3.66 \(\int \cos ^5(a+b x) \sin ^3(a+b x) \, dx\)

Optimal. Leaf size=31 \[ \frac{\cos ^8(a+b x)}{8 b}-\frac{\cos ^6(a+b x)}{6 b} \]

[Out]

-Cos[a + b*x]^6/(6*b) + Cos[a + b*x]^8/(8*b)

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Rubi [A]  time = 0.0334502, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2565, 14} \[ \frac{\cos ^8(a+b x)}{8 b}-\frac{\cos ^6(a+b x)}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^5*Sin[a + b*x]^3,x]

[Out]

-Cos[a + b*x]^6/(6*b) + Cos[a + b*x]^8/(8*b)

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \cos ^5(a+b x) \sin ^3(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int x^5 \left (1-x^2\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \left (x^5-x^7\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{\cos ^6(a+b x)}{6 b}+\frac{\cos ^8(a+b x)}{8 b}\\ \end{align*}

Mathematica [A]  time = 0.129495, size = 48, normalized size = 1.55 \[ \frac{-72 \cos (2 (a+b x))-12 \cos (4 (a+b x))+8 \cos (6 (a+b x))+3 \cos (8 (a+b x))}{3072 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^5*Sin[a + b*x]^3,x]

[Out]

(-72*Cos[2*(a + b*x)] - 12*Cos[4*(a + b*x)] + 8*Cos[6*(a + b*x)] + 3*Cos[8*(a + b*x)])/(3072*b)

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Maple [A]  time = 0.01, size = 34, normalized size = 1.1 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{6} \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{8}}-{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{6}}{24}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^5*sin(b*x+a)^3,x)

[Out]

1/b*(-1/8*cos(b*x+a)^6*sin(b*x+a)^2-1/24*cos(b*x+a)^6)

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Maxima [A]  time = 0.970098, size = 49, normalized size = 1.58 \begin{align*} \frac{3 \, \sin \left (b x + a\right )^{8} - 8 \, \sin \left (b x + a\right )^{6} + 6 \, \sin \left (b x + a\right )^{4}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^5*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

1/24*(3*sin(b*x + a)^8 - 8*sin(b*x + a)^6 + 6*sin(b*x + a)^4)/b

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Fricas [A]  time = 1.69465, size = 62, normalized size = 2. \begin{align*} \frac{3 \, \cos \left (b x + a\right )^{8} - 4 \, \cos \left (b x + a\right )^{6}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^5*sin(b*x+a)^3,x, algorithm="fricas")

[Out]

1/24*(3*cos(b*x + a)^8 - 4*cos(b*x + a)^6)/b

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Sympy [A]  time = 12.2605, size = 63, normalized size = 2.03 \begin{align*} \begin{cases} \frac{\sin ^{8}{\left (a + b x \right )}}{24 b} + \frac{\sin ^{6}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{6 b} + \frac{\sin ^{4}{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{4 b} & \text{for}\: b \neq 0 \\x \sin ^{3}{\left (a \right )} \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*sin(b*x+a)**3,x)

[Out]

Piecewise((sin(a + b*x)**8/(24*b) + sin(a + b*x)**6*cos(a + b*x)**2/(6*b) + sin(a + b*x)**4*cos(a + b*x)**4/(4
*b), Ne(b, 0)), (x*sin(a)**3*cos(a)**5, True))

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Giac [A]  time = 1.16947, size = 36, normalized size = 1.16 \begin{align*} \frac{\cos \left (b x + a\right )^{8}}{8 \, b} - \frac{\cos \left (b x + a\right )^{6}}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^5*sin(b*x+a)^3,x, algorithm="giac")

[Out]

1/8*cos(b*x + a)^8/b - 1/6*cos(b*x + a)^6/b