Integrand size = 20, antiderivative size = 49 \[ \int \cos ^2(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {x}{4}-\frac {\cos (2 a+2 b x) \sin (2 a+2 b x)}{8 b}+\frac {\sin ^3(2 a+2 b x)}{12 b} \]
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Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4370, 2715, 8, 2644, 30} \[ \int \cos ^2(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {\sin ^3(2 a+2 b x)}{12 b}-\frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{8 b}+\frac {x}{4} \]
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Rule 8
Rule 30
Rule 2644
Rule 2715
Rule 4370
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \sin ^2(2 a+2 b x) \, dx+\frac {1}{2} \int \cos (2 a+2 b x) \sin ^2(2 a+2 b x) \, dx \\ & = -\frac {\cos (2 a+2 b x) \sin (2 a+2 b x)}{8 b}+\frac {\int 1 \, dx}{4}+\frac {\text {Subst}\left (\int x^2 \, dx,x,\sin (2 a+2 b x)\right )}{4 b} \\ & = \frac {x}{4}-\frac {\cos (2 a+2 b x) \sin (2 a+2 b x)}{8 b}+\frac {\sin ^3(2 a+2 b x)}{12 b} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.82 \[ \int \cos ^2(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {-12 b x-3 \sin (2 (a+b x))+3 \sin (4 (a+b x))+\sin (6 (a+b x))}{48 b} \]
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Time = 0.54 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90
method | result | size |
parallelrisch | \(\frac {12 x b +3 \sin \left (2 x b +2 a \right )-3 \sin \left (4 x b +4 a \right )-\sin \left (6 x b +6 a \right )}{48 b}\) | \(44\) |
default | \(\frac {x}{4}+\frac {\sin \left (2 x b +2 a \right )}{16 b}-\frac {\sin \left (4 x b +4 a \right )}{16 b}-\frac {\sin \left (6 x b +6 a \right )}{48 b}\) | \(47\) |
risch | \(\frac {x}{4}+\frac {\sin \left (2 x b +2 a \right )}{16 b}-\frac {\sin \left (4 x b +4 a \right )}{16 b}-\frac {\sin \left (6 x b +6 a \right )}{48 b}\) | \(47\) |
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Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.96 \[ \int \cos ^2(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {3 \, b x - {\left (8 \, \cos \left (b x + a\right )^{5} - 2 \, \cos \left (b x + a\right )^{3} - 3 \, \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{12 \, b} \]
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Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (41) = 82\).
Time = 0.82 (sec) , antiderivative size = 231, normalized size of antiderivative = 4.71 \[ \int \cos ^2(a+b x) \sin ^2(2 a+2 b x) \, dx=\begin {cases} \frac {x \sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )}}{4} + \frac {x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{4} + \frac {x \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} + \frac {x \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{4} + \frac {\sin ^{2}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{24 b} + \frac {\sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{6 b} + \frac {\sin {\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{3 b} - \frac {7 \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{24 b} & \text {for}\: b \neq 0 \\x \sin ^{2}{\left (2 a \right )} \cos ^{2}{\left (a \right )} & \text {otherwise} \end {cases} \]
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Time = 0.22 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.88 \[ \int \cos ^2(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {12 \, b x - \sin \left (6 \, b x + 6 \, a\right ) - 3 \, \sin \left (4 \, b x + 4 \, a\right ) + 3 \, \sin \left (2 \, b x + 2 \, a\right )}{48 \, b} \]
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Time = 0.31 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.94 \[ \int \cos ^2(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {12 \, b x + 12 \, a - \sin \left (6 \, b x + 6 \, a\right ) - 3 \, \sin \left (4 \, b x + 4 \, a\right ) + 3 \, \sin \left (2 \, b x + 2 \, a\right )}{48 \, b} \]
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Time = 19.74 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.88 \[ \int \cos ^2(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {x}{4}-\frac {\frac {\sin \left (4\,a+4\,b\,x\right )}{16}-\frac {\sin \left (2\,a+2\,b\,x\right )}{16}+\frac {\sin \left (6\,a+6\,b\,x\right )}{48}}{b} \]
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