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

Optimal. Leaf size=213 \[ -\frac {a^2 (10 A+9 B+8 C) \sin ^3(c+d x)}{15 d}+\frac {a^2 (10 A+9 B+8 C) \sin (c+d x)}{5 d}+\frac {a^2 (10 A+12 B+9 C) \sin (c+d x) \cos ^3(c+d x)}{40 d}+\frac {a^2 (14 A+12 B+11 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} a^2 x (14 A+12 B+11 C)+\frac {(3 B+C) \sin (c+d x) \cos ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{15 d}+\frac {C \sin (c+d x) \cos ^3(c+d x) (a \cos (c+d x)+a)^2}{6 d} \]

[Out]

1/16*a^2*(14*A+12*B+11*C)*x+1/5*a^2*(10*A+9*B+8*C)*sin(d*x+c)/d+1/16*a^2*(14*A+12*B+11*C)*cos(d*x+c)*sin(d*x+c
)/d+1/40*a^2*(10*A+12*B+9*C)*cos(d*x+c)^3*sin(d*x+c)/d+1/6*C*cos(d*x+c)^3*(a+a*cos(d*x+c))^2*sin(d*x+c)/d+1/15
*(3*B+C)*cos(d*x+c)^3*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d-1/15*a^2*(10*A+9*B+8*C)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.50, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {3045, 2976, 2968, 3023, 2748, 2635, 8, 2633} \[ -\frac {a^2 (10 A+9 B+8 C) \sin ^3(c+d x)}{15 d}+\frac {a^2 (10 A+9 B+8 C) \sin (c+d x)}{5 d}+\frac {a^2 (10 A+12 B+9 C) \sin (c+d x) \cos ^3(c+d x)}{40 d}+\frac {a^2 (14 A+12 B+11 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} a^2 x (14 A+12 B+11 C)+\frac {(3 B+C) \sin (c+d x) \cos ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{15 d}+\frac {C \sin (c+d x) \cos ^3(c+d x) (a \cos (c+d x)+a)^2}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(14*A + 12*B + 11*C)*x)/16 + (a^2*(10*A + 9*B + 8*C)*Sin[c + d*x])/(5*d) + (a^2*(14*A + 12*B + 11*C)*Cos[
c + d*x]*Sin[c + d*x])/(16*d) + (a^2*(10*A + 12*B + 9*C)*Cos[c + d*x]^3*Sin[c + d*x])/(40*d) + (C*Cos[c + d*x]
^3*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(6*d) + ((3*B + C)*Cos[c + d*x]^3*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x
])/(15*d) - (a^2*(10*A + 9*B + 8*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 2968

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

Rule 2976

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 - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&
NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[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 3045

