\(\int \cos ^2(c+d x) (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx\) [223]

Optimal result

Integrand size = 31, antiderivative size = 189 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {1}{8} \left (4 a^2 A+3 A b^2+6 a b B\right ) x+\frac {\left (4 b^2 B+5 a (2 A b+a B)\right ) \sin (c+d x)}{5 d}+\frac {\left (4 a^2 A+3 A b^2+6 a b B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b (5 A b+6 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {b B \cos ^3(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{5 d}-\frac {\left (4 b^2 B+5 a (2 A b+a B)\right ) \sin ^3(c+d x)}{15 d} \]

[Out]

1/8*(4*A*a^2+3*A*b^2+6*B*a*b)*x+1/5*(4*b^2*B+5*a*(2*A*b+B*a))*sin(d*x+c)/d+1/8*(4*A*a^2+3*A*b^2+6*B*a*b)*cos(d
*x+c)*sin(d*x+c)/d+1/20*b*(5*A*b+6*B*a)*cos(d*x+c)^3*sin(d*x+c)/d+1/5*b*B*cos(d*x+c)^3*(a+b*cos(d*x+c))*sin(d*
x+c)/d-1/15*(4*b^2*B+5*a*(2*A*b+B*a))*sin(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3069, 3102, 2827, 2715, 8, 2713} \[ \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {\left (4 a^2 A+6 a b B+3 A b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (4 a^2 A+6 a b B+3 A b^2\right )-\frac {\left (5 a (a B+2 A b)+4 b^2 B\right ) \sin ^3(c+d x)}{15 d}+\frac {\left (5 a (a B+2 A b)+4 b^2 B\right ) \sin (c+d x)}{5 d}+\frac {b (6 a B+5 A b) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {b B \sin (c+d x) \cos ^3(c+d x) (a+b \cos (c+d x))}{5 d} \]

[In]

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

[Out]

((4*a^2*A + 3*A*b^2 + 6*a*b*B)*x)/8 + ((4*b^2*B + 5*a*(2*A*b + a*B))*Sin[c + d*x])/(5*d) + ((4*a^2*A + 3*A*b^2
+ 6*a*b*B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (b*(5*A*b + 6*a*B)*Cos[c + d*x]^3*Sin[c + d*x])/(20*d) + (b*B*C
os[c + d*x]^3*(a + b*Cos[c + d*x])*Sin[c + d*x])/(5*d) - ((4*b^2*B + 5*a*(2*A*b + a*B))*Sin[c + d*x]^3)/(15*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 3069

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*
x])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f
*x])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c
- b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m
, 1] &&  !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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 {b B \cos ^3(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{5 d}+\frac {1}{5} \int \cos ^2(c+d x) \left (a (5 a A+3 b B)+\left (4 b^2 B+5 a (2 A b+a B)\right ) \cos (c+d x)+b (5 A b+6 a B) \cos ^2(c+d x)\right ) \, dx \\ & = \frac {b (5 A b+6 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {b B \cos ^3(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{5 d}+\frac {1}{20} \int \cos ^2(c+d x) \left (5 \left (4 a^2 A+3 A b^2+6 a b B\right )+4 \left (4 b^2 B+5 a (2 A b+a B)\right ) \cos (c+d x)\right ) \, dx \\ & = \frac {b (5 A b+6 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {b B \cos ^3(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{5 d}+\frac {1}{4} \left (4 a^2 A+3 A b^2+6 a b B\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{5} \left (4 b^2 B+5 a (2 A b+a B)\right ) \int \cos ^3(c+d x) \, dx \\ & = \frac {\left (4 a^2 A+3 A b^2+6 a b B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b (5 A b+6 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {b B \cos ^3(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{5 d}+\frac {1}{8} \left (4 a^2 A+3 A b^2+6 a b B\right ) \int 1 \, dx-\frac {\left (4 b^2 B+5 a (2 A b+a B)\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d} \\ & = \frac {1}{8} \left (4 a^2 A+3 A b^2+6 a b B\right ) x+\frac {\left (4 b^2 B+5 a (2 A b+a B)\right ) \sin (c+d x)}{5 d}+\frac {\left (4 a^2 A+3 A b^2+6 a b B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b (5 A b+6 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {b B \cos ^3(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{5 d}-\frac {\left (4 b^2 B+5 a (2 A b+a B)\right ) \sin ^3(c+d x)}{15 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.15 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.77 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {60 \left (4 a^2 A+3 A b^2+6 a b B\right ) (c+d x)+60 \left (12 a A b+6 a^2 B+5 b^2 B\right ) \sin (c+d x)+120 \left (a^2 A+A b^2+2 a b B\right ) \sin (2 (c+d x))+10 \left (8 a A b+4 a^2 B+5 b^2 B\right ) \sin (3 (c+d x))+15 b (A b+2 a B) \sin (4 (c+d x))+6 b^2 B \sin (5 (c+d x))}{480 d} \]

