\(\int \csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx\) [107]

Optimal result

Integrand size = 22, antiderivative size = 75 \[ \int \csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\frac {2 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{b}-\frac {2 \cos (2 a+2 b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{3 b}+\frac {\csc ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{3 b} \]

[Out]

-2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))/b-2/3*cos(2*b*x+2*a)*sin
(2*b*x+2*a)^(3/2)/b+1/3*csc(b*x+a)^2*sin(2*b*x+2*a)^(7/2)/b

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 75, 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, 2719} \[ \int \csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\frac {2 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{b}-\frac {2 \sin ^{\frac {3}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{3 b}+\frac {\sin ^{\frac {7}{2}}(2 a+2 b x) \csc ^2(a+b x)}{3 b} \]

[In]

Int[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(5/2),x]

[Out]

(2*EllipticE[a - Pi/4 + b*x, 2])/b - (2*Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x]^(3/2))/(3*b) + (Csc[a + b*x]^2*Sin[2
*a + 2*b*x]^(7/2))/(3*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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 {7}{2}}(2 a+2 b x)}{3 b}+\frac {10}{3} \int \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx \\ & = -\frac {2 \cos (2 a+2 b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{3 b}+\frac {\csc ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{3 b}+2 \int \sqrt {\sin (2 a+2 b x)} \, dx \\ & = \frac {2 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{b}-\frac {2 \cos (2 a+2 b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{3 b}+\frac {\csc ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{3 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.45 \[ \int \csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\frac {2 \left (3 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )+\sin ^{\frac {3}{2}}(2 (a+b x))\right )}{3 b} \]

[In]

Integrate[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(5/2),x]

[Out]

(2*(3*EllipticE[a - Pi/4 + b*x, 2] + Sin[2*(a + b*x)]^(3/2)))/(3*b)

Maple [A] (verified)

Time = 19.63 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.83

method	result	size
default	\(\frac {2 \sqrt {2}\, \left (\frac {\sqrt {2}\, \sin \left (2 x b +2 a \right )^{\frac {3}{2}}}{6}-\frac {\sqrt {2}\, \sqrt {\sin \left (2 x b +2 a \right )+1}\, \sqrt {-2 \sin \left (2 x b +2 a \right )+2}\, \sqrt {-\sin \left (2 x b +2 a \right )}\, \left (2 \operatorname {EllipticE}\left (\sqrt {\sin \left (2 x b +2 a \right )+1}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {\sin \left (2 x b +2 a \right )+1}, \frac {\sqrt {2}}{2}\right )\right )}{4 \cos \left (2 x b +2 a \right ) \sqrt {\sin \left (2 x b +2 a \right )}}\right )}{b}\)	\(137\)

[In]

int(csc(b*x+a)^2*sin(2*b*x+2*a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2*2^(1/2)*(1/6*2^(1/2)*sin(2*b*x+2*a)^(3/2)-1/4*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)*(2*EllipticE((sin(2*b*x+2*a)+1)^(1/2),1/2*2^(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]

\[ \int \csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^(5/2),x, algorithm="fricas")

[Out]

integral(-(cos(2*b*x + 2*a)^2 - 1)*csc(b*x + a)^2*sqrt(sin(2*b*x + 2*a)), x)

Sympy [F(-1)]

Timed out. \[ \int \csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\text {Timed out} \]

[In]

integrate(csc(b*x+a)**2*sin(2*b*x+2*a)**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^(5/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)^2*sin(2*b*x + 2*a)^(5/2), x)

Giac [F]

\[ \int \csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^(5/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)^2*sin(2*b*x + 2*a)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int \frac {{\sin \left (2\,a+2\,b\,x\right )}^{5/2}}{{\sin \left (a+b\,x\right )}^2} \,d x \]

[In]

int(sin(2*a + 2*b*x)^(5/2)/sin(a + b*x)^2,x)

[Out]

int(sin(2*a + 2*b*x)^(5/2)/sin(a + b*x)^2, x)