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

Optimal result

Integrand size = 10, antiderivative size = 50 \[ \int \sqrt {a \sin ^3(x)} \, dx=-\frac {2}{3} \cot (x) \sqrt {a \sin ^3(x)}-\frac {2 \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {x}{2},2\right ) \sqrt {a \sin ^3(x)}}{3 \sin ^{\frac {3}{2}}(x)} \]

[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)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3286, 2715, 2720} \[ \int \sqrt {a \sin ^3(x)} \, dx=-\frac {2}{3} \cot (x) \sqrt {a \sin ^3(x)}-\frac {2 \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {x}{2},2\right ) \sqrt {a \sin ^3(x)}}{3 \sin ^{\frac {3}{2}}(x)} \]

[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*} \text {integral}& = \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 \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {x}{2},2\right ) \sqrt {a \sin ^3(x)}}{3 \sin ^{\frac {3}{2}}(x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.82 \[ \int \sqrt {a \sin ^3(x)} \, dx=-\frac {2 \left (\operatorname {EllipticF}\left (\frac {1}{4} (\pi -2 x),2\right )+\cos (x) \sqrt {\sin (x)}\right ) \sqrt {a \sin ^3(x)}}{3 \sin ^{\frac {3}{2}}(x)} \]

[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))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.41 (sec) , antiderivative size = 163, normalized size of antiderivative = 3.26

method	result	size
default	\(\frac {\left (-i \sqrt {-i \left (i-\cot \left (x \right )+\csc \left (x \right )\right )}\, \sqrt {-i \left (i+\cot \left (x \right )-\csc \left (x \right )\right )}\, \sqrt {i \left (\csc \left (x \right )-\cot \left (x \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (x \right )+\csc \left (x \right )\right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (x \right )-i \sqrt {-i \left (i-\cot \left (x \right )+\csc \left (x \right )\right )}\, \sqrt {-i \left (i+\cot \left (x \right )-\csc \left (x \right )\right )}\, \sqrt {i \left (\csc \left (x \right )-\cot \left (x \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (x \right )+\csc \left (x \right )\right )}, \frac {\sqrt {2}}{2}\right )+\cos \left (x \right ) \sin \left (x \right ) \sqrt {2}\right ) \sqrt {a \left (\sin ^{3}\left (x \right )\right )}\, \sqrt {8}}{6 \left (\cos \left (x \right )-1\right ) \left (\cos \left (x \right )+1\right )}\)	\(163\)

[In]

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

[Out]

1/6*(-I*(-I*(I-cot(x)+csc(x)))^(1/2)*(-I*(I+cot(x)-csc(x)))^(1/2)*(I*(csc(x)-cot(x)))^(1/2)*EllipticF((-I*(I-c
ot(x)+csc(x)))^(1/2),1/2*2^(1/2))*cos(x)-I*(-I*(I-cot(x)+csc(x)))^(1/2)*(-I*(I+cot(x)-csc(x)))^(1/2)*(I*(csc(x
)-cot(x)))^(1/2)*EllipticF((-I*(I-cot(x)+csc(x)))^(1/2),1/2*2^(1/2))+cos(x)*sin(x)*2^(1/2))*(a*sin(x)^3)^(1/2)
/(cos(x)-1)/(cos(x)+1)*8^(1/2)

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.38 \[ \int \sqrt {a \sin ^3(x)} \, dx=\frac {\sqrt {2} \sqrt {-i \, a} \sin \left (x\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (x\right ) + i \, \sin \left (x\right )\right ) + \sqrt {2} \sqrt {i \, a} \sin \left (x\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (x\right ) - i \, \sin \left (x\right )\right ) - 2 \, \sqrt {-{\left (a \cos \left (x\right )^{2} - a\right )} \sin \left (x\right )} \cos \left (x\right )}{3 \, \sin \left (x\right )} \]

[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)

Sympy [F]

\[ \int \sqrt {a \sin ^3(x)} \, dx=\int \sqrt {a \sin ^{3}{\left (x \right )}}\, dx \]

[In]

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

[Out]

Integral(sqrt(a*sin(x)**3), x)

Maxima [F]

\[ \int \sqrt {a \sin ^3(x)} \, dx=\int { \sqrt {a \sin \left (x\right )^{3}} \,d x } \]

[In]

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

[Out]

integrate(sqrt(a*sin(x)^3), x)

Giac [F]

\[ \int \sqrt {a \sin ^3(x)} \, dx=\int { \sqrt {a \sin \left (x\right )^{3}} \,d x } \]

[In]

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

[Out]

integrate(sqrt(a*sin(x)^3), x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a \sin ^3(x)} \, dx=\int \sqrt {a\,{\sin \left (x\right )}^3} \,d x \]

[In]

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

[Out]

int((a*sin(x)^3)^(1/2), x)