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/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x
])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (C*(a*d*m - b*c*(m + 1)) + b*B*
d*(m + n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {\int \cos ^2(c+d x) (a+a \cos (c+d x))^2 (3 a (2 A+C)+2 a (3 B+C) \cos (c+d x)) \, dx}{6 a}\\ &=\frac {C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(3 B+C) \cos ^3(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {\int \cos ^2(c+d x) (a+a \cos (c+d x)) \left (3 a^2 (10 A+6 B+7 C)+3 a^2 (10 A+12 B+9 C) \cos (c+d x)\right ) \, dx}{30 a}\\ &=\frac {C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(3 B+C) \cos ^3(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {\int \cos ^2(c+d x) \left (3 a^3 (10 A+6 B+7 C)+\left (3 a^3 (10 A+6 B+7 C)+3 a^3 (10 A+12 B+9 C)\right ) \cos (c+d x)+3 a^3 (10 A+12 B+9 C) \cos ^2(c+d x)\right ) \, dx}{30 a}\\ &=\frac {a^2 (10 A+12 B+9 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(3 B+C) \cos ^3(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {\int \cos ^2(c+d x) \left (15 a^3 (14 A+12 B+11 C)+24 a^3 (10 A+9 B+8 C) \cos (c+d x)\right ) \, dx}{120 a}\\ &=\frac {a^2 (10 A+12 B+9 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(3 B+C) \cos ^3(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{5} \left (a^2 (10 A+9 B+8 C)\right ) \int \cos ^3(c+d x) \, dx+\frac {1}{8} \left (a^2 (14 A+12 B+11 C)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {a^2 (14 A+12 B+11 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^2 (10 A+12 B+9 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(3 B+C) \cos ^3(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{16} \left (a^2 (14 A+12 B+11 C)\right ) \int 1 \, dx-\frac {\left (a^2 (10 A+9 B+8 C)\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {1}{16} a^2 (14 A+12 B+11 C) x+\frac {a^2 (10 A+9 B+8 C) \sin (c+d x)}{5 d}+\frac {a^2 (14 A+12 B+11 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^2 (10 A+12 B+9 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(3 B+C) \cos ^3(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d}-\frac {a^2 (10 A+9 B+8 C) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.73, size = 171, normalized size = 0.80 \[ \frac {a^2 (120 (12 A+11 B+10 C) \sin (c+d x)+15 (32 A+32 B+31 C) \sin (2 (c+d x))+160 A \sin (3 (c+d x))+30 A \sin (4 (c+d x))+840 A d x+180 B \sin (3 (c+d x))+60 B \sin (4 (c+d x))+12 B \sin (5 (c+d x))+720 B c+720 B d x+200 C \sin (3 (c+d x))+75 C \sin (4 (c+d x))+24 C \sin (5 (c+d x))+5 C \sin (6 (c+d x))+420 c C+660 C d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(720*B*c + 420*c*C + 840*A*d*x + 720*B*d*x + 660*C*d*x + 120*(12*A + 11*B + 10*C)*Sin[c + d*x] + 15*(32*A
+ 32*B + 31*C)*Sin[2*(c + d*x)] + 160*A*Sin[3*(c + d*x)] + 180*B*Sin[3*(c + d*x)] + 200*C*Sin[3*(c + d*x)] +
30*A*Sin[4*(c + d*x)] + 60*B*Sin[4*(c + d*x)] + 75*C*Sin[4*(c + d*x)] + 12*B*Sin[5*(c + d*x)] + 24*C*Sin[5*(c
+ d*x)] + 5*C*Sin[6*(c + d*x)]))/(960*d)

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fricas [A]  time = 0.45, size = 145, normalized size = 0.68 \[ \frac {15 \, {\left (14 \, A + 12 \, B + 11 \, C\right )} a^{2} d x + {\left (40 \, C a^{2} \cos \left (d x + c\right )^{5} + 48 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 10 \, {\left (6 \, A + 12 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 16 \, {\left (10 \, A + 9 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (14 \, A + 12 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right ) + 32 \, {\left (10 \, A + 9 \, B + 8 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(14*A + 12*B + 11*C)*a^2*d*x + (40*C*a^2*cos(d*x + c)^5 + 48*(B + 2*C)*a^2*cos(d*x + c)^4 + 10*(6*A
+ 12*B + 11*C)*a^2*cos(d*x + c)^3 + 16*(10*A + 9*B + 8*C)*a^2*cos(d*x + c)^2 + 15*(14*A + 12*B + 11*C)*a^2*cos
(d*x + c) + 32*(10*A + 9*B + 8*C)*a^2)*sin(d*x + c))/d

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giac [A]  time = 0.52, size = 196, normalized size = 0.92 \[ \frac {C a^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {1}{16} \, {\left (14 \, A a^{2} + 12 \, B a^{2} + 11 \, C a^{2}\right )} x + \frac {{\left (B a^{2} + 2 \, C a^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (2 \, A a^{2} + 4 \, B a^{2} + 5 \, C a^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (8 \, A a^{2} + 9 \, B a^{2} + 10 \, C a^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (32 \, A a^{2} + 32 \, B a^{2} + 31 \, C a^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (12 \, A a^{2} + 11 \, B a^{2} + 10 \, C a^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*C*a^2*sin(6*d*x + 6*c)/d + 1/16*(14*A*a^2 + 12*B*a^2 + 11*C*a^2)*x + 1/80*(B*a^2 + 2*C*a^2)*sin(5*d*x +
5*c)/d + 1/64*(2*A*a^2 + 4*B*a^2 + 5*C*a^2)*sin(4*d*x + 4*c)/d + 1/48*(8*A*a^2 + 9*B*a^2 + 10*C*a^2)*sin(3*d*x
+ 3*c)/d + 1/64*(32*A*a^2 + 32*B*a^2 + 31*C*a^2)*sin(2*d*x + 2*c)/d + 1/8*(12*A*a^2 + 11*B*a^2 + 10*C*a^2)*si
n(d*x + c)/d

