\(\int \csc ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx\) [119]

Optimal result

Integrand size = 22, antiderivative size = 28 \[ \int \csc ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=-\frac {\csc ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{3 b} \]

[Out]

-1/3*csc(b*x+a)^3*sin(2*b*x+2*a)^(3/2)/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {4377} \[ \int \csc ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b} \]

[In]

Int[Csc[a + b*x]^3*Sqrt[Sin[2*a + 2*b*x]],x]

[Out]

-1/3*(Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(3/2))/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 ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{3 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \csc ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=-\frac {\csc ^3(a+b x) \sin ^{\frac {3}{2}}(2 (a+b x))}{3 b} \]

[In]

Integrate[Csc[a + b*x]^3*Sqrt[Sin[2*a + 2*b*x]],x]

[Out]

-1/3*(Csc[a + b*x]^3*Sin[2*(a + b*x)]^(3/2))/b

Maple [C] (verified)

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

Time = 13.02 (sec) , antiderivative size = 192, normalized size of antiderivative = 6.86

method	result	size
default	\(\frac {\sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1}}\, \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right ) \left (4 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}-1\right )}{3 \tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, b}\)	\(192\)

[In]

int(csc(b*x+a)^3*sin(2*b*x+2*a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/3*(-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)^2-1)/tan(1/2*a+1/2*x*b)*(4*(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)+tan(1/2*a+1/2*x*b)^4-1)/(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)^3-tan(1/2*a+1/2*x*b))^(1/2)/b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (24) = 48\).

Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.89 \[ \int \csc ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\frac {2 \, {\left (\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \cos \left (b x + a\right ) + \cos \left (b x + a\right )^{2} - 1\right )}}{3 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}} \]

[In]

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^(1/2),x, algorithm="fricas")

[Out]

2/3*(sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a))*cos(b*x + a) + cos(b*x + a)^2 - 1)/(b*cos(b*x + a)^2 - b)

Sympy [F(-1)]

Timed out. \[ \int \csc ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\text {Timed out} \]

[In]

integrate(csc(b*x+a)**3*sin(2*b*x+2*a)**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \csc ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int { \csc \left (b x + a\right )^{3} \sqrt {\sin \left (2 \, b x + 2 \, a\right )} \,d x } \]

[In]

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^(1/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)^3*sqrt(sin(2*b*x + 2*a)), x)

Giac [F]

\[ \int \csc ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int { \csc \left (b x + a\right )^{3} \sqrt {\sin \left (2 \, b x + 2 \, a\right )} \,d x } \]

[In]

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^(1/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)^3*sqrt(sin(2*b*x + 2*a)), x)

Mupad [B] (verification not implemented)

Time = 21.34 (sec) , antiderivative size = 95, normalized size of antiderivative = 3.39 \[ \int \csc ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\frac {4\,\sqrt {\sin \left (2\,a+2\,b\,x\right )}\,\left (4\,{\sin \left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-6\,{\sin \left (\frac {3\,a}{2}+\frac {3\,b\,x}{2}\right )}^2+2\,{\sin \left (\frac {5\,a}{2}+\frac {5\,b\,x}{2}\right )}^2\right )}{3\,b\,\left (30\,{\sin \left (a+b\,x\right )}^2-12\,{\sin \left (2\,a+2\,b\,x\right )}^2+2\,{\sin \left (3\,a+3\,b\,x\right )}^2\right )} \]

[In]

int(sin(2*a + 2*b*x)^(1/2)/sin(a + b*x)^3,x)

[Out]

(4*sin(2*a + 2*b*x)^(1/2)*(4*sin(a/2 + (b*x)/2)^2 - 6*sin((3*a)/2 + (3*b*x)/2)^2 + 2*sin((5*a)/2 + (5*b*x)/2)^
2))/(3*b*(2*sin(3*a + 3*b*x)^2 - 12*sin(2*a + 2*b*x)^2 + 30*sin(a + b*x)^2))