[In]

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

[Out]

(60*(4*a^2*A + 3*A*b^2 + 6*a*b*B)*(c + d*x) + 60*(12*a*A*b + 6*a^2*B + 5*b^2*B)*Sin[c + d*x] + 120*(a^2*A + A*
b^2 + 2*a*b*B)*Sin[2*(c + d*x)] + 10*(8*a*A*b + 4*a^2*B + 5*b^2*B)*Sin[3*(c + d*x)] + 15*b*(A*b + 2*a*B)*Sin[4
*(c + d*x)] + 6*b^2*B*Sin[5*(c + d*x)])/(480*d)

Maple [A] (verified)

Time = 4.00 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.78

method	result	size
parallelrisch	\(\frac {120 \left (A \,a^{2}+A \,b^{2}+2 B a b \right ) \sin \left (2 d x +2 c \right )+10 \left (8 A a b +4 B \,a^{2}+5 B \,b^{2}\right ) \sin \left (3 d x +3 c \right )+15 \left (A \,b^{2}+2 B a b \right ) \sin \left (4 d x +4 c \right )+6 B \,b^{2} \sin \left (5 d x +5 c \right )+60 \left (12 A a b +6 B \,a^{2}+5 B \,b^{2}\right ) \sin \left (d x +c \right )+240 x d \left (A \,a^{2}+\frac {3}{4} A \,b^{2}+\frac {3}{2} B a b \right )}{480 d}\)	\(147\)
parts	\(\frac {\left (A \,b^{2}+2 B a b \right ) \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {\left (2 A a b +B \,a^{2}\right ) \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {A \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {B \,b^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5 d}\)	\(147\)
derivativedivides	\(\frac {\frac {B \,b^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+A \,b^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 B a b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 A a b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {B \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(184\)
default	\(\frac {\frac {B \,b^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+A \,b^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 B a b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 A a b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {B \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(184\)
risch	\(\frac {a^{2} x A}{2}+\frac {3 x A \,b^{2}}{8}+\frac {3 x B a b}{4}+\frac {3 \sin \left (d x +c \right ) A a b}{2 d}+\frac {3 \sin \left (d x +c \right ) B \,a^{2}}{4 d}+\frac {5 b^{2} B \sin \left (d x +c \right )}{8 d}+\frac {B \,b^{2} \sin \left (5 d x +5 c \right )}{80 d}+\frac {\sin \left (4 d x +4 c \right ) A \,b^{2}}{32 d}+\frac {\sin \left (4 d x +4 c \right ) B a b}{16 d}+\frac {\sin \left (3 d x +3 c \right ) A a b}{6 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{2}}{12 d}+\frac {5 \sin \left (3 d x +3 c \right ) B \,b^{2}}{48 d}+\frac {\sin \left (2 d x +2 c \right ) A \,a^{2}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) A \,b^{2}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B a b}{2 d}\)	\(225\)
norman	\(\frac {\left (\frac {1}{2} A \,a^{2}+\frac {3}{8} A \,b^{2}+\frac {3}{4} B a b \right ) x +\left (5 A \,a^{2}+\frac {15}{4} A \,b^{2}+\frac {15}{2} B a b \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (5 A \,a^{2}+\frac {15}{4} A \,b^{2}+\frac {15}{2} B a b \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} A \,a^{2}+\frac {3}{8} A \,b^{2}+\frac {3}{4} B a b \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {5}{2} A \,a^{2}+\frac {15}{8} A \,b^{2}+\frac {15}{4} B a b \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {5}{2} A \,a^{2}+\frac {15}{8} A \,b^{2}+\frac {15}{4} B a b \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4 \left (50 A a b +25 B \,a^{2}+29 B \,b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {\left (4 A \,a^{2}-16 A a b +5 A \,b^{2}-8 B \,a^{2}+10 B a b -8 B \,b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (4 A \,a^{2}+16 A a b +5 A \,b^{2}+8 B \,a^{2}+10 B a b +8 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (12 A \,a^{2}-64 A a b +3 A \,b^{2}-32 B \,a^{2}+6 B a b -16 B \,b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {\left (12 A \,a^{2}+64 A a b +3 A \,b^{2}+32 B \,a^{2}+6 B a b +16 B \,b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\)	\(429\)

