3.965 \(\int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=375 \[ \frac {\sin (c+d x) \left (-4 a^3 C+24 a^2 b B+a b^2 (70 A+53 C)+32 b^3 B\right ) (a+b \cos (c+d x))^2}{120 b d}+\frac {\sin (c+d x) \cos (c+d x) \left (-8 a^4 C+48 a^3 b B+2 a^2 b^2 (130 A+89 C)+232 a b^3 B+15 b^4 (6 A+5 C)\right )}{240 d}+\frac {1}{16} x \left (8 a^4 (2 A+C)+32 a^3 b B+12 a^2 b^2 (4 A+3 C)+24 a b^3 B+b^4 (6 A+5 C)\right )+\frac {\sin (c+d x) \left (-4 a^5 C+24 a^4 b B+a^3 b^2 (190 A+121 C)+224 a^2 b^3 B+32 a b^4 (5 A+4 C)+32 b^5 B\right )}{60 b d}+\frac {\sin (c+d x) \left (4 a (6 b B-a C)+5 b^2 (6 A+5 C)\right ) (a+b \cos (c+d x))^3}{120 b d}+\frac {(6 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^4}{30 b d}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^5}{6 b d} \]

[Out]

1/16*(32*a^3*b*B+24*a*b^3*B+8*a^4*(2*A+C)+12*a^2*b^2*(4*A+3*C)+b^4*(6*A+5*C))*x+1/60*(24*a^4*b*B+224*a^2*b^3*B
+32*b^5*B-4*a^5*C+32*a*b^4*(5*A+4*C)+a^3*b^2*(190*A+121*C))*sin(d*x+c)/b/d+1/240*(48*a^3*b*B+232*a*b^3*B-8*a^4
*C+15*b^4*(6*A+5*C)+2*a^2*b^2*(130*A+89*C))*cos(d*x+c)*sin(d*x+c)/d+1/120*(24*a^2*b*B+32*b^3*B-4*a^3*C+a*b^2*(
70*A+53*C))*(a+b*cos(d*x+c))^2*sin(d*x+c)/b/d+1/120*(5*b^2*(6*A+5*C)+4*a*(6*B*b-C*a))*(a+b*cos(d*x+c))^3*sin(d
*x+c)/b/d+1/30*(6*B*b-C*a)*(a+b*cos(d*x+c))^4*sin(d*x+c)/b/d+1/6*C*(a+b*cos(d*x+c))^5*sin(d*x+c)/b/d

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Rubi [A]  time = 0.68, antiderivative size = 375, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3023, 2753, 2734} \[ \frac {\sin (c+d x) \left (a^3 b^2 (190 A+121 C)+224 a^2 b^3 B+24 a^4 b B-4 a^5 C+32 a b^4 (5 A+4 C)+32 b^5 B\right )}{60 b d}+\frac {\sin (c+d x) \left (24 a^2 b B-4 a^3 C+a b^2 (70 A+53 C)+32 b^3 B\right ) (a+b \cos (c+d x))^2}{120 b d}+\frac {\sin (c+d x) \cos (c+d x) \left (2 a^2 b^2 (130 A+89 C)+48 a^3 b B-8 a^4 C+232 a b^3 B+15 b^4 (6 A+5 C)\right )}{240 d}+\frac {1}{16} x \left (12 a^2 b^2 (4 A+3 C)+8 a^4 (2 A+C)+32 a^3 b B+24 a b^3 B+b^4 (6 A+5 C)\right )+\frac {\sin (c+d x) \left (4 a (6 b B-a C)+5 b^2 (6 A+5 C)\right ) (a+b \cos (c+d x))^3}{120 b d}+\frac {(6 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^4}{30 b d}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^5}{6 b d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

