Integrand size = 18, antiderivative size = 15 \[ \int \cos ^3(a+b x) \sin (2 a+2 b x) \, dx=-\frac {2 \cos ^5(a+b x)}{5 b} \]
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Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4372, 2645, 30} \[ \int \cos ^3(a+b x) \sin (2 a+2 b x) \, dx=-\frac {2 \cos ^5(a+b x)}{5 b} \]
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Rule 30
Rule 2645
Rule 4372
Rubi steps \begin{align*} \text {integral}& = 2 \int \cos ^4(a+b x) \sin (a+b x) \, dx \\ & = -\frac {2 \text {Subst}\left (\int x^4 \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {2 \cos ^5(a+b x)}{5 b} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \cos ^3(a+b x) \sin (2 a+2 b x) \, dx=-\frac {2 \cos ^5(a+b x)}{5 b} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(40\) vs. \(2(13)=26\).
Time = 0.45 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.73
method | result | size |
default | \(-\frac {\cos \left (x b +a \right )}{4 b}-\frac {\cos \left (3 x b +3 a \right )}{8 b}-\frac {\cos \left (5 x b +5 a \right )}{40 b}\) | \(41\) |
risch | \(-\frac {\cos \left (x b +a \right )}{4 b}-\frac {\cos \left (3 x b +3 a \right )}{8 b}-\frac {\cos \left (5 x b +5 a \right )}{40 b}\) | \(41\) |
parallelrisch | \(\frac {-\frac {4}{5}+4 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{5} \tan \left (x b +a \right )+4 \left (-\tan \left (x b +a \right )^{2}+1\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}-8 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3} \tan \left (x b +a \right )-\frac {12 \tan \left (x b +a \right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{5}+\frac {4 \tan \left (x b +a \right )^{2}}{5}}{b \left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right )^{3} \left (1+\tan \left (x b +a \right )^{2}\right )}\) | \(123\) |
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none
Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \cos ^3(a+b x) \sin (2 a+2 b x) \, dx=-\frac {2 \, \cos \left (b x + a\right )^{5}}{5 \, b} \]
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Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (14) = 28\).
Time = 0.79 (sec) , antiderivative size = 117, normalized size of antiderivative = 7.80 \[ \int \cos ^3(a+b x) \sin (2 a+2 b x) \, dx=\begin {cases} - \frac {2 \sin ^{3}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )}}{5 b} - \frac {4 \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{5 b} + \frac {\sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{5 b} - \frac {2 \cos ^{3}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{5 b} & \text {for}\: b \neq 0 \\x \sin {\left (2 a \right )} \cos ^{3}{\left (a \right )} & \text {otherwise} \end {cases} \]
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Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (13) = 26\).
Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.27 \[ \int \cos ^3(a+b x) \sin (2 a+2 b x) \, dx=-\frac {\cos \left (5 \, b x + 5 \, a\right ) + 5 \, \cos \left (3 \, b x + 3 \, a\right ) + 10 \, \cos \left (b x + a\right )}{40 \, b} \]
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Time = 0.31 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \cos ^3(a+b x) \sin (2 a+2 b x) \, dx=-\frac {2 \, \cos \left (b x + a\right )^{5}}{5 \, b} \]
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Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \cos ^3(a+b x) \sin (2 a+2 b x) \, dx=-\frac {2\,{\cos \left (a+b\,x\right )}^5}{5\,b} \]
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