\(\int \cos (c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx\) [3]

Optimal result

Integrand size = 27, antiderivative size = 77 \[ \int \cos (c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {1}{2} a (A+B) x+\frac {a (3 A+2 B) \sin (c+d x)}{3 d}+\frac {a (A+B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a B \cos ^2(c+d x) \sin (c+d x)}{3 d} \]

[Out]

1/2*a*(A+B)*x+1/3*a*(3*A+2*B)*sin(d*x+c)/d+1/2*a*(A+B)*cos(d*x+c)*sin(d*x+c)/d+1/3*a*B*cos(d*x+c)^2*sin(d*x+c)
/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3047, 3102, 2813} \[ \int \cos (c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {a (3 A+2 B) \sin (c+d x)}{3 d}+\frac {a (A+B) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} a x (A+B)+\frac {a B \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

[In]

Int[Cos[c + d*x]*(a + a*Cos[c + d*x])*(A + B*Cos[c + d*x]),x]

[Out]

(a*(A + B)*x)/2 + (a*(3*A + 2*B)*Sin[c + d*x])/(3*d) + (a*(A + B)*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (a*B*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 3047

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]

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}& = \int \cos (c+d x) \left (a A+(a A+a B) \cos (c+d x)+a B \cos ^2(c+d x)\right ) \, dx \\ & = \frac {a B \cos ^2(c+d x) \sin (c+d x)}{3 d}+\frac {1}{3} \int \cos (c+d x) (a (3 A+2 B)+3 a (A+B) \cos (c+d x)) \, dx \\ & = \frac {1}{2} a (A+B) x+\frac {a (3 A+2 B) \sin (c+d x)}{3 d}+\frac {a (A+B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a B \cos ^2(c+d x) \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.84 \[ \int \cos (c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {a (6 A c+6 B c+6 A d x+6 B d x+3 (4 A+3 B) \sin (c+d x)+3 (A+B) \sin (2 (c+d x))+B \sin (3 (c+d x)))}{12 d} \]

[In]

Integrate[Cos[c + d*x]*(a + a*Cos[c + d*x])*(A + B*Cos[c + d*x]),x]

[Out]

(a*(6*A*c + 6*B*c + 6*A*d*x + 6*B*d*x + 3*(4*A + 3*B)*Sin[c + d*x] + 3*(A + B)*Sin[2*(c + d*x)] + B*Sin[3*(c +
d*x)]))/(12*d)

Maple [A] (verified)

Time = 1.81 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(\frac {a \left (\frac {\left (A +B \right ) \sin \left (2 d x +2 c \right )}{2}+\frac {B \sin \left (3 d x +3 c \right )}{6}+\left (2 A +\frac {3 B}{2}\right ) \sin \left (d x +c \right )+\left (A +B \right ) x d \right )}{2 d}\)	\(54\)
parts	\(\frac {\left (a A +B a \right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\sin \left (d x +c \right ) a A}{d}+\frac {B a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}\)	\(70\)
derivativedivides	\(\frac {\frac {B a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a A \sin \left (d x +c \right )}{d}\)	\(85\)
default	\(\frac {\frac {B a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a A \sin \left (d x +c \right )}{d}\)	\(85\)
risch	\(\frac {a x A}{2}+\frac {a B x}{2}+\frac {\sin \left (d x +c \right ) a A}{d}+\frac {3 a B \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (3 d x +3 c \right ) B a}{12 d}+\frac {\sin \left (2 d x +2 c \right ) a A}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B a}{4 d}\)	\(85\)
norman	\(\frac {\frac {a \left (A +B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a \left (A +B \right ) x}{2}+\frac {3 a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {3 a \left (A +B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {3 a \left (A +B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {a \left (A +B \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {4 a \left (3 A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(138\)

[In]

int(cos(d*x+c)*(a+cos(d*x+c)*a)*(A+B*cos(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/2*a*(1/2*(A+B)*sin(2*d*x+2*c)+1/6*B*sin(3*d*x+3*c)+(2*A+3/2*B)*sin(d*x+c)+(A+B)*x*d)/d

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.73 \[ \int \cos (c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {3 \, {\left (A + B\right )} a d x + {\left (2 \, B a \cos \left (d x + c\right )^{2} + 3 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 2 \, {\left (3 \, A + 2 \, B\right )} a\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

1/6*(3*(A + B)*a*d*x + (2*B*a*cos(d*x + c)^2 + 3*(A + B)*a*cos(d*x + c) + 2*(3*A + 2*B)*a)*sin(d*x + c))/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (70) = 140\).

Time = 0.13 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.18 \[ \int \cos (c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\begin {cases} \frac {A a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {A a \sin {\left (c + d x \right )}}{d} + \frac {B a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {2 B a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )}\right ) \left (a \cos {\left (c \right )} + a\right ) \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((A*a*x*sin(c + d*x)**2/2 + A*a*x*cos(c + d*x)**2/2 + A*a*sin(c + d*x)*cos(c + d*x)/(2*d) + A*a*sin(c
+ d*x)/d + 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)*cos(c + d*x)/(2*d), Ne(d, 0)), (x*(A + B*cos(c))*(a*cos(c) + a)*cos(c),
True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.03 \[ \int \cos (c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a + 12 \, A a \sin \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a + 3*(2*d*x + 2*c + sin(
2*d*x + 2*c))*B*a + 12*A*a*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.88 \[ \int \cos (c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {1}{2} \, {\left (A a + B a\right )} x + \frac {B a \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (A a + B a\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, A a + 3 \, B a\right )} \sin \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

1/2*(A*a + B*a)*x + 1/12*B*a*sin(3*d*x + 3*c)/d + 1/4*(A*a + B*a)*sin(2*d*x + 2*c)/d + 1/4*(4*A*a + 3*B*a)*sin
(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.09 \[ \int \cos (c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {A\,a\,x}{2}+\frac {B\,a\,x}{2}+\frac {A\,a\,\sin \left (c+d\,x\right )}{d}+\frac {3\,B\,a\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \]

[In]

int(cos(c + d*x)*(A + B*cos(c + d*x))*(a + a*cos(c + d*x)),x)

[Out]

(A*a*x)/2 + (B*a*x)/2 + (A*a*sin(c + d*x))/d + (3*B*a*sin(c + d*x))/(4*d) + (A*a*sin(2*c + 2*d*x))/(4*d) + (B*
a*sin(2*c + 2*d*x))/(4*d) + (B*a*sin(3*c + 3*d*x))/(12*d)