3.14 \(\int (a \sin ^4(x))^{3/2} \, dx\)

Optimal. Leaf size=78 \[ -\frac {5}{24} a \sin (x) \cos (x) \sqrt {a \sin ^4(x)}-\frac {1}{6} a \sin ^3(x) \cos (x) \sqrt {a \sin ^4(x)}-\frac {5}{16} a \cot (x) \sqrt {a \sin ^4(x)}+\frac {5}{16} a x \csc ^2(x) \sqrt {a \sin ^4(x)} \]

[Out]

-5/16*a*cot(x)*(a*sin(x)^4)^(1/2)+5/16*a*x*csc(x)^2*(a*sin(x)^4)^(1/2)-5/24*a*cos(x)*sin(x)*(a*sin(x)^4)^(1/2)
-1/6*a*cos(x)*sin(x)^3*(a*sin(x)^4)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3207, 2635, 8} \[ -\frac {1}{6} a \sin ^3(x) \cos (x) \sqrt {a \sin ^4(x)}-\frac {5}{24} a \sin (x) \cos (x) \sqrt {a \sin ^4(x)}-\frac {5}{16} a \cot (x) \sqrt {a \sin ^4(x)}+\frac {5}{16} a x \csc ^2(x) \sqrt {a \sin ^4(x)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sin[x]^4)^(3/2),x]

[Out]

(-5*a*Cot[x]*Sqrt[a*Sin[x]^4])/16 + (5*a*x*Csc[x]^2*Sqrt[a*Sin[x]^4])/16 - (5*a*Cos[x]*Sin[x]*Sqrt[a*Sin[x]^4]
)/24 - (a*Cos[x]*Sin[x]^3*Sqrt[a*Sin[x]^4])/6

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2635

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] && Integer
Q[2*n]

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 ^4(x)\right )^{3/2} \, dx &=\left (a \csc ^2(x) \sqrt {a \sin ^4(x)}\right ) \int \sin ^6(x) \, dx\\ &=-\frac {1}{6} a \cos (x) \sin ^3(x) \sqrt {a \sin ^4(x)}+\frac {1}{6} \left (5 a \csc ^2(x) \sqrt {a \sin ^4(x)}\right ) \int \sin ^4(x) \, dx\\ &=-\frac {5}{24} a \cos (x) \sin (x) \sqrt {a \sin ^4(x)}-\frac {1}{6} a \cos (x) \sin ^3(x) \sqrt {a \sin ^4(x)}+\frac {1}{8} \left (5 a \csc ^2(x) \sqrt {a \sin ^4(x)}\right ) \int \sin ^2(x) \, dx\\ &=-\frac {5}{16} a \cot (x) \sqrt {a \sin ^4(x)}-\frac {5}{24} a \cos (x) \sin (x) \sqrt {a \sin ^4(x)}-\frac {1}{6} a \cos (x) \sin ^3(x) \sqrt {a \sin ^4(x)}+\frac {1}{16} \left (5 a \csc ^2(x) \sqrt {a \sin ^4(x)}\right ) \int 1 \, dx\\ &=-\frac {5}{16} a \cot (x) \sqrt {a \sin ^4(x)}+\frac {5}{16} a x \csc ^2(x) \sqrt {a \sin ^4(x)}-\frac {5}{24} a \cos (x) \sin (x) \sqrt {a \sin ^4(x)}-\frac {1}{6} a \cos (x) \sin ^3(x) \sqrt {a \sin ^4(x)}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 38, normalized size = 0.49 \[ -\frac {1}{192} (-60 x+45 \sin (2 x)-9 \sin (4 x)+\sin (6 x)) \csc ^6(x) \left (a \sin ^4(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sin[x]^4)^(3/2),x]

[Out]

-1/192*(Csc[x]^6*(a*Sin[x]^4)^(3/2)*(-60*x + 45*Sin[2*x] - 9*Sin[4*x] + Sin[6*x]))

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fricas [A]  time = 0.44, size = 56, normalized size = 0.72 \[ -\frac {\sqrt {a \cos \relax (x)^{4} - 2 \, a \cos \relax (x)^{2} + a} {\left (15 \, a x - {\left (8 \, a \cos \relax (x)^{5} - 26 \, a \cos \relax (x)^{3} + 33 \, a \cos \relax (x)\right )} \sin \relax (x)\right )}}{48 \, {\left (\cos \relax (x)^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(x)^4)^(3/2),x, algorithm="fricas")

[Out]

-1/48*sqrt(a*cos(x)^4 - 2*a*cos(x)^2 + a)*(15*a*x - (8*a*cos(x)^5 - 26*a*cos(x)^3 + 33*a*cos(x))*sin(x))/(cos(
x)^2 - 1)

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giac [A]  time = 0.13, size = 27, normalized size = 0.35 \[ \frac {1}{192} \, a^{\frac {3}{2}} {\left (60 \, x - \sin \left (6 \, x\right ) + 9 \, \sin \left (4 \, x\right ) - 45 \, \sin \left (2 \, x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(x)^4)^(3/2),x, algorithm="giac")

[Out]

1/192*a^(3/2)*(60*x - sin(6*x) + 9*sin(4*x) - 45*sin(2*x))

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maple [A]  time = 0.28, size = 47, normalized size = 0.60 \[ -\frac {\left (a \left (1-\left (\cos ^{2}\relax (x )\right )\right )^{2}\right )^{\frac {3}{2}} \left (8 \left (\cos ^{5}\relax (x )\right ) \sin \relax (x )-26 \left (\cos ^{3}\relax (x )\right ) \sin \relax (x )+33 \sin \relax (x ) \cos \relax (x )-15 x \right )}{48 \sin \relax (x )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(x)^4)^(3/2),x)

[Out]

-1/48*(a*(1-cos(x)^2)^2)^(3/2)*(8*cos(x)^5*sin(x)-26*cos(x)^3*sin(x)+33*sin(x)*cos(x)-15*x)/sin(x)^6

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maxima [A]  time = 0.61, size = 55, normalized size = 0.71 \[ \frac {5}{16} \, a^{\frac {3}{2}} x - \frac {33 \, a^{\frac {3}{2}} \tan \relax (x)^{5} + 40 \, a^{\frac {3}{2}} \tan \relax (x)^{3} + 15 \, a^{\frac {3}{2}} \tan \relax (x)}{48 \, {\left (\tan \relax (x)^{6} + 3 \, \tan \relax (x)^{4} + 3 \, \tan \relax (x)^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(x)^4)^(3/2),x, algorithm="maxima")

[Out]

5/16*a^(3/2)*x - 1/48*(33*a^(3/2)*tan(x)^5 + 40*a^(3/2)*tan(x)^3 + 15*a^(3/2)*tan(x))/(tan(x)^6 + 3*tan(x)^4 +
3*tan(x)^2 + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a\,{\sin \relax (x)}^4\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(x)^4)^(3/2),x)

[Out]

int((a*sin(x)^4)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \sin ^{4}{\relax (x )}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(x)**4)**(3/2),x)

[Out]

Integral((a*sin(x)**4)**(3/2), x)

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