3.1.7 \(\int (a \sin ^3(x))^{5/2} \, dx\) [7]

Optimal. Leaf size=123 \[ -\frac {26}{77} a^2 \cot (x) \sqrt {a \sin ^3(x)}-\frac {26 a^2 F\left (\left .\frac {\pi }{4}-\frac {x}{2}\right |2\right ) \sqrt {a \sin ^3(x)}}{77 \sin ^{\frac {3}{2}}(x)}-\frac {78}{385} a^2 \cos (x) \sin (x) \sqrt {a \sin ^3(x)}-\frac {26}{165} a^2 \cos (x) \sin ^3(x) \sqrt {a \sin ^3(x)}-\frac {2}{15} a^2 \cos (x) \sin ^5(x) \sqrt {a \sin ^3(x)} \]

[Out]

-26/77*a^2*cot(x)*(a*sin(x)^3)^(1/2)-26/77*a^2*(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)-78/385*a^2*cos(x)*sin(x)*(a*sin(x)^3)^(1/2)-26/165*a^2*cos
(x)*sin(x)^3*(a*sin(x)^3)^(1/2)-2/15*a^2*cos(x)*sin(x)^5*(a*sin(x)^3)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {26}{165} a^2 \sin ^3(x) \cos (x) \sqrt {a \sin ^3(x)}-\frac {78}{385} a^2 \sin (x) \cos (x) \sqrt {a \sin ^3(x)}-\frac {2}{15} a^2 \sin ^5(x) \cos (x) \sqrt {a \sin ^3(x)}-\frac {26}{77} a^2 \cot (x) \sqrt {a \sin ^3(x)}-\frac {26 a^2 F\left (\left .\frac {\pi }{4}-\frac {x}{2}\right |2\right ) \sqrt {a \sin ^3(x)}}{77 \sin ^{\frac {3}{2}}(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*Sin[x]^3)^(5/2),x]

[Out]

(-26*a^2*Cot[x]*Sqrt[a*Sin[x]^3])/77 - (26*a^2*EllipticF[Pi/4 - x/2, 2]*Sqrt[a*Sin[x]^3])/(77*Sin[x]^(3/2)) -
(78*a^2*Cos[x]*Sin[x]*Sqrt[a*Sin[x]^3])/385 - (26*a^2*Cos[x]*Sin[x]^3*Sqrt[a*Sin[x]^3])/165 - (2*a^2*Cos[x]*Si
n[x]^5*Sqrt[a*Sin[x]^3])/15

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 \left (a \sin ^3(x)\right )^{5/2} \, dx &=\frac {\left (a^2 \sqrt {a \sin ^3(x)}\right ) \int \sin ^{\frac {15}{2}}(x) \, dx}{\sin ^{\frac {3}{2}}(x)}\\ &=-\frac {2}{15} a^2 \cos (x) \sin ^5(x) \sqrt {a \sin ^3(x)}+\frac {\left (13 a^2 \sqrt {a \sin ^3(x)}\right ) \int \sin ^{\frac {11}{2}}(x) \, dx}{15 \sin ^{\frac {3}{2}}(x)}\\ &=-\frac {26}{165} a^2 \cos (x) \sin ^3(x) \sqrt {a \sin ^3(x)}-\frac {2}{15} a^2 \cos (x) \sin ^5(x) \sqrt {a \sin ^3(x)}+\frac {\left (39 a^2 \sqrt {a \sin ^3(x)}\right ) \int \sin ^{\frac {7}{2}}(x) \, dx}{55 \sin ^{\frac {3}{2}}(x)}\\ &=-\frac {78}{385} a^2 \cos (x) \sin (x) \sqrt {a \sin ^3(x)}-\frac {26}{165} a^2 \cos (x) \sin ^3(x) \sqrt {a \sin ^3(x)}-\frac {2}{15} a^2 \cos (x) \sin ^5(x) \sqrt {a \sin ^3(x)}+\frac {\left (39 a^2 \sqrt {a \sin ^3(x)}\right ) \int \sin ^{\frac {3}{2}}(x) \, dx}{77 \sin ^{\frac {3}{2}}(x)}\\ &=-\frac {26}{77} a^2 \cot (x) \sqrt {a \sin ^3(x)}-\frac {78}{385} a^2 \cos (x) \sin (x) \sqrt {a \sin ^3(x)}-\frac {26}{165} a^2 \cos (x) \sin ^3(x) \sqrt {a \sin ^3(x)}-\frac {2}{15} a^2 \cos (x) \sin ^5(x) \sqrt {a \sin ^3(x)}+\frac {\left (13 a^2 \sqrt {a \sin ^3(x)}\right ) \int \frac {1}{\sqrt {\sin (x)}} \, dx}{77 \sin ^{\frac {3}{2}}(x)}\\ &=-\frac {26}{77} a^2 \cot (x) \sqrt {a \sin ^3(x)}-\frac {26 a^2 F\left (\left .\frac {\pi }{4}-\frac {x}{2}\right |2\right ) \sqrt {a \sin ^3(x)}}{77 \sin ^{\frac {3}{2}}(x)}-\frac {78}{385} a^2 \cos (x) \sin (x) \sqrt {a \sin ^3(x)}-\frac {26}{165} a^2 \cos (x) \sin ^3(x) \sqrt {a \sin ^3(x)}-\frac {2}{15} a^2 \cos (x) \sin ^5(x) \sqrt {a \sin ^3(x)}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 65, normalized size = 0.53 \begin {gather*} \frac {a \left (-12480 F\left (\left .\frac {1}{4} (\pi -2 x)\right |2\right )+(-15465 \cos (x)+3657 \cos (3 x)-749 \cos (5 x)+77 \cos (7 x)) \sqrt {\sin (x)}\right ) \left (a \sin ^3(x)\right )^{3/2}}{36960 \sin ^{\frac {9}{2}}(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*Sin[x]^3)^(5/2),x]

