, b - d\right )} x - 2 \, a\right ) - 2 \, {\left (4 \, b^{2} \sin \relax (c) - d^{2} \sin \relax (c)\right )} \sin \left (d x + 2 \, c\right ) + 2 \, {\left (4 \, b^{2} \sin \relax (c) - d^{2} \sin \relax (c)\right )} \sin \left (d x\right )}{8 \, {\left ({\left (\cos \relax (c)^{2} + \sin \relax (c)^{2}\right )} d^{3} - 4 \, {\left (b^{2} \cos \relax (c)^{2} + b^{2} \sin \relax (c)^{2}\right )} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.75, size = 105, normalized size = 1.54 \[ -\frac {d^2\,\cos \left (2\,a+c+2\,b\,x+d\,x\right )-b\,\left (2\,d\,\cos \left (2\,a+c+2\,b\,x+d\,x\right )-2\,d\,\cos \left (2\,a-c+2\,b\,x-d\,x\right )\right )+d^2\,\cos \left (2\,a-c+2\,b\,x-d\,x\right )}{16\,b^2\,d-4\,d^3}-\frac {\cos \left (c+d\,x\right )}{2\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 6.51, size = 408, normalized size = 6.00 \[ \begin {cases} x \sin ^{2}{\relax (a )} \sin {\relax (c )} & \text {for}\: b = 0 \wedge d = 0 \\\frac {x \sin ^{2}{\left (a - \frac {d x}{2} \right )} \sin {\left (c + d x \right )}}{4} - \frac {x \sin {\left (a - \frac {d x}{2} \right )} \cos {\left (a - \frac {d x}{2} \right )} \cos {\left (c + d x \right )}}{2} - \frac {x \sin {\left (c + d x \right )} \cos ^{2}{\left (a - \frac {d x}{2} \right )}}{4} - \frac {\sin ^{2}{\left (a - \frac {d x}{2} \right )} \cos {\left (c + d x \right )}}{d} - \frac {\sin {\left (a - \frac {d x}{2} \right )} \sin {\left (c + d x \right )} \cos {\left (a - \frac {d x}{2} \right )}}{2 d} & \text {for}\: b = - \frac {d}{2} \\\frac {x \sin ^{2}{\left (a + \frac {d x}{2} \right )} \sin {\left (c + d x \right )}}{4} + \frac {x \sin {\left (a + \frac {d x}{2} \right )} \cos {\left (a + \frac {d x}{2} \right )} \cos {\left (c + d x \right )}}{2} - \frac {x \sin {\left (c + d x \right )} \cos ^{2}{\left (a + \frac {d x}{2} \right )}}{4} - \frac {\sin ^{2}{\left (a + \frac {d x}{2} \right )} \cos {\left (c + d x \right )}}{d} + \frac {\sin {\left (a + \frac {d x}{2} \right )} \sin {\left (c + d x \right )} \cos {\left (a + \frac {d x}{2} \right )}}{2 d} & \text {for}\: b = \frac {d}{2} \\\left (\frac {x \sin ^{2}{\left (a + b x \right )}}{2} + \frac {x \cos ^{2}{\left (a + b x \right )}}{2} - \frac {\sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b}\right ) \sin {\relax (c )} & \text {for}\: d = 0 \\- \frac {2 b^{2} \sin ^{2}{\left (a + b x \right )} \cos {\left (c + d x \right )}}{4 b^{2} d - d^{3}} - \frac {2 b^{2} \cos ^{2}{\left (a + b x \right )} \cos {\left (c + d x \right )}}{4 b^{2} d - d^{3}} - \frac {2 b d \sin {\left (a + b x \right )} \sin {\left (c + d x \right )} \cos {\left (a + b x \right )}}{4 b^{2} d - d^{3}} + \frac {d^{2} \sin ^{2}{\left (a + b x \right )} \cos {\left (c + d x \right )}}{4 b^{2} d - d^{3}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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