Integrand size = 20, antiderivative size = 44 \[ \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {4 \sin ^8(a+b x)}{b}-\frac {32 \sin ^{10}(a+b x)}{5 b}+\frac {8 \sin ^{12}(a+b x)}{3 b} \]
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Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4373, 2644, 272, 45} \[ \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {8 \sin ^{12}(a+b x)}{3 b}-\frac {32 \sin ^{10}(a+b x)}{5 b}+\frac {4 \sin ^8(a+b x)}{b} \]
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Rule 45
Rule 272
Rule 2644
Rule 4373
Rubi steps \begin{align*} \text {integral}& = 32 \int \cos ^5(a+b x) \sin ^7(a+b x) \, dx \\ & = \frac {32 \text {Subst}\left (\int x^7 \left (1-x^2\right )^2 \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {16 \text {Subst}\left (\int (1-x)^2 x^3 \, dx,x,\sin ^2(a+b x)\right )}{b} \\ & = \frac {16 \text {Subst}\left (\int \left (x^3-2 x^4+x^5\right ) \, dx,x,\sin ^2(a+b x)\right )}{b} \\ & = \frac {4 \sin ^8(a+b x)}{b}-\frac {32 \sin ^{10}(a+b x)}{5 b}+\frac {8 \sin ^{12}(a+b x)}{3 b} \\ \end{align*}
Time = 0.38 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.55 \[ \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {-600 \cos (2 (a+b x))+75 \cos (4 (a+b x))+100 \cos (6 (a+b x))-30 \cos (8 (a+b x))-12 \cos (10 (a+b x))+5 \cos (12 (a+b x))}{3840 b} \]
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Time = 3.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.68
method | result | size |
parallelrisch | \(\frac {-600 \cos \left (2 x b +2 a \right )+462+5 \cos \left (12 x b +12 a \right )-12 \cos \left (10 x b +10 a \right )-30 \cos \left (8 x b +8 a \right )+100 \cos \left (6 x b +6 a \right )+75 \cos \left (4 x b +4 a \right )}{3840 b}\) | \(74\) |
default | \(-\frac {5 \cos \left (2 x b +2 a \right )}{32 b}+\frac {5 \cos \left (4 x b +4 a \right )}{256 b}+\frac {5 \cos \left (6 x b +6 a \right )}{192 b}-\frac {\cos \left (8 x b +8 a \right )}{128 b}-\frac {\cos \left (10 x b +10 a \right )}{320 b}+\frac {\cos \left (12 x b +12 a \right )}{768 b}\) | \(86\) |
risch | \(-\frac {5 \cos \left (2 x b +2 a \right )}{32 b}+\frac {5 \cos \left (4 x b +4 a \right )}{256 b}+\frac {5 \cos \left (6 x b +6 a \right )}{192 b}-\frac {\cos \left (8 x b +8 a \right )}{128 b}-\frac {\cos \left (10 x b +10 a \right )}{320 b}+\frac {\cos \left (12 x b +12 a \right )}{768 b}\) | \(86\) |
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Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {4 \, {\left (10 \, \cos \left (b x + a\right )^{12} - 36 \, \cos \left (b x + a\right )^{10} + 45 \, \cos \left (b x + a\right )^{8} - 20 \, \cos \left (b x + a\right )^{6}\right )}}{15 \, b} \]
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Leaf count of result is larger than twice the leaf count of optimal. 593 vs. \(2 (37) = 74\).
Time = 11.20 (sec) , antiderivative size = 593, normalized size of antiderivative = 13.48 \[ \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=\begin {cases} \frac {5 x \sin ^{2}{\left (a + b x \right )} \sin ^{5}{\left (2 a + 2 b x \right )}}{32} + \frac {5 x \sin ^{2}{\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{16} + \frac {5 x \sin ^{2}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{32} + \frac {5 x \sin {\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{16} + \frac {5 x \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{8} + \frac {5 x \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{16} - \frac {5 x \sin ^{5}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{32} - \frac {5 x \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{16} - \frac {5 x \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{32} - \frac {65 \sin ^{2}{\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{128 b} - \frac {2 \sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{3 b} - \frac {167 \sin ^{2}{\left (a + b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{640 b} + \frac {11 \sin {\left (a + b x \right )} \sin ^{5}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{64 b} + \frac {\sin {\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{4 b} + \frac {19 \sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{192 b} + \frac {\sin ^{4}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{128 b} - \frac {11 \cos ^{2}{\left (a + b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{1920 b} & \text {for}\: b \neq 0 \\x \sin ^{2}{\left (a \right )} \sin ^{5}{\left (2 a \right )} & \text {otherwise} \end {cases} \]
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Time = 0.23 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.64 \[ \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {5 \, \cos \left (12 \, b x + 12 \, a\right ) - 12 \, \cos \left (10 \, b x + 10 \, a\right ) - 30 \, \cos \left (8 \, b x + 8 \, a\right ) + 100 \, \cos \left (6 \, b x + 6 \, a\right ) + 75 \, \cos \left (4 \, b x + 4 \, a\right ) - 600 \, \cos \left (2 \, b x + 2 \, a\right )}{3840 \, b} \]
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Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {4 \, {\left (10 \, \sin \left (b x + a\right )^{12} - 24 \, \sin \left (b x + a\right )^{10} + 15 \, \sin \left (b x + a\right )^{8}\right )}}{15 \, b} \]
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Time = 19.38 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {-\frac {8\,{\cos \left (a+b\,x\right )}^{12}}{3}+\frac {48\,{\cos \left (a+b\,x\right )}^{10}}{5}-12\,{\cos \left (a+b\,x\right )}^8+\frac {16\,{\cos \left (a+b\,x\right )}^6}{3}}{b} \]
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