((32*a^3*b*B + 24*a*b^3*B + 8*a^4*(2*A + C) + 12*a^2*b^2*(4*A + 3*C) + b^4*(6*A + 5*C))*x)/16 + ((24*a^4*b*B +
 224*a^2*b^3*B + 32*b^5*B - 4*a^5*C + 32*a*b^4*(5*A + 4*C) + a^3*b^2*(190*A + 121*C))*Sin[c + d*x])/(60*b*d) +
 ((48*a^3*b*B + 232*a*b^3*B - 8*a^4*C + 15*b^4*(6*A + 5*C) + 2*a^2*b^2*(130*A + 89*C))*Cos[c + d*x]*Sin[c + d*
x])/(240*d) + ((24*a^2*b*B + 32*b^3*B - 4*a^3*C + a*b^2*(70*A + 53*C))*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(1
20*b*d) + ((5*b^2*(6*A + 5*C) + 4*a*(6*b*B - a*C))*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(120*b*d) + ((6*b*B -
a*C)*(a + b*Cos[c + d*x])^4*Sin[c + d*x])/(30*b*d) + (C*(a + b*Cos[c + d*x])^5*Sin[c + d*x])/(6*b*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 2753

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

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]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {C (a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d}+\frac {\int (a+b \cos (c+d x))^4 (b (6 A+5 C)+(6 b B-a C) \cos (c+d x)) \, dx}{6 b}\\ &=\frac {(6 b B-a C) (a+b \cos (c+d x))^4 \sin (c+d x)}{30 b d}+\frac {C (a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d}+\frac {\int (a+b \cos (c+d x))^3 \left (3 b (10 a A+8 b B+7 a C)+\left (5 b^2 (6 A+5 C)+4 a (6 b B-a C)\right ) \cos (c+d x)\right ) \, dx}{30 b}\\ &=\frac {\left (5 b^2 (6 A+5 C)+4 a (6 b B-a C)\right ) (a+b \cos (c+d x))^3 \sin (c+d x)}{120 b d}+\frac {(6 b B-a C) (a+b \cos (c+d x))^4 \sin (c+d x)}{30 b d}+\frac {C (a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d}+\frac {\int (a+b \cos (c+d x))^2 \left (3 b \left (56 a b B+8 a^2 (5 A+3 C)+5 b^2 (6 A+5 C)\right )+3 \left (24 a^2 b B+32 b^3 B-4 a^3 C+a b^2 (70 A+53 C)\right ) \cos (c+d x)\right ) \, dx}{120 b}\\ &=\frac {\left (24 a^2 b B+32 b^3 B-4 a^3 C+a b^2 (70 A+53 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{120 b d}+\frac {\left (5 b^2 (6 A+5 C)+4 a (6 b B-a C)\right ) (a+b \cos (c+d x))^3 \sin (c+d x)}{120 b d}+\frac {(6 b B-a C) (a+b \cos (c+d x))^4 \sin (c+d x)}{30 b d}+\frac {C (a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d}+\frac {\int (a+b \cos (c+d x)) \left (3 b \left (216 a^2 b B+64 b^3 B+8 a^3 (15 A+8 C)+a b^2 (230 A+181 C)\right )+3 \left (48 a^3 b B+232 a b^3 B-8 a^4 C+15 b^4 (6 A+5 C)+2 a^2 b^2 (130 A+89 C)\right ) \cos (c+d x)\right ) \, dx}{360 b}\\ &=\frac {1}{16} \left (32 a^3 b B+24 a b^3 B+8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) x+\frac {\left (24 a^4 b B+224 a^2 b^3 B+32 b^5 B-4 a^5 C+32 a b^4 (5 A+4 C)+a^3 b^2 (190 A+121 C)\right ) \sin (c+d x)}{60 b d}+\frac {\left (48 a^3 b B+232 a b^3 B-8 a^4 C+15 b^4 (6 A+5 C)+2 a^2 b^2 (130 A+89 C)\right ) \cos (c+d x) \sin (c+d x)}{240 d}+\frac {\left (24 a^2 b B+32 b^3 B-4 a^3 C+a b^2 (70 A+53 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{120 b d}+\frac {\left (5 b^2 (6 A+5 C)+4 a (6 b B-a C)\right ) (a+b \cos (c+d x))^3 \sin (c+d x)}{120 b d}+\frac {(6 b B-a C) (a+b \cos (c+d x))^4 \sin (c+d x)}{30 b d}+\frac {C (a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d}\\ \end {align*}

