3.1.9 \(\int \sqrt {a \sin ^3(x)} \, dx\) [9]

Optimal. Leaf size=50 \[ -\frac {2}{3} \cot (x) \sqrt {a \sin ^3(x)}-\frac {2 F\left (\left .\frac {\pi }{4}-\frac {x}{2}\right |2\right ) \sqrt {a \sin ^3(x)}}{3 \sin ^{\frac {3}{2}}(x)} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 50, 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 = {3286, 2715, 2720} \begin {gather*} -\frac {2}{3} \cot (x) \sqrt {a \sin ^3(x)}-\frac {2 F\left (\left .\frac {\pi }{4}-\frac {x}{2}\right |2\right ) \sqrt {a \sin ^3(x)}}{3 \sin ^{\frac {3}{2}}(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*Sin[x]^3],x]

[Out]

(-2*Cot[x]*Sqrt[a*Sin[x]^3])/3 - (2*EllipticF[Pi/4 - x/2, 2]*Sqrt[a*Sin[x]^3])/(3*Sin[x]^(3/2))

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]

Rule 3286

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 41, normalized size = 0.82 \begin {gather*} -\frac {2 \left (F\left (\left .\frac {1}{4} (\pi -2 x)\right |2\right )+\cos (x) \sqrt {\sin (x)}\right ) \sqrt {a \sin ^3(x)}}{3 \sin ^{\frac {3}{2}}(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*Sin[x]^3],x]

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.54, size = 124, normalized size = 2.48

method result size
default \(-\frac {\left (i \sqrt {-\frac {i \left (-1+\cos \left (x \right )\right )}{\sin \left (x \right )}}\, \sin \left (x \right ) \sqrt {\frac {i \cos \left (x \right )+\sin \left (x \right )-i}{\sin \left (x \right )}}\, \sqrt {-\frac {i \cos \left (x \right )-\sin \left (x \right )-i}{\sin \left (x \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (x \right )+\sin \left (x \right )-i}{\sin \left (x \right )}}, \frac {\sqrt {2}}{2}\right )+\left (\cos ^{2}\left (x \right )\right ) \sqrt {2}-\cos \left (x \right ) \sqrt {2}\right ) \sqrt {a \left (1-\left (\cos ^{2}\left (x \right )\right )\right ) \sin \left (x \right )}\, \sqrt {8}}{6 \sin \left (x \right ) \left (-1+\cos \left (x \right )\right )}\) \(124\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(x)^3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/6*(I*(-I*(-1+cos(x))/sin(x))^(1/2)*sin(x)*((I*cos(x)+sin(x)-I)/sin(x))^(1/2)*(-(I*cos(x)-sin(x)-I)/sin(x))^
(1/2)*EllipticF(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))+cos(x)^2*2^(1/2)-cos(x)*2^(1/2))*(a*(1-cos(x)^
2)*sin(x))^(1/2)/sin(x)/(-1+cos(x))*8^(1/2)

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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((a*sin(x)^3)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sin(x)^3), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 69, normalized size = 1.38 \begin {gather*} \frac {\sqrt {2} \sqrt {-i \, a} \sin \left (x\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (x\right ) + i \, \sin \left (x\right )\right ) + \sqrt {2} \sqrt {i \, a} \sin \left (x\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (x\right ) - i \, \sin \left (x\right )\right ) - 2 \, \sqrt {-{\left (a \cos \left (x\right )^{2} - a\right )} \sin \left (x\right )} \cos \left (x\right )}{3 \, \sin \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(x)^3)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(x)**3)**(1/2),x)

[Out]

Integral(sqrt(a*sin(x)**3), 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((a*sin(x)^3)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(a*sin(x)^3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \sqrt {a\,{\sin \left (x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(x)^3)^(1/2),x)

[Out]

int((a*sin(x)^3)^(1/2), x)

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