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

Optimal. Leaf size=214 \[ \frac {\left (2 a^2 C+b^2 (6 A+5 C)\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {\left (a^2 (8 A+6 C)+b^2 (6 A+5 C)\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x \left (a^2 (8 A+6 C)+b^2 (6 A+5 C)\right )-\frac {2 a b (5 A+4 C) \sin ^3(c+d x)}{15 d}+\frac {2 a b (5 A+4 C) \sin (c+d x)}{5 d}+\frac {a b C \sin (c+d x) \cos ^4(c+d x)}{15 d}+\frac {C \sin (c+d x) \cos ^3(c+d x) (a+b \cos (c+d x))^2}{6 d} \]

[Out]

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

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Rubi [A]  time = 0.49, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3050, 3033, 3023, 2748, 2635, 8, 2633} \[ \frac {\left (2 a^2 C+b^2 (6 A+5 C)\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {\left (a^2 (8 A+6 C)+b^2 (6 A+5 C)\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x \left (a^2 (8 A+6 C)+b^2 (6 A+5 C)\right )-\frac {2 a b (5 A+4 C) \sin ^3(c+d x)}{15 d}+\frac {2 a b (5 A+4 C) \sin (c+d x)}{5 d}+\frac {a b C \sin (c+d x) \cos ^4(c+d x)}{15 d}+\frac {C \sin (c+d x) \cos ^3(c+d x) (a+b \cos (c+d x))^2}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2*(6*A + 5*C) + a^2*(8*A + 6*C))*x)/16 + (2*a*b*(5*A + 4*C)*Sin[c + d*x])/(5*d) + ((b^2*(6*A + 5*C) + a^2*
(8*A + 6*C))*Cos[c + d*x]*Sin[c + d*x])/(16*d) + ((2*a^2*C + b^2*(6*A + 5*C))*Cos[c + d*x]^3*Sin[c + d*x])/(24
*d) + (a*b*C*Cos[c + d*x]^4*Sin[c + d*x])/(15*d) + (C*Cos[c + d*x]^3*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(6*d
) - (2*a*b*(5*A + 4*C)*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 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 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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 ^2(c+d x) (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac {C \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+C)+b (6 A+5 C) \cos (c+d x)+2 a C \cos ^2(c+d x)\right ) \, dx\\ &=\frac {a b C \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac {C \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+C)+12 a b (5 A+4 C) \cos (c+d x)+5 \left (2 a^2 C+b^2 (6 A+5 C)\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {\left (2 a^2 C+b^2 (6 A+5 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a b C \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac {C \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 (b^2 (6 A+5 C)+a^2 (8 A+6 C)\right )+48 a b (5 A+4 C) \cos (c+d x)\right ) \, dx\\ &=\frac {\left (2 a^2 C+b^2 (6 A+5 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a b C \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac {C \cos ^3(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {1}{5} (2 a b (5 A+4 C)) \int \cos ^3(c+d x) \, dx+\frac {1}{8} \left (b^2 (6 A+5 C)+a^2 (8 A+6 C)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {\left (b^2 (6 A+5 C)+a^2 (8 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {\left (2 a^2 C+b^2 (6 A+5 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a b C \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac {C \cos ^3(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {1}{16} \left (b^2 (6 A+5 C)+a^2 (8 A+6 C)\right ) \int 1 \, dx-\frac {(2 a b (5 A+4 C)) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {1}{16} \left (b^2 (6 A+5 C)+a^2 (8 A+6 C)\right ) x+\frac {2 a b (5 A+4 C) \sin (c+d x)}{5 d}+\frac {\left (b^2 (6 A+5 C)+a^2 (8 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {\left (2 a^2 C+b^2 (6 A+5 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a b C \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac {C \cos ^3(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}-\frac {2 a b (5 A+4 C) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.72, size = 160, normalized size = 0.75 \[ \frac {60 (c+d x) \left (a^2 (8 A+6 C)+b^2 (6 A+5 C)\right )+15 \left (16 a^2 (A+C)+b^2 (16 A+15 C)\right ) \sin (2 (c+d x))+15 \left (2 a^2 C+2 A b^2+3 b^2 C\right ) \sin (4 (c+d x))+240 a b (6 A+5 C) \sin (c+d x)+40 a b (4 A+5 C) \sin (3 (c+d x))+24 a b C \sin (5 (c+d x))+5 b^2 C \sin (6 (c+d x))}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*(b^2*(6*A + 5*C) + a^2*(8*A + 6*C))*(c + d*x) + 240*a*b*(6*A + 5*C)*Sin[c + d*x] + 15*(16*a^2*(A + C) + b^
2*(16*A + 15*C))*Sin[2*(c + d*x)] + 40*a*b*(4*A + 5*C)*Sin[3*(c + d*x)] + 15*(2*A*b^2 + 2*a^2*C + 3*b^2*C)*Sin
[4*(c + d*x)] + 24*a*b*C*Sin[5*(c + d*x)] + 5*b^2*C*Sin[6*(c + d*x)])/(960*d)

