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

Optimal. Leaf size=79 \[ \frac{8 \sin (a+b x)}{15 b \sin ^{\frac{3}{2}}(2 a+2 b x)}-\frac{2 \cos (a+b x)}{5 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{16 \cos (a+b x)}{15 b \sqrt{\sin (2 a+2 b x)}} \]

[Out]

(-2*Cos[a + b*x])/(5*b*Sin[2*a + 2*b*x]^(5/2)) + (8*Sin[a + b*x])/(15*b*Sin[2*a + 2*b*x]^(3/2)) - (16*Cos[a +
b*x])/(15*b*Sqrt[Sin[2*a + 2*b*x]])

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Rubi [A]  time = 0.0860161, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4308, 4303, 4304, 4291} \[ \frac{8 \sin (a+b x)}{15 b \sin ^{\frac{3}{2}}(2 a+2 b x)}-\frac{2 \cos (a+b x)}{5 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{16 \cos (a+b x)}{15 b \sqrt{\sin (2 a+2 b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]/Sin[2*a + 2*b*x]^(5/2),x]

[Out]

(-2*Cos[a + b*x])/(5*b*Sin[2*a + 2*b*x]^(5/2)) + (8*Sin[a + b*x])/(15*b*Sin[2*a + 2*b*x]^(3/2)) - (16*Cos[a +
b*x])/(15*b*Sqrt[Sin[2*a + 2*b*x]])

Rule 4308

Int[((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_)/sin[(a_.) + (b_.)*(x_)], x_Symbol] :> Dist[2*g, Int[Cos[a + b*x]*(g*S
in[c + d*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, g, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ
[p] && IntegerQ[2*p]

Rule 4303

Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(Cos[a + b*x]*(g*Sin[c + d
*x])^(p + 1))/(2*b*g*(p + 1)), x] + Dist[(2*p + 3)/(2*g*(p + 1)), Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p + 1), x
], x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && LtQ[p, -1] && Integ
erQ[2*p]

Rule 4304

Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> -Simp[(Sin[a + b*x]*(g*Sin[c +
d*x])^(p + 1))/(2*b*g*(p + 1)), x] + Dist[(2*p + 3)/(2*g*(p + 1)), Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p + 1),
x], x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && LtQ[p, -1] && Inte
gerQ[2*p]

Rule 4291

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> -Simp[((e*Cos[a +
b*x])^m*(g*Sin[c + d*x])^(p + 1))/(b*g*m), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && E
qQ[d/b, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rubi steps

\begin{align*} \int \frac{\csc (a+b x)}{\sin ^{\frac{5}{2}}(2 a+2 b x)} \, dx &=2 \int \frac{\cos (a+b x)}{\sin ^{\frac{7}{2}}(2 a+2 b x)} \, dx\\ &=-\frac{2 \cos (a+b x)}{5 b \sin ^{\frac{5}{2}}(2 a+2 b x)}+\frac{8}{5} \int \frac{\sin (a+b x)}{\sin ^{\frac{5}{2}}(2 a+2 b x)} \, dx\\ &=-\frac{2 \cos (a+b x)}{5 b \sin ^{\frac{5}{2}}(2 a+2 b x)}+\frac{8 \sin (a+b x)}{15 b \sin ^{\frac{3}{2}}(2 a+2 b x)}+\frac{16}{15} \int \frac{\cos (a+b x)}{\sin ^{\frac{3}{2}}(2 a+2 b x)} \, dx\\ &=-\frac{2 \cos (a+b x)}{5 b \sin ^{\frac{5}{2}}(2 a+2 b x)}+\frac{8 \sin (a+b x)}{15 b \sin ^{\frac{3}{2}}(2 a+2 b x)}-\frac{16 \cos (a+b x)}{15 b \sqrt{\sin (2 a+2 b x)}}\\ \end{align*}

Mathematica [A]  time = 0.124077, size = 52, normalized size = 0.66 \[ -\frac{\sqrt{\sin (2 (a+b x))} \left (3 \csc ^3(a+b x)+27 \csc (a+b x)-5 \tan (a+b x) \sec (a+b x)\right )}{60 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]/Sin[2*a + 2*b*x]^(5/2),x]

[Out]

-(Sqrt[Sin[2*(a + b*x)]]*(27*Csc[a + b*x] + 3*Csc[a + b*x]^3 - 5*Sec[a + b*x]*Tan[a + b*x]))/(60*b)

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Maple [C]  time = 5.486, size = 481, normalized size = 6.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/80/b*(-tan(1/2*b*x+1/2*a)/(tan(1/2*b*x+1/2*a)^2-1))^(1/2)/tan(1/2*b*x+1/2*a)^3*(24*(tan(1/2*b*x+1/2*a)*(tan
(1/2*b*x+1/2*a)^2-1))^(1/2)*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))
^(1/2)*EllipticE((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))*tan(1/2*b*x+1/2*a)^2-12*(tan(1/2*b*x+1/2*a)*(tan(1/
2*b*x+1/2*a)^2-1))^(1/2)*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1
/2)*EllipticF((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))*tan(1/2*b*x+1/2*a)^2+(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+
1/2*a)^2-1))^(1/2)*tan(1/2*b*x+1/2*a)^6-(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)^2-1))^(1/2)*tan(1/2*b*x+1/2*a)
^4+12*(tan(1/2*b*x+1/2*a)^3-tan(1/2*b*x+1/2*a))^(1/2)*tan(1/2*b*x+1/2*a)^4-(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/
2*a)^2-1))^(1/2)*tan(1/2*b*x+1/2*a)^2-12*(tan(1/2*b*x+1/2*a)^3-tan(1/2*b*x+1/2*a))^(1/2)*tan(1/2*b*x+1/2*a)^2+
(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)^2-1))^(1/2))/(tan(1/2*b*x+1/2*a)^3-tan(1/2*b*x+1/2*a))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (b x + a\right )}{\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)/sin(2*b*x+2*a)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.514205, size = 267, normalized size = 3.38 \begin{align*} -\frac{\sqrt{2}{\left (32 \, \cos \left (b x + a\right )^{4} - 40 \, \cos \left (b x + a\right )^{2} + 5\right )} \sqrt{\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \,{\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{60 \,{\left (b \cos \left (b x + a\right )^{4} - b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/sin(2*b*x+2*a)^(5/2),x, algorithm="fricas")

[Out]

-1/60*(sqrt(2)*(32*cos(b*x + a)^4 - 40*cos(b*x + a)^2 + 5)*sqrt(cos(b*x + a)*sin(b*x + a)) + 32*(cos(b*x + a)^
4 - cos(b*x + a)^2)*sin(b*x + a))/((b*cos(b*x + a)^4 - b*cos(b*x + a)^2)*sin(b*x + a))

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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)/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 \frac{\csc \left (b x + a\right )}{\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)/sin(2*b*x+2*a)^(5/2),x, algorithm="giac")

[Out]

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