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maple [A]  time = 0.31, size = 304, normalized size = 1.43 \[ \frac {a^{2} A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{2} B \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a^{2} C \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {2 a^{2} A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 B \,a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 a^{2} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a^{2} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {B \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a^{2} C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.47, size = 296, normalized size = 1.39 \[ -\frac {640 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} - 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{2} + 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} - 60 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 128 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a^{2} + 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} - 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*(640*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2 - 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c
))*A*a^2 - 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 - 64*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x
+ c))*B*a^2 + 320*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^2 - 60*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x
+ 2*c))*B*a^2 - 128*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*a^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*a^2 - 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*s
in(2*d*x + 2*c))*C*a^2)/d

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mupad [B]  time = 2.77, size = 366, normalized size = 1.72 \[ \frac {\left (\frac {7\,A\,a^2}{4}+\frac {3\,B\,a^2}{2}+\frac {11\,C\,a^2}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {119\,A\,a^2}{12}+\frac {17\,B\,a^2}{2}+\frac {187\,C\,a^2}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {43\,A\,a^2}{2}+\frac {107\,B\,a^2}{5}+\frac {331\,C\,a^2}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {53\,A\,a^2}{2}+\frac {117\,B\,a^2}{5}+\frac {501\,C\,a^2}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {233\,A\,a^2}{12}+\frac {31\,B\,a^2}{2}+\frac {87\,C\,a^2}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {25\,A\,a^2}{4}+\frac {13\,B\,a^2}{2}+\frac {53\,C\,a^2}{8}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (14\,A+12\,B+11\,C\right )}{8\,\left (\frac {7\,A\,a^2}{4}+\frac {3\,B\,a^2}{2}+\frac {11\,C\,a^2}{8}\right )}\right )\,\left (14\,A+12\,B+11\,C\right )}{8\,d}-\frac {a^2\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )\,\left (14\,A+12\,B+11\,C\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^11*((7*A*a^2)/4 + (3*B*a^2)/2 + (11*C*a^2)/8) + tan(c/2 + (d*x)/2)^9*((119*A*a^2)/12 + (17
*B*a^2)/2 + (187*C*a^2)/24) + tan(c/2 + (d*x)/2)^3*((233*A*a^2)/12 + (31*B*a^2)/2 + (87*C*a^2)/8) + tan(c/2 +
(d*x)/2)^7*((43*A*a^2)/2 + (107*B*a^2)/5 + (331*C*a^2)/20) + tan(c/2 + (d*x)/2)^5*((53*A*a^2)/2 + (117*B*a^2)/
5 + (501*C*a^2)/20) + tan(c/2 + (d*x)/2)*((25*A*a^2)/4 + (13*B*a^2)/2 + (53*C*a^2)/8))/(d*(6*tan(c/2 + (d*x)/2
)^2 + 15*tan(c/2 + (d*x)/2)^4 + 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 + 6*tan(c/2 + (d*x)/2)^10 +
tan(c/2 + (d*x)/2)^12 + 1)) + (a^2*atan((a^2*tan(c/2 + (d*x)/2)*(14*A + 12*B + 11*C))/(8*((7*A*a^2)/4 + (3*B*a
^2)/2 + (11*C*a^2)/8)))*(14*A + 12*B + 11*C))/(8*d) - (a^2*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2)*(14*A + 12*B +
11*C))/(8*d)

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sympy [A]  time = 4.90, size = 821, normalized size = 3.85 \[ \begin {cases} \frac {3 A a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {A a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {A a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {4 A a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 A a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {2 A a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {3 B a^{2} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {3 B a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a^{2} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {8 B a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 B a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {3 B a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {2 B a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 B a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {B a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {5 C a^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 C a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 C a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {15 C a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 C a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {5 C a^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 C a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {5 C a^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {16 C a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {5 C a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {8 C a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {3 C a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {11 C a^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac {2 C a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 C a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + a\right )^{2} \left (A + B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \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+a*cos(d*x+c))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

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

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