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

Optimal. Leaf size=106 \[ \frac {30 F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{77 b}-\frac {30 \cos (2 a+2 b x)}{77 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {18 \cos (2 a+2 b x)}{77 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {\csc ^2(a+b x)}{11 b \sin ^{\frac {7}{2}}(2 a+2 b x)} \]

[Out]

-30/77*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))/b-18/77*cos(2*b*x+2*
a)/b/sin(2*b*x+2*a)^(7/2)-1/11*csc(b*x+a)^2/b/sin(2*b*x+2*a)^(7/2)-30/77*cos(2*b*x+2*a)/b/sin(2*b*x+2*a)^(3/2)

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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, 2636, 2641} \[ \frac {30 F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{77 b}-\frac {30 \cos (2 a+2 b x)}{77 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {18 \cos (2 a+2 b x)}{77 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {\csc ^2(a+b x)}{11 b \sin ^{\frac {7}{2}}(2 a+2 b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(30*EllipticF[a - Pi/4 + b*x, 2])/(77*b) - (18*Cos[2*a + 2*b*x])/(77*b*Sin[2*a + 2*b*x]^(7/2)) - Csc[a + b*x]^
2/(11*b*Sin[2*a + 2*b*x]^(7/2)) - (30*Cos[2*a + 2*b*x])/(77*b*Sin[2*a + 2*b*x]^(3/2))

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[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]

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 \frac {\csc ^2(a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx &=-\frac {\csc ^2(a+b x)}{11 b \sin ^{\frac {7}{2}}(2 a+2 b x)}+\frac {18}{11} \int \frac {1}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx\\ &=-\frac {18 \cos (2 a+2 b x)}{77 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {\csc ^2(a+b x)}{11 b \sin ^{\frac {7}{2}}(2 a+2 b x)}+\frac {90}{77} \int \frac {1}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\\ &=-\frac {18 \cos (2 a+2 b x)}{77 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {\csc ^2(a+b x)}{11 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {30 \cos (2 a+2 b x)}{77 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {30}{77} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx\\ &=\frac {30 F\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{77 b}-\frac {18 \cos (2 a+2 b x)}{77 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {\csc ^2(a+b x)}{11 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {30 \cos (2 a+2 b x)}{77 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 86, normalized size = 0.81 \[ \frac {480 F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )+\sqrt {\sin (2 (a+b x))} \left (-7 \csc ^6(a+b x)-32 \csc ^4(a+b x)-141 \csc ^2(a+b x)+11 \sec ^2(a+b x) \left (\sec ^2(a+b x)+9\right )\right )}{1232 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(480*EllipticF[a - Pi/4 + b*x, 2] + (-141*Csc[a + b*x]^2 - 32*Csc[a + b*x]^4 - 7*Csc[a + b*x]^6 + 11*Sec[a + b
*x]^2*(9 + Sec[a + b*x]^2))*Sqrt[Sin[2*(a + b*x)]])/(1232*b)

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\csc \left (b x + a\right )^{2}}{{\left (\cos \left (2 \, b x + 2 \, a\right )^{4} - 2 \, \cos \left (2 \, b x + 2 \, a\right )^{2} + 1\right )} \sqrt {\sin \left (2 \, b x + 2 \, a\right )}}, x\right ) \]

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(csc(b*x + a)^2/((cos(2*b*x + 2*a)^4 - 2*cos(2*b*x + 2*a)^2 + 1)*sqrt(sin(2*b*x + 2*a))), x)

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

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="giac")

[Out]

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

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{2}\left (b x +a \right )}{\sin \left (2 b x +2 a \right )^{\frac {9}{2}}}\, dx \]

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]

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

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

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 \[ \int \frac {1}{{\sin \left (a+b\,x\right )}^2\,{\sin \left (2\,a+2\,b\,x\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(a + b*x)^2*sin(2*a + 2*b*x)^(9/2)),x)

[Out]

int(1/(sin(a + b*x)^2*sin(2*a + 2*b*x)^(9/2)), 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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