Integrand size = 20, antiderivative size = 53 \[ \int \csc (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{b}+\frac {\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{b} \] Output:
-arcsin(cos(b*x+a)-sin(b*x+a))/b+ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x+2*a)^( 1/2))/b
Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.98 \[ \int \csc (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{b}+\frac {\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )}{b} \] Input:
Integrate[Csc[a + b*x]*Sqrt[Sin[2*a + 2*b*x]],x]
Output:
-(ArcSin[Cos[a + b*x] - Sin[a + b*x]]/b) + Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*(a + b*x)]]]/b
Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4796, 3042, 4793}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\sin (2 a+2 b x)}}{\sin (a+b x)}dx\) |
\(\Big \downarrow \) 4796 |
\(\displaystyle 2 \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx\) |
\(\Big \downarrow \) 4793 |
\(\displaystyle 2 \left (\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}\right )\) |
Input:
Int[Csc[a + b*x]*Sqrt[Sin[2*a + 2*b*x]],x]
Output:
2*(-1/2*ArcSin[Cos[a + b*x] - Sin[a + b*x]]/b + Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]/(2*b))
Int[cos[(a_.) + (b_.)*(x_)]/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Sim p[-ArcSin[Cos[a + b*x] - Sin[a + b*x]]/d, x] + Simp[Log[Cos[a + b*x] + Sin[ a + b*x] + Sqrt[Sin[c + d*x]]]/d, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2]
Int[((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_)/sin[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[2*g Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; FreeQ[{ a, b, c, d, g, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !IntegerQ[p] && IntegerQ[2*p]
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 4.74 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.96
method | result | size |
default | \(\frac {2 \sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}{\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-1}}\, \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-1\right ) \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}}\) | \(157\) |
Input:
int(csc(b*x+a)*sin(2*b*x+2*a)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/b*(-tan(1/2*a+1/2*b*x)/(tan(1/2*a+1/2*b*x)^2-1))^(1/2)*(tan(1/2*a+1/2*b* x)^2-1)/(tan(1/2*a+1/2*b*x)*(tan(1/2*a+1/2*b*x)^2-1))^(1/2)*(tan(1/2*a+1/2 *b*x)+1)^(1/2)*(-2*tan(1/2*a+1/2*b*x)+2)^(1/2)*(-tan(1/2*a+1/2*b*x))^(1/2) /(tan(1/2*a+1/2*b*x)^3-tan(1/2*a+1/2*b*x))^(1/2)*EllipticF((tan(1/2*a+1/2* b*x)+1)^(1/2),1/2*2^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 242 vs. \(2 (51) = 102\).
Time = 0.09 (sec) , antiderivative size = 242, normalized size of antiderivative = 4.57 \[ \int \csc (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\frac {2 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) - 2 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) - \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{4 \, b} \] Input:
integrate(csc(b*x+a)*sin(2*b*x+2*a)^(1/2),x, algorithm="fricas")
Output:
1/4*(2*arctan(-(sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a))*(cos(b*x + a) - si n(b*x + a)) + cos(b*x + a)*sin(b*x + a))/(cos(b*x + a)^2 + 2*cos(b*x + a)* sin(b*x + a) - 1)) - 2*arctan(-(2*sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a)) - cos(b*x + a) - sin(b*x + a))/(cos(b*x + a) - sin(b*x + a))) - log(-32*co s(b*x + a)^4 + 4*sqrt(2)*(4*cos(b*x + a)^3 - (4*cos(b*x + a)^2 + 1)*sin(b* x + a) - 5*cos(b*x + a))*sqrt(cos(b*x + a)*sin(b*x + a)) + 32*cos(b*x + a) ^2 + 16*cos(b*x + a)*sin(b*x + a) + 1))/b
Timed out. \[ \int \csc (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\text {Timed out} \] Input:
integrate(csc(b*x+a)*sin(2*b*x+2*a)**(1/2),x)
Output:
Timed out
\[ \int \csc (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int { \csc \left (b x + a\right ) \sqrt {\sin \left (2 \, b x + 2 \, a\right )} \,d x } \] Input:
integrate(csc(b*x+a)*sin(2*b*x+2*a)^(1/2),x, algorithm="maxima")
Output:
integrate(csc(b*x + a)*sqrt(sin(2*b*x + 2*a)), x)
\[ \int \csc (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int { \csc \left (b x + a\right ) \sqrt {\sin \left (2 \, b x + 2 \, a\right )} \,d x } \] Input:
integrate(csc(b*x+a)*sin(2*b*x+2*a)^(1/2),x, algorithm="giac")
Output:
integrate(csc(b*x + a)*sqrt(sin(2*b*x + 2*a)), x)
Timed out. \[ \int \csc (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int \frac {\sqrt {\sin \left (2\,a+2\,b\,x\right )}}{\sin \left (a+b\,x\right )} \,d x \] Input:
int(sin(2*a + 2*b*x)^(1/2)/sin(a + b*x),x)
Output:
int(sin(2*a + 2*b*x)^(1/2)/sin(a + b*x), x)
\[ \int \csc (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int \sqrt {\sin \left (2 b x +2 a \right )}\, \csc \left (b x +a \right )d x \] Input:
int(csc(b*x+a)*sin(2*b*x+2*a)^(1/2),x)
Output:
int(sqrt(sin(2*a + 2*b*x))*csc(a + b*x),x)