Integrand size = 22, antiderivative size = 70 \[ \int \csc ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\frac {2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )}{b}-\frac {2 \cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{b}+\frac {\csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{b} \] Output:
2*InverseJacobiAM(a-1/4*Pi+b*x,2^(1/2))/b-2*cos(2*b*x+2*a)*sin(2*b*x+2*a)^ (1/2)/b+csc(b*x+a)^2*sin(2*b*x+2*a)^(5/2)/b
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.74 \[ \int \csc ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\frac {2 \left (1+\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)}\right ) \sqrt {\sin (2 (a+b x))}}{b} \] Input:
Integrate[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(3/2),x]
Output:
(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
Time = 0.34 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 4788, 3042, 3115, 3042, 3120}
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.
\(\displaystyle \int \sin ^{\frac {3}{2}}(2 a+2 b x) \csc ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (2 a+2 b x)^{3/2}}{\sin (a+b x)^2}dx\) |
\(\Big \downarrow \) 4788 |
\(\displaystyle 6 \int \sin ^{\frac {3}{2}}(2 a+2 b x)dx+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \csc ^2(a+b x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 6 \int \sin (2 a+2 b x)^{3/2}dx+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \csc ^2(a+b x)}{b}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle 6 \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{3 b}\right )+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \csc ^2(a+b x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 6 \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{3 b}\right )+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \csc ^2(a+b x)}{b}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \csc ^2(a+b x)}{b}+6 \left (\frac {\operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{3 b}-\frac {\sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{3 b}\right )\) |
Input:
Int[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(3/2),x]
Output:
6*(EllipticF[a - Pi/4 + b*x, 2]/(3*b) - (Cos[2*a + 2*b*x]*Sqrt[Sin[2*a + 2 *b*x]])/(3*b)) + (Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(5/2))/b
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] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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]
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] + Simp[(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]
Time = 7.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.59
method | result | size |
default | \(\frac {\sqrt {2}\, \left (\sqrt {2}\, \sqrt {\sin \left (2 b x +2 a \right )}+\frac {\sqrt {2}\, \sqrt {\sin \left (2 b x +2 a \right )+1}\, \sqrt {-2 \sin \left (2 b x +2 a \right )+2}\, \sqrt {-\sin \left (2 b x +2 a \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (2 b x +2 a \right )+1}, \frac {\sqrt {2}}{2}\right )}{2 \cos \left (2 b x +2 a \right ) \sqrt {\sin \left (2 b x +2 a \right )}}\right )}{b}\) | \(111\) |
Input:
int(csc(b*x+a)^2*sin(2*b*x+2*a)^(3/2),x,method=_RETURNVERBOSE)
Output:
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
\[ \int \csc ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^(3/2),x, algorithm="fricas")
Output:
integral(csc(b*x + a)^2*sin(2*b*x + 2*a)^(3/2), x)
Timed out. \[ \int \csc ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\text {Timed out} \] Input:
integrate(csc(b*x+a)**2*sin(2*b*x+2*a)**(3/2),x)
Output:
Timed out
\[ \int \csc ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^(3/2),x, algorithm="maxima")
Output:
integrate(csc(b*x + a)^2*sin(2*b*x + 2*a)^(3/2), x)
\[ \int \csc ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^(3/2),x, algorithm="giac")
Output:
integrate(csc(b*x + a)^2*sin(2*b*x + 2*a)^(3/2), x)
Timed out. \[ \int \csc ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int \frac {{\sin \left (2\,a+2\,b\,x\right )}^{3/2}}{{\sin \left (a+b\,x\right )}^2} \,d x \] Input:
int(sin(2*a + 2*b*x)^(3/2)/sin(a + b*x)^2,x)
Output:
int(sin(2*a + 2*b*x)^(3/2)/sin(a + b*x)^2, x)
\[ \int \csc ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int \sqrt {\sin \left (2 b x +2 a \right )}\, \csc \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )d x \] Input:
int(csc(b*x+a)^2*sin(2*b*x+2*a)^(3/2),x)
Output:
int(sqrt(sin(2*a + 2*b*x))*csc(a + b*x)**2*sin(2*a + 2*b*x),x)