3.1.6 \(\int \cos ^6(a+b x) \, dx\) [6]

Optimal. Leaf size=67 \[ \frac {5 x}{16}+\frac {5 \cos (a+b x) \sin (a+b x)}{16 b}+\frac {5 \cos ^3(a+b x) \sin (a+b x)}{24 b}+\frac {\cos ^5(a+b x) \sin (a+b x)}{6 b} \]

[Out]

5/16*x+5/16*cos(b*x+a)*sin(b*x+a)/b+5/24*cos(b*x+a)^3*sin(b*x+a)/b+1/6*cos(b*x+a)^5*sin(b*x+a)/b

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Rubi [A]
time = 0.02, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2715, 8} \begin {gather*} \frac {\sin (a+b x) \cos ^5(a+b x)}{6 b}+\frac {5 \sin (a+b x) \cos ^3(a+b x)}{24 b}+\frac {5 \sin (a+b x) \cos (a+b x)}{16 b}+\frac {5 x}{16} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^6,x]

[Out]

(5*x)/16 + (5*Cos[a + b*x]*Sin[a + b*x])/(16*b) + (5*Cos[a + b*x]^3*Sin[a + b*x])/(24*b) + (Cos[a + b*x]^5*Sin
[a + b*x])/(6*b)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rubi steps

\begin {align*} \int \cos ^6(a+b x) \, dx &=\frac {\cos ^5(a+b x) \sin (a+b x)}{6 b}+\frac {5}{6} \int \cos ^4(a+b x) \, dx\\ &=\frac {5 \cos ^3(a+b x) \sin (a+b x)}{24 b}+\frac {\cos ^5(a+b x) \sin (a+b x)}{6 b}+\frac {5}{8} \int \cos ^2(a+b x) \, dx\\ &=\frac {5 \cos (a+b x) \sin (a+b x)}{16 b}+\frac {5 \cos ^3(a+b x) \sin (a+b x)}{24 b}+\frac {\cos ^5(a+b x) \sin (a+b x)}{6 b}+\frac {5 \int 1 \, dx}{16}\\ &=\frac {5 x}{16}+\frac {5 \cos (a+b x) \sin (a+b x)}{16 b}+\frac {5 \cos ^3(a+b x) \sin (a+b x)}{24 b}+\frac {\cos ^5(a+b x) \sin (a+b x)}{6 b}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 43, normalized size = 0.64 \begin {gather*} \frac {60 a+60 b x+45 \sin (2 (a+b x))+9 \sin (4 (a+b x))+\sin (6 (a+b x))}{192 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^6,x]

[Out]

(60*a + 60*b*x + 45*Sin[2*(a + b*x)] + 9*Sin[4*(a + b*x)] + Sin[6*(a + b*x)])/(192*b)

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Maple [A]
time = 0.09, size = 48, normalized size = 0.72

method	result	size
risch	\(\frac {5 x}{16}+\frac {\sin \left (6 b x +6 a \right )}{192 b}+\frac {3 \sin \left (4 b x +4 a \right )}{64 b}+\frac {15 \sin \left (2 b x +2 a \right )}{64 b}\)	\(47\)
derivativedivides	\(\frac {\frac {\left (\cos ^{5}\left (b x +a \right )+\frac {5 \left (\cos ^{3}\left (b x +a \right )\right )}{4}+\frac {15 \cos \left (b x +a \right )}{8}\right ) \sin \left (b x +a \right )}{6}+\frac {5 b x}{16}+\frac {5 a}{16}}{b}\)	\(48\)
default	\(\frac {\frac {\left (\cos ^{5}\left (b x +a \right )+\frac {5 \left (\cos ^{3}\left (b x +a \right )\right )}{4}+\frac {15 \cos \left (b x +a \right )}{8}\right ) \sin \left (b x +a \right )}{6}+\frac {5 b x}{16}+\frac {5 a}{16}}{b}\)	\(48\)
norman	\(\frac {\frac {5 x}{16}+\frac {11 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{8 b}-\frac {5 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{24 b}+\frac {15 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}-\frac {15 \left (\tan ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {5 \left (\tan ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{24 b}-\frac {11 \left (\tan ^{11}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}+\frac {15 x \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8}+\frac {75 x \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{16}+\frac {25 x \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4}+\frac {75 x \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{16}+\frac {15 x \left (\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8}+\frac {5 x \left (\tan ^{12}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{16}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{6}}\)	\(199\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^6,x,method=_RETURNVERBOSE)

