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

Optimal. Leaf size=75 \[ \frac{2 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{b}-\frac{2 \sin ^{\frac{3}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{3 b}+\frac{\sin ^{\frac{7}{2}}(2 a+2 b x) \csc ^2(a+b x)}{3 b} \]

[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)

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Rubi [A]  time = 0.0469874, antiderivative size = 75, 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, 2639} \[ \frac{2 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{b}-\frac{2 \sin ^{\frac{3}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{3 b}+\frac{\sin ^{\frac{7}{2}}(2 a+2 b x) \csc ^2(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[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 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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(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{5}{2}}(2 a+2 b x) \, dx &=\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]  time = 0.0784348, size = 34, normalized size = 0.45 \[ \frac{2 \left (\sin ^{\frac{3}{2}}(2 (a+b x))+3 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )\right )}{3 b} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 3.971, size = 137, normalized size = 1.8 \begin{align*} 2\,{\frac{\sqrt{2}}{b} \left ( 1/6\,\sqrt{2} \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{3/2}-1/4\,{\frac{\sqrt{2}\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 ) } \left ( 2\,{\it EllipticE} \left ( \sqrt{\sin \left ( 2\,bx+2\,a \right ) +1},1/2\,\sqrt{2} \right ) -{\it EllipticF} \left ( \sqrt{\sin \left ( 2\,bx+2\,a \right ) +1},1/2\,\sqrt{2} \right ) \right ) }{\cos \left ( 2\,bx+2\,a \right ) \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)^(5/2),x)

[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

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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{5}{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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (\cos \left (2 \, b x + 2 \, a\right )^{2} - 1\right )} \csc \left (b x + a\right )^{2} \sqrt{\sin \left (2 \, b x + 2 \, a\right )}, 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)^(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)

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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)**(5/2),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{5}{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)^(5/2),x, algorithm="giac")

[Out]

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