Integrand size = 19, antiderivative size = 114 \[ \int \cos ^5(c+d x) (a+b \cos (c+d x)) \, dx=\frac {5 b x}{16}+\frac {a \sin (c+d x)}{d}+\frac {5 b \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 b \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {b \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d} \] Output:
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*s in(d*x+c)^5/d
Time = 0.14 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.68 \[ \int \cos ^5(c+d x) (a+b \cos (c+d x)) \, dx=\frac {960 a \sin (c+d x)-640 a \sin ^3(c+d x)+192 a \sin ^5(c+d x)+5 b (60 c+60 d x+45 \sin (2 (c+d x))+9 \sin (4 (c+d x))+\sin (6 (c+d x)))}{960 d} \] Input:
Integrate[Cos[c + d*x]^5*(a + b*Cos[c + d*x]),x]
Output:
(960*a*Sin[c + d*x] - 640*a*Sin[c + d*x]^3 + 192*a*Sin[c + d*x]^5 + 5*b*(6 0*c + 60*d*x + 45*Sin[2*(c + d*x)] + 9*Sin[4*(c + d*x)] + Sin[6*(c + d*x)] ))/(960*d)
Time = 0.48 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.579, Rules used = {3042, 3227, 3042, 3113, 2009, 3115, 3042, 3115, 3042, 3115, 24}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \cos ^5(c+d x) (a+b \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^5 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle a \int \cos ^5(c+d x)dx+b \int \cos ^6(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^5dx+b \int \sin \left (c+d x+\frac {\pi }{2}\right )^6dx\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle b \int \sin \left (c+d x+\frac {\pi }{2}\right )^6dx-\frac {a \int \left (\sin ^4(c+d x)-2 \sin ^2(c+d x)+1\right )d(-\sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle b \int \sin \left (c+d x+\frac {\pi }{2}\right )^6dx-\frac {a \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle b \left (\frac {5}{6} \int \cos ^4(c+d x)dx+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b \left (\frac {5}{6} \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle b \left (\frac {5}{6} \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b \left (\frac {5}{6} \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle b \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle b \left (\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5}{6} \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )\right )-\frac {a \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
Input:
Int[Cos[c + d*x]^5*(a + b*Cos[c + d*x]),x]
Output:
-((a*(-Sin[c + d*x] + (2*Sin[c + d*x]^3)/3 - Sin[c + d*x]^5/5))/d) + b*((C os[c + d*x]^5*Sin[c + d*x])/(6*d) + (5*((Cos[c + d*x]^3*Sin[c + d*x])/(4*d ) + (3*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4))/6)
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
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] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Time = 10.16 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.70
| method | result | size |
| derivativedivides | \(\frac {b \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {a \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) | \(80\) |
| default | \(\frac {b \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {a \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) | \(80\) |
| parallelrisch | \(\frac {300 d x b +5 b \sin \left (6 d x +6 c \right )+12 a \sin \left (5 d x +5 c \right )+45 b \sin \left (4 d x +4 c \right )+100 a \sin \left (3 d x +3 c \right )+225 b \sin \left (2 d x +2 c \right )+600 a \sin \left (d x +c \right )}{960 d}\) | \(81\) |
| parts | \(\frac {a \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5 d}+\frac {b \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\) | \(82\) |
| 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 | \(\frac {\frac {5 b x}{16}+\frac {15 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {75 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{16}+\frac {25 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4}+\frac {75 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{16}+\frac {15 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {5 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{16}+\frac {\left (16 a -11 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 d}+\frac {\left (16 a +11 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {\left (112 a -5 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d}+\frac {\left (112 a +5 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 d}+\frac {\left (208 a -75 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 d}+\frac {\left (208 a +75 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}\) | \(248\) |
| orering | \(\text {Expression too large to display}\) | \(1678\) |
Input:
int(cos(d*x+c)^5*(a+cos(d*x+c)*b),x,method=_RETURNVERBOSE)
Output:
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)+5/1 6*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))
Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.66 \[ \int \cos ^5(c+d x) (a+b \cos (c+d x)) \, dx=\frac {75 \, b d x + {\left (40 \, b \cos \left (d x + c\right )^{5} + 48 \, a \cos \left (d x + c\right )^{4} + 50 \, b \cos \left (d x + c\right )^{3} + 64 \, a \cos \left (d x + c\right )^{2} + 75 \, b \cos \left (d x + c\right ) + 128 \, a\right )} \sin \left (d x + c\right )}{240 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+b*cos(d*x+c)),x, algorithm="fricas")
Output:
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
Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (107) = 214\).
Time = 0.34 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.89 \[ \int \cos ^5(c+d x) (a+b \cos (c+d x)) \, dx=\begin {cases} \frac {8 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 b x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 b x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 b x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 b x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 b \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 b \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 b \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\left (c \right )}\right ) \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**5*(a+b*cos(d*x+c)),x)
Output:
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)/(1 6*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))
Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.74 \[ \int \cos ^5(c+d x) (a+b \cos (c+d x)) \, dx=\frac {64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b}{960 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+b*cos(d*x+c)),x, algorithm="maxima")
Output:
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
Time = 0.52 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.81 \[ \int \cos ^5(c+d x) (a+b \cos (c+d x)) \, dx=\frac {5}{16} \, b x + \frac {b \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {3 \, b \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {5 \, a \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {15 \, b \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {5 \, a \sin \left (d x + c\right )}{8 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+b*cos(d*x+c)),x, algorithm="giac")
Output:
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
Time = 39.95 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.01 \[ \int \cos ^5(c+d x) (a+b \cos (c+d x)) \, dx=\frac {5\,b\,x}{16}+\frac {8\,a\,\sin \left (c+d\,x\right )}{15\,d}+\frac {5\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{16\,d}+\frac {4\,a\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{15\,d}+\frac {a\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{5\,d}+\frac {5\,b\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{24\,d}+\frac {b\,{\cos \left (c+d\,x\right )}^5\,\sin \left (c+d\,x\right )}{6\,d} \] Input:
int(cos(c + d*x)^5*(a + b*cos(c + d*x)),x)
Output:
(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)
Time = 0.15 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.80 \[ \int \cos ^5(c+d x) (a+b \cos (c+d x)) \, dx=\frac {40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} b -130 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b +165 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +48 \sin \left (d x +c \right )^{5} a -160 \sin \left (d x +c \right )^{3} a +240 \sin \left (d x +c \right ) a +75 b d x}{240 d} \] Input:
int(cos(d*x+c)^5*(a+b*cos(d*x+c)),x)
Output:
(40*cos(c + d*x)*sin(c + d*x)**5*b - 130*cos(c + d*x)*sin(c + d*x)**3*b + 165*cos(c + d*x)*sin(c + d*x)*b + 48*sin(c + d*x)**5*a - 160*sin(c + d*x)* *3*a + 240*sin(c + d*x)*a + 75*b*d*x)/(240*d)