Integrand size = 20, antiderivative size = 44 \[ \int \csc ^2(a+b x) \sin ^7(2 a+2 b x) \, dx=-\frac {16 \cos ^8(a+b x)}{b}+\frac {128 \cos ^{10}(a+b x)}{5 b}-\frac {32 \cos ^{12}(a+b x)}{3 b} \]
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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, 2645, 272, 45} \[ \int \csc ^2(a+b x) \sin ^7(2 a+2 b x) \, dx=-\frac {32 \cos ^{12}(a+b x)}{3 b}+\frac {128 \cos ^{10}(a+b x)}{5 b}-\frac {16 \cos ^8(a+b x)}{b} \]
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Rule 45
Rule 272
Rule 2645
Rule 4373
Rubi steps \begin{align*} \text {integral}& = 128 \int \cos ^7(a+b x) \sin ^5(a+b x) \, dx \\ & = -\frac {128 \text {Subst}\left (\int x^7 \left (1-x^2\right )^2 \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {64 \text {Subst}\left (\int (1-x)^2 x^3 \, dx,x,\cos ^2(a+b x)\right )}{b} \\ & = -\frac {64 \text {Subst}\left (\int \left (x^3-2 x^4+x^5\right ) \, dx,x,\cos ^2(a+b x)\right )}{b} \\ & = -\frac {16 \cos ^8(a+b x)}{b}+\frac {128 \cos ^{10}(a+b x)}{5 b}-\frac {32 \cos ^{12}(a+b x)}{3 b} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.34 \[ \int \csc ^2(a+b x) \sin ^7(2 a+2 b x) \, dx=\frac {64 \sin ^6(a+b x)}{3 b}-\frac {48 \sin ^8(a+b x)}{b}+\frac {192 \sin ^{10}(a+b x)}{5 b}-\frac {32 \sin ^{12}(a+b x)}{3 b} \]
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Time = 28.90 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.07
method | result | size |
default | \(-\frac {128 \left (\frac {\sin \left (x b +a \right )^{12}}{12}-\frac {3 \sin \left (x b +a \right )^{10}}{10}+\frac {3 \sin \left (x b +a \right )^{8}}{8}-\frac {\sin \left (x b +a \right )^{6}}{6}\right )}{b}\) | \(47\) |
risch | \(-\frac {\cos \left (12 x b +12 a \right )}{192 b}-\frac {\cos \left (10 x b +10 a \right )}{80 b}+\frac {\cos \left (8 x b +8 a \right )}{32 b}+\frac {5 \cos \left (6 x b +6 a \right )}{48 b}-\frac {5 \cos \left (4 x b +4 a \right )}{64 b}-\frac {5 \cos \left (2 x b +2 a \right )}{8 b}\) | \(86\) |
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Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int \csc ^2(a+b x) \sin ^7(2 a+2 b x) \, dx=-\frac {16 \, {\left (10 \, \cos \left (b x + a\right )^{12} - 24 \, \cos \left (b x + a\right )^{10} + 15 \, \cos \left (b x + a\right )^{8}\right )}}{15 \, b} \]
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Timed out. \[ \int \csc ^2(a+b x) \sin ^7(2 a+2 b x) \, dx=\text {Timed out} \]
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Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.64 \[ \int \csc ^2(a+b x) \sin ^7(2 a+2 b x) \, dx=-\frac {5 \, \cos \left (12 \, b x + 12 \, a\right ) + 12 \, \cos \left (10 \, b x + 10 \, a\right ) - 30 \, \cos \left (8 \, b x + 8 \, a\right ) - 100 \, \cos \left (6 \, b x + 6 \, a\right ) + 75 \, \cos \left (4 \, b x + 4 \, a\right ) + 600 \, \cos \left (2 \, b x + 2 \, a\right )}{960 \, b} \]
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Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \csc ^2(a+b x) \sin ^7(2 a+2 b x) \, dx=-\frac {16 \, {\left (10 \, \sin \left (b x + a\right )^{12} - 36 \, \sin \left (b x + a\right )^{10} + 45 \, \sin \left (b x + a\right )^{8} - 20 \, \sin \left (b x + a\right )^{6}\right )}}{15 \, b} \]
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Time = 19.91 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.80 \[ \int \csc ^2(a+b x) \sin ^7(2 a+2 b x) \, dx=-\frac {16\,{\cos \left (a+b\,x\right )}^8\,\left (10\,{\cos \left (a+b\,x\right )}^4-24\,{\cos \left (a+b\,x\right )}^2+15\right )}{15\,b} \]
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