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

Optimal. Leaf size=327 \[ \frac {b \sin (c+d x) \cos ^3(c+d x) \left (6 a^2 C+42 a b B+30 A b^2+25 b^2 C\right )}{120 d}+\frac {\sin (c+d x) \left (5 a^3 (3 A+2 C)+30 a^2 b B+6 a b^2 (5 A+4 C)+8 b^3 B\right )}{15 d}+\frac {\sin (c+d x) \cos ^2(c+d x) \left (a^3 C+12 a^2 b B+3 a b^2 (5 A+4 C)+4 b^3 B\right )}{15 d}+\frac {\sin (c+d x) \cos (c+d x) \left (8 a^3 B+6 a^2 b (4 A+3 C)+18 a b^2 B+b^3 (6 A+5 C)\right )}{16 d}+\frac {1}{16} x \left (8 a^3 B+6 a^2 b (4 A+3 C)+18 a b^2 B+b^3 (6 A+5 C)\right )+\frac {(a C+2 b B) \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{10 d}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^3}{6 d} \]

[Out]

1/16*(8*a^3*B+18*a*b^2*B+6*a^2*b*(4*A+3*C)+b^3*(6*A+5*C))*x+1/15*(30*a^2*b*B+8*b^3*B+5*a^3*(3*A+2*C)+6*a*b^2*(
5*A+4*C))*sin(d*x+c)/d+1/16*(8*a^3*B+18*a*b^2*B+6*a^2*b*(4*A+3*C)+b^3*(6*A+5*C))*cos(d*x+c)*sin(d*x+c)/d+1/15*
(12*a^2*b*B+4*b^3*B+a^3*C+3*a*b^2*(5*A+4*C))*cos(d*x+c)^2*sin(d*x+c)/d+1/120*b*(30*A*b^2+42*B*a*b+6*C*a^2+25*C
*b^2)*cos(d*x+c)^3*sin(d*x+c)/d+1/10*(2*B*b+C*a)*cos(d*x+c)^2*(a+b*cos(d*x+c))^2*sin(d*x+c)/d+1/6*C*cos(d*x+c)
^2*(a+b*cos(d*x+c))^3*sin(d*x+c)/d

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Rubi [A]  time = 0.61, antiderivative size = 327, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3049, 3033, 3023, 2734} \[ \frac {\sin (c+d x) \left (5 a^3 (3 A+2 C)+30 a^2 b B+6 a b^2 (5 A+4 C)+8 b^3 B\right )}{15 d}+\frac {b \sin (c+d x) \cos ^3(c+d x) \left (6 a^2 C+42 a b B+30 A b^2+25 b^2 C\right )}{120 d}+\frac {\sin (c+d x) \cos ^2(c+d x) \left (12 a^2 b B+a^3 C+3 a b^2 (5 A+4 C)+4 b^3 B\right )}{15 d}+\frac {\sin (c+d x) \cos (c+d x) \left (6 a^2 b (4 A+3 C)+8 a^3 B+18 a b^2 B+b^3 (6 A+5 C)\right )}{16 d}+\frac {1}{16} x \left (6 a^2 b (4 A+3 C)+8 a^3 B+18 a b^2 B+b^3 (6 A+5 C)\right )+\frac {(a C+2 b B) \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{10 d}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^3}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^3*B + 18*a*b^2*B + 6*a^2*b*(4*A + 3*C) + b^3*(6*A + 5*C))*x)/16 + ((30*a^2*b*B + 8*b^3*B + 5*a^3*(3*A +
2*C) + 6*a*b^2*(5*A + 4*C))*Sin[c + d*x])/(15*d) + ((8*a^3*B + 18*a*b^2*B + 6*a^2*b*(4*A + 3*C) + b^3*(6*A + 5
*C))*Cos[c + d*x]*Sin[c + d*x])/(16*d) + ((12*a^2*b*B + 4*b^3*B + a^3*C + 3*a*b^2*(5*A + 4*C))*Cos[c + d*x]^2*
Sin[c + d*x])/(15*d) + (b*(30*A*b^2 + 42*a*b*B + 6*a^2*C + 25*b^2*C)*Cos[c + d*x]^3*Sin[c + d*x])/(120*d) + ((
2*b*B + a*C)*Cos[c + d*x]^2*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(10*d) + (C*Cos[c + d*x]^2*(a + b*Cos[c + d*x
])^3*Sin[c + d*x])/(6*d)

