3.33 \(\int (a-a \sin ^2(x))^3 \, dx\)

Optimal. Leaf size=46 \[ \frac{5 a^3 x}{16}+\frac{1}{6} a^3 \sin (x) \cos ^5(x)+\frac{5}{24} a^3 \sin (x) \cos ^3(x)+\frac{5}{16} a^3 \sin (x) \cos (x) \]

[Out]

(5*a^3*x)/16 + (5*a^3*Cos[x]*Sin[x])/16 + (5*a^3*Cos[x]^3*Sin[x])/24 + (a^3*Cos[x]^5*Sin[x])/6

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Rubi [A]  time = 0.0324165, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {3175, 2635, 8} \[ \frac{5 a^3 x}{16}+\frac{1}{6} a^3 \sin (x) \cos ^5(x)+\frac{5}{24} a^3 \sin (x) \cos ^3(x)+\frac{5}{16} a^3 \sin (x) \cos (x) \]

Antiderivative was successfully verified.

[In]

Int[(a - a*Sin[x]^2)^3,x]

[Out]

(5*a^3*x)/16 + (5*a^3*Cos[x]*Sin[x])/16 + (5*a^3*Cos[x]^3*Sin[x])/24 + (a^3*Cos[x]^5*Sin[x])/6

Rule 3175

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Dist[a^p, Int[ActivateTrig[u*cos[e + f*x
]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]

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 8

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

Rubi steps

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

Mathematica [A]  time = 0.0027158, size = 34, normalized size = 0.74 \[ a^3 \left (\frac{5 x}{16}+\frac{15}{64} \sin (2 x)+\frac{3}{64} \sin (4 x)+\frac{1}{192} \sin (6 x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a - a*Sin[x]^2)^3,x]

[Out]

a^3*((5*x)/16 + (15*Sin[2*x])/64 + (3*Sin[4*x])/64 + Sin[6*x]/192)

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Maple [A]  time = 0.02, size = 72, normalized size = 1.6 \begin{align*} -{a}^{3} \left ( -{\frac{\cos \left ( x \right ) }{6} \left ( \left ( \sin \left ( x \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( x \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( x \right ) }{8}} \right ) }+{\frac{5\,x}{16}} \right ) +3\,{a}^{3} \left ( -1/4\, \left ( \left ( \sin \left ( x \right ) \right ) ^{3}+3/2\,\sin \left ( x \right ) \right ) \cos \left ( x \right ) +3/8\,x \right ) -3\,{a}^{3} \left ( -1/2\,\sin \left ( x \right ) \cos \left ( x \right ) +x/2 \right ) +{a}^{3}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a-a*sin(x)^2)^3,x)

[Out]

-a^3*(-1/6*(sin(x)^5+5/4*sin(x)^3+15/8*sin(x))*cos(x)+5/16*x)+3*a^3*(-1/4*(sin(x)^3+3/2*sin(x))*cos(x)+3/8*x)-
3*a^3*(-1/2*sin(x)*cos(x)+1/2*x)+a^3*x

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Maxima [A]  time = 0.949992, size = 93, normalized size = 2.02 \begin{align*} -\frac{1}{192} \,{\left (4 \, \sin \left (2 \, x\right )^{3} + 60 \, x + 9 \, \sin \left (4 \, x\right ) - 48 \, \sin \left (2 \, x\right )\right )} a^{3} + \frac{3}{32} \, a^{3}{\left (12 \, x + \sin \left (4 \, x\right ) - 8 \, \sin \left (2 \, x\right )\right )} - \frac{3}{4} \, a^{3}{\left (2 \, x - \sin \left (2 \, x\right )\right )} + a^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(4*sin(2*x)^3 + 60*x + 9*sin(4*x) - 48*sin(2*x))*a^3 + 3/32*a^3*(12*x + sin(4*x) - 8*sin(2*x)) - 3/4*a^
3*(2*x - sin(2*x)) + a^3*x

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Fricas [A]  time = 1.62099, size = 104, normalized size = 2.26 \begin{align*} \frac{5}{16} \, a^{3} x + \frac{1}{48} \,{\left (8 \, a^{3} \cos \left (x\right )^{5} + 10 \, a^{3} \cos \left (x\right )^{3} + 15 \, a^{3} \cos \left (x\right )\right )} \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/16*a^3*x + 1/48*(8*a^3*cos(x)^5 + 10*a^3*cos(x)^3 + 15*a^3*cos(x))*sin(x)

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Sympy [B]  time = 4.28107, size = 233, normalized size = 5.07 \begin{align*} - \frac{5 a^{3} x \sin ^{6}{\left (x \right )}}{16} - \frac{15 a^{3} x \sin ^{4}{\left (x \right )} \cos ^{2}{\left (x \right )}}{16} + \frac{9 a^{3} x \sin ^{4}{\left (x \right )}}{8} - \frac{15 a^{3} x \sin ^{2}{\left (x \right )} \cos ^{4}{\left (x \right )}}{16} + \frac{9 a^{3} x \sin ^{2}{\left (x \right )} \cos ^{2}{\left (x \right )}}{4} - \frac{3 a^{3} x \sin ^{2}{\left (x \right )}}{2} - \frac{5 a^{3} x \cos ^{6}{\left (x \right )}}{16} + \frac{9 a^{3} x \cos ^{4}{\left (x \right )}}{8} - \frac{3 a^{3} x \cos ^{2}{\left (x \right )}}{2} + a^{3} x + \frac{11 a^{3} \sin ^{5}{\left (x \right )} \cos{\left (x \right )}}{16} + \frac{5 a^{3} \sin ^{3}{\left (x \right )} \cos ^{3}{\left (x \right )}}{6} - \frac{15 a^{3} \sin ^{3}{\left (x \right )} \cos{\left (x \right )}}{8} + \frac{5 a^{3} \sin{\left (x \right )} \cos ^{5}{\left (x \right )}}{16} - \frac{9 a^{3} \sin{\left (x \right )} \cos ^{3}{\left (x \right )}}{8} + \frac{3 a^{3} \sin{\left (x \right )} \cos{\left (x \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*sin(x)**2)**3,x)

[Out]

-5*a**3*x*sin(x)**6/16 - 15*a**3*x*sin(x)**4*cos(x)**2/16 + 9*a**3*x*sin(x)**4/8 - 15*a**3*x*sin(x)**2*cos(x)*
*4/16 + 9*a**3*x*sin(x)**2*cos(x)**2/4 - 3*a**3*x*sin(x)**2/2 - 5*a**3*x*cos(x)**6/16 + 9*a**3*x*cos(x)**4/8 -
3*a**3*x*cos(x)**2/2 + a**3*x + 11*a**3*sin(x)**5*cos(x)/16 + 5*a**3*sin(x)**3*cos(x)**3/6 - 15*a**3*sin(x)**
3*cos(x)/8 + 5*a**3*sin(x)*cos(x)**5/16 - 9*a**3*sin(x)*cos(x)**3/8 + 3*a**3*sin(x)*cos(x)/2

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Giac [A]  time = 1.12124, size = 46, normalized size = 1. \begin{align*} \frac{5}{16} \, a^{3} x + \frac{1}{192} \, a^{3} \sin \left (6 \, x\right ) + \frac{3}{64} \, a^{3} \sin \left (4 \, x\right ) + \frac{15}{64} \, a^{3} \sin \left (2 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/16*a^3*x + 1/192*a^3*sin(6*x) + 3/64*a^3*sin(4*x) + 15/64*a^3*sin(2*x)