3.2.14 \(\int \sin ^3(a+b x) \, dx\) [114]

Optimal. Leaf size=27 \[ -\frac {\cos (a+b x)}{b}+\frac {\cos ^3(a+b x)}{3 b} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 27, 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 {\cos ^3(a+b x)}{3 b}-\frac {\cos (a+b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3,x]

[Out]

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

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Mathematica [A]
time = 0.01, size = 29, normalized size = 1.07 \begin {gather*} -\frac {3 \cos (a+b x)}{4 b}+\frac {\cos (3 (a+b x))}{12 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^3,x]

[Out]

(-3*Cos[a + b*x])/(4*b) + Cos[3*(a + b*x)]/(12*b)

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Maple [A]
time = 0.04, size = 22, normalized size = 0.81

method	result	size
derivativedivides	\(-\frac {\left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{3 b}\)	\(22\)
default	\(-\frac {\left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{3 b}\)	\(22\)
risch	\(-\frac {3 \cos \left (b x +a \right )}{4 b}+\frac {\cos \left (3 b x +3 a \right )}{12 b}\)	\(27\)
norman	\(\frac {-\frac {4 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {4}{3 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{3}}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 2.36, size = 22, normalized size = 0.81 \begin {gather*} \frac {\cos \left (b x + a\right )^{3} - 3 \, \cos \left (b x + a\right )}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.79, size = 22, normalized size = 0.81 \begin {gather*} \frac {\cos \left (b x + a\right )^{3} - 3 \, \cos \left (b x + a\right )}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.11, size = 37, normalized size = 1.37 \begin {gather*} \begin {cases} - \frac {\sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{b} - \frac {2 \cos ^{3}{\left (a + b x \right )}}{3 b} & \text {for}\: b \neq 0 \\x \sin ^{3}{\left (a \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.72, size = 25, normalized size = 0.93 \begin {gather*} \frac {\cos \left (b x + a\right )^{3}}{3 \, b} - \frac {\cos \left (b x + a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*cos(b*x + a)^3/b - cos(b*x + a)/b

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Mupad [B]
time = 0.15, size = 24, normalized size = 0.89 \begin {gather*} -\frac {3\,\cos \left (a+b\,x\right )-{\cos \left (a+b\,x\right )}^3}{3\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^3,x)

[Out]

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

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