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

Optimal. Leaf size=277 \[ \frac {\sin (c+d x) \cos (c+d x) \left (-6 a^3 C+30 a^2 b B+a b^2 (100 A+71 C)+45 b^3 B\right )}{120 d}+\frac {1}{8} x \left (4 a^3 (2 A+C)+12 a^2 b B+3 a b^2 (4 A+3 C)+3 b^3 B\right )+\frac {\sin (c+d x) \left (-3 a^4 C+15 a^3 b B+4 a^2 b^2 (20 A+13 C)+60 a b^3 B+4 b^4 (5 A+4 C)\right )}{30 b d}+\frac {\sin (c+d x) \left (3 a (5 b B-a C)+4 b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^2}{60 b d}+\frac {(5 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^3}{20 b d}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 b d} \]

[Out]

1/8*(12*a^2*b*B+3*b^3*B+4*a^3*(2*A+C)+3*a*b^2*(4*A+3*C))*x+1/30*(15*a^3*b*B+60*a*b^3*B-3*a^4*C+4*b^4*(5*A+4*C)
+4*a^2*b^2*(20*A+13*C))*sin(d*x+c)/b/d+1/120*(30*a^2*b*B+45*b^3*B-6*a^3*C+a*b^2*(100*A+71*C))*cos(d*x+c)*sin(d
*x+c)/d+1/60*(4*b^2*(5*A+4*C)+3*a*(5*B*b-C*a))*(a+b*cos(d*x+c))^2*sin(d*x+c)/b/d+1/20*(5*B*b-C*a)*(a+b*cos(d*x
+c))^3*sin(d*x+c)/b/d+1/5*C*(a+b*cos(d*x+c))^4*sin(d*x+c)/b/d

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Rubi [A]  time = 0.42, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3023, 2753, 2734} \[ \frac {\sin (c+d x) \left (4 a^2 b^2 (20 A+13 C)+15 a^3 b B-3 a^4 C+60 a b^3 B+4 b^4 (5 A+4 C)\right )}{30 b d}+\frac {\sin (c+d x) \cos (c+d x) \left (30 a^2 b B-6 a^3 C+a b^2 (100 A+71 C)+45 b^3 B\right )}{120 d}+\frac {1}{8} x \left (4 a^3 (2 A+C)+12 a^2 b B+3 a b^2 (4 A+3 C)+3 b^3 B\right )+\frac {\sin (c+d x) \left (3 a (5 b B-a C)+4 b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^2}{60 b d}+\frac {(5 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^3}{20 b d}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((12*a^2*b*B + 3*b^3*B + 4*a^3*(2*A + C) + 3*a*b^2*(4*A + 3*C))*x)/8 + ((15*a^3*b*B + 60*a*b^3*B - 3*a^4*C + 4
*b^4*(5*A + 4*C) + 4*a^2*b^2*(20*A + 13*C))*Sin[c + d*x])/(30*b*d) + ((30*a^2*b*B + 45*b^3*B - 6*a^3*C + a*b^2
*(100*A + 71*C))*Cos[c + d*x]*Sin[c + d*x])/(120*d) + ((4*b^2*(5*A + 4*C) + 3*a*(5*b*B - a*C))*(a + b*Cos[c +
d*x])^2*Sin[c + d*x])/(60*b*d) + ((5*b*B - a*C)*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(20*b*d) + (C*(a + b*Cos[
c + d*x])^4*Sin[c + d*x])/(5*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))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}+\frac {\int (a+b \cos (c+d x))^3 (b (5 A+4 C)+(5 b B-a C) \cos (c+d x)) \, dx}{5 b}\\ &=\frac {(5 b B-a C) (a+b \cos (c+d x))^3 \sin (c+d x)}{20 b d}+\frac {C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}+\frac {\int (a+b \cos (c+d x))^2 \left (b (20 a A+15 b B+13 a C)+\left (4 b^2 (5 A+4 C)+3 a (5 b B-a C)\right ) \cos (c+d x)\right ) \, dx}{20 b}\\ &=\frac {\left (4 b^2 (5 A+4 C)+3 a (5 b B-a C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{60 b d}+\frac {(5 b B-a C) (a+b \cos (c+d x))^3 \sin (c+d x)}{20 b d}+\frac {C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}+\frac {\int (a+b \cos (c+d x)) \left (b \left (75 a b B+8 b^2 (5 A+4 C)+a^2 (60 A+33 C)\right )+\left (30 a^2 b B+45 b^3 B-6 a^3 C+a b^2 (100 A+71 C)\right ) \cos (c+d x)\right ) \, dx}{60 b}\\ &=\frac {1}{8} \left (12 a^2 b B+3 b^3 B+4 a^3 (2 A+C)+3 a b^2 (4 A+3 C)\right ) x+\frac {\left (15 a^3 b B+60 a b^3 B-3 a^4 C+4 b^4 (5 A+4 C)+4 a^2 b^2 (20 A+13 C)\right ) \sin (c+d x)}{30 b d}+\frac {\left (30 a^2 b B+45 b^3 B-6 a^3 C+a b^2 (100 A+71 C)\right ) \cos (c+d x) \sin (c+d x)}{120 d}+\frac {\left (4 b^2 (5 A+4 C)+3 a (5 b B-a C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{60 b d}+\frac {(5 b B-a C) (a+b \cos (c+d x))^3 \sin (c+d x)}{20 b d}+\frac {C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}\\ \end {align*}