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fricas [A]  time = 0.78, size = 159, normalized size = 0.74 \[ \frac {15 \, {\left (2 \, {\left (4 \, A + 3 \, C\right )} a^{2} + {\left (6 \, A + 5 \, C\right )} b^{2}\right )} d x + {\left (40 \, C b^{2} \cos \left (d x + c\right )^{5} + 96 \, C a b \cos \left (d x + c\right )^{4} + 32 \, {\left (5 \, A + 4 \, C\right )} a b \cos \left (d x + c\right )^{2} + 10 \, {\left (6 \, C a^{2} + {\left (6 \, A + 5 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{3} + 64 \, {\left (5 \, A + 4 \, C\right )} a b + 15 \, {\left (2 \, {\left (4 \, A + 3 \, C\right )} a^{2} + {\left (6 \, A + 5 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(2*(4*A + 3*C)*a^2 + (6*A + 5*C)*b^2)*d*x + (40*C*b^2*cos(d*x + c)^5 + 96*C*a*b*cos(d*x + c)^4 + 32*
(5*A + 4*C)*a*b*cos(d*x + c)^2 + 10*(6*C*a^2 + (6*A + 5*C)*b^2)*cos(d*x + c)^3 + 64*(5*A + 4*C)*a*b + 15*(2*(4
*A + 3*C)*a^2 + (6*A + 5*C)*b^2)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.39, size = 183, normalized size = 0.86 \[ \frac {C b^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {C a b \sin \left (5 \, d x + 5 \, c\right )}{40 \, d} + \frac {1}{16} \, {\left (8 \, A a^{2} + 6 \, C a^{2} + 6 \, A b^{2} + 5 \, C b^{2}\right )} x + \frac {{\left (2 \, C a^{2} + 2 \, A b^{2} + 3 \, C b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (4 \, A a b + 5 \, C a b\right )} \sin \left (3 \, d x + 3 \, c\right )}{24 \, d} + \frac {{\left (16 \, A a^{2} + 16 \, C a^{2} + 16 \, A b^{2} + 15 \, C b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (6 \, A a b + 5 \, C a b\right )} \sin \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*C*b^2*sin(6*d*x + 6*c)/d + 1/40*C*a*b*sin(5*d*x + 5*c)/d + 1/16*(8*A*a^2 + 6*C*a^2 + 6*A*b^2 + 5*C*b^2)*
x + 1/64*(2*C*a^2 + 2*A*b^2 + 3*C*b^2)*sin(4*d*x + 4*c)/d + 1/24*(4*A*a*b + 5*C*a*b)*sin(3*d*x + 3*c)/d + 1/64
*(16*A*a^2 + 16*C*a^2 + 16*A*b^2 + 15*C*b^2)*sin(2*d*x + 2*c)/d + 1/4*(6*A*a*b + 5*C*a*b)*sin(d*x + c)/d

