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

Optimal result

Integrand size = 29, antiderivative size = 97 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {1}{8} a (4 A+3 B) x+\frac {a (A+B) \sin (c+d x)}{d}+\frac {a (4 A+3 B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a B \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {a (A+B) \sin ^3(c+d x)}{3 d} \]

[Out]

1/8*a*(4*A+3*B)*x+a*(A+B)*sin(d*x+c)/d+1/8*a*(4*A+3*B)*cos(d*x+c)*sin(d*x+c)/d+1/4*a*B*cos(d*x+c)^3*sin(d*x+c)
/d-1/3*a*(A+B)*sin(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3047, 3102, 2827, 2715, 8, 2713} \[ \int \cos ^2(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=-\frac {a (A+B) \sin ^3(c+d x)}{3 d}+\frac {a (A+B) \sin (c+d x)}{d}+\frac {a (4 A+3 B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} a x (4 A+3 B)+\frac {a B \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

[In]

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

[Out]

(a*(4*A + 3*B)*x)/8 + (a*(A + B)*Sin[c + d*x])/d + (a*(4*A + 3*B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*B*Cos[
c + d*x]^3*Sin[c + d*x])/(4*d) - (a*(A + B)*Sin[c + d*x]^3)/(3*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2715

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] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 ^2(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 ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{4} \int \cos ^2(c+d x) (a (4 A+3 B)+4 a (A+B) \cos (c+d x)) \, dx \\ & = \frac {a B \cos ^3(c+d x) \sin (c+d x)}{4 d}+(a (A+B)) \int \cos ^3(c+d x) \, dx+\frac {1}{4} (a (4 A+3 B)) \int \cos ^2(c+d x) \, dx \\ & = \frac {a (4 A+3 B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a B \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{8} (a (4 A+3 B)) \int 1 \, dx-\frac {(a (A+B)) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d} \\ & = \frac {1}{8} a (4 A+3 B) x+\frac {a (A+B) \sin (c+d x)}{d}+\frac {a (4 A+3 B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a B \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {a (A+B) \sin ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.77 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {a \left (48 A c+36 B c+48 A d x+36 B d x+96 (A+B) \sin (c+d x)-32 (A+B) \sin ^3(c+d x)+24 (A+B) \sin (2 (c+d x))+3 B \sin (4 (c+d x))\right )}{96 d} \]

[In]

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

[Out]

(a*(48*A*c + 36*B*c + 48*A*d*x + 36*B*d*x + 96*(A + B)*Sin[c + d*x] - 32*(A + B)*Sin[c + d*x]^3 + 24*(A + B)*S
in[2*(c + d*x)] + 3*B*Sin[4*(c + d*x)]))/(96*d)

Maple [A] (verified)

Time = 2.52 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.69

method	result	size
parallelrisch	\(\frac {\left (\frac {\left (A +B \right ) \sin \left (2 d x +2 c \right )}{2}+\frac {\left (A +B \right ) \sin \left (3 d x +3 c \right )}{6}+\frac {\sin \left (4 d x +4 c \right ) B}{16}+\frac {3 \left (A +B \right ) \sin \left (d x +c \right )}{2}+d x \left (A +\frac {3 B}{4}\right )\right ) a}{2 d}\)	\(67\)
parts	\(\frac {\left (a A +B a \right ) \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {B a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(97\)
derivativedivides	\(\frac {B a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {B a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(107\)
default	\(\frac {B a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {B a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(107\)
risch	\(\frac {a x A}{2}+\frac {3 a B x}{8}+\frac {3 \sin \left (d x +c \right ) a A}{4 d}+\frac {3 a B \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (4 d x +4 c \right ) B a}{32 d}+\frac {\sin \left (3 d x +3 c \right ) a A}{12 d}+\frac {\sin \left (3 d x +3 c \right ) B a}{12 d}+\frac {\sin \left (2 d x +2 c \right ) a A}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B a}{4 d}\)	\(118\)
norman	\(\frac {\frac {a \left (4 A +3 B \right ) x}{8}+\frac {a \left (4 A +3 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a \left (4 A +3 B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {3 a \left (4 A +3 B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {a \left (4 A +3 B \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {a \left (4 A +3 B \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {7 a \left (4 A +7 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {a \left (12 A +13 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (52 A +31 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(211\)

