3.1.33 \(\int (a-a \sin ^2(x))^3 \, dx\) [33]

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

[Out]

5/16*a^3*x+5/16*a^3*cos(x)*sin(x)+5/24*a^3*cos(x)^3*sin(x)+1/6*a^3*cos(x)^5*sin(x)

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Rubi [A]
time = 0.02, antiderivative size = 46, normalized size of antiderivative = 1.00, 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 = {3254, 2715, 8} \begin {gather*} \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) \end {gather*}

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 8

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

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 3254

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]

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*}

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Mathematica [A]
time = 0.00, size = 34, normalized size = 0.74 \begin {gather*} 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 ) \end {gather*}

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 = 0.22, size = 72, normalized size = 1.57

method	result	size
risch	\(\frac {5 a^{3} x}{16}+\frac {a^{3} \sin \left (6 x \right )}{192}+\frac {3 a^{3} \sin \left (4 x \right )}{64}+\frac {15 a^{3} \sin \left (2 x \right )}{64}\)	\(35\)
default	\(-a^{3} \left (-\frac {\left (\sin ^{5}\left (x \right )+\frac {5 \left (\sin ^{3}\left (x \right )\right )}{4}+\frac {15 \sin \left (x \right )}{8}\right ) \cos \left (x \right )}{6}+\frac {5 x}{16}\right )+3 a^{3} \left (-\frac {\left (\sin ^{3}\left (x \right )+\frac {3 \sin \left (x \right )}{2}\right ) \cos \left (x \right )}{4}+\frac {3 x}{8}\right )-3 a^{3} \left (-\frac {\sin \left (x \right ) \cos \left (x \right )}{2}+\frac {x}{2}\right )+a^{3} x\)	\(72\)
norman	\(\frac {\frac {5 a^{3} x}{16}+\frac {11 a^{3} \tan \left (\frac {x}{2}\right )}{8}-\frac {5 a^{3} \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{24}+\frac {15 a^{3} \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{4}-\frac {15 a^{3} \left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{4}+\frac {5 a^{3} \left (\tan ^{9}\left (\frac {x}{2}\right )\right )}{24}-\frac {11 a^{3} \left (\tan ^{11}\left (\frac {x}{2}\right )\right )}{8}+\frac {15 a^{3} x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{8}+\frac {75 a^{3} x \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{16}+\frac {25 a^{3} x \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{4}+\frac {75 a^{3} x \left (\tan ^{8}\left (\frac {x}{2}\right )\right )}{16}+\frac {15 a^{3} x \left (\tan ^{10}\left (\frac {x}{2}\right )\right )}{8}+\frac {5 a^{3} x \left (\tan ^{12}\left (\frac {x}{2}\right )\right )}{16}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{6}}\)	\(155\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.30, size = 69, normalized size = 1.50 \begin {gather*} -\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 {gather*}

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 = 0.41, size = 37, normalized size = 0.80 \begin {gather*} \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (49) = 98\).
time = 0.32, size = 233, normalized size = 5.07 \begin {gather*} - \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 {gather*}

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 = 0.44, size = 34, normalized size = 0.74 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 13.68, size = 42, normalized size = 0.91 \begin {gather*} \frac {11\,a^3\,{\cos \left (x\right )}^5\,\sin \left (x\right )}{16}+\frac {5\,a^3\,{\cos \left (x\right )}^3\,{\sin \left (x\right )}^3}{6}+\frac {5\,a^3\,\cos \left (x\right )\,{\sin \left (x\right )}^5}{16}+\frac {5\,x\,a^3}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - a*sin(x)^2)^3,x)

[Out]

(5*a^3*x)/16 + (5*a^3*cos(x)*sin(x)^5)/16 + (11*a^3*cos(x)^5*sin(x))/16 + (5*a^3*cos(x)^3*sin(x)^3)/6

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