[Out]

(a*(-12480*EllipticF[(Pi - 2*x)/4, 2] + (-15465*Cos[x] + 3657*Cos[3*x] - 749*Cos[5*x] + 77*Cos[7*x])*Sqrt[Sin[
x]])*(a*Sin[x]^3)^(3/2))/(36960*Sin[x]^(9/2))

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Maple [C] Result contains complex when optimal does not.
time = 0.54, size = 152, normalized size = 1.24

method result size
default \(-\frac {\left (-154 \left (\cos ^{8}\left (x \right )\right )+195 i \sqrt {2}\, \sin \left (x \right ) \sqrt {\frac {i \cos \left (x \right )+\sin \left (x \right )-i}{\sin \left (x \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (x \right )\right )}{\sin \left (x \right )}}\, \sqrt {\frac {-i \cos \left (x \right )+\sin \left (x \right )+i}{\sin \left (x \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (x \right )+\sin \left (x \right )-i}{\sin \left (x \right )}}, \frac {\sqrt {2}}{2}\right )+154 \left (\cos ^{7}\left (x \right )\right )+644 \left (\cos ^{6}\left (x \right )\right )-644 \left (\cos ^{5}\left (x \right )\right )-1060 \left (\cos ^{4}\left (x \right )\right )+1060 \left (\cos ^{3}\left (x \right )\right )+960 \left (\cos ^{2}\left (x \right )\right )-960 \cos \left (x \right )\right ) \left (a \left (1-\left (\cos ^{2}\left (x \right )\right )\right ) \sin \left (x \right )\right )^{\frac {5}{2}}}{1155 \sin \left (x \right )^{7} \left (-1+\cos \left (x \right )\right )}\) \(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(x)^3)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/1155*(-154*cos(x)^8+195*I*2^(1/2)*sin(x)*((I*cos(x)+sin(x)-I)/sin(x))^(1/2)*(-I*(-1+cos(x))/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))+154*cos(x)^7+644*
cos(x)^6-644*cos(x)^5-1060*cos(x)^4+1060*cos(x)^3+960*cos(x)^2-960*cos(x))*(a*(1-cos(x)^2)*sin(x))^(5/2)/sin(x
)^7/(-1+cos(x))

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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)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(x)^3)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 110, normalized size = 0.89 \begin {gather*} \frac {195 \, \sqrt {2} \sqrt {-i \, a} a^{2} \sin \left (x\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (x\right ) + i \, \sin \left (x\right )\right ) + 195 \, \sqrt {2} \sqrt {i \, a} a^{2} \sin \left (x\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (x\right ) - i \, \sin \left (x\right )\right ) + 2 \, {\left (77 \, a^{2} \cos \left (x\right )^{7} - 322 \, a^{2} \cos \left (x\right )^{5} + 530 \, a^{2} \cos \left (x\right )^{3} - 480 \, a^{2} \cos \left (x\right )\right )} \sqrt {-{\left (a \cos \left (x\right )^{2} - a\right )} \sin \left (x\right )}}{1155 \, \sin \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(x)^3)^(5/2),x, algorithm="fricas")

[Out]

1/1155*(195*sqrt(2)*sqrt(-I*a)*a^2*sin(x)*weierstrassPInverse(4, 0, cos(x) + I*sin(x)) + 195*sqrt(2)*sqrt(I*a)
*a^2*sin(x)*weierstrassPInverse(4, 0, cos(x) - I*sin(x)) + 2*(77*a^2*cos(x)^7 - 322*a^2*cos(x)^5 + 530*a^2*cos
(x)^3 - 480*a^2*cos(x))*sqrt(-(a*cos(x)^2 - a)*sin(x)))/sin(x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(x)**3)**(5/2),x)

[Out]

Timed out

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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)^(5/2),x, algorithm="giac")

[Out]

integrate((a*sin(x)^3)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a\,{\sin \left (x\right )}^3\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(x)^3)^(5/2),x)

[Out]

int((a*sin(x)^3)^(5/2), x)

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