3.1.10 \(\int \csc ^{\frac {5}{2}}(a+b x) \, dx\) [10]

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

[Out]

-2/3*cos(b*x+a)*csc(b*x+a)^(3/2)/b-2/3*(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)*EllipticF
(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 = 0.02, 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 = {3853, 3856, 2720} \begin {gather*} \frac {2 \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} F\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{3 b}-\frac {2 \cos (a+b x) \csc ^{\frac {3}{2}}(a+b x)}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^(5/2),x]

[Out]

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

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 3853

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

Rule 3856

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

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Mathematica [A]
time = 0.09, size = 50, normalized size = 0.75 \begin {gather*} -\frac {2 \csc ^{\frac {3}{2}}(a+b x) \left (\cos (a+b x)+F\left (\left .\frac {1}{4} (-2 a+\pi -2 b x)\right |2\right ) \sin ^{\frac {3}{2}}(a+b x)\right )}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^(5/2),x]

[Out]

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

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Maple [A]
time = 0.09, size = 88, normalized size = 1.31

method	result	size
default	\(\frac {\sqrt {\sin \left (x b +a \right )+1}\, \sqrt {-2 \sin \left (x b +a \right )+2}\, \sqrt {-\sin \left (x b +a \right )}\, \EllipticF \left (\sqrt {\sin \left (x b +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \sin \left (x b +a \right )-2 \left (\cos ^{2}\left (x b +a \right )\right )}{3 \sin \left (x b +a \right )^{\frac {3}{2}} \cos \left (x b +a \right ) b}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3/sin(b*x+a)^(3/2)*((sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*EllipticF((sin(b*x+a)+1
)^(1/2),1/2*2^(1/2))*sin(b*x+a)-2*cos(b*x+a)^2)/cos(b*x+a)/b

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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)^(5/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.72, size = 88, normalized size = 1.31 \begin {gather*} \frac {-i \, \sqrt {2 i} \sin \left (b x + a\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + i \, \sqrt {-2 i} \sin \left (b x + a\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - \frac {2 \, \cos \left (b x + a\right )}{\sqrt {\sin \left (b x + a\right )}}}{3 \, b \sin \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-I*sqrt(2*I)*sin(b*x + a)*weierstrassPInverse(4, 0, cos(b*x + a) + I*sin(b*x + a)) + I*sqrt(-2*I)*sin(b*x
+ a)*weierstrassPInverse(4, 0, cos(b*x + a) - I*sin(b*x + a)) - 2*cos(b*x + a)/sqrt(sin(b*x + a)))/(b*sin(b*x
+ a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(a + b*x)**(5/2), x)

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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)^(5/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/sin(a + b*x))^(5/2),x)

[Out]

int((1/sin(a + b*x))^(5/2), x)

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