[In]

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

[Out]

1/2*(1/2*(A+B)*sin(2*d*x+2*c)+1/6*(A+B)*sin(3*d*x+3*c)+1/16*sin(4*d*x+4*c)*B+3/2*(A+B)*sin(d*x+c)+d*x*(A+3/4*B
))*a/d

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.76 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {3 \, {\left (4 \, A + 3 \, B\right )} a d x + {\left (6 \, B a \cos \left (d x + c\right )^{3} + 8 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 3 \, B\right )} a \cos \left (d x + c\right ) + 16 \, {\left (A + B\right )} a\right )} \sin \left (d x + c\right )}{24 \, d} \]

[In]

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

[Out]

1/24*(3*(4*A + 3*B)*a*d*x + (6*B*a*cos(d*x + c)^3 + 8*(A + B)*a*cos(d*x + c)^2 + 3*(4*A + 3*B)*a*cos(d*x + c)
+ 16*(A + 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. 252 vs. \(2 (88) = 176\).

Time = 0.18 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.60 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\begin {cases} \frac {A a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {2 A a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {A a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {3 B a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 B a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 B a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 B a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 B a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {B a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )}\right ) \left (a \cos {\left (c \right )} + a\right ) \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**2*(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 + 2*A*a*sin(c + d*x)**3/(3*d) + A*a*sin(c + d*x)*
cos(c + d*x)**2/d + A*a*sin(c + d*x)*cos(c + d*x)/(2*d) + 3*B*a*x*sin(c + d*x)**4/8 + 3*B*a*x*sin(c + d*x)**2*
cos(c + d*x)**2/4 + 3*B*a*x*cos(c + d*x)**4/8 + 3*B*a*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 2*B*a*sin(c + d*x)*
*3/(3*d) + 5*B*a*sin(c + d*x)*cos(c + d*x)**3/(8*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)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.04 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=-\frac {32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a}{96 \, d} \]

[In]

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

[Out]

-1/96*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a + 32*(sin(d*x + c)^3
- 3*sin(d*x + c))*B*a - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a)/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.92 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {1}{8} \, {\left (4 \, A a + 3 \, B a\right )} x + \frac {B a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (A a + B a\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (A a + B a\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {3 \, {\left (A a + B a\right )} \sin \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

1/8*(4*A*a + 3*B*a)*x + 1/32*B*a*sin(4*d*x + 4*c)/d + 1/12*(A*a + B*a)*sin(3*d*x + 3*c)/d + 1/4*(A*a + B*a)*si
n(2*d*x + 2*c)/d + 3/4*(A*a + B*a)*sin(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 0.77 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.19 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {\left (A\,a+\frac {3\,B\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {7\,A\,a}{3}+\frac {49\,B\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {13\,A\,a}{3}+\frac {31\,B\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A\,a+\frac {13\,B\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,A+3\,B\right )}{4\,\left (A\,a+\frac {3\,B\,a}{4}\right )}\right )\,\left (4\,A+3\,B\right )}{4\,d}-\frac {a\,\left (4\,A+3\,B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(3*A*a + (13*B*a)/4) + tan(c/2 + (d*x)/2)^7*(A*a + (3*B*a)/4) + tan(c/2 + (d*x)/2)^3*((13*
A*a)/3 + (31*B*a)/12) + tan(c/2 + (d*x)/2)^5*((7*A*a)/3 + (49*B*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*atan((a*tan(c/2 + (d*x)/2)*(4*A + 3*B
))/(4*(A*a + (3*B*a)/4)))*(4*A + 3*B))/(4*d) - (a*(4*A + 3*B)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(4*d)