Integrand size = 14, antiderivative size = 27 \[ \int \sin (c-b x) \sin (a+b x) \, dx=-\frac {1}{2} x \cos (a+c)+\frac {\sin (a-c+2 b x)}{4 b} \]
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Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4665, 2717} \[ \int \sin (c-b x) \sin (a+b x) \, dx=\frac {\sin (a+2 b x-c)}{4 b}-\frac {1}{2} x \cos (a+c) \]
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Rule 2717
Rule 4665
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {1}{2} \cos (a+c)+\frac {1}{2} \cos (a-c+2 b x)\right ) \, dx \\ & = -\frac {1}{2} x \cos (a+c)+\frac {1}{2} \int \cos (a-c+2 b x) \, dx \\ & = -\frac {1}{2} x \cos (a+c)+\frac {\sin (a-c+2 b x)}{4 b} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \sin (c-b x) \sin (a+b x) \, dx=\frac {-2 b x \cos (a+c)+\sin (a-c+2 b x)}{4 b} \]
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Time = 0.47 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89
method | result | size |
default | \(-\frac {x \cos \left (a +c \right )}{2}+\frac {\sin \left (2 x b +a -c \right )}{4 b}\) | \(24\) |
risch | \(-\frac {x \cos \left (a +c \right )}{2}+\frac {\sin \left (2 x b +a -c \right )}{4 b}\) | \(24\) |
parallelrisch | \(-\frac {2 x \cos \left (a +c \right ) b -\sin \left (2 x b +a -c \right )+\sin \left (a +c \right )}{4 b}\) | \(31\) |
norman | \(\frac {\frac {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{b}-\frac {x}{2}+\frac {x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{2}+\frac {x \tan \left (\frac {x b}{2}-\frac {c}{2}\right )^{2}}{2}-2 x \tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \tan \left (\frac {x b}{2}-\frac {c}{2}\right )-\frac {x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2} \tan \left (\frac {x b}{2}-\frac {c}{2}\right )^{2}}{2}-\frac {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \tan \left (\frac {x b}{2}-\frac {c}{2}\right )^{2}}{b}}{\left (1+\tan \left (\frac {x b}{2}-\frac {c}{2}\right )^{2}\right ) \left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right )}\) | \(148\) |
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Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int \sin (c-b x) \sin (a+b x) \, dx=-\frac {b x \cos \left (a + c\right ) - \cos \left (b x + a\right ) \cos \left (a + c\right ) \sin \left (b x + a\right ) + \cos \left (b x + a\right )^{2} \sin \left (a + c\right )}{2 \, b} \]
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Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.26 \[ \int \sin (c-b x) \sin (a+b x) \, dx=- \begin {cases} \frac {x \sin {\left (a + b x \right )} \sin {\left (b x - c \right )}}{2} + \frac {x \cos {\left (a + b x \right )} \cos {\left (b x - c \right )}}{2} - \frac {\sin {\left (a + b x \right )} \cos {\left (b x - c \right )}}{2 b} & \text {for}\: b \neq 0 \\- x \sin {\left (a \right )} \sin {\left (c \right )} & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \sin (c-b x) \sin (a+b x) \, dx=-\frac {1}{2} \, x \cos \left (a + c\right ) + \frac {\sin \left (2 \, b x + a - c\right )}{4 \, b} \]
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Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \sin (c-b x) \sin (a+b x) \, dx=-\frac {1}{2} \, x \cos \left (a + c\right ) + \frac {\sin \left (2 \, b x + a - c\right )}{4 \, b} \]
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Time = 27.34 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \sin (c-b x) \sin (a+b x) \, dx=\left \{\begin {array}{cl} x\,\sin \left (a\right )\,\sin \left (c\right ) & \text {\ if\ \ }b=0\\ \frac {\sin \left (a-c+2\,b\,x\right )}{4\,b}-\frac {x\,\cos \left (a+c\right )}{2} & \text {\ if\ \ }b\neq 0 \end {array}\right . \]
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