\(\int \cos ^4(c+d x) (a+a \cos (c+d x)) \, dx\) [2]

Optimal result

Integrand size = 19, antiderivative size = 92 \[ \int \cos ^4(c+d x) (a+a \cos (c+d x)) \, dx=\frac {3 a x}{8}+\frac {a \sin (c+d x)}{d}+\frac {3 a \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d} \]

[Out]

3/8*a*x+a*sin(d*x+c)/d+3/8*a*cos(d*x+c)*sin(d*x+c)/d+1/4*a*cos(d*x+c)^3*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] (verified)

Time = 0.07 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2827, 2715, 8, 2713} \[ \int \cos ^4(c+d x) (a+a \cos (c+d x)) \, dx=\frac {a \sin ^5(c+d x)}{5 d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}+\frac {a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 a x}{8} \]

[In]

Int[Cos[c + d*x]^4*(a + a*Cos[c + d*x]),x]

[Out]

(3*a*x)/8 + (a*Sin[c + d*x])/d + (3*a*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*Cos[c + d*x]^3*Sin[c + d*x])/(4*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*} \text {integral}& = a \int \cos ^4(c+d x) \, dx+a \int \cos ^5(c+d x) \, dx \\ & = \frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{4} (3 a) \int \cos ^2(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 {3 a \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d}+\frac {1}{8} (3 a) \int 1 \, dx \\ & = \frac {3 a x}{8}+\frac {a \sin (c+d x)}{d}+\frac {3 a \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.71 \[ \int \cos ^4(c+d x) (a+a \cos (c+d x)) \, dx=\frac {a \left (480 \sin (c+d x)-320 \sin ^3(c+d x)+96 \sin ^5(c+d x)+15 (12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x)))\right )}{480 d} \]

[In]

Integrate[Cos[c + d*x]^4*(a + a*Cos[c + d*x]),x]

[Out]

(a*(480*Sin[c + d*x] - 320*Sin[c + d*x]^3 + 96*Sin[c + d*x]^5 + 15*(12*(c + d*x) + 8*Sin[2*(c + d*x)] + Sin[4*
(c + d*x)])))/(480*d)

Maple [A] (verified)

Time = 2.25 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(\frac {a \left (180 d x +300 \sin \left (d x +c \right )+6 \sin \left (5 d x +5 c \right )+15 \sin \left (4 d x +4 c \right )+50 \sin \left (3 d x +3 c \right )+120 \sin \left (2 d x +2 c \right )\right )}{480 d}\)	\(64\)
derivativedivides	\(\frac {\frac {a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(70\)
default	\(\frac {\frac {a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(70\)
parts	\(\frac {a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5 d}\)	\(72\)
risch	\(\frac {3 a x}{8}+\frac {5 a \sin \left (d x +c \right )}{8 d}+\frac {a \sin \left (5 d x +5 c \right )}{80 d}+\frac {a \sin \left (4 d x +4 c \right )}{32 d}+\frac {5 a \sin \left (3 d x +3 c \right )}{48 d}+\frac {a \sin \left (2 d x +2 c \right )}{4 d}\)	\(78\)
norman	\(\frac {\frac {3 a x}{8}+\frac {13 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {19 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {116 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {13 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {3 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {15 a x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {15 a x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {15 a x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {15 a x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {3 a x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\)	\(180\)

[In]

int(cos(d*x+c)^4*(a+cos(d*x+c)*a),x,method=_RETURNVERBOSE)

[Out]

1/480*a*(180*d*x+300*sin(d*x+c)+6*sin(5*d*x+5*c)+15*sin(4*d*x+4*c)+50*sin(3*d*x+3*c)+120*sin(2*d*x+2*c))/d

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.70 \[ \int \cos ^4(c+d x) (a+a \cos (c+d x)) \, dx=\frac {45 \, a d x + {\left (24 \, a \cos \left (d x + c\right )^{4} + 30 \, a \cos \left (d x + c\right )^{3} + 32 \, a \cos \left (d x + c\right )^{2} + 45 \, a \cos \left (d x + c\right ) + 64 \, a\right )} \sin \left (d x + c\right )}{120 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

1/120*(45*a*d*x + (24*a*cos(d*x + c)^4 + 30*a*cos(d*x + c)^3 + 32*a*cos(d*x + c)^2 + 45*a*cos(d*x + c) + 64*a)
*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.83 \[ \int \cos ^4(c+d x) (a+a \cos (c+d x)) \, dx=\begin {cases} \frac {3 a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {8 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {3 a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + a\right ) \cos ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**4*(a+a*cos(d*x+c)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.75 \[ \int \cos ^4(c+d x) (a+a \cos (c+d x)) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a}{480 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+a*cos(d*x+c)),x, algorithm="maxima")

[Out]

1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) +
8*sin(2*d*x + 2*c))*a)/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.84 \[ \int \cos ^4(c+d x) (a+a \cos (c+d x)) \, dx=\frac {3}{8} \, a x + \frac {a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {5 \, a \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {a \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {5 \, a \sin \left (d x + c\right )}{8 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+a*cos(d*x+c)),x, algorithm="giac")

[Out]

3/8*a*x + 1/80*a*sin(5*d*x + 5*c)/d + 1/32*a*sin(4*d*x + 4*c)/d + 5/48*a*sin(3*d*x + 3*c)/d + 1/4*a*sin(2*d*x
+ 2*c)/d + 5/8*a*sin(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 16.63 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.01 \[ \int \cos ^4(c+d x) (a+a \cos (c+d x)) \, dx=\frac {3\,a\,x}{8}+\frac {\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+\frac {13\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{6}+\frac {116\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}+\frac {19\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {13\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]

[In]

int(cos(c + d*x)^4*(a + a*cos(c + d*x)),x)

[Out]

(3*a*x)/8 + ((13*a*tan(c/2 + (d*x)/2))/4 + (19*a*tan(c/2 + (d*x)/2)^3)/6 + (116*a*tan(c/2 + (d*x)/2)^5)/15 + (
13*a*tan(c/2 + (d*x)/2)^7)/6 + (3*a*tan(c/2 + (d*x)/2)^9)/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^5)