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

Optimal. Leaf size=201 \[ -\frac {a^3 (19 B+17 C) \sin ^3(c+d x)}{15 d}+\frac {a^3 (19 B+17 C) \sin (c+d x)}{5 d}+\frac {a^3 (22 B+21 C) \sin (c+d x) \cos ^3(c+d x)}{40 d}+\frac {(3 B+4 C) \sin (c+d x) \cos ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{15 d}+\frac {a^3 (26 B+23 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} a^3 x (26 B+23 C)+\frac {a C \sin (c+d x) \cos ^3(c+d x) (a \cos (c+d x)+a)^2}{6 d} \]

[Out]

1/16*a^3*(26*B+23*C)*x+1/5*a^3*(19*B+17*C)*sin(d*x+c)/d+1/16*a^3*(26*B+23*C)*cos(d*x+c)*sin(d*x+c)/d+1/40*a^3*
(22*B+21*C)*cos(d*x+c)^3*sin(d*x+c)/d+1/6*a*C*cos(d*x+c)^3*(a+a*cos(d*x+c))^2*sin(d*x+c)/d+1/15*(3*B+4*C)*cos(
d*x+c)^3*(a^3+a^3*cos(d*x+c))*sin(d*x+c)/d-1/15*a^3*(19*B+17*C)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.49, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {3029, 2976, 2968, 3023, 2748, 2635, 8, 2633} \[ -\frac {a^3 (19 B+17 C) \sin ^3(c+d x)}{15 d}+\frac {a^3 (19 B+17 C) \sin (c+d x)}{5 d}+\frac {a^3 (22 B+21 C) \sin (c+d x) \cos ^3(c+d x)}{40 d}+\frac {(3 B+4 C) \sin (c+d x) \cos ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{15 d}+\frac {a^3 (26 B+23 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} a^3 x (26 B+23 C)+\frac {a C \sin (c+d x) \cos ^3(c+d x) (a \cos (c+d x)+a)^2}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(26*B + 23*C)*x)/16 + (a^3*(19*B + 17*C)*Sin[c + d*x])/(5*d) + (a^3*(26*B + 23*C)*Cos[c + d*x]*Sin[c + d*
x])/(16*d) + (a^3*(22*B + 21*C)*Cos[c + d*x]^3*Sin[c + d*x])/(40*d) + (a*C*Cos[c + d*x]^3*(a + a*Cos[c + d*x])
^2*Sin[c + d*x])/(6*d) + ((3*B + 4*C)*Cos[c + d*x]^3*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(15*d) - (a^3*(19*
B + 17*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 3029

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

Rubi steps

\begin {align*} \int \cos (c+d x) (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\int \cos ^2(c+d x) (a+a \cos (c+d x))^3 (B+C \cos (c+d x)) \, dx\\ &=\frac {a C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {1}{6} \int \cos ^2(c+d x) (a+a \cos (c+d x))^2 (3 a (2 B+C)+2 a (3 B+4 C) \cos (c+d x)) \, dx\\ &=\frac {a C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(3 B+4 C) \cos ^3(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{30} \int \cos ^2(c+d x) (a+a \cos (c+d x)) \left (3 a^2 (16 B+13 C)+3 a^2 (22 B+21 C) \cos (c+d x)\right ) \, dx\\ &=\frac {a C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(3 B+4 C) \cos ^3(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{30} \int \cos ^2(c+d x) \left (3 a^3 (16 B+13 C)+\left (3 a^3 (16 B+13 C)+3 a^3 (22 B+21 C)\right ) \cos (c+d x)+3 a^3 (22 B+21 C) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {a^3 (22 B+21 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {a C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(3 B+4 C) \cos ^3(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{120} \int \cos ^2(c+d x) \left (15 a^3 (26 B+23 C)+24 a^3 (19 B+17 C) \cos (c+d x)\right ) \, dx\\ &=\frac {a^3 (22 B+21 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {a C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(3 B+4 C) \cos ^3(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{5} \left (a^3 (19 B+17 C)\right ) \int \cos ^3(c+d x) \, dx+\frac {1}{8} \left (a^3 (26 B+23 C)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {a^3 (26 B+23 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^3 (22 B+21 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {a C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(3 B+4 C) \cos ^3(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{16} \left (a^3 (26 B+23 C)\right ) \int 1 \, dx-\frac {\left (a^3 (19 B+17 C)\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {1}{16} a^3 (26 B+23 C) x+\frac {a^3 (19 B+17 C) \sin (c+d x)}{5 d}+\frac {a^3 (26 B+23 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^3 (22 B+21 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {a C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(3 B+4 C) \cos ^3(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}-\frac {a^3 (19 B+17 C) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.42, size = 130, normalized size = 0.65 \[ \frac {a^3 (120 (23 B+21 C) \sin (c+d x)+15 (64 B+63 C) \sin (2 (c+d x))+340 B \sin (3 (c+d x))+90 B \sin (4 (c+d x))+12 B \sin (5 (c+d x))+1560 B d x+380 C \sin (3 (c+d x))+135 C \sin (4 (c+d x))+36 C \sin (5 (c+d x))+5 C \sin (6 (c+d x))+1380 C d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1560*B*d*x + 1380*C*d*x + 120*(23*B + 21*C)*Sin[c + d*x] + 15*(64*B + 63*C)*Sin[2*(c + d*x)] + 340*B*Sin
[3*(c + d*x)] + 380*C*Sin[3*(c + d*x)] + 90*B*Sin[4*(c + d*x)] + 135*C*Sin[4*(c + d*x)] + 12*B*Sin[5*(c + d*x)
] + 36*C*Sin[5*(c + d*x)] + 5*C*Sin[6*(c + d*x)]))/(960*d)

