3.2.16 \(\int \cos ^3(a+b x) \, dx\) [116]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.00, 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 = {2713} \begin {gather*} \frac {\sin (a+b x)}{b}-\frac {\sin ^3(a+b x)}{3 b} \end {gather*}

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 2713

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 {\text {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*}

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Mathematica [A]
time = 0.01, size = 26, normalized size = 1.00 \begin {gather*} \frac {\sin (a+b x)}{b}-\frac {\sin ^3(a+b x)}{3 b} \end {gather*}

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 = 0.04, size = 22, normalized size = 0.85

method	result	size
derivativedivides	\(\frac {\left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{3 b}\)	\(22\)
default	\(\frac {\left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{3 b}\)	\(22\)
risch	\(\frac {3 \sin \left (b x +a \right )}{4 b}+\frac {\sin \left (3 b x +3 a \right )}{12 b}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 3.53, size = 22, normalized size = 0.85 \begin {gather*} -\frac {\sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )}{3 \, b} \end {gather*}

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 = 1.19, size = 21, normalized size = 0.81 \begin {gather*} \frac {{\left (\cos \left (b x + a\right )^{2} + 2\right )} \sin \left (b x + a\right )}{3 \, b} \end {gather*}

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 = 0.11, size = 36, normalized size = 1.38 \begin {gather*} \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 {gather*}

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 = 1.04, size = 22, normalized size = 0.85 \begin {gather*} -\frac {\sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )}{3 \, b} \end {gather*}

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

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Mupad [B]
time = 0.03, size = 24, normalized size = 0.92 \begin {gather*} \frac {3\,\sin \left (a+b\,x\right )-{\sin \left (a+b\,x\right )}^3}{3\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^3,x)

[Out]

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

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