Integrand size = 20, antiderivative size = 46 \[ \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {4 \cos ^3(a+b x)}{3 b}+\frac {8 \cos ^5(a+b x)}{5 b}-\frac {4 \cos ^7(a+b x)}{7 b} \]
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Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4373, 2645, 276} \[ \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {4 \cos ^7(a+b x)}{7 b}+\frac {8 \cos ^5(a+b x)}{5 b}-\frac {4 \cos ^3(a+b x)}{3 b} \]
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Rule 276
Rule 2645
Rule 4373
Rubi steps \begin{align*} \text {integral}& = 4 \int \cos ^2(a+b x) \sin ^5(a+b x) \, dx \\ & = -\frac {4 \text {Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {4 \text {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {4 \cos ^3(a+b x)}{3 b}+\frac {8 \cos ^5(a+b x)}{5 b}-\frac {4 \cos ^7(a+b x)}{7 b} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {\cos ^3(a+b x) (-157+108 \cos (2 (a+b x))-15 \cos (4 (a+b x)))}{210 b} \]
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Time = 1.51 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.20
method | result | size |
default | \(-\frac {5 \cos \left (x b +a \right )}{16 b}-\frac {\cos \left (3 x b +3 a \right )}{48 b}+\frac {3 \cos \left (5 x b +5 a \right )}{80 b}-\frac {\cos \left (7 x b +7 a \right )}{112 b}\) | \(55\) |
risch | \(-\frac {5 \cos \left (x b +a \right )}{16 b}-\frac {\cos \left (3 x b +3 a \right )}{48 b}+\frac {3 \cos \left (5 x b +5 a \right )}{80 b}-\frac {\cos \left (7 x b +7 a \right )}{112 b}\) | \(55\) |
parallelrisch | \(\frac {24+\left (64 \tan \left (x b +a \right )^{4}+176 \tan \left (x b +a \right )^{2}+88\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}-192 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{5} \tan \left (x b +a \right )^{3}+\left (384 \tan \left (x b +a \right )^{4}-48 \tan \left (x b +a \right )^{2}+264\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}+\left (768 \tan \left (x b +a \right )^{3}-768 \tan \left (x b +a \right )\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}+\left (576 \tan \left (x b +a \right )^{2}-120\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}+192 \tan \left (x b +a \right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{105 b \left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right )^{3} \left (1+\tan \left (x b +a \right )^{2}\right )^{2}}\) | \(196\) |
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Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {4 \, {\left (15 \, \cos \left (b x + a\right )^{7} - 42 \, \cos \left (b x + a\right )^{5} + 35 \, \cos \left (b x + a\right )^{3}\right )}}{105 \, b} \]
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Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (39) = 78\).
Time = 2.07 (sec) , antiderivative size = 202, normalized size of antiderivative = 4.39 \[ \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=\begin {cases} - \frac {12 \sin ^{3}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{35 b} - \frac {11 \sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{35 b} - \frac {24 \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{35 b} + \frac {8 \sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{35 b} - \frac {38 \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )}}{105 b} - \frac {32 \cos ^{3}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{105 b} & \text {for}\: b \neq 0 \\x \sin ^{3}{\left (a \right )} \sin ^{2}{\left (2 a \right )} & \text {otherwise} \end {cases} \]
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Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02 \[ \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {15 \, \cos \left (7 \, b x + 7 \, a\right ) - 63 \, \cos \left (5 \, b x + 5 \, a\right ) + 35 \, \cos \left (3 \, b x + 3 \, a\right ) + 525 \, \cos \left (b x + a\right )}{1680 \, b} \]
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Time = 0.32 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02 \[ \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {15 \, \cos \left (7 \, b x + 7 \, a\right ) - 63 \, \cos \left (5 \, b x + 5 \, a\right ) + 35 \, \cos \left (3 \, b x + 3 \, a\right ) + 525 \, \cos \left (b x + a\right )}{1680 \, b} \]
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Time = 19.23 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {4\,\left (15\,{\cos \left (a+b\,x\right )}^7-42\,{\cos \left (a+b\,x\right )}^5+35\,{\cos \left (a+b\,x\right )}^3\right )}{105\,b} \]
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