3.5.5 \(\int \cos ^5(c+d x) (a+b \cos (c+d x)) \, dx\) [405]

Optimal. Leaf size=114 \[ \frac {5 b x}{16}+\frac {a \sin (c+d x)}{d}+\frac {5 b \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 b \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {b \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d} \]

[Out]

5/16*b*x+a*sin(d*x+c)/d+5/16*b*cos(d*x+c)*sin(d*x+c)/d+5/24*b*cos(d*x+c)^3*sin(d*x+c)/d+1/6*b*cos(d*x+c)^5*sin
(d*x+c)/d-2/3*a*sin(d*x+c)^3/d+1/5*a*sin(d*x+c)^5/d

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2827, 2713, 2715, 8} \begin {gather*} \frac {a \sin ^5(c+d x)}{5 d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}+\frac {b \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5 b \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {5 b \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 b x}{16} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + b*Cos[c + d*x]),x]

[Out]

(5*b*x)/16 + (a*Sin[c + d*x])/d + (5*b*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (5*b*Cos[c + d*x]^3*Sin[c + d*x])/(
24*d) + (b*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (2*a*Sin[c + d*x]^3)/(3*d) + (a*Sin[c + d*x]^5)/(5*d)

Rule 8

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

Rule 2713

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]

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]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.09, size = 81, normalized size = 0.71 \begin {gather*} \frac {300 b c+300 b d x+600 a \sin (c+d x)+225 b \sin (2 (c+d x))+100 a \sin (3 (c+d x))+45 b \sin (4 (c+d x))+12 a \sin (5 (c+d x))+5 b \sin (6 (c+d x))}{960 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + b*Cos[c + d*x]),x]

[Out]

(300*b*c + 300*b*d*x + 600*a*Sin[c + d*x] + 225*b*Sin[2*(c + d*x)] + 100*a*Sin[3*(c + d*x)] + 45*b*Sin[4*(c +
d*x)] + 12*a*Sin[5*(c + d*x)] + 5*b*Sin[6*(c + d*x)])/(960*d)

________________________________________________________________________________________

Maple [A]
time = 0.14, size = 80, normalized size = 0.70

method	result	size
derivativedivides	\(\frac {b \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(80\)
default	\(\frac {b \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(80\)
risch	\(\frac {5 b x}{16}+\frac {5 a \sin \left (d x +c \right )}{8 d}+\frac {b \sin \left (6 d x +6 c \right )}{192 d}+\frac {a \sin \left (5 d x +5 c \right )}{80 d}+\frac {3 b \sin \left (4 d x +4 c \right )}{64 d}+\frac {5 a \sin \left (3 d x +3 c \right )}{48 d}+\frac {15 b \sin \left (2 d x +2 c \right )}{64 d}\)	\(93\)
norman	\(\frac {\frac {5 b x}{16}+\frac {15 b x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {75 b x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {25 b x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {75 b x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {15 b x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {5 b x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {\left (16 a -11 b \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {\left (16 a +11 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {\left (112 a -5 b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {\left (112 a +5 b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {\left (208 a -75 b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d}+\frac {\left (208 a +75 b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\)	\(248\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+b*cos(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(b*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c)+1/5*a*(8/3+cos(d*x+c)^
4+4/3*cos(d*x+c)^2)*sin(d*x+c))

________________________________________________________________________________________

Maxima [A]
time = 0.28, size = 84, normalized size = 0.74 \begin {gather*} \frac {64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b}{960 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(64*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a - 5*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c
- 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*b)/d

________________________________________________________________________________________

Fricas [A]
time = 0.39, size = 75, normalized size = 0.66 \begin {gather*} \frac {75 \, b d x + {\left (40 \, b \cos \left (d x + c\right )^{5} + 48 \, a \cos \left (d x + c\right )^{4} + 50 \, b \cos \left (d x + c\right )^{3} + 64 \, a \cos \left (d x + c\right )^{2} + 75 \, b \cos \left (d x + c\right ) + 128 \, a\right )} \sin \left (d x + c\right )}{240 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(75*b*d*x + (40*b*cos(d*x + c)^5 + 48*a*cos(d*x + c)^4 + 50*b*cos(d*x + c)^3 + 64*a*cos(d*x + c)^2 + 75*
b*cos(d*x + c) + 128*a)*sin(d*x + c))/d

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (107) = 214\).
time = 0.44, size = 216, normalized size = 1.89 \begin {gather*} \begin {cases} \frac {8 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 b x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 b x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 b x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 b x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 b \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 b \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 b \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\left (c \right )}\right ) \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+b*cos(d*x+c)),x)

[Out]

Piecewise((8*a*sin(c + d*x)**5/(15*d) + 4*a*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + a*sin(c + d*x)*cos(c + d*x
)**4/d + 5*b*x*sin(c + d*x)**6/16 + 15*b*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*b*x*sin(c + d*x)**2*cos(c +
d*x)**4/16 + 5*b*x*cos(c + d*x)**6/16 + 5*b*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*b*sin(c + d*x)**3*cos(c +
d*x)**3/(6*d) + 11*b*sin(c + d*x)*cos(c + d*x)**5/(16*d), Ne(d, 0)), (x*(a + b*cos(c))*cos(c)**5, True))

________________________________________________________________________________________

Giac [A]
time = 0.43, size = 92, normalized size = 0.81 \begin {gather*} \frac {5}{16} \, b x + \frac {b \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {3 \, b \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {5 \, a \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {15 \, b \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {5 \, a \sin \left (d x + c\right )}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/16*b*x + 1/192*b*sin(6*d*x + 6*c)/d + 1/80*a*sin(5*d*x + 5*c)/d + 3/64*b*sin(4*d*x + 4*c)/d + 5/48*a*sin(3*d
*x + 3*c)/d + 15/64*b*sin(2*d*x + 2*c)/d + 5/8*a*sin(d*x + c)/d

________________________________________________________________________________________

Mupad [B]
time = 0.66, size = 115, normalized size = 1.01 \begin {gather*} \frac {5\,b\,x}{16}+\frac {8\,a\,\sin \left (c+d\,x\right )}{15\,d}+\frac {5\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{16\,d}+\frac {4\,a\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{15\,d}+\frac {a\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{5\,d}+\frac {5\,b\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{24\,d}+\frac {b\,{\cos \left (c+d\,x\right )}^5\,\sin \left (c+d\,x\right )}{6\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*b*x)/16 + (8*a*sin(c + d*x))/(15*d) + (5*b*cos(c + d*x)*sin(c + d*x))/(16*d) + (4*a*cos(c + d*x)^2*sin(c +
d*x))/(15*d) + (a*cos(c + d*x)^4*sin(c + d*x))/(5*d) + (5*b*cos(c + d*x)^3*sin(c + d*x))/(24*d) + (b*cos(c + d
*x)^5*sin(c + d*x))/(6*d)

________________________________________________________________________________________