Integrand size = 27, antiderivative size = 69 \[ \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {B x}{2}+\frac {(3 A+2 C) \sin (c+d x)}{3 d}+\frac {B \cos (c+d x) \sin (c+d x)}{2 d}+\frac {C \cos ^2(c+d x) \sin (c+d x)}{3 d} \] Output:
1/2*B*x+1/3*(3*A+2*C)*sin(d*x+c)/d+1/2*B*cos(d*x+c)*sin(d*x+c)/d+1/3*C*cos (d*x+c)^2*sin(d*x+c)/d
Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77 \[ \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {6 B c+6 B d x+3 (4 A+3 C) \sin (c+d x)+3 B \sin (2 (c+d x))+C \sin (3 (c+d x))}{12 d} \] Input:
Integrate[Cos[c + d*x]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]
Output:
(6*B*c + 6*B*d*x + 3*(4*A + 3*C)*Sin[c + d*x] + 3*B*Sin[2*(c + d*x)] + C*S in[3*(c + d*x)])/(12*d)
Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3042, 3502, 3042, 3213}
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.
\(\displaystyle \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {1}{3} \int \cos (c+d x) (3 A+2 C+3 B \cos (c+d x))dx+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (3 A+2 C+3 B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle \frac {1}{3} \left (\frac {(3 A+2 C) \sin (c+d x)}{d}+\frac {3 B \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3 B x}{2}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 d}\) |
Input:
Int[Cos[c + d*x]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]
Output:
(C*Cos[c + d*x]^2*Sin[c + d*x])/(3*d) + ((3*B*x)/2 + ((3*A + 2*C)*Sin[c + d*x])/d + (3*B*Cos[c + d*x]*Sin[c + d*x])/(2*d))/3
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, 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]
Time = 1.00 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.71
method | result | size |
parallelrisch | \(\frac {3 B \sin \left (2 d x +2 c \right )+C \sin \left (3 d x +3 c \right )+\left (12 A +9 C \right ) \sin \left (d x +c \right )+6 B x d}{12 d}\) | \(49\) |
derivativedivides | \(\frac {\frac {C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \sin \left (d x +c \right )}{d}\) | \(57\) |
default | \(\frac {\frac {C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \sin \left (d x +c \right )}{d}\) | \(57\) |
risch | \(\frac {B x}{2}+\frac {\sin \left (d x +c \right ) A}{d}+\frac {3 C \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (3 d x +3 c \right ) C}{12 d}+\frac {B \sin \left (2 d x +2 c \right )}{4 d}\) | \(59\) |
parts | \(\frac {\sin \left (d x +c \right ) A}{d}+\frac {B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3 d}\) | \(62\) |
norman | \(\frac {\frac {\left (2 A -B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {\left (2 A +B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {B x}{2}+\frac {3 B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {3 B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}+\frac {B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}+\frac {4 \left (3 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\) | \(134\) |
orering | \(\text {Expression too large to display}\) | \(1158\) |
Input:
int(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/12*(3*B*sin(2*d*x+2*c)+C*sin(3*d*x+3*c)+(12*A+9*C)*sin(d*x+c)+6*B*x*d)/d
Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.65 \[ \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 \, B d x + {\left (2 \, C \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 6 \, A + 4 \, C\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas" )
Output:
1/6*(3*B*d*x + (2*C*cos(d*x + c)^2 + 3*B*cos(d*x + c) + 6*A + 4*C)*sin(d*x + c))/d
Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.55 \[ \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\begin {cases} \frac {A \sin {\left (c + d x \right )}}{d} + \frac {B x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 C \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) \cos {\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)
Output:
Piecewise((A*sin(c + d*x)/d + B*x*sin(c + d*x)**2/2 + B*x*cos(c + d*x)**2/ 2 + B*sin(c + d*x)*cos(c + d*x)/(2*d) + 2*C*sin(c + d*x)**3/(3*d) + C*sin( c + d*x)*cos(c + d*x)**2/d, Ne(d, 0)), (x*(A + B*cos(c) + C*cos(c)**2)*cos (c), True))
Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.80 \[ \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C + 12 \, A \sin \left (d x + c\right )}{12 \, d} \] Input:
integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima" )
Output:
1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*C + 12*A*sin(d*x + c))/d
Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77 \[ \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{2} \, B x + \frac {C \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {B \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, A + 3 \, C\right )} \sin \left (d x + c\right )}{4 \, d} \] Input:
integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")
Output:
1/2*B*x + 1/12*C*sin(3*d*x + 3*c)/d + 1/4*B*sin(2*d*x + 2*c)/d + 1/4*(4*A + 3*C)*sin(d*x + c)/d
Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {B\,x}{2}+\frac {A\,\sin \left (c+d\,x\right )}{d}+\frac {2\,C\,\sin \left (c+d\,x\right )}{3\,d}+\frac {B\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {C\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \] Input:
int(cos(c + d*x)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)
Output:
(B*x)/2 + (A*sin(c + d*x))/d + (2*C*sin(c + d*x))/(3*d) + (B*cos(c + d*x)* sin(c + d*x))/(2*d) + (C*cos(c + d*x)^2*sin(c + d*x))/(3*d)
Time = 0.16 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.80 \[ \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -2 \sin \left (d x +c \right )^{3} c +6 \sin \left (d x +c \right ) a +6 \sin \left (d x +c \right ) c +3 b d x}{6 d} \] Input:
int(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)
Output:
(3*cos(c + d*x)*sin(c + d*x)*b - 2*sin(c + d*x)**3*c + 6*sin(c + d*x)*a + 6*sin(c + d*x)*c + 3*b*d*x)/(6*d)