3.1.15 \(\int \frac {1}{\sin ^{\frac {5}{2}}(b x)} \, dx\) [15]

Optimal. Leaf size=41 \[ -\frac {2 F\left (\left .\frac {\pi }{4}-\frac {b x}{2}\right |2\right )}{3 b}-\frac {2 \cos (b x)}{3 b \sin ^{\frac {3}{2}}(b x)} \]

[Out]

-2/3*(sin(1/4*Pi+1/2*b*x)^2)^(1/2)/sin(1/4*Pi+1/2*b*x)*EllipticF(cos(1/4*Pi+1/2*b*x),2^(1/2))/b-2/3*cos(b*x)/b
/sin(b*x)^(3/2)

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Rubi [A]
time = 0.01, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2716, 2720} \begin {gather*} -\frac {2 F\left (\left .\frac {\pi }{4}-\frac {b x}{2}\right |2\right )}{3 b}-\frac {2 \cos (b x)}{3 b \sin ^{\frac {3}{2}}(b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[b*x]^(-5/2),x]

[Out]

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

Rubi steps

\begin {align*} \int \frac {1}{\sin ^{\frac {5}{2}}(b x)} \, dx &=-\frac {2 \cos (b x)}{3 b \sin ^{\frac {3}{2}}(b x)}+\frac {1}{3} \int \frac {1}{\sqrt {\sin (b x)}} \, dx\\ &=-\frac {2 F\left (\left .\frac {\pi }{4}-\frac {b x}{2}\right |2\right )}{3 b}-\frac {2 \cos (b x)}{3 b \sin ^{\frac {3}{2}}(b x)}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 33, normalized size = 0.80 \begin {gather*} -\frac {2 \left (F\left (\left .\frac {1}{4} (\pi -2 b x)\right |2\right )+\frac {\cos (b x)}{\sin ^{\frac {3}{2}}(b x)}\right )}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[b*x]^(-5/2),x]

[Out]

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

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Maple [A]
time = 0.05, size = 72, normalized size = 1.76

method	result	size
default	\(\frac {\sqrt {\sin \left (b x \right )+1}\, \sqrt {-2 \sin \left (b x \right )+2}\, \sqrt {-\sin \left (b x \right )}\, \EllipticF \left (\sqrt {\sin \left (b x \right )+1}, \frac {\sqrt {2}}{2}\right ) \sin \left (b x \right )-2 \left (\cos ^{2}\left (b x \right )\right )}{3 \sin \left (b x \right )^{\frac {3}{2}} \cos \left (b x \right ) b}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sin(b*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3/sin(b*x)^(3/2)*((sin(b*x)+1)^(1/2)*(-2*sin(b*x)+2)^(1/2)*(-sin(b*x))^(1/2)*EllipticF((sin(b*x)+1)^(1/2),1/
2*2^(1/2))*sin(b*x)-2*cos(b*x)^2)/cos(b*x)/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(1/sin(b*x)^(5/2),x, algorithm="maxima")

[Out]

integrate(sin(b*x)^(-5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 97, normalized size = 2.37 \begin {gather*} \frac {\sqrt {-i} {\left (\sqrt {2} \cos \left (b x\right )^{2} - \sqrt {2}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x\right ) + i \, \sin \left (b x\right )\right ) + \sqrt {i} {\left (\sqrt {2} \cos \left (b x\right )^{2} - \sqrt {2}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x\right ) - i \, \sin \left (b x\right )\right ) + 2 \, \cos \left (b x\right ) \sqrt {\sin \left (b x\right )}}{3 \, {\left (b \cos \left (b x\right )^{2} - b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sin ^{\frac {5}{2}}{\left (b x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(b*x)**(5/2),x)

[Out]

Integral(sin(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(1/sin(b*x)^(5/2),x, algorithm="giac")

[Out]

integrate(sin(b*x)^(-5/2), x)

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Mupad [B]
time = 0.59, size = 34, normalized size = 0.83 \begin {gather*} -\frac {\cos \left (b\,x\right )\,{\left ({\sin \left (b\,x\right )}^2\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {3}{2};\ {\cos \left (b\,x\right )}^2\right )}{b\,{\sin \left (b\,x\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sin(b*x)^(5/2),x)

[Out]

-(cos(b*x)*(sin(b*x)^2)^(3/4)*hypergeom([1/2, 7/4], 3/2, cos(b*x)^2))/(b*sin(b*x)^(3/2))

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