Integrand size = 15, antiderivative size = 62 \[ \int \sin (a+b x) \sin ^2(c+d x) \, dx=-\frac {\cos (a+b x)}{2 b}+\frac {\cos (a-2 c+(b-2 d) x)}{4 (b-2 d)}+\frac {\cos (a+2 c+(b+2 d) x)}{4 (b+2 d)} \]
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Time = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4665, 2718} \[ \int \sin (a+b x) \sin ^2(c+d x) \, dx=\frac {\cos (a+x (b-2 d)-2 c)}{4 (b-2 d)}+\frac {\cos (a+x (b+2 d)+2 c)}{4 (b+2 d)}-\frac {\cos (a+b x)}{2 b} \]
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Rule 2718
Rule 4665
Rubi steps \begin{align*} \text {integral}& = \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 \\ & = -\left (\frac {1}{4} \int \sin (a-2 c+(b-2 d) x) \, dx\right )-\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*}
Time = 0.81 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.11 \[ \int \sin (a+b x) \sin ^2(c+d x) \, dx=\frac {1}{4} \left (-\frac {2 \cos (a) \cos (b x)}{b}+\frac {\cos (a-2 c+b x-2 d x)}{b-2 d}+\frac {\cos (a+2 c+b x+2 d x)}{b+2 d}+\frac {2 \sin (a) \sin (b x)}{b}\right ) \]
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Time = 0.53 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92
method | result | size |
default | \(-\frac {\cos \left (x b +a \right )}{2 b}+\frac {\cos \left (a -2 c +\left (b -2 d \right ) x \right )}{4 b -8 d}+\frac {\cos \left (a +2 c +\left (b +2 d \right ) x \right )}{4 b +8 d}\) | \(57\) |
risch | \(-\frac {\cos \left (x b +a \right )}{2 b}+\frac {\cos \left (x b -2 d x +a -2 c \right )}{4 b -8 d}+\frac {\cos \left (x b +2 d x +a +2 c \right )}{4 b +8 d}\) | \(57\) |
parallelrisch | \(\frac {b \left (b +2 d \right ) \cos \left (a -2 c +\left (b -2 d \right ) x \right )+b \left (b -2 d \right ) \cos \left (a +2 c +\left (b +2 d \right ) x \right )+\left (-2 b^{2}+8 d^{2}\right ) \cos \left (x b +a \right )-8 d^{2}}{4 b^{3}-16 b \,d^{2}}\) | \(80\) |
norman | \(\frac {\frac {4 d^{2}}{b \left (b^{2}-4 d^{2}\right )}+\frac {4 d^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{b \left (b^{2}-4 d^{2}\right )}+\frac {8 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{2}-4 d^{2}}-\frac {8 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{b^{2}-4 d^{2}}+\frac {4 b \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b^{2}-4 d^{2}}+\frac {2 \left (-2 b^{2}+4 d^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b \left (b^{2}-4 d^{2}\right )}}{\left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}\) | \(222\) |
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Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.15 \[ \int \sin (a+b x) \sin ^2(c+d x) \, dx=\frac {b^{2} \cos \left (b x + a\right ) \cos \left (d x + c\right )^{2} + 2 \, b d \cos \left (d x + c\right ) \sin \left (b x + a\right ) \sin \left (d x + c\right ) - {\left (b^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right )}{b^{3} - 4 \, b d^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 408 vs. \(2 (49) = 98\).
Time = 0.67 (sec) , antiderivative size = 408, normalized size of antiderivative = 6.58 \[ \int \sin (a+b x) \sin ^2(c+d x) \, dx=\begin {cases} x \sin {\left (a \right )} \sin ^{2}{\left (c \right )} & \text {for}\: b = 0 \wedge d = 0 \\\left (\frac {x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {x \cos ^{2}{\left (c + d x \right )}}{2} - \frac {\sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d}\right ) \sin {\left (a \right )} & \text {for}\: b = 0 \\\frac {x \sin {\left (a - 2 d x \right )} \sin ^{2}{\left (c + d x \right )}}{4} - \frac {x \sin {\left (a - 2 d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} - \frac {x \sin {\left (c + d x \right )} \cos {\left (a - 2 d x \right )} \cos {\left (c + d x \right )}}{2} - \frac {3 \sin {\left (a - 2 d x \right )} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {\cos {\left (a - 2 d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: b = - 2 d \\\frac {x \sin {\left (a + 2 d x \right )} \sin ^{2}{\left (c + d x \right )}}{4} - \frac {x \sin {\left (a + 2 d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {x \sin {\left (c + d x \right )} \cos {\left (a + 2 d x \right )} \cos {\left (c + d x \right )}}{2} - \frac {3 \sin {\left (a + 2 d x \right )} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} - \frac {\cos {\left (a + 2 d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: b = 2 d \\- \frac {b^{2} \sin ^{2}{\left (c + d x \right )} \cos {\left (a + b x \right )}}{b^{3} - 4 b d^{2}} + \frac {2 b d \sin {\left (a + b x \right )} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{b^{3} - 4 b d^{2}} + \frac {2 d^{2} \sin ^{2}{\left (c + d x \right )} \cos {\left (a + b x \right )}}{b^{3} - 4 b d^{2}} + \frac {2 d^{2} \cos {\left (a + b x \right )} \cos ^{2}{\left (c + d x \right )}}{b^{3} - 4 b d^{2}} & \text {otherwise} \end {cases} \]
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Leaf count of result is larger than twice the leaf count of optimal. 414 vs. \(2 (56) = 112\).
