3.1.1 \(\int \cos ^5(c+d x) (a+a \cos (c+d x)) \, dx\) [1]

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

[Out]

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

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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 {a \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5 a \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {5 a \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a x}{16} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a*x)/16 + (a*Sin[c + d*x])/d + (5*a*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (5*a*Cos[c + d*x]^3*Sin[c + d*x])/(
24*d) + (a*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+a \cos (c+d x)) \, dx &=a \int \cos ^5(c+d x) \, dx+a \int \cos ^6(c+d x) \, dx\\ &=\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {1}{6} (5 a) \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 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a \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 a) \int \cos ^2(c+d x) \, dx\\ &=\frac {a \sin (c+d x)}{d}+\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a \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 a) \int 1 \, dx\\ &=\frac {5 a x}{16}+\frac {a \sin (c+d x)}{d}+\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a \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*}

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Mathematica [A]
time = 0.11, size = 74, normalized size = 0.65 \begin {gather*} \frac {a (300 c+300 d x+600 \sin (c+d x)+225 \sin (2 (c+d x))+100 \sin (3 (c+d x))+45 \sin (4 (c+d x))+12 \sin (5 (c+d x))+5 \sin (6 (c+d x)))}{960 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(300*c + 300*d*x + 600*Sin[c + d*x] + 225*Sin[2*(c + d*x)] + 100*Sin[3*(c + d*x)] + 45*Sin[4*(c + d*x)] + 1
2*Sin[5*(c + d*x)] + 5*Sin[6*(c + d*x)]))/(960*d)

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Maple [A]
time = 0.16, size = 80, normalized size = 0.70

method	result	size
derivativedivides	\(\frac {a \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(80\)
default	\(\frac {a \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(80\)
risch	\(\frac {5 a x}{16}+\frac {5 a \sin \left (d x +c \right )}{8 d}+\frac {a \sin \left (6 d x +6 c \right )}{192 d}+\frac {a \sin \left (5 d x +5 c \right )}{80 d}+\frac {3 a \sin \left (4 d x +4 c \right )}{64 d}+\frac {5 a \sin \left (3 d x +3 c \right )}{48 d}+\frac {15 a \sin \left (2 d x +2 c \right )}{64 d}\)	\(93\)
norman	\(\frac {\frac {5 a x}{16}+\frac {27 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {107 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {283 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d}+\frac {133 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d}+\frac {39 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {5 a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {15 a x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {75 a x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {25 a x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {75 a x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {15 a x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {5 a x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\)	\(212\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(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))

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Maxima [A]
time = 0.27, 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 )} a}{960 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*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))*a)/d

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Fricas [A]
time = 0.41, size = 75, normalized size = 0.66 \begin {gather*} \frac {75 \, a d x + {\left (40 \, a \cos \left (d x + c\right )^{5} + 48 \, a \cos \left (d x + c\right )^{4} + 50 \, a \cos \left (d x + c\right )^{3} + 64 \, a \cos \left (d x + c\right )^{2} + 75 \, a \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+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (107) = 214\).
time = 0.43, size = 216, normalized size = 1.89 \begin {gather*} \begin {cases} \frac {5 a x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 a x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 a x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 a x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 a \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {8 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {5 a \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {4 a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {11 a \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac {a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + a\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+a*cos(d*x+c)),x)

[Out]

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

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Giac [A]
time = 0.45, size = 92, normalized size = 0.81 \begin {gather*} \frac {5}{16} \, a x + \frac {a \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {3 \, a \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 \, a \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+a*cos(d*x+c)),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 2.98, size = 107, normalized size = 0.94 \begin {gather*} \frac {5\,a\,x}{16}+\frac {\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {39\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{8}+\frac {133\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{20}+\frac {283\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {107\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {27\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a*x)/16 + ((27*a*tan(c/2 + (d*x)/2))/8 + (107*a*tan(c/2 + (d*x)/2)^3)/24 + (283*a*tan(c/2 + (d*x)/2)^5)/20
+ (133*a*tan(c/2 + (d*x)/2)^7)/20 + (39*a*tan(c/2 + (d*x)/2)^9)/8 + (5*a*tan(c/2 + (d*x)/2)^11)/8)/(d*(tan(c/2
+ (d*x)/2)^2 + 1)^6)

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