3.105 \(\int \csc ^2(a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x) \, dx\)

Optimal. Leaf size=106 \[ \frac {6 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{5 b}-\frac {2 \sin ^{\frac {7}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{7 b}-\frac {2 \sin ^{\frac {3}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{5 b}+\frac {\sin ^{\frac {11}{2}}(2 a+2 b x) \csc ^2(a+b x)}{7 b} \]

[Out]

-6/5*(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-2/5*cos(2*b*x+2*a)*s
in(2*b*x+2*a)^(3/2)/b-2/7*cos(2*b*x+2*a)*sin(2*b*x+2*a)^(7/2)/b+1/7*csc(b*x+a)^2*sin(2*b*x+2*a)^(11/2)/b

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Rubi [A]  time = 0.06, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4300, 2635, 2639} \[ \frac {6 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{5 b}-\frac {2 \sin ^{\frac {7}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{7 b}-\frac {2 \sin ^{\frac {3}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{5 b}+\frac {\sin ^{\frac {11}{2}}(2 a+2 b x) \csc ^2(a+b x)}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(9/2),x]

[Out]

(6*EllipticE[a - Pi/4 + b*x, 2])/(5*b) - (2*Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x]^(3/2))/(5*b) - (2*Cos[2*a + 2*b*
x]*Sin[2*a + 2*b*x]^(7/2))/(7*b) + (Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(11/2))/(7*b)

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]

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]

Rubi steps

\begin {align*} \int \csc ^2(a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x) \, dx &=\frac {\csc ^2(a+b x) \sin ^{\frac {11}{2}}(2 a+2 b x)}{7 b}+\frac {18}{7} \int \sin ^{\frac {9}{2}}(2 a+2 b x) \, dx\\ &=-\frac {2 \cos (2 a+2 b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{7 b}+\frac {\csc ^2(a+b x) \sin ^{\frac {11}{2}}(2 a+2 b x)}{7 b}+2 \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)}{5 b}-\frac {2 \cos (2 a+2 b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{7 b}+\frac {\csc ^2(a+b x) \sin ^{\frac {11}{2}}(2 a+2 b x)}{7 b}+\frac {6}{5} \int \sqrt {\sin (2 a+2 b x)} \, dx\\ &=\frac {6 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{5 b}-\frac {2 \cos (2 a+2 b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{5 b}-\frac {2 \cos (2 a+2 b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{7 b}+\frac {\csc ^2(a+b x) \sin ^{\frac {11}{2}}(2 a+2 b x)}{7 b}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 66, normalized size = 0.62 \[ \frac {\sqrt {\sin (2 (a+b x))} (15 \sin (2 (a+b x))-14 \sin (4 (a+b x))-5 \sin (6 (a+b x)))+84 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{70 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(9/2),x]

[Out]

(84*EllipticE[a - Pi/4 + b*x, 2] + Sqrt[Sin[2*(a + b*x)]]*(15*Sin[2*(a + b*x)] - 14*Sin[4*(a + b*x)] - 5*Sin[6
*(a + b*x)]))/(70*b)

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (\cos \left (2 \, b x + 2 \, a\right )^{4} - 2 \, \cos \left (2 \, b x + 2 \, a\right )^{2} + 1\right )} \csc \left (b x + a\right )^{2} \sqrt {\sin \left (2 \, b x + 2 \, a\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^(9/2),x, algorithm="fricas")

[Out]

integral((cos(2*b*x + 2*a)^4 - 2*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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^(9/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 29.08, size = 204, normalized size = 1.92 \[ \frac {8 \sqrt {2}\, \left (\frac {\sqrt {2}\, \left (\sin ^{\frac {7}{2}}\left (2 b x +2 a \right )\right )}{56}-\frac {\sqrt {2}\, \left (6 \sqrt {1+\sin \left (2 b x +2 a \right )}\, \sqrt {-2 \sin \left (2 b x +2 a \right )+2}\, \sqrt {-\sin \left (2 b x +2 a \right )}\, \EllipticE \left (\sqrt {1+\sin \left (2 b x +2 a \right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {1+\sin \left (2 b x +2 a \right )}\, \sqrt {-2 \sin \left (2 b x +2 a \right )+2}\, \sqrt {-\sin \left (2 b x +2 a \right )}\, \EllipticF \left (\sqrt {1+\sin \left (2 b x +2 a \right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{4}\left (2 b x +2 a \right )\right )+2 \left (\sin ^{2}\left (2 b x +2 a \right )\right )\right )}{80 \cos \left (2 b x +2 a \right ) \sqrt {\sin \left (2 b x +2 a \right )}}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^2*sin(2*b*x+2*a)^(9/2),x)

[Out]

8*2^(1/2)*(1/56*2^(1/2)*sin(2*b*x+2*a)^(7/2)-1/80*2^(1/2)*(6*(1+sin(2*b*x+2*a))^(1/2)*(-2*sin(2*b*x+2*a)+2)^(1
/2)*(-sin(2*b*x+2*a))^(1/2)*EllipticE((1+sin(2*b*x+2*a))^(1/2),1/2*2^(1/2))-3*(1+sin(2*b*x+2*a))^(1/2)*(-2*sin
(2*b*x+2*a)+2)^(1/2)*(-sin(2*b*x+2*a))^(1/2)*EllipticF((1+sin(2*b*x+2*a))^(1/2),1/2*2^(1/2))-2*sin(2*b*x+2*a)^
4+2*sin(2*b*x+2*a)^2)/cos(2*b*x+2*a)/sin(2*b*x+2*a)^(1/2))/b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \csc \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac {9}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^(9/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)^2*sin(2*b*x + 2*a)^(9/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\sin \left (2\,a+2\,b\,x\right )}^{9/2}}{{\sin \left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*a + 2*b*x)^(9/2)/sin(a + b*x)^2,x)

[Out]

int(sin(2*a + 2*b*x)^(9/2)/sin(a + b*x)^2, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**2*sin(2*b*x+2*a)**(9/2),x)

[Out]

Timed out

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