Rule 2734

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 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]

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rubi steps

\begin {align*} \int \cos (c+d x) (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {C \cos ^2(c+d x) (a+b \cos (c+d x))^3 \sin (c+d x)}{6 d}+\frac {1}{6} \int \cos (c+d x) (a+b \cos (c+d x))^2 \left (2 a (3 A+C)+(6 A b+6 a B+5 b C) \cos (c+d x)+3 (2 b B+a C) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {(2 b B+a C) \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{10 d}+\frac {C \cos ^2(c+d x) (a+b \cos (c+d x))^3 \sin (c+d x)}{6 d}+\frac {1}{30} \int \cos (c+d x) (a+b \cos (c+d x)) \left (2 a (15 a A+6 b B+8 a C)+\left (30 a^2 B+24 b^2 B+a b (60 A+47 C)\right ) \cos (c+d x)+\left (30 A b^2+42 a b B+6 a^2 C+25 b^2 C\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {b \left (30 A b^2+42 a b B+6 a^2 C+25 b^2 C\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac {(2 b B+a C) \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{10 d}+\frac {C \cos ^2(c+d x) (a+b \cos (c+d x))^3 \sin (c+d x)}{6 d}+\frac {1}{120} \int \cos (c+d x) \left (8 a^2 (15 a A+6 b B+8 a C)+15 \left (8 a^3 B+18 a b^2 B+6 a^2 b (4 A+3 C)+b^3 (6 A+5 C)\right ) \cos (c+d x)+24 \left (12 a^2 b B+4 b^3 B+a^3 C+3 a b^2 (5 A+4 C)\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {\left (12 a^2 b B+4 b^3 B+a^3 C+3 a b^2 (5 A+4 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{15 d}+\frac {b \left (30 A b^2+42 a b B+6 a^2 C+25 b^2 C\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac {(2 b B+a C) \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{10 d}+\frac {C \cos ^2(c+d x) (a+b \cos (c+d x))^3 \sin (c+d x)}{6 d}+\frac {1}{360} \int \cos (c+d x) \left (24 \left (30 a^2 b B+8 b^3 B+5 a^3 (3 A+2 C)+6 a b^2 (5 A+4 C)\right )+45 \left (8 a^3 B+18 a b^2 B+6 a^2 b (4 A+3 C)+b^3 (6 A+5 C)\right ) \cos (c+d x)\right ) \, dx\\ &=\frac {1}{16} \left (8 a^3 B+18 a b^2 B+6 a^2 b (4 A+3 C)+b^3 (6 A+5 C)\right ) x+\frac {\left (30 a^2 b B+8 b^3 B+5 a^3 (3 A+2 C)+6 a b^2 (5 A+4 C)\right ) \sin (c+d x)}{15 d}+\frac {\left (8 a^3 B+18 a b^2 B+6 a^2 b (4 A+3 C)+b^3 (6 A+5 C)\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {\left (12 a^2 b B+4 b^3 B+a^3 C+3 a b^2 (5 A+4 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{15 d}+\frac {b \left (30 A b^2+42 a b B+6 a^2 C+25 b^2 C\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac {(2 b B+a C) \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{10 d}+\frac {C \cos ^2(c+d x) (a+b \cos (c+d x))^3 \sin (c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]  time = 1.22, size = 368, normalized size = 1.13 \[ \frac {480 a^3 B c+480 a^3 B d x+80 a^3 C \sin (3 (c+d x))+1440 a^2 A b c+1440 a^2 A b d x+240 a^2 b B \sin (3 (c+d x))+90 a^2 b C \sin (4 (c+d x))+1080 a^2 b c C+1080 a^2 b C d x+120 \sin (c+d x) \left (a^3 (8 A+6 C)+18 a^2 b B+3 a b^2 (6 A+5 C)+5 b^3 B\right )+15 \sin (2 (c+d x)) \left (16 a^3 B+48 a^2 b (A+C)+48 a b^2 B+b^3 (16 A+15 C)\right )+240 a A b^2 \sin (3 (c+d x))+90 a b^2 B \sin (4 (c+d x))+1080 a b^2 B c+1080 a b^2 B d x+300 a b^2 C \sin (3 (c+d x))+36 a b^2 C \sin (5 (c+d x))+30 A b^3 \sin (4 (c+d x))+360 A b^3 c+360 A b^3 d x+100 b^3 B \sin (3 (c+d x))+12 b^3 B \sin (5 (c+d x))+45 b^3 C \sin (4 (c+d x))+5 b^3 C \sin (6 (c+d x))+300 b^3 c C+300 b^3 C d x}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1440*a^2*A*b*c + 360*A*b^3*c + 480*a^3*B*c + 1080*a*b^2*B*c + 1080*a^2*b*c*C + 300*b^3*c*C + 1440*a^2*A*b*d*x
+ 360*A*b^3*d*x + 480*a^3*B*d*x + 1080*a*b^2*B*d*x + 1080*a^2*b*C*d*x + 300*b^3*C*d*x + 120*(18*a^2*b*B + 5*b
^3*B + 3*a*b^2*(6*A + 5*C) + a^3*(8*A + 6*C))*Sin[c + d*x] + 15*(16*a^3*B + 48*a*b^2*B + 48*a^2*b*(A + C) + b^
3*(16*A + 15*C))*Sin[2*(c + d*x)] + 240*a*A*b^2*Sin[3*(c + d*x)] + 240*a^2*b*B*Sin[3*(c + d*x)] + 100*b^3*B*Si
n[3*(c + d*x)] + 80*a^3*C*Sin[3*(c + d*x)] + 300*a*b^2*C*Sin[3*(c + d*x)] + 30*A*b^3*Sin[4*(c + d*x)] + 90*a*b
^2*B*Sin[4*(c + d*x)] + 90*a^2*b*C*Sin[4*(c + d*x)] + 45*b^3*C*Sin[4*(c + d*x)] + 12*b^3*B*Sin[5*(c + d*x)] +
36*a*b^2*C*Sin[5*(c + d*x)] + 5*b^3*C*Sin[6*(c + d*x)])/(960*d)

