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

Optimal. Leaf size=41 \[ \frac{\sin ^5(a+b x)}{5 b}-\frac{2 \sin ^3(a+b x)}{3 b}+\frac{\sin (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.0129965, antiderivative size = 41, 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} \[ \frac{\sin ^5(a+b x)}{5 b}-\frac{2 \sin ^3(a+b x)}{3 b}+\frac{\sin (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^5,x]

[Out]

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

Mathematica [A]  time = 0.0139859, size = 41, normalized size = 1. \[ \frac{\sin ^5(a+b x)}{5 b}-\frac{2 \sin ^3(a+b x)}{3 b}+\frac{\sin (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^5,x]

[Out]

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

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Maple [A]  time = 0.027, size = 32, normalized size = 0.8 \begin{align*}{\frac{\sin \left ( bx+a \right ) }{5\,b} \left ({\frac{8}{3}}+ \left ( \cos \left ( bx+a \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.49008, size = 46, normalized size = 1.12 \begin{align*} \frac{3 \, \sin \left (b x + a\right )^{5} - 10 \, \sin \left (b x + a\right )^{3} + 15 \, \sin \left (b x + a\right )}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.64212, size = 85, normalized size = 2.07 \begin{align*} \frac{{\left (3 \, \cos \left (b x + a\right )^{4} + 4 \, \cos \left (b x + a\right )^{2} + 8\right )} \sin \left (b x + a\right )}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*cos(b*x + a)^4 + 4*cos(b*x + a)^2 + 8)*sin(b*x + a)/b

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Sympy [A]  time = 2.00431, size = 58, normalized size = 1.41 \begin{align*} \begin{cases} \frac{8 \sin ^{5}{\left (a + b x \right )}}{15 b} + \frac{4 \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b} + \frac{\sin{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{b} & \text{for}\: b \neq 0 \\x \cos ^{5}{\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)**5,x)

[Out]

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

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Giac [A]  time = 1.26137, size = 46, normalized size = 1.12 \begin{align*} \frac{3 \, \sin \left (b x + a\right )^{5} - 10 \, \sin \left (b x + a\right )^{3} + 15 \, \sin \left (b x + a\right )}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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