3.231 \(\int \cos ^2(c+d x) (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=269 \[ \frac {b \left (14 a^2 B+18 a A b+5 b^2 B\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}-\frac {\left (5 a^3 B+15 a^2 A b+12 a b^2 B+4 A b^3\right ) \sin ^3(c+d x)}{15 d}+\frac {\left (5 a^3 B+15 a^2 A b+12 a b^2 B+4 A b^3\right ) \sin (c+d x)}{5 d}+\frac {\left (8 a^3 A+18 a^2 b B+18 a A b^2+5 b^3 B\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x \left (8 a^3 A+18 a^2 b B+18 a A b^2+5 b^3 B\right )+\frac {b^2 (4 a B+3 A b) \sin (c+d x) \cos ^4(c+d x)}{15 d}+\frac {b B \sin (c+d x) \cos ^3(c+d x) (a+b \cos (c+d x))^2}{6 d} \]

[Out]

1/16*(8*A*a^3+18*A*a*b^2+18*B*a^2*b+5*B*b^3)*x+1/5*(15*A*a^2*b+4*A*b^3+5*B*a^3+12*B*a*b^2)*sin(d*x+c)/d+1/16*(
8*A*a^3+18*A*a*b^2+18*B*a^2*b+5*B*b^3)*cos(d*x+c)*sin(d*x+c)/d+1/24*b*(18*A*a*b+14*B*a^2+5*B*b^2)*cos(d*x+c)^3
*sin(d*x+c)/d+1/15*b^2*(3*A*b+4*B*a)*cos(d*x+c)^4*sin(d*x+c)/d+1/6*b*B*cos(d*x+c)^3*(a+b*cos(d*x+c))^2*sin(d*x
+c)/d-1/15*(15*A*a^2*b+4*A*b^3+5*B*a^3+12*B*a*b^2)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.51, antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2990, 3033, 3023, 2748, 2635, 8, 2633} \[ -\frac {\left (15 a^2 A b+5 a^3 B+12 a b^2 B+4 A b^3\right ) \sin ^3(c+d x)}{15 d}+\frac {\left (15 a^2 A b+5 a^3 B+12 a b^2 B+4 A b^3\right ) \sin (c+d x)}{5 d}+\frac {b \left (14 a^2 B+18 a A b+5 b^2 B\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {\left (8 a^3 A+18 a^2 b B+18 a A b^2+5 b^3 B\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x \left (8 a^3 A+18 a^2 b B+18 a A b^2+5 b^3 B\right )+\frac {b^2 (4 a B+3 A b) \sin (c+d x) \cos ^4(c+d x)}{15 d}+\frac {b B \sin (c+d x) \cos ^3(c+d x) (a+b \cos (c+d x))^2}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x]),x]

[Out]

((8*a^3*A + 18*a*A*b^2 + 18*a^2*b*B + 5*b^3*B)*x)/16 + ((15*a^2*A*b + 4*A*b^3 + 5*a^3*B + 12*a*b^2*B)*Sin[c +
d*x])/(5*d) + ((8*a^3*A + 18*a*A*b^2 + 18*a^2*b*B + 5*b^3*B)*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (b*(18*a*A*b
+ 14*a^2*B + 5*b^2*B)*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) + (b^2*(3*A*b + 4*a*B)*Cos[c + d*x]^4*Sin[c + d*x])/
(15*d) + (b*B*Cos[c + d*x]^3*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(6*d) - ((15*a^2*A*b + 4*A*b^3 + 5*a^3*B + 1
2*a*b^2*B)*Sin[c + d*x]^3)/(15*d)

Rule 8

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

Rule 2633

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 2635

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] && Integer
Q[2*n]

Rule 2748

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 2990

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] /; F
reeQ[{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 3023

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]