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fricas [A]  time = 0.45, size = 256, normalized size = 0.78 \[ \frac {15 \, {\left (8 \, B a^{3} + 6 \, {\left (4 \, A + 3 \, C\right )} a^{2} b + 18 \, B a b^{2} + {\left (6 \, A + 5 \, C\right )} b^{3}\right )} d x + {\left (40 \, C b^{3} \cos \left (d x + c\right )^{5} + 48 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{4} + 80 \, {\left (3 \, A + 2 \, C\right )} a^{3} + 480 \, B a^{2} b + 96 \, {\left (5 \, A + 4 \, C\right )} a b^{2} + 128 \, B b^{3} + 10 \, {\left (18 \, C a^{2} b + 18 \, B a b^{2} + {\left (6 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{3} + 16 \, {\left (5 \, C a^{3} + 15 \, B a^{2} b + 3 \, {\left (5 \, A + 4 \, C\right )} a b^{2} + 4 \, B b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (8 \, B a^{3} + 6 \, {\left (4 \, A + 3 \, C\right )} a^{2} b + 18 \, B a b^{2} + {\left (6 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(8*B*a^3 + 6*(4*A + 3*C)*a^2*b + 18*B*a*b^2 + (6*A + 5*C)*b^3)*d*x + (40*C*b^3*cos(d*x + c)^5 + 48*(
3*C*a*b^2 + B*b^3)*cos(d*x + c)^4 + 80*(3*A + 2*C)*a^3 + 480*B*a^2*b + 96*(5*A + 4*C)*a*b^2 + 128*B*b^3 + 10*(
18*C*a^2*b + 18*B*a*b^2 + (6*A + 5*C)*b^3)*cos(d*x + c)^3 + 16*(5*C*a^3 + 15*B*a^2*b + 3*(5*A + 4*C)*a*b^2 + 4
*B*b^3)*cos(d*x + c)^2 + 15*(8*B*a^3 + 6*(4*A + 3*C)*a^2*b + 18*B*a*b^2 + (6*A + 5*C)*b^3)*cos(d*x + c))*sin(d
*x + c))/d

