Integrand size = 36, antiderivative size = 97 \[ \int \cos (c+d x) (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{8} a (4 B+3 C) x+\frac {a (B+C) \sin (c+d x)}{d}+\frac {a (4 B+3 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a C \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {a (B+C) \sin ^3(c+d x)}{3 d} \]
1/8*a*(4*B+3*C)*x+a*(B+C)*sin(d*x+c)/d+1/8*a*(4*B+3*C)*cos(d*x+c)*sin(d*x+ c)/d+1/4*a*C*cos(d*x+c)^3*sin(d*x+c)/d-1/3*a*(B+C)*sin(d*x+c)^3/d
Time = 0.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.78 \[ \int \cos (c+d x) (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {a (48 B d x+36 C d x+72 (B+C) \sin (c+d x)+24 (B+C) \sin (2 (c+d x))+8 B \sin (3 (c+d x))+8 C \sin (3 (c+d x))+3 C \sin (4 (c+d x)))}{96 d} \]
(a*(48*B*d*x + 36*C*d*x + 72*(B + C)*Sin[c + d*x] + 24*(B + C)*Sin[2*(c + d*x)] + 8*B*Sin[3*(c + d*x)] + 8*C*Sin[3*(c + d*x)] + 3*C*Sin[4*(c + d*x)] ))/(96*d)
Time = 0.62 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.96, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.361, Rules used = {3042, 3508, 3042, 3447, 3042, 3502, 3042, 3227, 3042, 3113, 2009, 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.
| \(\displaystyle \int \cos (c+d x) (a \cos (c+d x)+a) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \cos ^2(c+d x) (a \cos (c+d x)+a) (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \int \cos ^2(c+d x) \left ((a B+a C) \cos (c+d x)+a B+a C \cos ^2(c+d x)\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left ((a B+a C) \sin \left (c+d x+\frac {\pi }{2}\right )+a B+a C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{4} \int \cos ^2(c+d x) (a (4 B+3 C)+4 a (B+C) \cos (c+d x))dx+\frac {a C \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a (4 B+3 C)+4 a (B+C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a C \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{4} \left (4 a (B+C) \int \cos ^3(c+d x)dx+a (4 B+3 C) \int \cos ^2(c+d x)dx\right )+\frac {a C \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (a (4 B+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+4 a (B+C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {a C \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {1}{4} \left (a (4 B+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {4 a (B+C) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a C \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} \left (a (4 B+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {4 a (B+C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a C \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{4} \left (a (4 B+3 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {4 a (B+C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a C \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{4} \left (a (4 B+3 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {4 a (B+C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a C \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
(a*C*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (a*(4*B + 3*C)*(x/2 + (Cos[c + d *x]*Sin[c + d*x])/(2*d)) - (4*a*(B + C)*(-Sin[c + d*x] + Sin[c + d*x]^3/3) )/d)/4
3.3.27.3.1 Defintions of rubi rules used
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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 4.76 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.69
| method | result | size |
| parallelrisch | \(\frac {\left (\frac {\left (B +C \right ) \sin \left (2 d x +2 c \right )}{2}+\frac {\left (B +C \right ) \sin \left (3 d x +3 c \right )}{6}+\frac {\sin \left (4 d x +4 c \right ) C}{16}+\frac {3 \left (B +C \right ) \sin \left (d x +c \right )}{2}+d x \left (B +\frac {3 C}{4}\right )\right ) a}{2 d}\) | \(67\) |
| parts | \(\frac {\left (B a +a C \right ) \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(97\) |
| derivativedivides | \(\frac {a C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(107\) |
| default | \(\frac {a C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(107\) |
| risch | \(\frac {a B x}{2}+\frac {3 a C x}{8}+\frac {3 a B \sin \left (d x +c \right )}{4 d}+\frac {3 a C \sin \left (d x +c \right )}{4 d}+\frac {a C \sin \left (4 d x +4 c \right )}{32 d}+\frac {\sin \left (3 d x +3 c \right ) B a}{12 d}+\frac {\sin \left (3 d x +3 c \right ) a C}{12 d}+\frac {\sin \left (2 d x +2 c \right ) B a}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a C}{4 d}\) | \(118\) |
| norman | \(\frac {\frac {a \left (4 B +3 C \right ) x}{8}+\frac {a \left (4 B +3 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a \left (4 B +3 C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {3 a \left (4 B +3 C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {a \left (4 B +3 C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {a \left (4 B +3 C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {7 a \left (4 B +7 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {a \left (12 B +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (52 B +31 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\) | \(211\) |
1/2*(1/2*(B+C)*sin(2*d*x+2*c)+1/6*(B+C)*sin(3*d*x+3*c)+1/16*sin(4*d*x+4*c) *C+3/2*(B+C)*sin(d*x+c)+d*x*(B+3/4*C))*a/d
Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.76 \[ \int \cos (c+d x) (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (4 \, B + 3 \, C\right )} a d x + {\left (6 \, C a \cos \left (d x + c\right )^{3} + 8 \, {\left (B + C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, B + 3 \, C\right )} a \cos \left (d x + c\right ) + 16 \, {\left (B + C\right )} a\right )} \sin \left (d x + c\right )}{24 \, d} \]
1/24*(3*(4*B + 3*C)*a*d*x + (6*C*a*cos(d*x + c)^3 + 8*(B + C)*a*cos(d*x + c)^2 + 3*(4*B + 3*C)*a*cos(d*x + c) + 16*(B + C)*a)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (88) = 176\).
