3.1 \(\int (a \sin ^2(x))^{5/2} \, dx\)

Optimal. Leaf size=53 \[ -\frac {8}{15} a^2 \cot (x) \sqrt {a \sin ^2(x)}-\frac {1}{5} \cot (x) \left (a \sin ^2(x)\right )^{5/2}-\frac {4}{15} a \cot (x) \left (a \sin ^2(x)\right )^{3/2} \]

[Out]

-4/15*a*cot(x)*(a*sin(x)^2)^(3/2)-1/5*cot(x)*(a*sin(x)^2)^(5/2)-8/15*a^2*cot(x)*(a*sin(x)^2)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3203, 3207, 2638} \[ -\frac {8}{15} a^2 \cot (x) \sqrt {a \sin ^2(x)}-\frac {1}{5} \cot (x) \left (a \sin ^2(x)\right )^{5/2}-\frac {4}{15} a \cot (x) \left (a \sin ^2(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a^2*Cot[x]*Sqrt[a*Sin[x]^2])/15 - (4*a*Cot[x]*(a*Sin[x]^2)^(3/2))/15 - (Cot[x]*(a*Sin[x]^2)^(5/2))/5

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3203

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> -Simp[(Cot[e + f*x]*(b*Sin[e + f*x]^2)^p)/(2*f*p), x]
+ Dist[(b*(2*p - 1))/(2*p), Int[(b*Sin[e + f*x]^2)^(p - 1), x], x] /; FreeQ[{b, e, f}, x] &&  !IntegerQ[p] &&
GtQ[p, 1]

Rule 3207

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 ^2(x)\right )^{5/2} \, dx &=-\frac {1}{5} \cot (x) \left (a \sin ^2(x)\right )^{5/2}+\frac {1}{5} (4 a) \int \left (a \sin ^2(x)\right )^{3/2} \, dx\\ &=-\frac {4}{15} a \cot (x) \left (a \sin ^2(x)\right )^{3/2}-\frac {1}{5} \cot (x) \left (a \sin ^2(x)\right )^{5/2}+\frac {1}{15} \left (8 a^2\right ) \int \sqrt {a \sin ^2(x)} \, dx\\ &=-\frac {4}{15} a \cot (x) \left (a \sin ^2(x)\right )^{3/2}-\frac {1}{5} \cot (x) \left (a \sin ^2(x)\right )^{5/2}+\frac {1}{15} \left (8 a^2 \csc (x) \sqrt {a \sin ^2(x)}\right ) \int \sin (x) \, dx\\ &=-\frac {8}{15} a^2 \cot (x) \sqrt {a \sin ^2(x)}-\frac {4}{15} a \cot (x) \left (a \sin ^2(x)\right )^{3/2}-\frac {1}{5} \cot (x) \left (a \sin ^2(x)\right )^{5/2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 36, normalized size = 0.68 \[ -\frac {1}{240} a^2 (150 \cos (x)-25 \cos (3 x)+3 \cos (5 x)) \csc (x) \sqrt {a \sin ^2(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/240*(a^2*(150*Cos[x] - 25*Cos[3*x] + 3*Cos[5*x])*Csc[x]*Sqrt[a*Sin[x]^2])

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fricas [A]  time = 0.43, size = 43, normalized size = 0.81 \[ -\frac {{\left (3 \, a^{2} \cos \relax (x)^{5} - 10 \, a^{2} \cos \relax (x)^{3} + 15 \, a^{2} \cos \relax (x)\right )} \sqrt {-a \cos \relax (x)^{2} + a}}{15 \, \sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(3*a^2*cos(x)^5 - 10*a^2*cos(x)^3 + 15*a^2*cos(x))*sqrt(-a*cos(x)^2 + a)/sin(x)

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giac [A]  time = 3.74, size = 45, normalized size = 0.85 \[ \frac {1}{15} \, {\left (8 \, a^{2} \mathrm {sgn}\left (\sin \relax (x)\right ) - {\left (3 \, a^{2} \cos \relax (x)^{5} - 10 \, a^{2} \cos \relax (x)^{3} + 15 \, a^{2} \cos \relax (x)\right )} \mathrm {sgn}\left (\sin \relax (x)\right )\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(8*a^2*sgn(sin(x)) - (3*a^2*cos(x)^5 - 10*a^2*cos(x)^3 + 15*a^2*cos(x))*sgn(sin(x)))*sqrt(a)

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maple [A]  time = 0.71, size = 32, normalized size = 0.60 \[ -\frac {a^{3} \sin \relax (x ) \cos \relax (x ) \left (3 \left (\sin ^{4}\relax (x )\right )+4 \left (\sin ^{2}\relax (x )\right )+8\right )}{15 \sqrt {a \left (\sin ^{2}\relax (x )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*a^3*sin(x)*cos(x)*(3*sin(x)^4+4*sin(x)^2+8)/(a*sin(x)^2)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \sin \relax (x)^{2}\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sin(x)^2)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (a\,{\sin \relax (x)}^2\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a*sin(x)^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \sin ^{2}{\relax (x )}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*sin(x)**2)**(5/2), x)

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