Rule 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx &=\frac {b B \cos ^3(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {1}{6} \int \cos ^2(c+d x) (a+b \cos (c+d x)) \left (3 a (2 a A+b B)+\left (5 b^2 B+6 a (2 A b+a B)\right ) \cos (c+d x)+2 b (3 A b+4 a B) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {b^2 (3 A b+4 a B) \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac {b B \cos ^3(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {1}{30} \int \cos ^2(c+d x) \left (15 a^2 (2 a A+b B)+6 \left (15 a^2 A b+4 A b^3+5 a^3 B+12 a b^2 B\right ) \cos (c+d x)+5 b \left (18 a A b+14 a^2 B+5 b^2 B\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {b \left (18 a A b+14 a^2 B+5 b^2 B\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {b^2 (3 A b+4 a B) \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac {b B \cos ^3(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {1}{120} \int \cos ^2(c+d x) \left (15 \left (8 a^3 A+18 a A b^2+18 a^2 b B+5 b^3 B\right )+24 \left (15 a^2 A b+4 A b^3+5 a^3 B+12 a b^2 B\right ) \cos (c+d x)\right ) \, dx\\ &=\frac {b \left (18 a A b+14 a^2 B+5 b^2 B\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {b^2 (3 A b+4 a B) \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac {b B \cos ^3(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {1}{5} \left (15 a^2 A b+4 A b^3+5 a^3 B+12 a b^2 B\right ) \int \cos ^3(c+d x) \, dx+\frac {1}{8} \left (8 a^3 A+18 a A b^2+18 a^2 b B+5 b^3 B\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {\left (8 a^3 A+18 a A b^2+18 a^2 b B+5 b^3 B\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {b \left (18 a A b+14 a^2 B+5 b^2 B\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {b^2 (3 A b+4 a B) \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac {b B \cos ^3(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {1}{16} \left (8 a^3 A+18 a A b^2+18 a^2 b B+5 b^3 B\right ) \int 1 \, dx-\frac {\left (15 a^2 A b+4 A b^3+5 a^3 B+12 a b^2 B\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {1}{16} \left (8 a^3 A+18 a A b^2+18 a^2 b B+5 b^3 B\right ) x+\frac {\left (15 a^2 A b+4 A b^3+5 a^3 B+12 a b^2 B\right ) \sin (c+d x)}{5 d}+\frac {\left (8 a^3 A+18 a A b^2+18 a^2 b B+5 b^3 B\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {b \left (18 a A b+14 a^2 B+5 b^2 B\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {b^2 (3 A b+4 a B) \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac {b B \cos ^3(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}-\frac {\left (15 a^2 A b+4 A b^3+5 a^3 B+12 a b^2 B\right ) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.69, size = 289, normalized size = 1.07 \[ \frac {480 a^3 A c+480 a^3 A d x+80 a^3 B \sin (3 (c+d x))+240 a^2 A b \sin (3 (c+d x))+90 a^2 b B \sin (4 (c+d x))+1080 a^2 b B c+1080 a^2 b B d x+120 \left (6 a^3 B+18 a^2 A b+15 a b^2 B+5 A b^3\right ) \sin (c+d x)+15 \left (16 a^3 A+48 a^2 b B+48 a A b^2+15 b^3 B\right ) \sin (2 (c+d x))+90 a A b^2 \sin (4 (c+d x))+1080 a A b^2 c+1080 a A b^2 d x+300 a b^2 B \sin (3 (c+d x))+36 a b^2 B \sin (5 (c+d x))+100 A b^3 \sin (3 (c+d x))+12 A b^3 \sin (5 (c+d x))+45 b^3 B \sin (4 (c+d x))+5 b^3 B \sin (6 (c+d x))+300 b^3 B c+300 b^3 B d x}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x]),x]

[Out]

(480*a^3*A*c + 1080*a*A*b^2*c + 1080*a^2*b*B*c + 300*b^3*B*c + 480*a^3*A*d*x + 1080*a*A*b^2*d*x + 1080*a^2*b*B
*d*x + 300*b^3*B*d*x + 120*(18*a^2*A*b + 5*A*b^3 + 6*a^3*B + 15*a*b^2*B)*Sin[c + d*x] + 15*(16*a^3*A + 48*a*A*
b^2 + 48*a^2*b*B + 15*b^3*B)*Sin[2*(c + d*x)] + 240*a^2*A*b*Sin[3*(c + d*x)] + 100*A*b^3*Sin[3*(c + d*x)] + 80
*a^3*B*Sin[3*(c + d*x)] + 300*a*b^2*B*Sin[3*(c + d*x)] + 90*a*A*b^2*Sin[4*(c + d*x)] + 90*a^2*b*B*Sin[4*(c + d
*x)] + 45*b^3*B*Sin[4*(c + d*x)] + 12*A*b^3*Sin[5*(c + d*x)] + 36*a*b^2*B*Sin[5*(c + d*x)] + 5*b^3*B*Sin[6*(c
+ d*x)])/(960*d)

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fricas [A]  time = 0.96, size = 211, normalized size = 0.78 \[ \frac {15 \, {\left (8 \, A a^{3} + 18 \, B a^{2} b + 18 \, A a b^{2} + 5 \, B b^{3}\right )} d x + {\left (40 \, B b^{3} \cos \left (d x + c\right )^{5} + 48 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{4} + 160 \, B a^{3} + 480 \, A a^{2} b + 384 \, B a b^{2} + 128 \, A b^{3} + 10 \, {\left (18 \, B a^{2} b + 18 \, A a b^{2} + 5 \, B b^{3}\right )} \cos \left (d x + c\right )^{3} + 16 \, {\left (5 \, B a^{3} + 15 \, A a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (8 \, A a^{3} + 18 \, B a^{2} b + 18 \, A a b^{2} + 5 \, B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