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Mathematica [A]  time = 1.39, size = 432, normalized size = 1.15 \[ \frac {960 a^4 A c+960 a^4 A d x+480 a^4 c C+480 a^4 C d x+1920 a^3 b B c+1920 a^3 b B d x+320 a^3 b C \sin (3 (c+d x))+2880 a^2 A b^2 c+2880 a^2 A b^2 d x+480 a^2 b^2 B \sin (3 (c+d x))+180 a^2 b^2 C \sin (4 (c+d x))+2160 a^2 b^2 c C+2160 a^2 b^2 C d x+120 \sin (c+d x) \left (8 a^4 B+8 a^3 b (4 A+3 C)+36 a^2 b^2 B+4 a b^3 (6 A+5 C)+5 b^4 B\right )+15 \sin (2 (c+d x)) \left (16 a^4 C+64 a^3 b B+96 a^2 b^2 (A+C)+64 a b^3 B+b^4 (16 A+15 C)\right )+320 a A b^3 \sin (3 (c+d x))+120 a b^3 B \sin (4 (c+d x))+1440 a b^3 B c+1440 a b^3 B d x+400 a b^3 C \sin (3 (c+d x))+48 a b^3 C \sin (5 (c+d x))+30 A b^4 \sin (4 (c+d x))+360 A b^4 c+360 A b^4 d x+100 b^4 B \sin (3 (c+d x))+12 b^4 B \sin (5 (c+d x))+45 b^4 C \sin (4 (c+d x))+5 b^4 C \sin (6 (c+d x))+300 b^4 c C+300 b^4 C d x}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(960*a^4*A*c + 2880*a^2*A*b^2*c + 360*A*b^4*c + 1920*a^3*b*B*c + 1440*a*b^3*B*c + 480*a^4*c*C + 2160*a^2*b^2*c
*C + 300*b^4*c*C + 960*a^4*A*d*x + 2880*a^2*A*b^2*d*x + 360*A*b^4*d*x + 1920*a^3*b*B*d*x + 1440*a*b^3*B*d*x +
480*a^4*C*d*x + 2160*a^2*b^2*C*d*x + 300*b^4*C*d*x + 120*(8*a^4*B + 36*a^2*b^2*B + 5*b^4*B + 8*a^3*b*(4*A + 3*
C) + 4*a*b^3*(6*A + 5*C))*Sin[c + d*x] + 15*(64*a^3*b*B + 64*a*b^3*B + 16*a^4*C + 96*a^2*b^2*(A + C) + b^4*(16
*A + 15*C))*Sin[2*(c + d*x)] + 320*a*A*b^3*Sin[3*(c + d*x)] + 480*a^2*b^2*B*Sin[3*(c + d*x)] + 100*b^4*B*Sin[3
*(c + d*x)] + 320*a^3*b*C*Sin[3*(c + d*x)] + 400*a*b^3*C*Sin[3*(c + d*x)] + 30*A*b^4*Sin[4*(c + d*x)] + 120*a*
b^3*B*Sin[4*(c + d*x)] + 180*a^2*b^2*C*Sin[4*(c + d*x)] + 45*b^4*C*Sin[4*(c + d*x)] + 12*b^4*B*Sin[5*(c + d*x)
] + 48*a*b^3*C*Sin[5*(c + d*x)] + 5*b^4*C*Sin[6*(c + d*x)])/(960*d)