Time = 0.18 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.63 \[ \int \cos (c+d x) (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\begin {cases} \frac {B a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {2 B a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {3 C a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 C a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 C a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 C a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 C a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {C a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (B \cos {\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) \left (a \cos {\left (c \right )} + a\right ) \cos {\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((B*a*x*sin(c + d*x)**2/2 + B*a*x*cos(c + d*x)**2/2 + 2*B*a*sin(c + d*x)**3/(3*d) + B*a*sin(c + d*x)*cos(c + d*x)**2/d + B*a*sin(c + d*x)*c os(c + d*x)/(2*d) + 3*C*a*x*sin(c + d*x)**4/8 + 3*C*a*x*sin(c + d*x)**2*co s(c + d*x)**2/4 + 3*C*a*x*cos(c + d*x)**4/8 + 3*C*a*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 2*C*a*sin(c + d*x)**3/(3*d) + 5*C*a*sin(c + d*x)*cos(c + d* x)**3/(8*d) + C*a*sin(c + d*x)*cos(c + d*x)**2/d, Ne(d, 0)), (x*(B*cos(c) + C*cos(c)**2)*(a*cos(c) + a)*cos(c), True))
Time = 0.22 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.04 \[ \int \cos (c+d x) (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a}{96 \, d} \]
-1/96*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a - 24*(2*d*x + 2*c + sin(2* d*x + 2*c))*B*a + 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a - 3*(12*d*x + 1 2*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a)/d
Time = 0.31 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.92 \[ \int \cos (c+d x) (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{8} \, {\left (4 \, B a + 3 \, C a\right )} x + \frac {C a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (B a + C a\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (B a + C a\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {3 \, {\left (B a + C a\right )} \sin \left (d x + c\right )}{4 \, d} \]
1/8*(4*B*a + 3*C*a)*x + 1/32*C*a*sin(4*d*x + 4*c)/d + 1/12*(B*a + C*a)*sin (3*d*x + 3*c)/d + 1/4*(B*a + C*a)*sin(2*d*x + 2*c)/d + 3/4*(B*a + C*a)*sin (d*x + c)/d
Time = 2.33 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.19 \[ \int \cos (c+d x) (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\left (B\,a+\frac {3\,C\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {7\,B\,a}{3}+\frac {49\,C\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {13\,B\,a}{3}+\frac {31\,C\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,B\,a+\frac {13\,C\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,B+3\,C\right )}{4\,\left (B\,a+\frac {3\,C\,a}{4}\right )}\right )\,\left (4\,B+3\,C\right )}{4\,d}-\frac {a\,\left (4\,B+3\,C\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d} \]
(tan(c/2 + (d*x)/2)*(3*B*a + (13*C*a)/4) + tan(c/2 + (d*x)/2)^7*(B*a + (3* C*a)/4) + tan(c/2 + (d*x)/2)^3*((13*B*a)/3 + (31*C*a)/12) + tan(c/2 + (d*x )/2)^5*((7*B*a)/3 + (49*C*a)/12))/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (a*ata n((a*tan(c/2 + (d*x)/2)*(4*B + 3*C))/(4*(B*a + (3*C*a)/4)))*(4*B + 3*C))/( 4*d) - (a*(4*B + 3*C)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(4*d)