1/240*(15*(8*A*a^3 + 18*B*a^2*b + 18*A*a*b^2 + 5*B*b^3)*d*x + (40*B*b^3*cos(d*x + c)^5 + 48*(3*B*a*b^2 + A*b^3
)*cos(d*x + c)^4 + 160*B*a^3 + 480*A*a^2*b + 384*B*a*b^2 + 128*A*b^3 + 10*(18*B*a^2*b + 18*A*a*b^2 + 5*B*b^3)*
cos(d*x + c)^3 + 16*(5*B*a^3 + 15*A*a^2*b + 12*B*a*b^2 + 4*A*b^3)*cos(d*x + c)^2 + 15*(8*A*a^3 + 18*B*a^2*b +
18*A*a*b^2 + 5*B*b^3)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.47, size = 230, normalized size = 0.86 \[ \frac {B b^{3} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {1}{16} \, {\left (8 \, A a^{3} + 18 \, B a^{2} b + 18 \, A a b^{2} + 5 \, B b^{3}\right )} x + \frac {{\left (3 \, B a b^{2} + A b^{3}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {3 \, {\left (2 \, B a^{2} b + 2 \, A a b^{2} + B b^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (4 \, B a^{3} + 12 \, A a^{2} b + 15 \, B a b^{2} + 5 \, A b^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (16 \, A a^{3} + 48 \, B a^{2} b + 48 \, A a b^{2} + 15 \, B b^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (6 \, B a^{3} + 18 \, A a^{2} b + 15 \, B a b^{2} + 5 \, A b^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

1/192*B*b^3*sin(6*d*x + 6*c)/d + 1/16*(8*A*a^3 + 18*B*a^2*b + 18*A*a*b^2 + 5*B*b^3)*x + 1/80*(3*B*a*b^2 + A*b^
3)*sin(5*d*x + 5*c)/d + 3/64*(2*B*a^2*b + 2*A*a*b^2 + B*b^3)*sin(4*d*x + 4*c)/d + 1/48*(4*B*a^3 + 12*A*a^2*b +
15*B*a*b^2 + 5*A*b^3)*sin(3*d*x + 3*c)/d + 1/64*(16*A*a^3 + 48*B*a^2*b + 48*A*a*b^2 + 15*B*b^3)*sin(2*d*x + 2
*c)/d + 1/8*(6*B*a^3 + 18*A*a^2*b + 15*B*a*b^2 + 5*A*b^3)*sin(d*x + c)/d

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maple [A]  time = 0.10, size = 270, normalized size = 1.00 \[ \frac {A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {a^{3} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \,a^{2} b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 a^{2} b B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 A a \,b^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {3 B a \,b^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {A \,b^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+b^{3} B \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)),x)

[Out]

1/d*(A*a^3*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+1/3*a^3*B*(2+cos(d*x+c)^2)*sin(d*x+c)+A*a^2*b*(2+cos(d*x+
c)^2)*sin(d*x+c)+3*a^2*b*B*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+3*A*a*b^2*(1/4*(cos(d*
x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+3/5*B*a*b^2*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+1/
5*A*b^3*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+b^3*B*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+
c))*sin(d*x+c)+5/16*d*x+5/16*c))

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maxima [A]  time = 0.57, size = 266, normalized size = 0.99 \[ \frac {240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 960 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} b + 90 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} b + 90 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a b^{2} + 192 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a b^{2} + 64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A b^{3} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{3}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

1/960*(240*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 - 320*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 - 960*(sin(d*x
+ c)^3 - 3*sin(d*x + c))*A*a^2*b + 90*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^2*b + 90*(1
2*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a*b^2 + 192*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15
*sin(d*x + c))*B*a*b^2 + 64*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*b^3 - 5*(4*sin(2*d*x +
2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*B*b^3)/d

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mupad [B]  time = 1.11, size = 352, normalized size = 1.31 \[ \frac {A\,a^3\,x}{2}+\frac {5\,B\,b^3\,x}{16}+\frac {9\,A\,a\,b^2\,x}{8}+\frac {9\,B\,a^2\,b\,x}{8}+\frac {5\,A\,b^3\,\sin \left (c+d\,x\right )}{8\,d}+\frac {3\,B\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {5\,A\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {A\,b^3\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {15\,B\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{64\,d}+\frac {3\,B\,b^3\,\sin \left (4\,c+4\,d\,x\right )}{64\,d}+\frac {B\,b^3\,\sin \left (6\,c+6\,d\,x\right )}{192\,d}+\frac {3\,A\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,A\,a\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,B\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {5\,B\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{16\,d}+\frac {3\,B\,a^2\,b\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,B\,a\,b^2\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {9\,A\,a^2\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {15\,B\,a\,b^2\,\sin \left (c+d\,x\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(A + B*cos(c + d*x))*(a + b*cos(c + d*x))^3,x)

