\(\int \csc ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx\) [108]

Optimal result

Integrand size = 22, antiderivative size = 70 \[ \int \csc ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\frac {2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )}{b}-\frac {2 \cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{b}+\frac {\csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{b} \]

[Out]

-2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))/b+csc(b*x+a)^2*sin(2*b*x
+2*a)^(5/2)/b-2*cos(2*b*x+2*a)*sin(2*b*x+2*a)^(1/2)/b

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4385, 2715, 2720} \[ \int \csc ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\frac {2 \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{b}-\frac {2 \sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{b}+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \csc ^2(a+b x)}{b} \]

[In]

Int[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(3/2),x]

[Out]

(2*EllipticF[a - Pi/4 + b*x, 2])/b - (2*Cos[2*a + 2*b*x]*Sqrt[Sin[2*a + 2*b*x]])/b + (Csc[a + b*x]^2*Sin[2*a +
2*b*x]^(5/2))/b

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 4385

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)/(2*b*g*(m + p + 1))), x] + Dist[(m + 2*p + 2)/(e^2*(m + p + 1)), Int[(e*Sin[a
+ b*x])^(m + 2)*(g*Sin[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, e, g, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b,
2] &&  !IntegerQ[p] && LtQ[m, -1] && NeQ[m + 2*p + 2, 0] && NeQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]

Rubi steps \begin{align*} \text {integral}& = \frac {\csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{b}+6 \int \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx \\ & = -\frac {2 \cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{b}+\frac {\csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{b}+2 \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx \\ & = \frac {2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )}{b}-\frac {2 \cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{b}+\frac {\csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{b} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.42 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.74 \[ \int \csc ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\frac {2 \left (1+\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)}\right ) \sqrt {\sin (2 (a+b x))}}{b} \]

[In]

Integrate[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(3/2),x]

[Out]

(2*(1 + Hypergeometric2F1[1/4, 1/2, 5/4, -Tan[a + b*x]^2]*Sqrt[Sec[a + b*x]^2])*Sqrt[Sin[2*(a + b*x)]])/b

Maple [A] (verified)

Time = 4.72 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.59

method	result	size
default	\(\frac {\sqrt {2}\, \left (\sqrt {2}\, \sqrt {\sin \left (2 x b +2 a \right )}+\frac {\sqrt {2}\, \sqrt {\sin \left (2 x b +2 a \right )+1}\, \sqrt {-2 \sin \left (2 x b +2 a \right )+2}\, \sqrt {-\sin \left (2 x b +2 a \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (2 x b +2 a \right )+1}, \frac {\sqrt {2}}{2}\right )}{2 \cos \left (2 x b +2 a \right ) \sqrt {\sin \left (2 x b +2 a \right )}}\right )}{b}\)	\(111\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(csc(b*x + a)^2*sin(2*b*x + 2*a)^(3/2), x)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \csc ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int \frac {{\sin \left (2\,a+2\,b\,x\right )}^{3/2}}{{\sin \left (a+b\,x\right )}^2} \,d x \]

[In]

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

[Out]

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