Integrand size = 22, antiderivative size = 48 \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {\operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )}{6 b}+\frac {\sin ^2(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]
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Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4381, 2720} \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {\sin ^2(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {\operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{6 b} \]
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Rule 2720
Rule 4381
Rubi steps \begin{align*} \text {integral}& = \frac {\sin ^2(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {1}{6} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx \\ & = \frac {\operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )}{6 b}+\frac {\sin ^2(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \\ \end{align*}
Time = 0.41 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.73 \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {\sec ^2(a+b x) \sqrt {\sin (2 (a+b x))}-\frac {\sqrt {2} \operatorname {EllipticF}\left (\arcsin (\cos (a+b x)-\sin (a+b x)),\frac {1}{2}\right ) (\cos (a+b x)+\sin (a+b x))}{\sqrt {1+\sin (2 (a+b x))}}}{12 b} \]
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Time = 28.15 (sec) , antiderivative size = 123, normalized size of antiderivative = 2.56
method | result | size |
default | \(\frac {\sqrt {\sin \left (2 x b +2 a \right )+1}\, \sqrt {-2 \sin \left (2 x b +2 a \right )+2}\, \sqrt {-\sin \left (2 x b +2 a \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (2 x b +2 a \right )+1}, \frac {\sqrt {2}}{2}\right ) \sin \left (2 x b +2 a \right )-2 \cos \left (2 x b +2 a \right )^{2}+2 \cos \left (2 x b +2 a \right )}{12 \sin \left (2 x b +2 a \right )^{\frac {3}{2}} \cos \left (2 x b +2 a \right ) b}\) | \(123\) |
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Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.96 \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {2 i} \cos \left (b x + a\right )^{2} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + \sqrt {-2 i} \cos \left (b x + a\right )^{2} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{12 \, b \cos \left (b x + a\right )^{2}} \]
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Timed out. \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]
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\[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int { \frac {\sin \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int { \frac {\sin \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int \frac {{\sin \left (a+b\,x\right )}^2}{{\sin \left (2\,a+2\,b\,x\right )}^{5/2}} \,d x \]
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