Integrand size = 22, antiderivative size = 45 \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=-\frac {E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{2 b}+\frac {\sin ^2(a+b x)}{b \sqrt {\sin (2 a+2 b x)}} \]
[Out]
Time = 0.05 (sec) , antiderivative size = 45, 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, 2719} \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {\sin ^2(a+b x)}{b \sqrt {\sin (2 a+2 b x)}}-\frac {E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{2 b} \]
[In]
[Out]
Rule 2719
Rule 4381
Rubi steps \begin{align*} \text {integral}& = \frac {\sin ^2(a+b x)}{b \sqrt {\sin (2 a+2 b x)}}-\frac {1}{2} \int \sqrt {\sin (2 a+2 b x)} \, dx \\ & = -\frac {E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{2 b}+\frac {\sin ^2(a+b x)}{b \sqrt {\sin (2 a+2 b x)}} \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.91 \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {-E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )+\sqrt {\sin (2 (a+b x))} \tan (a+b x)}{2 b} \]
[In]
[Out]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 26.14 (sec) , antiderivative size = 108031867, normalized size of antiderivative = 2400708.16
[In]
[Out]
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 156, normalized size of antiderivative = 3.47 \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {-i \, \sqrt {2 i} \cos \left (b x + a\right ) E(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + i \, \sqrt {-2 i} \cos \left (b x + a\right ) E(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) + i \, \sqrt {2 i} \cos \left (b x + a\right ) F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) - i \, \sqrt {-2 i} \cos \left (b x + a\right ) F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) + 2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \sin \left (b x + a\right )}{4 \, b \cos \left (b x + a\right )} \]
[In]
[Out]
Timed out. \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int { \frac {\sin \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}} \,d x } \]
[In]
[Out]
\[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int { \frac {\sin \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int \frac {{\sin \left (a+b\,x\right )}^2}{{\sin \left (2\,a+2\,b\,x\right )}^{3/2}} \,d x \]
[In]
[Out]