[Out]

(A*a^3*x)/2 + (5*B*b^3*x)/16 + (9*A*a*b^2*x)/8 + (9*B*a^2*b*x)/8 + (5*A*b^3*sin(c + d*x))/(8*d) + (3*B*a^3*sin
(c + d*x))/(4*d) + (A*a^3*sin(2*c + 2*d*x))/(4*d) + (5*A*b^3*sin(3*c + 3*d*x))/(48*d) + (B*a^3*sin(3*c + 3*d*x
))/(12*d) + (A*b^3*sin(5*c + 5*d*x))/(80*d) + (15*B*b^3*sin(2*c + 2*d*x))/(64*d) + (3*B*b^3*sin(4*c + 4*d*x))/
(64*d) + (B*b^3*sin(6*c + 6*d*x))/(192*d) + (3*A*a*b^2*sin(2*c + 2*d*x))/(4*d) + (A*a^2*b*sin(3*c + 3*d*x))/(4
*d) + (3*A*a*b^2*sin(4*c + 4*d*x))/(32*d) + (3*B*a^2*b*sin(2*c + 2*d*x))/(4*d) + (5*B*a*b^2*sin(3*c + 3*d*x))/
(16*d) + (3*B*a^2*b*sin(4*c + 4*d*x))/(32*d) + (3*B*a*b^2*sin(5*c + 5*d*x))/(80*d) + (9*A*a^2*b*sin(c + d*x))/
(4*d) + (15*B*a*b^2*sin(c + d*x))/(8*d)

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sympy [A]  time = 4.62, size = 721, normalized size = 2.68 \[ \begin {cases} \frac {A a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 A a^{2} b \sin ^{3}{\left (c + d x \right )}}{d} + \frac {3 A a^{2} b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {9 A a b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 A a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {9 A a b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {9 A a b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {15 A a b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {8 A b^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 A b^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {A b^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {2 B a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {9 B a^{2} b x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 B a^{2} b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {9 B a^{2} b x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {9 B a^{2} b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {15 B a^{2} b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {8 B a b^{2} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {4 B a b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 B a b^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 B b^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 B b^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 B b^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 B b^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 B b^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 B b^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 B b^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\relax (c )}\right ) \left (a + b \cos {\relax (c )}\right )^{3} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+b*cos(d*x+c))**3*(A+B*cos(d*x+c)),x)

[Out]

Piecewise((A*a**3*x*sin(c + d*x)**2/2 + A*a**3*x*cos(c + d*x)**2/2 + A*a**3*sin(c + d*x)*cos(c + d*x)/(2*d) +
2*A*a**2*b*sin(c + d*x)**3/d + 3*A*a**2*b*sin(c + d*x)*cos(c + d*x)**2/d + 9*A*a*b**2*x*sin(c + d*x)**4/8 + 9*
A*a*b**2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 9*A*a*b**2*x*cos(c + d*x)**4/8 + 9*A*a*b**2*sin(c + d*x)**3*cos
(c + d*x)/(8*d) + 15*A*a*b**2*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 8*A*b**3*sin(c + d*x)**5/(15*d) + 4*A*b**3*
sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + A*b**3*sin(c + d*x)*cos(c + d*x)**4/d + 2*B*a**3*sin(c + d*x)**3/(3*d)
+ B*a**3*sin(c + d*x)*cos(c + d*x)**2/d + 9*B*a**2*b*x*sin(c + d*x)**4/8 + 9*B*a**2*b*x*sin(c + d*x)**2*cos(c
+ d*x)**2/4 + 9*B*a**2*b*x*cos(c + d*x)**4/8 + 9*B*a**2*b*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 15*B*a**2*b*si
n(c + d*x)*cos(c + d*x)**3/(8*d) + 8*B*a*b**2*sin(c + d*x)**5/(5*d) + 4*B*a*b**2*sin(c + d*x)**3*cos(c + d*x)*
*2/d + 3*B*a*b**2*sin(c + d*x)*cos(c + d*x)**4/d + 5*B*b**3*x*sin(c + d*x)**6/16 + 15*B*b**3*x*sin(c + d*x)**4
*cos(c + d*x)**2/16 + 15*B*b**3*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 5*B*b**3*x*cos(c + d*x)**6/16 + 5*B*b**
3*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*B*b**3*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) + 11*B*b**3*sin(c + d*x
)*cos(c + d*x)**5/(16*d), Ne(d, 0)), (x*(A + B*cos(c))*(a + b*cos(c))**3*cos(c)**2, True))

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