Integrand size = 20, antiderivative size = 42 \[ \int \cos ^2(a+b x) \csc ^4(2 a+2 b x) \, dx=-\frac {\cot (a+b x)}{8 b}-\frac {\cot ^3(a+b x)}{48 b}+\frac {\tan (a+b x)}{16 b} \]
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Time = 0.08 (sec) , antiderivative size = 42, 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 = {4372, 2700, 276} \[ \int \cos ^2(a+b x) \csc ^4(2 a+2 b x) \, dx=\frac {\tan (a+b x)}{16 b}-\frac {\cot ^3(a+b x)}{48 b}-\frac {\cot (a+b x)}{8 b} \]
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Rule 276
Rule 2700
Rule 4372
Rubi steps \begin{align*} \text {integral}& = \frac {1}{16} \int \csc ^4(a+b x) \sec ^2(a+b x) \, dx \\ & = \frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^4} \, dx,x,\tan (a+b x)\right )}{16 b} \\ & = \frac {\text {Subst}\left (\int \left (1+\frac {1}{x^4}+\frac {2}{x^2}\right ) \, dx,x,\tan (a+b x)\right )}{16 b} \\ & = -\frac {\cot (a+b x)}{8 b}-\frac {\cot ^3(a+b x)}{48 b}+\frac {\tan (a+b x)}{16 b} \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.14 \[ \int \cos ^2(a+b x) \csc ^4(2 a+2 b x) \, dx=-\frac {5 \cot (a+b x)}{48 b}-\frac {\cot (a+b x) \csc ^2(a+b x)}{48 b}+\frac {\tan (a+b x)}{16 b} \]
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Result contains complex when optimal does not.
Time = 5.42 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.10
method | result | size |
risch | \(\frac {i \left (2 \,{\mathrm e}^{2 i \left (x b +a \right )}-1\right )}{3 b \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}\) | \(46\) |
default | \(\frac {-\frac {1}{3 \cos \left (x b +a \right ) \sin \left (x b +a \right )^{3}}+\frac {4}{3 \sin \left (x b +a \right ) \cos \left (x b +a \right )}-\frac {8 \cot \left (x b +a \right )}{3}}{16 b}\) | \(51\) |
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Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.29 \[ \int \cos ^2(a+b x) \csc ^4(2 a+2 b x) \, dx=-\frac {8 \, \cos \left (b x + a\right )^{4} - 12 \, \cos \left (b x + a\right )^{2} + 3}{48 \, {\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )} \]
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Timed out. \[ \int \cos ^2(a+b x) \csc ^4(2 a+2 b x) \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (36) = 72\).
Time = 0.23 (sec) , antiderivative size = 308, normalized size of antiderivative = 7.33 \[ \int \cos ^2(a+b x) \csc ^4(2 a+2 b x) \, dx=\frac {{\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) - 1\right )} \sin \left (8 \, b x + 8 \, a\right ) - 2 \, {\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) - 1\right )} \sin \left (6 \, b x + 6 \, a\right ) - 2 \, \cos \left (8 \, b x + 8 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 4 \, \cos \left (6 \, b x + 6 \, a\right ) \sin \left (2 \, b x + 2 \, a\right )}{3 \, {\left (b \cos \left (8 \, b x + 8 \, a\right )^{2} + 4 \, b \cos \left (6 \, b x + 6 \, a\right )^{2} + 4 \, b \cos \left (2 \, b x + 2 \, a\right )^{2} + b \sin \left (8 \, b x + 8 \, a\right )^{2} + 4 \, b \sin \left (6 \, b x + 6 \, a\right )^{2} - 8 \, b \sin \left (6 \, b x + 6 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 4 \, b \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, {\left (2 \, b \cos \left (6 \, b x + 6 \, a\right ) - 2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )} \cos \left (8 \, b x + 8 \, a\right ) - 4 \, {\left (2 \, b \cos \left (2 \, b x + 2 \, a\right ) - b\right )} \cos \left (6 \, b x + 6 \, a\right ) - 4 \, b \cos \left (2 \, b x + 2 \, a\right ) - 4 \, {\left (b \sin \left (6 \, b x + 6 \, a\right ) - b \sin \left (2 \, b x + 2 \, a\right )\right )} \sin \left (8 \, b x + 8 \, a\right ) + b\right )}} \]
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Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.83 \[ \int \cos ^2(a+b x) \csc ^4(2 a+2 b x) \, dx=-\frac {\frac {6 \, \tan \left (b x + a\right )^{2} + 1}{\tan \left (b x + a\right )^{3}} - 3 \, \tan \left (b x + a\right )}{48 \, b} \]
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Time = 19.60 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int \cos ^2(a+b x) \csc ^4(2 a+2 b x) \, dx=\frac {\mathrm {tan}\left (a+b\,x\right )}{16\,b}-\frac {\frac {{\mathrm {tan}\left (a+b\,x\right )}^2}{8}+\frac {1}{48}}{b\,{\mathrm {tan}\left (a+b\,x\right )}^3} \]
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