3.5 \(\int \sin ^5(a+b x) \, dx\)

Optimal. Leaf size=42 \[ -\frac{\cos ^5(a+b x)}{5 b}+\frac{2 \cos ^3(a+b x)}{3 b}-\frac{\cos (a+b x)}{b} \]

[Out]

-(Cos[a + b*x]/b) + (2*Cos[a + b*x]^3)/(3*b) - Cos[a + b*x]^5/(5*b)

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Rubi [A]  time = 0.0130112, antiderivative size = 42, 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} \[ -\frac{\cos ^5(a+b x)}{5 b}+\frac{2 \cos ^3(a+b x)}{3 b}-\frac{\cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^5,x]

[Out]

-(Cos[a + b*x]/b) + (2*Cos[a + b*x]^3)/(3*b) - Cos[a + b*x]^5/(5*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 \sin ^5(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{\cos (a+b x)}{b}+\frac{2 \cos ^3(a+b x)}{3 b}-\frac{\cos ^5(a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0121988, size = 44, normalized size = 1.05 \[ -\frac{5 \cos (a+b x)}{8 b}+\frac{5 \cos (3 (a+b x))}{48 b}-\frac{\cos (5 (a+b x))}{80 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^5,x]

[Out]

(-5*Cos[a + b*x])/(8*b) + (5*Cos[3*(a + b*x)])/(48*b) - Cos[5*(a + b*x)]/(80*b)

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Maple [A]  time = 0.037, size = 32, normalized size = 0.8 \begin{align*} -{\frac{\cos \left ( bx+a \right ) }{5\,b} \left ({\frac{8}{3}}+ \left ( \sin \left ( bx+a \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^5,x)

[Out]

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

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Maxima [A]  time = 0.995149, size = 46, normalized size = 1.1 \begin{align*} -\frac{3 \, \cos \left (b x + a\right )^{5} - 10 \, \cos \left (b x + a\right )^{3} + 15 \, \cos \left (b x + a\right )}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^5,x, algorithm="maxima")

[Out]

-1/15*(3*cos(b*x + a)^5 - 10*cos(b*x + a)^3 + 15*cos(b*x + a))/b

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Fricas [A]  time = 2.1845, size = 89, normalized size = 2.12 \begin{align*} -\frac{3 \, \cos \left (b x + a\right )^{5} - 10 \, \cos \left (b x + a\right )^{3} + 15 \, \cos \left (b x + a\right )}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^5,x, algorithm="fricas")

[Out]

-1/15*(3*cos(b*x + a)^5 - 10*cos(b*x + a)^3 + 15*cos(b*x + a))/b

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Sympy [A]  time = 1.83787, size = 60, normalized size = 1.43 \begin{align*} \begin{cases} - \frac{\sin ^{4}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{b} - \frac{4 \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac{8 \cos ^{5}{\left (a + b x \right )}}{15 b} & \text{for}\: b \neq 0 \\x \sin ^{5}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)**5,x)

[Out]

Piecewise((-sin(a + b*x)**4*cos(a + b*x)/b - 4*sin(a + b*x)**2*cos(a + b*x)**3/(3*b) - 8*cos(a + b*x)**5/(15*b
), Ne(b, 0)), (x*sin(a)**5, True))

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Giac [A]  time = 1.12195, size = 51, normalized size = 1.21 \begin{align*} -\frac{\cos \left (b x + a\right )^{5}}{5 \, b} + \frac{2 \, \cos \left (b x + a\right )^{3}}{3 \, b} - \frac{\cos \left (b x + a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^5,x, algorithm="giac")

[Out]

-1/5*cos(b*x + a)^5/b + 2/3*cos(b*x + a)^3/b - cos(b*x + a)/b