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Mathematica [A]  time = 0.97, size = 288, normalized size = 1.04 \[ \frac {480 a^3 A c+480 a^3 A d x+240 a^3 c C+240 a^3 C d x+720 a^2 b B c+720 a^2 b B d x+120 a^2 b C \sin (3 (c+d x))+60 \sin (c+d x) \left (8 a^3 B+6 a^2 b (4 A+3 C)+18 a b^2 B+b^3 (6 A+5 C)\right )+120 \sin (2 (c+d x)) \left (a^3 C+3 a^2 b B+3 a b^2 (A+C)+b^3 B\right )+720 a A b^2 c+720 a A b^2 d x+120 a b^2 B \sin (3 (c+d x))+45 a b^2 C \sin (4 (c+d x))+540 a b^2 c C+540 a b^2 C d x+40 A b^3 \sin (3 (c+d x))+15 b^3 B \sin (4 (c+d x))+180 b^3 B c+180 b^3 B d x+50 b^3 C \sin (3 (c+d x))+6 b^3 C \sin (5 (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(480*a^3*A*c + 720*a*A*b^2*c + 720*a^2*b*B*c + 180*b^3*B*c + 240*a^3*c*C + 540*a*b^2*c*C + 480*a^3*A*d*x + 720
*a*A*b^2*d*x + 720*a^2*b*B*d*x + 180*b^3*B*d*x + 240*a^3*C*d*x + 540*a*b^2*C*d*x + 60*(8*a^3*B + 18*a*b^2*B +
6*a^2*b*(4*A + 3*C) + b^3*(6*A + 5*C))*Sin[c + d*x] + 120*(3*a^2*b*B + b^3*B + a^3*C + 3*a*b^2*(A + C))*Sin[2*
(c + d*x)] + 40*A*b^3*Sin[3*(c + d*x)] + 120*a*b^2*B*Sin[3*(c + d*x)] + 120*a^2*b*C*Sin[3*(c + d*x)] + 50*b^3*
C*Sin[3*(c + d*x)] + 15*b^3*B*Sin[4*(c + d*x)] + 45*a*b^2*C*Sin[4*(c + d*x)] + 6*b^3*C*Sin[5*(c + d*x)])/(480*
d)

