3.1.66 \(\int \cos ^5(a+b x) \sin ^3(a+b x) \, dx\) [66]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.00, 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 = {2645, 14} \begin {gather*} \frac {\cos ^8(a+b x)}{8 b}-\frac {\cos ^6(a+b x)}{6 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^5*Sin[a + b*x]^3,x]

[Out]

-1/6*Cos[a + b*x]^6/b + Cos[a + b*x]^8/(8*b)

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]]

Rule 2645

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])

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 48, normalized size = 1.55 \begin {gather*} \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} \end {gather*}

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 = 0.10, size = 34, normalized size = 1.10

method	result	size
derivativedivides	\(\frac {-\frac {\left (\cos ^{6}\left (b x +a \right )\right ) \left (\sin ^{2}\left (b x +a \right )\right )}{8}-\frac {\left (\cos ^{6}\left (b x +a \right )\right )}{24}}{b}\)	\(34\)
default	\(\frac {-\frac {\left (\cos ^{6}\left (b x +a \right )\right ) \left (\sin ^{2}\left (b x +a \right )\right )}{8}-\frac {\left (\cos ^{6}\left (b x +a \right )\right )}{24}}{b}\)	\(34\)
risch	\(\frac {\cos \left (8 b x +8 a \right )}{1024 b}+\frac {\cos \left (6 b x +6 a \right )}{384 b}-\frac {\cos \left (4 b x +4 a \right )}{256 b}-\frac {3 \cos \left (2 b x +2 a \right )}{128 b}\)	\(58\)
norman	\(\frac {\frac {4 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {4 \left (\tan ^{12}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {16 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}-\frac {16 \left (\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}+\frac {40 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{8}}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.29, size = 36, normalized size = 1.16 \begin {gather*} \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 {gather*}

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 = 0.37, size = 26, normalized size = 0.84 \begin {gather*} \frac {3 \, \cos \left (b x + a\right )^{8} - 4 \, \cos \left (b x + a\right )^{6}}{24 \, b} \end {gather*}

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 = 0.86, size = 44, normalized size = 1.42 \begin {gather*} \begin {cases} - \frac {\sin ^{2}{\left (a + b x \right )} \cos ^{6}{\left (a + b x \right )}}{6 b} - \frac {\cos ^{8}{\left (a + b x \right )}}{24 b} & \text {for}\: b \neq 0 \\x \sin ^{3}{\left (a \right )} \cos ^{5}{\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)**5*sin(b*x+a)**3,x)

[Out]

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

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Giac [A]
time = 3.63, size = 27, normalized size = 0.87 \begin {gather*} \frac {\cos \left (b x + a\right )^{8}}{8 \, b} - \frac {\cos \left (b x + a\right )^{6}}{6 \, b} \end {gather*}

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

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Mupad [B]
time = 0.37, size = 25, normalized size = 0.81 \begin {gather*} \frac {{\cos \left (a+b\,x\right )}^6\,\left (3\,{\cos \left (a+b\,x\right )}^2-4\right )}{24\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^5*sin(a + b*x)^3,x)

[Out]

(cos(a + b*x)^6*(3*cos(a + b*x)^2 - 4))/(24*b)

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