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maple [A]  time = 0.32, size = 209, normalized size = 0.98 \[ \frac {a^{2} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{2} C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 A a b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {2 C a b \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+A \,b^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b^{2} C \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.57, size = 202, normalized size = 0.94 \[ \frac {240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} - 640 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b + 128 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a b + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{2} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{2}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(240*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))
*C*a^2 - 640*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a*b + 128*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x
+ c))*C*a*b + 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*b^2 - 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^2)/d

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mupad [B]  time = 1.66, size = 252, normalized size = 1.18 \[ \frac {A\,a^2\,x}{2}+\frac {3\,A\,b^2\,x}{8}+\frac {3\,C\,a^2\,x}{8}+\frac {5\,C\,b^2\,x}{16}+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {15\,C\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{64\,d}+\frac {3\,C\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{64\,d}+\frac {C\,b^2\,\sin \left (6\,c+6\,d\,x\right )}{192\,d}+\frac {3\,A\,a\,b\,\sin \left (c+d\,x\right )}{2\,d}+\frac {5\,C\,a\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,a\,b\,\sin \left (3\,c+3\,d\,x\right )}{6\,d}+\frac {5\,C\,a\,b\,\sin \left (3\,c+3\,d\,x\right )}{24\,d}+\frac {C\,a\,b\,\sin \left (5\,c+5\,d\,x\right )}{40\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*a^2*x)/2 + (3*A*b^2*x)/8 + (3*C*a^2*x)/8 + (5*C*b^2*x)/16 + (A*a^2*sin(2*c + 2*d*x))/(4*d) + (A*b^2*sin(2*c
+ 2*d*x))/(4*d) + (A*b^2*sin(4*c + 4*d*x))/(32*d) + (C*a^2*sin(2*c + 2*d*x))/(4*d) + (C*a^2*sin(4*c + 4*d*x))
/(32*d) + (15*C*b^2*sin(2*c + 2*d*x))/(64*d) + (3*C*b^2*sin(4*c + 4*d*x))/(64*d) + (C*b^2*sin(6*c + 6*d*x))/(1
92*d) + (3*A*a*b*sin(c + d*x))/(2*d) + (5*C*a*b*sin(c + d*x))/(4*d) + (A*a*b*sin(3*c + 3*d*x))/(6*d) + (5*C*a*
b*sin(3*c + 3*d*x))/(24*d) + (C*a*b*sin(5*c + 5*d*x))/(40*d)

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sympy [A]  time = 4.35, size = 592, normalized size = 2.77 \[ \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 {3 C a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 C a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 C a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 C a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {16 C a b \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {8 C a b \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {2 C a b \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 C b^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 C b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 C b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 C b^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 C b^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 C b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 C b^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\relax (c )}\right ) \left (a + b \cos {\relax (c )}\right )^{2} \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))**2*(A+C*cos(d*x+c)**2),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) + 3*C*a**2*x*sin(c + d*x)**4/8 + 3*C*a**2*x*sin(c + d*x)**
2*cos(c + d*x)**2/4 + 3*C*a**2*x*cos(c + d*x)**4/8 + 3*C*a**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 5*C*a**2*si
n(c + d*x)*cos(c + d*x)**3/(8*d) + 16*C*a*b*sin(c + d*x)**5/(15*d) + 8*C*a*b*sin(c + d*x)**3*cos(c + d*x)**2/(
3*d) + 2*C*a*b*sin(c + d*x)*cos(c + d*x)**4/d + 5*C*b**2*x*sin(c + d*x)**6/16 + 15*C*b**2*x*sin(c + d*x)**4*co
s(c + d*x)**2/16 + 15*C*b**2*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 5*C*b**2*x*cos(c + d*x)**6/16 + 5*C*b**2*s
in(c + d*x)**5*cos(c + d*x)/(16*d) + 5*C*b**2*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) + 11*C*b**2*sin(c + d*x)*c
os(c + d*x)**5/(16*d), Ne(d, 0)), (x*(A + C*cos(c)**2)*(a + b*cos(c))**2*cos(c)**2, True))

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