[In]

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

[Out]

1/480*(120*(A*a^2+A*b^2+2*B*a*b)*sin(2*d*x+2*c)+10*(8*A*a*b+4*B*a^2+5*B*b^2)*sin(3*d*x+3*c)+15*(A*b^2+2*B*a*b)
*sin(4*d*x+4*c)+6*B*b^2*sin(5*d*x+5*c)+60*(12*A*a*b+6*B*a^2+5*B*b^2)*sin(d*x+c)+240*x*d*(A*a^2+3/4*A*b^2+3/2*B
*a*b))/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.75 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {15 \, {\left (4 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} d x + {\left (24 \, B b^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{3} + 80 \, B a^{2} + 160 \, A a b + 64 \, B b^{2} + 8 \, {\left (5 \, B a^{2} + 10 \, A a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (4 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

[In]

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

[Out]

1/120*(15*(4*A*a^2 + 6*B*a*b + 3*A*b^2)*d*x + (24*B*b^2*cos(d*x + c)^4 + 30*(2*B*a*b + A*b^2)*cos(d*x + c)^3 +
80*B*a^2 + 160*A*a*b + 64*B*b^2 + 8*(5*B*a^2 + 10*A*a*b + 4*B*b^2)*cos(d*x + c)^2 + 15*(4*A*a^2 + 6*B*a*b + 3
*A*b^2)*cos(d*x + c))*sin(d*x + c))/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 459 vs. \(2 (184) = 368\).

Time = 0.29 (sec) , antiderivative size = 459, normalized size of antiderivative = 2.43 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\begin {cases} \frac {A a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {4 A a b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {2 A a b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 A b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 A b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 A b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 A b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 A b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {2 B a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 B a b x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {3 B a b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a b x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {3 B a b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {5 B a b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {8 B b^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 B b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {B b^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )}\right ) \left (a + b \cos {\left (c \right )}\right )^{2} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((A*a**2*x*sin(c + d*x)**2/2 + A*a**2*x*cos(c + d*x)**2/2 + A*a**2*sin(c + d*x)*cos(c + d*x)/(2*d) +
4*A*a*b*sin(c + d*x)**3/(3*d) + 2*A*a*b*sin(c + d*x)*cos(c + d*x)**2/d + 3*A*b**2*x*sin(c + d*x)**4/8 + 3*A*b*
*2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*A*b**2*x*cos(c + d*x)**4/8 + 3*A*b**2*sin(c + d*x)**3*cos(c + d*x)/
(8*d) + 5*A*b**2*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 2*B*a**2*sin(c + d*x)**3/(3*d) + B*a**2*sin(c + d*x)*cos
(c + d*x)**2/d + 3*B*a*b*x*sin(c + d*x)**4/4 + 3*B*a*b*x*sin(c + d*x)**2*cos(c + d*x)**2/2 + 3*B*a*b*x*cos(c +
d*x)**4/4 + 3*B*a*b*sin(c + d*x)**3*cos(c + d*x)/(4*d) + 5*B*a*b*sin(c + d*x)*cos(c + d*x)**3/(4*d) + 8*B*b**
2*sin(c + d*x)**5/(15*d) + 4*B*b**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + B*b**2*sin(c + d*x)*cos(c + d*x)**
4/d, Ne(d, 0)), (x*(A + B*cos(c))*(a + b*cos(c))**2*cos(c)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.93 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a b + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{2} + 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B b^{2}}{480 \, d} \]

