\(\int \frac {\csc (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx\) [101]

Optimal result

Integrand size = 20, antiderivative size = 24 \[ \int \frac {\csc (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {\csc (a+b x) \sqrt {\sin (2 a+2 b x)}}{b} \]

[Out]

-csc(b*x+a)*sin(2*b*x+2*a)^(1/2)/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {4377} \[ \int \frac {\csc (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {\sqrt {\sin (2 a+2 b x)} \csc (a+b x)}{b} \]

[In]

Int[Csc[a + b*x]/Sqrt[Sin[2*a + 2*b*x]],x]

[Out]

-((Csc[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/b)

Rule 4377

Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(e*Sin[a + b
*x])^m*((g*Sin[c + d*x])^(p + 1)/(b*g*m)), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && Eq
Q[d/b, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\csc (a+b x) \sqrt {\sin (2 a+2 b x)}}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {\csc (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {\csc (a+b x) \sqrt {\sin (2 (a+b x))}}{b} \]

[In]

Integrate[Csc[a + b*x]/Sqrt[Sin[2*a + 2*b*x]],x]

[Out]

-((Csc[a + b*x]*Sqrt[Sin[2*(a + b*x)]])/b)

Maple [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 3.38 (sec) , antiderivative size = 308, normalized size of antiderivative = 12.83

method	result	size
default	\(\frac {\sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1}}\, \left (2 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticE}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right )+\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\right )}{b \tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}}\)	\(308\)

[In]

int(csc(b*x+a)/sin(2*b*x+2*a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/b*(-tan(1/2*a+1/2*x*b)/(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*(2*(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2
)*(tan(1/2*a+1/2*x*b)+1)^(1/2)*(-2*tan(1/2*a+1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b))^(1/2)*EllipticE((tan(1/2*
a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))-(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*(tan(1/2*a+1/2*x*b)+1)^(1
/2)*(-2*tan(1/2*a+1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b))^(1/2)*EllipticF((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(
1/2))+(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1/2)*tan(1/2*a+1/2*x*b)^2-(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2
*x*b))^(1/2))/tan(1/2*a+1/2*x*b)/(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {\csc (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + \sin \left (b x + a\right )}{b \sin \left (b x + a\right )} \]

[In]

integrate(csc(b*x+a)/sin(2*b*x+2*a)^(1/2),x, algorithm="fricas")

[Out]

-(sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a)) + sin(b*x + a))/(b*sin(b*x + a))

Sympy [F(-1)]

Timed out. \[ \int \frac {\csc (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=\text {Timed out} \]

[In]

integrate(csc(b*x+a)/sin(2*b*x+2*a)**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\csc (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=\int { \frac {\csc \left (b x + a\right )}{\sqrt {\sin \left (2 \, b x + 2 \, a\right )}} \,d x } \]

[In]

integrate(csc(b*x+a)/sin(2*b*x+2*a)^(1/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)/sqrt(sin(2*b*x + 2*a)), x)

Giac [F]

\[ \int \frac {\csc (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=\int { \frac {\csc \left (b x + a\right )}{\sqrt {\sin \left (2 \, b x + 2 \, a\right )}} \,d x } \]

[In]

integrate(csc(b*x+a)/sin(2*b*x+2*a)^(1/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)/sqrt(sin(2*b*x + 2*a)), x)

Mupad [B] (verification not implemented)

Time = 21.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\csc (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {\sqrt {\sin \left (2\,a+2\,b\,x\right )}}{b\,\sin \left (a+b\,x\right )} \]

[In]

int(1/(sin(a + b*x)*sin(2*a + 2*b*x)^(1/2)),x)

[Out]

-sin(2*a + 2*b*x)^(1/2)/(b*sin(a + b*x))