Integrand size = 20, antiderivative size = 60 \[ \int \csc ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=x+\frac {\cos (a+b x) \sin (a+b x)}{b}+\frac {2 \cos ^3(a+b x) \sin (a+b x)}{3 b}-\frac {8 \cos ^5(a+b x) \sin (a+b x)}{3 b} \]
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Time = 0.10 (sec) , antiderivative size = 60, 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, 2648, 2715, 8} \[ \int \csc ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {8 \sin (a+b x) \cos ^5(a+b x)}{3 b}+\frac {2 \sin (a+b x) \cos ^3(a+b x)}{3 b}+\frac {\sin (a+b x) \cos (a+b x)}{b}+x \]
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Rule 8
Rule 2648
Rule 2715
Rule 4373
Rubi steps \begin{align*} \text {integral}& = 16 \int \cos ^4(a+b x) \sin ^2(a+b x) \, dx \\ & = -\frac {8 \cos ^5(a+b x) \sin (a+b x)}{3 b}+\frac {8}{3} \int \cos ^4(a+b x) \, dx \\ & = \frac {2 \cos ^3(a+b x) \sin (a+b x)}{3 b}-\frac {8 \cos ^5(a+b x) \sin (a+b x)}{3 b}+2 \int \cos ^2(a+b x) \, dx \\ & = \frac {\cos (a+b x) \sin (a+b x)}{b}+\frac {2 \cos ^3(a+b x) \sin (a+b x)}{3 b}-\frac {8 \cos ^5(a+b x) \sin (a+b x)}{3 b}+\int 1 \, dx \\ & = x+\frac {\cos (a+b x) \sin (a+b x)}{b}+\frac {2 \cos ^3(a+b x) \sin (a+b x)}{3 b}-\frac {8 \cos ^5(a+b x) \sin (a+b x)}{3 b} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.67 \[ \int \csc ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {-12 b x-3 \sin (2 (a+b x))+3 \sin (4 (a+b x))+\sin (6 (a+b x))}{12 b} \]
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Time = 3.56 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.75
method | result | size |
risch | \(x -\frac {\sin \left (6 x b +6 a \right )}{12 b}-\frac {\sin \left (4 x b +4 a \right )}{4 b}+\frac {\sin \left (2 x b +2 a \right )}{4 b}\) | \(45\) |
default | \(\frac {-\frac {8 \cos \left (x b +a \right )^{5} \sin \left (x b +a \right )}{3}+\frac {2 \left (\cos \left (x b +a \right )^{3}+\frac {3 \cos \left (x b +a \right )}{2}\right ) \sin \left (x b +a \right )}{3}+x b +a}{b}\) | \(55\) |
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Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.78 \[ \int \csc ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {3 \, b x - {\left (8 \, \cos \left (b x + a\right )^{5} - 2 \, \cos \left (b x + a\right )^{3} - 3 \, \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{3 \, b} \]
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Timed out. \[ \int \csc ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\text {Timed out} \]
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Time = 0.21 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.72 \[ \int \csc ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {12 \, b x - \sin \left (6 \, b x + 6 \, a\right ) - 3 \, \sin \left (4 \, b x + 4 \, a\right ) + 3 \, \sin \left (2 \, b x + 2 \, a\right )}{12 \, b} \]
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Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92 \[ \int \csc ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {3 \, b x + 3 \, a + \frac {3 \, \tan \left (b x + a\right )^{5} + 8 \, \tan \left (b x + a\right )^{3} - 3 \, \tan \left (b x + a\right )}{{\left (\tan \left (b x + a\right )^{2} + 1\right )}^{3}}}{3 \, b} \]
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Time = 20.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.08 \[ \int \csc ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=x+\frac {{\mathrm {tan}\left (a+b\,x\right )}^5+\frac {8\,{\mathrm {tan}\left (a+b\,x\right )}^3}{3}-\mathrm {tan}\left (a+b\,x\right )}{b\,\left ({\mathrm {tan}\left (a+b\,x\right )}^6+3\,{\mathrm {tan}\left (a+b\,x\right )}^4+3\,{\mathrm {tan}\left (a+b\,x\right )}^2+1\right )} \]
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