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

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

[Out]

(30*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))

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Rubi [A]  time = 0.059475, antiderivative size = 106, normalized size of antiderivative = 1., 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 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 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]

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*}

Mathematica [A]  time = 0.343713, size = 86, normalized size = 0.81 \[ \frac{480 \text{EllipticF}\left (a+b x-\frac{\pi }{4},2\right )+\sqrt{\sin (2 (a+b x))} \left (-7 \csc ^6(a+b x)-32 \csc ^4(a+b x)-141 \csc ^2(a+b x)+11 \sec ^2(a+b x) \left (\sec ^2(a+b x)+9\right )\right )}{1232 b} \]

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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Maple [A]  time = 47.27, size = 167, normalized size = 1.6 \begin{align*}{\frac{\sqrt{2}}{64\,b} \left ( -{\frac{64\,\sqrt{2}}{11} \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{-{\frac{11}{2}}}}+{\frac{32\,\sqrt{2}}{77\,\cos \left ( 2\,bx+2\,a \right ) } \left ( 15\,\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},1/2\,\sqrt{2} \right ) \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{5}+30\, \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{6}-12\, \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{4}-4\, \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{2}-14 \right ) \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{-{\frac{11}{2}}}} \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)^(9/2),x)

[Out]

1/64*2^(1/2)*(-64/11*2^(1/2)/sin(2*b*x+2*a)^(11/2)+32/77*2^(1/2)/sin(2*b*x+2*a)^(11/2)*(15*(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))*sin(
2*b*x+2*a)^5+30*sin(2*b*x+2*a)^6-12*sin(2*b*x+2*a)^4-4*sin(2*b*x+2*a)^2-14)/cos(2*b*x+2*a))/b

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Maxima [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)^(9/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \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)^(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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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)**(9/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac{9}{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)^(9/2),x, algorithm="giac")

[Out]

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