Time = 0.25 (sec) , antiderivative size = 414, normalized size of antiderivative = 6.68 \[ \int \sin (a+b x) \sin ^2(c+d x) \, dx=\frac {{\left (b^{2} \cos \left (2 \, c\right ) - 2 \, b d \cos \left (2 \, c\right )\right )} \cos \left ({\left (b + 2 \, d\right )} x + a + 4 \, c\right ) + {\left (b^{2} \cos \left (2 \, c\right ) - 2 \, b d \cos \left (2 \, c\right )\right )} \cos \left ({\left (b + 2 \, d\right )} x + a\right ) + {\left (b^{2} \cos \left (2 \, c\right ) + 2 \, b d \cos \left (2 \, c\right )\right )} \cos \left (-{\left (b - 2 \, d\right )} x - a + 4 \, c\right ) + {\left (b^{2} \cos \left (2 \, c\right ) + 2 \, b d \cos \left (2 \, c\right )\right )} \cos \left (-{\left (b - 2 \, d\right )} x - a\right ) - 2 \, {\left (b^{2} \cos \left (2 \, c\right ) - 4 \, d^{2} \cos \left (2 \, c\right )\right )} \cos \left (b x + a + 2 \, c\right ) - 2 \, {\left (b^{2} \cos \left (2 \, c\right ) - 4 \, d^{2} \cos \left (2 \, c\right )\right )} \cos \left (b x + a - 2 \, c\right ) + {\left (b^{2} \sin \left (2 \, c\right ) - 2 \, b d \sin \left (2 \, c\right )\right )} \sin \left ({\left (b + 2 \, d\right )} x + a + 4 \, c\right ) - {\left (b^{2} \sin \left (2 \, c\right ) - 2 \, b d \sin \left (2 \, c\right )\right )} \sin \left ({\left (b + 2 \, d\right )} x + a\right ) + {\left (b^{2} \sin \left (2 \, c\right ) + 2 \, b d \sin \left (2 \, c\right )\right )} \sin \left (-{\left (b - 2 \, d\right )} x - a + 4 \, c\right ) - {\left (b^{2} \sin \left (2 \, c\right ) + 2 \, b d \sin \left (2 \, c\right )\right )} \sin \left (-{\left (b - 2 \, d\right )} x - a\right ) - 2 \, {\left (b^{2} \sin \left (2 \, c\right ) - 4 \, d^{2} \sin \left (2 \, c\right )\right )} \sin \left (b x + a + 2 \, c\right ) + 2 \, {\left (b^{2} \sin \left (2 \, c\right ) - 4 \, d^{2} \sin \left (2 \, c\right )\right )} \sin \left (b x + a - 2 \, c\right )}{8 \, {\left (b^{3} \cos \left (2 \, c\right )^{2} + b^{3} \sin \left (2 \, c\right )^{2} - 4 \, {\left (b \cos \left (2 \, c\right )^{2} + b \sin \left (2 \, c\right )^{2}\right )} d^{2}\right )}} \]
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Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.90 \[ \int \sin (a+b x) \sin ^2(c+d x) \, dx=\frac {\cos \left (b x + 2 \, d x + a + 2 \, c\right )}{4 \, {\left (b + 2 \, d\right )}} + \frac {\cos \left (b x - 2 \, d x + a - 2 \, c\right )}{4 \, {\left (b - 2 \, d\right )}} - \frac {\cos \left (b x + a\right )}{2 \, b} \]
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Time = 20.80 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.58 \[ \int \sin (a+b x) \sin ^2(c+d x) \, dx=-\frac {d\,\left (2\,b\,\cos \left (a-2\,c+b\,x-2\,d\,x\right )-2\,b\,\cos \left (a+2\,c+b\,x+2\,d\,x\right )\right )+b^2\,\cos \left (a-2\,c+b\,x-2\,d\,x\right )+b^2\,\cos \left (a+2\,c+b\,x+2\,d\,x\right )}{16\,b\,d^2-4\,b^3}-\frac {\cos \left (a+b\,x\right )}{2\,b} \]
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