\(\int \cos (c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [293]

Optimal result

Integrand size = 27, antiderivative size = 69 \[ \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {B x}{2}+\frac {(3 A+2 C) \sin (c+d x)}{3 d}+\frac {B \cos (c+d x) \sin (c+d x)}{2 d}+\frac {C \cos ^2(c+d x) \sin (c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {3102, 2813} \[ \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {(3 A+2 C) \sin (c+d x)}{3 d}+\frac {B \sin (c+d x) \cos (c+d x)}{2 d}+\frac {B x}{2}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

[In]

Int[Cos[c + d*x]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(B*x)/2 + ((3*A + 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)

Rule 2813

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)*(Cos[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {C \cos ^2(c+d x) \sin (c+d x)}{3 d}+\frac {1}{3} \int \cos (c+d x) (3 A+2 C+3 B \cos (c+d x)) \, 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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77 \[ \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {6 B c+6 B d x+3 (4 A+3 C) \sin (c+d x)+3 B \sin (2 (c+d x))+C \sin (3 (c+d x))}{12 d} \]

[In]

Integrate[Cos[c + d*x]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(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)

Maple [A] (verified)

Time = 3.78 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.71

method	result	size
parallelrisch	\(\frac {3 \sin \left (2 d x +2 c \right ) B +\sin \left (3 d x +3 c \right ) C +\left (12 A +9 C \right ) \sin \left (d x +c \right )+6 d x B}{12 d}\)	\(49\)
derivativedivides	\(\frac {\frac {C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \sin \left (d x +c \right )}{d}\)	\(57\)
default	\(\frac {\frac {C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \sin \left (d x +c \right )}{d}\)	\(57\)
risch	\(\frac {x B}{2}+\frac {\sin \left (d x +c \right ) A}{d}+\frac {3 C \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (3 d x +3 c \right ) C}{12 d}+\frac {B \sin \left (2 d x +2 c \right )}{4 d}\)	\(59\)
parts	\(\frac {\sin \left (d x +c \right ) A}{d}+\frac {B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}\)	\(62\)
norman	\(\frac {\frac {\left (2 A -B +2 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (2 A +B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {x B}{2}+\frac {3 x B \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {3 x B \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {x B \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {4 \left (3 A +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(134\)

[In]

int(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/12*(3*sin(2*d*x+2*c)*B+sin(3*d*x+3*c)*C+(12*A+9*C)*sin(d*x+c)+6*d*x*B)/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.65 \[ \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 \, B d x + {\left (2 \, C \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 6 \, A + 4 \, C\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.55 \[ \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\begin {cases} \frac {A \sin {\left (c + d x \right )}}{d} + \frac {B x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 C \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

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)*c
os(c), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.80 \[ \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C + 12 \, A \sin \left (d x + c\right )}{12 \, d} \]

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77 \[ \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{2} \, B x + \frac {C \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {B \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, A + 3 \, C\right )} \sin \left (d x + c\right )}{4 \, d} \]

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.44 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {B\,x}{2}+\frac {A\,\sin \left (c+d\,x\right )}{d}+\frac {2\,C\,\sin \left (c+d\,x\right )}{3\,d}+\frac {B\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {C\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \]

[In]

int(cos(c + d*x)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)

[Out]

(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)