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fricas [A]  time = 0.46, size = 130, normalized size = 0.65 \[ \frac {15 \, {\left (26 \, B + 23 \, C\right )} a^{3} d x + {\left (40 \, C a^{3} \cos \left (d x + c\right )^{5} + 48 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 10 \, {\left (18 \, B + 23 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 16 \, {\left (19 \, B + 17 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 15 \, {\left (26 \, B + 23 \, C\right )} a^{3} \cos \left (d x + c\right ) + 32 \, {\left (19 \, B + 17 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(26*B + 23*C)*a^3*d*x + (40*C*a^3*cos(d*x + c)^5 + 48*(B + 3*C)*a^3*cos(d*x + c)^4 + 10*(18*B + 23*C
)*a^3*cos(d*x + c)^3 + 16*(19*B + 17*C)*a^3*cos(d*x + c)^2 + 15*(26*B + 23*C)*a^3*cos(d*x + c) + 32*(19*B + 17
*C)*a^3)*sin(d*x + c))/d

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giac [A]  time = 0.67, size = 166, normalized size = 0.83 \[ \frac {C a^{3} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {1}{16} \, {\left (26 \, B a^{3} + 23 \, C a^{3}\right )} x + \frac {{\left (B a^{3} + 3 \, C a^{3}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {3 \, {\left (2 \, B a^{3} + 3 \, C a^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (17 \, B a^{3} + 19 \, C a^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (64 \, B a^{3} + 63 \, C a^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (23 \, B a^{3} + 21 \, C a^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*C*a^3*sin(6*d*x + 6*c)/d + 1/16*(26*B*a^3 + 23*C*a^3)*x + 1/80*(B*a^3 + 3*C*a^3)*sin(5*d*x + 5*c)/d + 3/
64*(2*B*a^3 + 3*C*a^3)*sin(4*d*x + 4*c)/d + 1/48*(17*B*a^3 + 19*C*a^3)*sin(3*d*x + 3*c)/d + 1/64*(64*B*a^3 + 6
3*C*a^3)*sin(2*d*x + 2*c)/d + 1/8*(23*B*a^3 + 21*C*a^3)*sin(d*x + c)/d

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maple [A]  time = 0.30, size = 266, normalized size = 1.32 \[ \frac {C \,a^{3} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {a^{3} B \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {3 C \,a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 a^{3} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 C \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\frac {C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a^{3} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.76, size = 262, normalized size = 1.30 \[ \frac {64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{3} - 960 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} + 90 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 192 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a^{3} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 90 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.50, size = 315, normalized size = 1.57 \[ \frac {\left (\frac {13\,B\,a^3}{4}+\frac {23\,C\,a^3}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {221\,B\,a^3}{12}+\frac {391\,C\,a^3}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {429\,B\,a^3}{10}+\frac {759\,C\,a^3}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {499\,B\,a^3}{10}+\frac {969\,C\,a^3}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {419\,B\,a^3}{12}+\frac {211\,C\,a^3}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {51\,B\,a^3}{4}+\frac {105\,C\,a^3}{8}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^3\,\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (26\,B+23\,C\right )}{8\,\left (\frac {13\,B\,a^3}{4}+\frac {23\,C\,a^3}{8}\right )}\right )\,\left (26\,B+23\,C\right )}{8\,d}-\frac {a^3\,\left (26\,B+23\,C\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((51*B*a^3)/4 + (105*C*a^3)/8) + tan(c/2 + (d*x)/2)^11*((13*B*a^3)/4 + (23*C*a^3)/8) + tan
(c/2 + (d*x)/2)^3*((419*B*a^3)/12 + (211*C*a^3)/8) + tan(c/2 + (d*x)/2)^9*((221*B*a^3)/12 + (391*C*a^3)/24) +
tan(c/2 + (d*x)/2)^7*((429*B*a^3)/10 + (759*C*a^3)/20) + tan(c/2 + (d*x)/2)^5*((499*B*a^3)/10 + (969*C*a^3)/20
))/(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^3*atan((a^3*tan(c/2 + (d*x)/2)*(26*B + 23*C))/(8*((
13*B*a^3)/4 + (23*C*a^3)/8)))*(26*B + 23*C))/(8*d) - (a^3*(26*B + 23*C)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/
(8*d)

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sympy [A]  time = 4.58, size = 699, normalized size = 3.48 \[ \begin {cases} \frac {9 B a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 B a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {B a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {9 B a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {B a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {8 B a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 B a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {9 B a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 B a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {B a^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {15 B a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {3 B a^{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 )} \cos {\left (c + d x \right )}}{2 d} + \frac {5 C a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 C a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {9 C a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {15 C a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {9 C a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {5 C a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {9 C a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {5 C a^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {8 C a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {5 C a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {4 C a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {9 C a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 C a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {11 C a^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac {3 C a^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {15 C a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {C a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \left (a \cos {\relax (c )} + a\right )^{3} \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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