3.2.14 \(\int \frac {\csc ^2(a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx\) [114]

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

[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 = 0.04, 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 = {4385, 2716, 2720} \begin {gather*} \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)} \end {gather*}

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 2716

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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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*} \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 \begin {gather*} \frac {480 F\left (\left .a-\frac {\pi }{4}+b x\right |2\right )+\left (-141 \csc ^2(a+b x)-32 \csc ^4(a+b x)-7 \csc ^6(a+b x)+11 \sec ^2(a+b x) \left (9+\sec ^2(a+b x)\right )\right ) \sqrt {\sin (2 (a+b x))}}{1232 b} \end {gather*}

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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Maple [F(-1)]
time = 180.00, size = 0, normalized size = 0.00 hanged

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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [C] Result contains complex when optimal does not.
time = 0.94, size = 233, normalized size = 2.20 \begin {gather*} -\frac {240 \, \sqrt {2 i} {\left (\cos \left (b x + a\right )^{10} - 3 \, \cos \left (b x + a\right )^{8} + 3 \, \cos \left (b x + a\right )^{6} - \cos \left (b x + a\right )^{4}\right )} {\rm ellipticF}\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ), -1\right ) + 240 \, \sqrt {-2 i} {\left (\cos \left (b x + a\right )^{10} - 3 \, \cos \left (b x + a\right )^{8} + 3 \, \cos \left (b x + a\right )^{6} - \cos \left (b x + a\right )^{4}\right )} {\rm ellipticF}\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ), -1\right ) - \sqrt {2} {\left (240 \, \cos \left (b x + a\right )^{8} - 600 \, \cos \left (b x + a\right )^{6} + 444 \, \cos \left (b x + a\right )^{4} - 66 \, \cos \left (b x + a\right )^{2} - 11\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{1232 \, {\left (b \cos \left (b x + a\right )^{10} - 3 \, b \cos \left (b x + a\right )^{8} + 3 \, b \cos \left (b x + a\right )^{6} - b \cos \left (b x + a\right )^{4}\right )}} \end {gather*}

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]

-1/1232*(240*sqrt(2*I)*(cos(b*x + a)^10 - 3*cos(b*x + a)^8 + 3*cos(b*x + a)^6 - cos(b*x + a)^4)*ellipticF(cos(
b*x + a) + I*sin(b*x + a), -1) + 240*sqrt(-2*I)*(cos(b*x + a)^10 - 3*cos(b*x + a)^8 + 3*cos(b*x + a)^6 - cos(b
*x + a)^4)*ellipticF(cos(b*x + a) - I*sin(b*x + a), -1) - sqrt(2)*(240*cos(b*x + a)^8 - 600*cos(b*x + a)^6 + 4
44*cos(b*x + a)^4 - 66*cos(b*x + a)^2 - 11)*sqrt(cos(b*x + a)*sin(b*x + a)))/(b*cos(b*x + a)^10 - 3*b*cos(b*x
+ a)^8 + 3*b*cos(b*x + a)^6 - b*cos(b*x + a)^4)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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