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fricas [A]  time = 0.49, size = 292, normalized size = 0.78 \[ \frac {15 \, {\left (8 \, {\left (2 \, A + C\right )} a^{4} + 32 \, B a^{3} b + 12 \, {\left (4 \, A + 3 \, C\right )} a^{2} b^{2} + 24 \, B a b^{3} + {\left (6 \, A + 5 \, C\right )} b^{4}\right )} d x + {\left (40 \, C b^{4} \cos \left (d x + c\right )^{5} + 240 \, B a^{4} + 320 \, {\left (3 \, A + 2 \, C\right )} a^{3} b + 960 \, B a^{2} b^{2} + 128 \, {\left (5 \, A + 4 \, C\right )} a b^{3} + 128 \, B b^{4} + 48 \, {\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{4} + 10 \, {\left (36 \, C a^{2} b^{2} + 24 \, B a b^{3} + {\left (6 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{3} + 32 \, {\left (10 \, C a^{3} b + 15 \, B a^{2} b^{2} + 2 \, {\left (5 \, A + 4 \, C\right )} a b^{3} + 2 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (8 \, C a^{4} + 32 \, B a^{3} b + 12 \, {\left (4 \, A + 3 \, C\right )} a^{2} b^{2} + 24 \, B a b^{3} + {\left (6 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(15*(8*(2*A + C)*a^4 + 32*B*a^3*b + 12*(4*A + 3*C)*a^2*b^2 + 24*B*a*b^3 + (6*A + 5*C)*b^4)*d*x + (40*C*b
^4*cos(d*x + c)^5 + 240*B*a^4 + 320*(3*A + 2*C)*a^3*b + 960*B*a^2*b^2 + 128*(5*A + 4*C)*a*b^3 + 128*B*b^4 + 48
*(4*C*a*b^3 + B*b^4)*cos(d*x + c)^4 + 10*(36*C*a^2*b^2 + 24*B*a*b^3 + (6*A + 5*C)*b^4)*cos(d*x + c)^3 + 32*(10
*C*a^3*b + 15*B*a^2*b^2 + 2*(5*A + 4*C)*a*b^3 + 2*B*b^4)*cos(d*x + c)^2 + 15*(8*C*a^4 + 32*B*a^3*b + 12*(4*A +
 3*C)*a^2*b^2 + 24*B*a*b^3 + (6*A + 5*C)*b^4)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.36, size = 326, normalized size = 0.87 \[ \frac {C b^{4} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {1}{16} \, {\left (16 \, A a^{4} + 8 \, C a^{4} + 32 \, B a^{3} b + 48 \, A a^{2} b^{2} + 36 \, C a^{2} b^{2} + 24 \, B a b^{3} + 6 \, A b^{4} + 5 \, C b^{4}\right )} x + \frac {{\left (4 \, C a b^{3} + B b^{4}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (12 \, C a^{2} b^{2} + 8 \, B a b^{3} + 2 \, A b^{4} + 3 \, C b^{4}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (16 \, C a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 20 \, C a b^{3} + 5 \, B b^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (16 \, C a^{4} + 64 \, B a^{3} b + 96 \, A a^{2} b^{2} + 96 \, C a^{2} b^{2} + 64 \, B a b^{3} + 16 \, A b^{4} + 15 \, C b^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (8 \, B a^{4} + 32 \, A a^{3} b + 24 \, C a^{3} b + 36 \, B a^{2} b^{2} + 24 \, A a b^{3} + 20 \, C a b^{3} + 5 \, B b^{4}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

1/192*C*b^4*sin(6*d*x + 6*c)/d + 1/16*(16*A*a^4 + 8*C*a^4 + 32*B*a^3*b + 48*A*a^2*b^2 + 36*C*a^2*b^2 + 24*B*a*
b^3 + 6*A*b^4 + 5*C*b^4)*x + 1/80*(4*C*a*b^3 + B*b^4)*sin(5*d*x + 5*c)/d + 1/64*(12*C*a^2*b^2 + 8*B*a*b^3 + 2*
A*b^4 + 3*C*b^4)*sin(4*d*x + 4*c)/d + 1/48*(16*C*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3 + 20*C*a*b^3 + 5*B*b^4)*sin
(3*d*x + 3*c)/d + 1/64*(16*C*a^4 + 64*B*a^3*b + 96*A*a^2*b^2 + 96*C*a^2*b^2 + 64*B*a*b^3 + 16*A*b^4 + 15*C*b^4
)*sin(2*d*x + 2*c)/d + 1/8*(8*B*a^4 + 32*A*a^3*b + 24*C*a^3*b + 36*B*a^2*b^2 + 24*A*a*b^3 + 20*C*a*b^3 + 5*B*b
^4)*sin(d*x + c)/d

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maple [A]  time = 0.35, size = 431, normalized size = 1.15 \[ \frac {C \,b^{4} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {B \,b^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {4 C a \,b^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+A \,b^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+4 B a \,b^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+6 C \,a^{2} b^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 A a \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 B \,a^{2} b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\frac {4 a^{3} b C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+6 A \,a^{2} b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 B \,a^{3} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 A \,a^{3} b \sin \left (d x +c \right )+a^{4} B \sin \left (d x +c \right )+A \,a^{4} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

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

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maxima [A]  time = 0.34, size = 415, normalized size = 1.11 \[ \frac {960 \, {\left (d x + c\right )} A a^{4} + 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 960 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} b - 1280 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} b + 1440 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b^{2} - 1920 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} b^{2} + 180 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b^{2} - 1280 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b^{3} + 120 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{3} + 256 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a b^{3} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{4} + 64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B b^{4} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{4} + 960 \, B a^{4} \sin \left (d x + c\right ) + 3840 \, A a^{3} b \sin \left (d x + c\right )}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