[In]

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

[Out]

1/480*(120*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^2 - 320*(sin(d*x
+ c)^3 - 3*sin(d*x + c))*A*a*b + 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a*b + 15*(12*d*
x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*b^2 + 32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*
x + c))*B*b^2)/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.83 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {B b^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {1}{8} \, {\left (4 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} x + \frac {{\left (2 \, B a b + A b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (4 \, B a^{2} + 8 \, A a b + 5 \, B b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (6 \, B a^{2} + 12 \, A a b + 5 \, B b^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \]

[In]

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

[Out]

1/80*B*b^2*sin(5*d*x + 5*c)/d + 1/8*(4*A*a^2 + 6*B*a*b + 3*A*b^2)*x + 1/32*(2*B*a*b + A*b^2)*sin(4*d*x + 4*c)/
d + 1/48*(4*B*a^2 + 8*A*a*b + 5*B*b^2)*sin(3*d*x + 3*c)/d + 1/4*(A*a^2 + 2*B*a*b + A*b^2)*sin(2*d*x + 2*c)/d +
1/8*(6*B*a^2 + 12*A*a*b + 5*B*b^2)*sin(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 4.09 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.62 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {x\,\left (A\,a^2+\frac {3\,B\,a\,b}{2}+\frac {3\,A\,b^2}{4}\right )}{2}+\frac {\left (2\,B\,a^2-\frac {5\,A\,b^2}{4}-A\,a^2+2\,B\,b^2+4\,A\,a\,b-\frac {5\,B\,a\,b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {16\,B\,a^2}{3}-\frac {A\,b^2}{2}-2\,A\,a^2+\frac {8\,B\,b^2}{3}+\frac {32\,A\,a\,b}{3}-B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,B\,a^2}{3}+\frac {40\,A\,a\,b}{3}+\frac {116\,B\,b^2}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (2\,A\,a^2+\frac {A\,b^2}{2}+\frac {16\,B\,a^2}{3}+\frac {8\,B\,b^2}{3}+\frac {32\,A\,a\,b}{3}+B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A\,a^2+\frac {5\,A\,b^2}{4}+2\,B\,a^2+2\,B\,b^2+4\,A\,a\,b+\frac {5\,B\,a\,b}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

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

[Out]

(x*(A*a^2 + (3*A*b^2)/4 + (3*B*a*b)/2))/2 + (tan(c/2 + (d*x)/2)^5*((20*B*a^2)/3 + (116*B*b^2)/15 + (40*A*a*b)/
3) - tan(c/2 + (d*x)/2)^9*(A*a^2 + (5*A*b^2)/4 - 2*B*a^2 - 2*B*b^2 - 4*A*a*b + (5*B*a*b)/2) + tan(c/2 + (d*x)/
2)^3*(2*A*a^2 + (A*b^2)/2 + (16*B*a^2)/3 + (8*B*b^2)/3 + (32*A*a*b)/3 + B*a*b) - tan(c/2 + (d*x)/2)^7*(2*A*a^2
+ (A*b^2)/2 - (16*B*a^2)/3 - (8*B*b^2)/3 - (32*A*a*b)/3 + B*a*b) + tan(c/2 + (d*x)/2)*(A*a^2 + (5*A*b^2)/4 +
2*B*a^2 + 2*B*b^2 + 4*A*a*b + (5*B*a*b)/2))/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2
+ (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1))