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fricas [A]  time = 0.44, size = 207, normalized size = 0.75 \[ \frac {15 \, {\left (4 \, {\left (2 \, A + C\right )} a^{3} + 12 \, B a^{2} b + 3 \, {\left (4 \, A + 3 \, C\right )} a b^{2} + 3 \, B b^{3}\right )} d x + {\left (24 \, C b^{3} \cos \left (d x + c\right )^{4} + 120 \, B a^{3} + 120 \, {\left (3 \, A + 2 \, C\right )} a^{2} b + 240 \, B a b^{2} + 16 \, {\left (5 \, A + 4 \, C\right )} b^{3} + 30 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (15 \, C a^{2} b + 15 \, B a b^{2} + {\left (5 \, A + 4 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (4 \, C a^{3} + 12 \, B a^{2} b + 3 \, {\left (4 \, A + 3 \, C\right )} a b^{2} + 3 \, B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(4*(2*A + C)*a^3 + 12*B*a^2*b + 3*(4*A + 3*C)*a*b^2 + 3*B*b^3)*d*x + (24*C*b^3*cos(d*x + c)^4 + 120*
B*a^3 + 120*(3*A + 2*C)*a^2*b + 240*B*a*b^2 + 16*(5*A + 4*C)*b^3 + 30*(3*C*a*b^2 + B*b^3)*cos(d*x + c)^3 + 8*(
15*C*a^2*b + 15*B*a*b^2 + (5*A + 4*C)*b^3)*cos(d*x + c)^2 + 15*(4*C*a^3 + 12*B*a^2*b + 3*(4*A + 3*C)*a*b^2 + 3
*B*b^3)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.20, size = 227, normalized size = 0.82 \[ \frac {C b^{3} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {1}{8} \, {\left (8 \, A a^{3} + 4 \, C a^{3} + 12 \, B a^{2} b + 12 \, A a b^{2} + 9 \, C a b^{2} + 3 \, B b^{3}\right )} x + \frac {{\left (3 \, C a b^{2} + B b^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (12 \, C a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3} + 5 \, C b^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (C a^{3} + 3 \, B a^{2} b + 3 \, A a b^{2} + 3 \, C a b^{2} + B b^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (8 \, B a^{3} + 24 \, A a^{2} b + 18 \, C a^{2} b + 18 \, B a b^{2} + 6 \, A b^{3} + 5 \, C b^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*C*b^3*sin(5*d*x + 5*c)/d + 1/8*(8*A*a^3 + 4*C*a^3 + 12*B*a^2*b + 12*A*a*b^2 + 9*C*a*b^2 + 3*B*b^3)*x + 1/
32*(3*C*a*b^2 + B*b^3)*sin(4*d*x + 4*c)/d + 1/48*(12*C*a^2*b + 12*B*a*b^2 + 4*A*b^3 + 5*C*b^3)*sin(3*d*x + 3*c
)/d + 1/4*(C*a^3 + 3*B*a^2*b + 3*A*a*b^2 + 3*C*a*b^2 + B*b^3)*sin(2*d*x + 2*c)/d + 1/8*(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)*sin(d*x + c)/d

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maple [A]  time = 0.23, size = 301, normalized size = 1.09 \[ \frac {\frac {b^{3} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+b^{3} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 C a \,b^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B a \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+C \,a^{2} b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 A a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 A \,a^{2} b \sin \left (d x +c \right )+a^{3} B \sin \left (d x +c \right )+A \,a^{3} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 288, normalized size = 1.04 \[ \frac {480 \, {\left (d x + c\right )} A a^{3} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} b + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b^{2} - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b^{2} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{2} - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{3} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{3} + 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C b^{3} + 480 \, B a^{3} \sin \left (d x + c\right ) + 1440 \, A a^{2} b \sin \left (d x + c\right )}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(480*(d*x + c)*A*a^3 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 + 360*(2*d*x + 2*c + sin(2*d*x + 2*c))
*B*a^2*b - 480*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^2*b + 360*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a*b^2 - 480*
(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a*b^2 + 45*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a*b^2
- 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b^3 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*
b^3 + 32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*b^3 + 480*B*a^3*sin(d*x + c) + 1440*A*a^2*
b*sin(d*x + c))/d

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mupad [B]  time = 2.65, size = 359, normalized size = 1.30 \[ A\,a^3\,x+\frac {3\,B\,b^3\,x}{8}+\frac {C\,a^3\,x}{2}+\frac {3\,A\,a\,b^2\,x}{2}+\frac {3\,B\,a^2\,b\,x}{2}+\frac {9\,C\,a\,b^2\,x}{8}+\frac {3\,A\,b^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,C\,b^3\,\sin \left (c+d\,x\right )}{8\,d}+\frac {A\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,b^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {5\,C\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {C\,b^3\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {3\,A\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,B\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,C\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,C\,a\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,A\,a^2\,b\,\sin \left (c+d\,x\right )}{d}+\frac {9\,B\,a\,b^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {9\,C\,a^2\,b\,\sin \left (c+d\,x\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.51, size = 685, normalized size = 2.47 \[ \begin {cases} A a^{3} x + \frac {3 A a^{2} b \sin {\left (c + d x \right )}}{d} + \frac {3 A a b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a b^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 A b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A b^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a^{3} \sin {\left (c + d x \right )}}{d} + \frac {3 B a^{2} b x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a^{2} b x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a^{2} b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 B a b^{2} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {3 B a b^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 B b^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 B b^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 B b^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {C a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {C a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {C a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 C a^{2} b \sin ^{3}{\left (c + d x \right )}}{d} + \frac {3 C a^{2} b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {9 C a b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 C a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {9 C a b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {9 C a b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {15 C a b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {8 C b^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C b^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {C b^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\relax (c )}\right )^{3} \left (A + B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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