1/960*(960*(d*x + c)*A*a^4 + 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 + 960*(2*d*x + 2*c + sin(2*d*x + 2*c))
*B*a^3*b - 1280*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3*b + 1440*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2*b^2 -
1920*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^2*b^2 + 180*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))
*C*a^2*b^2 - 1280*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a*b^3 + 120*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*
d*x + 2*c))*B*a*b^3 + 256*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*a*b^3 + 30*(12*d*x + 12*c
 + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*b^4 + 64*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*
B*b^4 - 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^4 + 960*B*a^4*
sin(d*x + c) + 3840*A*a^3*b*sin(d*x + c))/d

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mupad [B]  time = 4.51, size = 534, normalized size = 1.42 \[ A\,a^4\,x+\frac {3\,A\,b^4\,x}{8}+\frac {C\,a^4\,x}{2}+\frac {5\,C\,b^4\,x}{16}+\frac {3\,B\,a\,b^3\,x}{2}+2\,B\,a^3\,b\,x+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {5\,B\,b^4\,\sin \left (c+d\,x\right )}{8\,d}+3\,A\,a^2\,b^2\,x+\frac {9\,C\,a^2\,b^2\,x}{4}+\frac {A\,b^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,b^4\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {C\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {5\,B\,b^4\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {B\,b^4\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {15\,C\,b^4\,\sin \left (2\,c+2\,d\,x\right )}{64\,d}+\frac {3\,C\,b^4\,\sin \left (4\,c+4\,d\,x\right )}{64\,d}+\frac {C\,b^4\,\sin \left (6\,c+6\,d\,x\right )}{192\,d}+\frac {A\,a\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{3\,d}+\frac {B\,a\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {B\,a^3\,b\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {B\,a\,b^3\,\sin \left (4\,c+4\,d\,x\right )}{8\,d}+\frac {9\,B\,a^2\,b^2\,\sin \left (c+d\,x\right )}{2\,d}+\frac {5\,C\,a\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,a^3\,b\,\sin \left (3\,c+3\,d\,x\right )}{3\,d}+\frac {C\,a\,b^3\,\sin \left (5\,c+5\,d\,x\right )}{20\,d}+\frac {3\,A\,a^2\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {B\,a^2\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{2\,d}+\frac {3\,C\,a^2\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {3\,C\,a^2\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{16\,d}+\frac {3\,A\,a\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {4\,A\,a^3\,b\,\sin \left (c+d\,x\right )}{d}+\frac {5\,C\,a\,b^3\,\sin \left (c+d\,x\right )}{2\,d}+\frac {3\,C\,a^3\,b\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))^4*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)

[Out]

A*a^4*x + (3*A*b^4*x)/8 + (C*a^4*x)/2 + (5*C*b^4*x)/16 + (3*B*a*b^3*x)/2 + 2*B*a^3*b*x + (B*a^4*sin(c + d*x))/
d + (5*B*b^4*sin(c + d*x))/(8*d) + 3*A*a^2*b^2*x + (9*C*a^2*b^2*x)/4 + (A*b^4*sin(2*c + 2*d*x))/(4*d) + (A*b^4
*sin(4*c + 4*d*x))/(32*d) + (C*a^4*sin(2*c + 2*d*x))/(4*d) + (5*B*b^4*sin(3*c + 3*d*x))/(48*d) + (B*b^4*sin(5*
c + 5*d*x))/(80*d) + (15*C*b^4*sin(2*c + 2*d*x))/(64*d) + (3*C*b^4*sin(4*c + 4*d*x))/(64*d) + (C*b^4*sin(6*c +
 6*d*x))/(192*d) + (A*a*b^3*sin(3*c + 3*d*x))/(3*d) + (B*a*b^3*sin(2*c + 2*d*x))/d + (B*a^3*b*sin(2*c + 2*d*x)
)/d + (B*a*b^3*sin(4*c + 4*d*x))/(8*d) + (9*B*a^2*b^2*sin(c + d*x))/(2*d) + (5*C*a*b^3*sin(3*c + 3*d*x))/(12*d
) + (C*a^3*b*sin(3*c + 3*d*x))/(3*d) + (C*a*b^3*sin(5*c + 5*d*x))/(20*d) + (3*A*a^2*b^2*sin(2*c + 2*d*x))/(2*d
) + (B*a^2*b^2*sin(3*c + 3*d*x))/(2*d) + (3*C*a^2*b^2*sin(2*c + 2*d*x))/(2*d) + (3*C*a^2*b^2*sin(4*c + 4*d*x))
/(16*d) + (3*A*a*b^3*sin(c + d*x))/d + (4*A*a^3*b*sin(c + d*x))/d + (5*C*a*b^3*sin(c + d*x))/(2*d) + (3*C*a^3*
b*sin(c + d*x))/d

