Integrand size = 20, antiderivative size = 53 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=-\frac {2 \cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {4 \sin (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}} \]
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Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4393, 4388, 4377} \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {4 \sin (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}}-\frac {2 \cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]
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Rule 4377
Rule 4388
Rule 4393
Rubi steps \begin{align*} \text {integral}& = 2 \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx \\ & = -\frac {2 \cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {4}{3} \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx \\ & = -\frac {2 \cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {4 \sin (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}} \\ \end{align*}
Time = 0.22 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.81 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {\left (-\frac {1}{6} \cot (a+b x) \csc (a+b x)+\frac {1}{2} \sec (a+b x)\right ) \sqrt {\sin (2 (a+b x))}}{b} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 5.35 (sec) , antiderivative size = 194, normalized size of antiderivative = 3.66
method | result | size |
default | \(-\frac {\sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1}}\, \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right ) \left (2 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}+1\right )}{12 b \tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}}\) | \(194\) |
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Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.40 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {4 \, \cos \left (b x + a\right )^{3} + \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{2} - 3\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - 4 \, \cos \left (b x + a\right )}{6 \, {\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )}} \]
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Timed out. \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]
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\[ \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}} \,d x } \]
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Time = 23.19 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.94 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {4\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,\left (1+{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}-{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\right )}{3\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}^2\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )} \]
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