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giac [A]  time = 0.71, size = 283, normalized size = 0.87 \[ \frac {C b^{3} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {1}{16} \, {\left (8 \, B a^{3} + 24 \, A a^{2} b + 18 \, C a^{2} b + 18 \, B a b^{2} + 6 \, A b^{3} + 5 \, C b^{3}\right )} x + \frac {{\left (3 \, C a b^{2} + B b^{3}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (6 \, C a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3} + 3 \, C b^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (4 \, C a^{3} + 12 \, B a^{2} b + 12 \, A a b^{2} + 15 \, C a b^{2} + 5 \, B b^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (16 \, B a^{3} + 48 \, A a^{2} b + 48 \, C a^{2} b + 48 \, B a b^{2} + 16 \, A b^{3} + 15 \, C b^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (8 \, A a^{3} + 6 \, C a^{3} + 18 \, B a^{2} b + 18 \, A a b^{2} + 15 \, C a b^{2} + 5 \, B b^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*C*b^3*sin(6*d*x + 6*c)/d + 1/16*(8*B*a^3 + 24*A*a^2*b + 18*C*a^2*b + 18*B*a*b^2 + 6*A*b^3 + 5*C*b^3)*x +
1/80*(3*C*a*b^2 + B*b^3)*sin(5*d*x + 5*c)/d + 1/64*(6*C*a^2*b + 6*B*a*b^2 + 2*A*b^3 + 3*C*b^3)*sin(4*d*x + 4*
c)/d + 1/48*(4*C*a^3 + 12*B*a^2*b + 12*A*a*b^2 + 15*C*a*b^2 + 5*B*b^3)*sin(3*d*x + 3*c)/d + 1/64*(16*B*a^3 + 4
8*A*a^2*b + 48*C*a^2*b + 48*B*a*b^2 + 16*A*b^3 + 15*C*b^3)*sin(2*d*x + 2*c)/d + 1/8*(8*A*a^3 + 6*C*a^3 + 18*B*
a^2*b + 18*A*a*b^2 + 15*C*a*b^2 + 5*B*b^3)*sin(d*x + c)/d

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maple [A]  time = 0.34, size = 370, normalized size = 1.13 \[ \frac {A \,a^{3} \sin \left (d x +c \right )+a^{3} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 A \,a^{2} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{2} b B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 C \,a^{2} b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+A a \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 B a \,b^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {3 C a \,b^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+A \,b^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {b^{3} B \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+b^{3} C \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(A*a^3*sin(d*x+c)+a^3*B*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+1/3*C*a^3*(2+cos(d*x+c)^2)*sin(d*x+c)+3*
A*a^2*b*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+a^2*b*B*(2+cos(d*x+c)^2)*sin(d*x+c)+3*C*a^2*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)+A*a*b^2*(2+cos(d*x+c)^2)*sin(d*x+c)+3*B*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*C*a*b^2*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+A*b^3*(
1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+1/5*b^3*B*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(
d*x+c)+b^3*C*(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 = 0.34, size = 360, normalized size = 1.10 \[ \frac {240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 720 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b - 960 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} b + 90 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b - 960 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b^{2} + 90 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{2} + 192 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a b^{2} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{3} + 64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B b^{3} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{3} + 960 \, A a^{3} \sin \left (d x + c\right )}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(240*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 - 320*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3 + 720*(2*d*x +
2*c + sin(2*d*x + 2*c))*A*a^2*b - 960*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^2*b + 90*(12*d*x + 12*c + sin(4*d
*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^2*b - 960*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a*b^2 + 90*(12*d*x + 12*c +
sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a*b^2 + 192*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*
C*a*b^2 + 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*b^3 + 64*(3*sin(d*x + c)^5 - 10*sin(d*x
+ c)^3 + 15*sin(d*x + c))*B*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))*C*b^3 + 960*A*a^3*sin(d*x + c))/d

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mupad [B]  time = 3.79, size = 471, normalized size = 1.44 \[ \frac {3\,A\,b^3\,x}{8}+\frac {B\,a^3\,x}{2}+\frac {5\,C\,b^3\,x}{16}+\frac {3\,A\,a^2\,b\,x}{2}+\frac {9\,B\,a\,b^2\,x}{8}+\frac {9\,C\,a^2\,b\,x}{8}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,B\,b^3\,\sin \left (c+d\,x\right )}{8\,d}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,b^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {5\,B\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {C\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,b^3\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {15\,C\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{64\,d}+\frac {3\,C\,b^3\,\sin \left (4\,c+4\,d\,x\right )}{64\,d}+\frac {C\,b^3\,\sin \left (6\,c+6\,d\,x\right )}{192\,d}+\frac {3\,A\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,B\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,B\,a\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,C\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {5\,C\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{16\,d}+\frac {3\,C\,a^2\,b\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,C\,a\,b^2\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {9\,A\,a\,b^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {9\,B\,a^2\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {15\,C\,a\,b^2\,\sin \left (c+d\,x\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 6.08, size = 966, normalized size = 2.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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