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

Optimal. Leaf size=70 \[ \frac{2 \text{EllipticF}\left (a+b x-\frac{\pi }{4},2\right )}{b}-\frac{2 \sqrt{\sin (2 a+2 b x)} \cos (2 a+2 b x)}{b}+\frac{\sin ^{\frac{5}{2}}(2 a+2 b x) \csc ^2(a+b x)}{b} \]

[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

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Rubi [A]  time = 0.0482702, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {4300, 2635, 2641} \[ \frac{2 F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{b}-\frac{2 \sqrt{\sin (2 a+2 b x)} \cos (2 a+2 b x)}{b}+\frac{\sin ^{\frac{5}{2}}(2 a+2 b x) \csc ^2(a+b x)}{b} \]

Antiderivative was successfully verified.

[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 4300

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]

Rule 2635

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] && Integer
Q[2*n]

Rule 2641

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

Rubi steps

\begin{align*} \int \csc ^2(a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x) \, dx &=\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 F\left (\left .a-\frac{\pi }{4}+b x\right |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 [A]  time = 0.859969, size = 73, normalized size = 1.04 \[ \frac{2 \sqrt{\sin (2 (a+b x))}-\frac{\sqrt{2} (\sin (a+b x)+\cos (a+b x)) \text{EllipticF}\left (\sin ^{-1}(\cos (a+b x)-\sin (a+b x)),\frac{1}{2}\right )}{\sqrt{\sin (2 (a+b x))+1}}}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[Sin[2*(a + b*x)]] - (Sqrt[2]*EllipticF[ArcSin[Cos[a + b*x] - Sin[a + b*x]], 1/2]*(Cos[a + b*x] + Sin[a
+ b*x]))/Sqrt[1 + Sin[2*(a + b*x)]])/b

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Maple [A]  time = 2.622, size = 111, normalized size = 1.6 \begin{align*}{\frac{\sqrt{2}}{b} \left ( \sqrt{2}\sqrt{\sin \left ( 2\,bx+2\,a \right ) }+{\frac{\sqrt{2}}{2\,\cos \left ( 2\,bx+2\,a \right ) }\sqrt{\sin \left ( 2\,bx+2\,a \right ) +1}\sqrt{-2\,\sin \left ( 2\,bx+2\,a \right ) +2}\sqrt{-\sin \left ( 2\,bx+2\,a \right ) }{\it EllipticF} \left ( \sqrt{\sin \left ( 2\,bx+2\,a \right ) +1},{\frac{\sqrt{2}}{2}} \right ){\frac{1}{\sqrt{\sin \left ( 2\,bx+2\,a \right ) }}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \csc \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\csc \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac{3}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \csc \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)