[Out]

1/b*(1/6*(cos(b*x+a)^5+5/4*cos(b*x+a)^3+15/8*cos(b*x+a))*sin(b*x+a)+5/16*b*x+5/16*a)

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Maxima [A]
time = 0.28, size = 48, normalized size = 0.72 \begin {gather*} -\frac {4 \, \sin \left (2 \, b x + 2 \, a\right )^{3} - 60 \, b x - 60 \, a - 9 \, \sin \left (4 \, b x + 4 \, a\right ) - 48 \, \sin \left (2 \, b x + 2 \, a\right )}{192 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^6,x, algorithm="maxima")

[Out]

-1/192*(4*sin(2*b*x + 2*a)^3 - 60*b*x - 60*a - 9*sin(4*b*x + 4*a) - 48*sin(2*b*x + 2*a))/b

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Fricas [A]
time = 0.38, size = 46, normalized size = 0.69 \begin {gather*} \frac {15 \, b x + {\left (8 \, \cos \left (b x + a\right )^{5} + 10 \, \cos \left (b x + a\right )^{3} + 15 \, \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{48 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^6,x, algorithm="fricas")

[Out]

1/48*(15*b*x + (8*cos(b*x + a)^5 + 10*cos(b*x + a)^3 + 15*cos(b*x + a))*sin(b*x + a))/b

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (61) = 122\).
time = 0.37, size = 139, normalized size = 2.07 \begin {gather*} \begin {cases} \frac {5 x \sin ^{6}{\left (a + b x \right )}}{16} + \frac {15 x \sin ^{4}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{16} + \frac {15 x \sin ^{2}{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{16} + \frac {5 x \cos ^{6}{\left (a + b x \right )}}{16} + \frac {5 \sin ^{5}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{16 b} + \frac {5 \sin ^{3}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{6 b} + \frac {11 \sin {\left (a + b x \right )} \cos ^{5}{\left (a + b x \right )}}{16 b} & \text {for}\: b \neq 0 \\x \cos ^{6}{\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)**6,x)

[Out]

Piecewise((5*x*sin(a + b*x)**6/16 + 15*x*sin(a + b*x)**4*cos(a + b*x)**2/16 + 15*x*sin(a + b*x)**2*cos(a + b*x
)**4/16 + 5*x*cos(a + b*x)**6/16 + 5*sin(a + b*x)**5*cos(a + b*x)/(16*b) + 5*sin(a + b*x)**3*cos(a + b*x)**3/(
6*b) + 11*sin(a + b*x)*cos(a + b*x)**5/(16*b), Ne(b, 0)), (x*cos(a)**6, True))

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Giac [A]
time = 0.45, size = 46, normalized size = 0.69 \begin {gather*} \frac {5}{16} \, x + \frac {\sin \left (6 \, b x + 6 \, a\right )}{192 \, b} + \frac {3 \, \sin \left (4 \, b x + 4 \, a\right )}{64 \, b} + \frac {15 \, \sin \left (2 \, b x + 2 \, a\right )}{64 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^6,x, algorithm="giac")

[Out]

5/16*x + 1/192*sin(6*b*x + 6*a)/b + 3/64*sin(4*b*x + 4*a)/b + 15/64*sin(2*b*x + 2*a)/b

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Mupad [B]
time = 0.22, size = 42, normalized size = 0.63 \begin {gather*} \frac {5\,x}{16}+\frac {\frac {15\,\sin \left (2\,a+2\,b\,x\right )}{64}+\frac {3\,\sin \left (4\,a+4\,b\,x\right )}{64}+\frac {\sin \left (6\,a+6\,b\,x\right )}{192}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^6,x)

[Out]

(5*x)/16 + ((15*sin(2*a + 2*b*x))/64 + (3*sin(4*a + 4*b*x))/64 + sin(6*a + 6*b*x)/192)/b

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