3.116 \(\int \cos ^3(a+b x) \, dx\)

Optimal. Leaf size=26 \[ \frac{\sin (a+b x)}{b}-\frac{\sin ^3(a+b x)}{3 b} \]

[Out]

Sin[a + b*x]/b - Sin[a + b*x]^3/(3*b)

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Rubi [A]  time = 0.0106401, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2633} \[ \frac{\sin (a+b x)}{b}-\frac{\sin ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3,x]

[Out]

Sin[a + b*x]/b - Sin[a + b*x]^3/(3*b)

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int \cos ^3(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (a+b x)\right )}{b}\\ &=\frac{\sin (a+b x)}{b}-\frac{\sin ^3(a+b x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.0074745, size = 26, normalized size = 1. \[ \frac{\sin (a+b x)}{b}-\frac{\sin ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3,x]

[Out]

Sin[a + b*x]/b - Sin[a + b*x]^3/(3*b)

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Maple [A]  time = 0.009, size = 22, normalized size = 0.9 \begin{align*}{\frac{ \left ( 2+ \left ( \cos \left ( bx+a \right ) \right ) ^{2} \right ) \sin \left ( bx+a \right ) }{3\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3,x)

[Out]

1/3/b*(2+cos(b*x+a)^2)*sin(b*x+a)

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Maxima [A]  time = 0.943241, size = 30, normalized size = 1.15 \begin{align*} -\frac{\sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/3*(sin(b*x + a)^3 - 3*sin(b*x + a))/b

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Fricas [A]  time = 1.61269, size = 55, normalized size = 2.12 \begin{align*} \frac{{\left (\cos \left (b x + a\right )^{2} + 2\right )} \sin \left (b x + a\right )}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3,x, algorithm="fricas")

[Out]

1/3*(cos(b*x + a)^2 + 2)*sin(b*x + a)/b

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Sympy [A]  time = 0.466847, size = 36, normalized size = 1.38 \begin{align*} \begin{cases} \frac{2 \sin ^{3}{\left (a + b x \right )}}{3 b} + \frac{\sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} & \text{for}\: b \neq 0 \\x \cos ^{3}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**3,x)

[Out]

Piecewise((2*sin(a + b*x)**3/(3*b) + sin(a + b*x)*cos(a + b*x)**2/b, Ne(b, 0)), (x*cos(a)**3, True))

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Giac [A]  time = 1.0726, size = 30, normalized size = 1.15 \begin{align*} -\frac{\sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3,x, algorithm="giac")

[Out]

-1/3*(sin(b*x + a)^3 - 3*sin(b*x + a))/b