3.15 \(\int \frac {1}{\csc ^{\frac {5}{2}}(a+b x)} \, dx\)

Optimal. Leaf size=67 \[ \frac {6 \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{5 b}-\frac {2 \cos (a+b x)}{5 b \csc ^{\frac {3}{2}}(a+b x)} \]

[Out]

-2/5*cos(b*x+a)/b/csc(b*x+a)^(3/2)-6/5*(sin(1/2*a+1/4*Pi+1/2*b*x)^2)^(1/2)/sin(1/2*a+1/4*Pi+1/2*b*x)*EllipticE
(cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/2))*csc(b*x+a)^(1/2)*sin(b*x+a)^(1/2)/b

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Rubi [A]  time = 0.03, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3769, 3771, 2639} \[ \frac {6 \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{5 b}-\frac {2 \cos (a+b x)}{5 b \csc ^{\frac {3}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^(-5/2),x]

[Out]

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

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 3769

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

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \frac {1}{\csc ^{\frac {5}{2}}(a+b x)} \, dx &=-\frac {2 \cos (a+b x)}{5 b \csc ^{\frac {3}{2}}(a+b x)}+\frac {3}{5} \int \frac {1}{\sqrt {\csc (a+b x)}} \, dx\\ &=-\frac {2 \cos (a+b x)}{5 b \csc ^{\frac {3}{2}}(a+b x)}+\frac {1}{5} \left (3 \sqrt {\csc (a+b x)} \sqrt {\sin (a+b x)}\right ) \int \sqrt {\sin (a+b x)} \, dx\\ &=-\frac {2 \cos (a+b x)}{5 b \csc ^{\frac {3}{2}}(a+b x)}+\frac {6 \sqrt {\csc (a+b x)} E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {\sin (a+b x)}}{5 b}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 60, normalized size = 0.90 \[ -\frac {2 \sqrt {\csc (a+b x)} \left (\sin ^2(a+b x) \cos (a+b x)+3 \sqrt {\sin (a+b x)} E\left (\left .\frac {1}{4} (-2 a-2 b x+\pi )\right |2\right )\right )}{5 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^(-5/2),x]

[Out]

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

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\csc \left (b x + a\right )^{\frac {5}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/csc(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

integral(csc(b*x + a)^(-5/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\csc \left (b x + a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/csc(b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)^(-5/2), x)

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maple [A]  time = 2.53, size = 142, normalized size = 2.12 \[ \frac {\frac {2 \left (\sin ^{4}\left (b x +a \right )\right )}{5}-\frac {2 \left (\sin ^{2}\left (b x +a \right )\right )}{5}-\frac {6 \sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, \EllipticE \left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )}{5}+\frac {3 \sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, \EllipticF \left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )}{5}}{\cos \left (b x +a \right ) \sqrt {\sin \left (b x +a \right )}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/csc(b*x+a)^(5/2),x)

[Out]

(2/5*sin(b*x+a)^4-2/5*sin(b*x+a)^2-6/5*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*Ellipt
icE((sin(b*x+a)+1)^(1/2),1/2*2^(1/2))+3/5*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*Ell
ipticF((sin(b*x+a)+1)^(1/2),1/2*2^(1/2)))/cos(b*x+a)/sin(b*x+a)^(1/2)/b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\csc \left (b x + a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/csc(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)^(-5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (\frac {1}{\sin \left (a+b\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1/sin(a + b*x))^(5/2),x)

[Out]

int(1/(1/sin(a + b*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\csc ^{\frac {5}{2}}{\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/csc(b*x+a)**(5/2),x)

[Out]

Integral(csc(a + b*x)**(-5/2), x)

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