3.1.11 \(\int \sin ^{\frac {3}{2}}(b x) \, dx\) [11]

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

[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)*s
in(b*x)^(1/2)/b

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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 = {2715, 2720} \begin {gather*} -\frac {2 F\left (\left .\frac {\pi }{4}-\frac {b x}{2}\right |2\right )}{3 b}-\frac {2 \sqrt {\sin (b x)} \cos (b x)}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[b*x]^(3/2),x]

[Out]

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

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[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 \sin ^{\frac {3}{2}}(b x) \, dx &=-\frac {2 \cos (b x) \sqrt {\sin (b x)}}{3 b}+\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) \sqrt {\sin (b x)}}{3 b}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 33, normalized size = 0.80 \begin {gather*} -\frac {2 \left (F\left (\left .\frac {1}{4} (\pi -2 b x)\right |2\right )+\cos (b x) \sqrt {\sin (b x)}\right )}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[b*x]^(3/2),x]

[Out]

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

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Maple [A]
time = 0.06, size = 72, normalized size = 1.76

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

[Out]

integrate(sin(b*x)^(3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 60, normalized size = 1.46 \begin {gather*} \frac {\sqrt {2} \sqrt {-i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x\right ) + i \, \sin \left (b x\right )\right ) + \sqrt {2} \sqrt {i} {\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 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin ^{\frac {3}{2}}{\left (b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x)**(3/2),x)

[Out]

Integral(sin(b*x)**(3/2), 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(sin(b*x)^(3/2),x, algorithm="giac")

[Out]

integrate(sin(b*x)^(3/2), x)

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Mupad [B]
time = 0.40, size = 34, normalized size = 0.83 \begin {gather*} -\frac {\cos \left (b\,x\right )\,{\sin \left (b\,x\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{2};\ {\cos \left (b\,x\right )}^2\right )}{b\,{\left ({\sin \left (b\,x\right )}^2\right )}^{5/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x)^(3/2),x)

[Out]

-(cos(b*x)*sin(b*x)^(5/2)*hypergeom([-1/4, 1/2], 3/2, cos(b*x)^2))/(b*(sin(b*x)^2)^(5/4))

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