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sympy [A]  time = 6.72, size = 1066, normalized size = 2.84 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**4*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Piecewise((A*a**4*x + 4*A*a**3*b*sin(c + d*x)/d + 3*A*a**2*b**2*x*sin(c + d*x)**2 + 3*A*a**2*b**2*x*cos(c + d*
x)**2 + 3*A*a**2*b**2*sin(c + d*x)*cos(c + d*x)/d + 8*A*a*b**3*sin(c + d*x)**3/(3*d) + 4*A*a*b**3*sin(c + d*x)
*cos(c + d*x)**2/d + 3*A*b**4*x*sin(c + d*x)**4/8 + 3*A*b**4*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*A*b**4*x*
cos(c + d*x)**4/8 + 3*A*b**4*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 5*A*b**4*sin(c + d*x)*cos(c + d*x)**3/(8*d)
+ B*a**4*sin(c + d*x)/d + 2*B*a**3*b*x*sin(c + d*x)**2 + 2*B*a**3*b*x*cos(c + d*x)**2 + 2*B*a**3*b*sin(c + d*x
)*cos(c + d*x)/d + 4*B*a**2*b**2*sin(c + d*x)**3/d + 6*B*a**2*b**2*sin(c + d*x)*cos(c + d*x)**2/d + 3*B*a*b**3
*x*sin(c + d*x)**4/2 + 3*B*a*b**3*x*sin(c + d*x)**2*cos(c + d*x)**2 + 3*B*a*b**3*x*cos(c + d*x)**4/2 + 3*B*a*b
**3*sin(c + d*x)**3*cos(c + d*x)/(2*d) + 5*B*a*b**3*sin(c + d*x)*cos(c + d*x)**3/(2*d) + 8*B*b**4*sin(c + d*x)
**5/(15*d) + 4*B*b**4*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + B*b**4*sin(c + d*x)*cos(c + d*x)**4/d + C*a**4*x
*sin(c + d*x)**2/2 + C*a**4*x*cos(c + d*x)**2/2 + C*a**4*sin(c + d*x)*cos(c + d*x)/(2*d) + 8*C*a**3*b*sin(c +
d*x)**3/(3*d) + 4*C*a**3*b*sin(c + d*x)*cos(c + d*x)**2/d + 9*C*a**2*b**2*x*sin(c + d*x)**4/4 + 9*C*a**2*b**2*
x*sin(c + d*x)**2*cos(c + d*x)**2/2 + 9*C*a**2*b**2*x*cos(c + d*x)**4/4 + 9*C*a**2*b**2*sin(c + d*x)**3*cos(c
+ d*x)/(4*d) + 15*C*a**2*b**2*sin(c + d*x)*cos(c + d*x)**3/(4*d) + 32*C*a*b**3*sin(c + d*x)**5/(15*d) + 16*C*a
*b**3*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 4*C*a*b**3*sin(c + d*x)*cos(c + d*x)**4/d + 5*C*b**4*x*sin(c + d
*x)**6/16 + 15*C*b**4*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*C*b**4*x*sin(c + d*x)**2*cos(c + d*x)**4/16 +
5*C*b**4*x*cos(c + d*x)**6/16 + 5*C*b**4*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*C*b**4*sin(c + d*x)**3*cos(c
+ d*x)**3/(6*d) + 11*C*b**4*sin(c + d*x)*cos(c + d*x)**5/(16*d), Ne(d, 0)), (x*(a + b*cos(c))**4*(A + B*cos(c)
 + C*cos(c)**2), True))

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