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

Optimal. Leaf size=188 \[ -\frac {a^3 (30 A+23 C) \sin ^3(c+d x)}{120 d}+\frac {a^3 (30 A+23 C) \sin (c+d x)}{10 d}+\frac {3 a^3 (30 A+23 C) \sin (c+d x) \cos (c+d x)}{80 d}+\frac {1}{16} a^3 x (30 A+23 C)+\frac {(30 A+7 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{120 d}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^3}{6 d}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{10 a d} \]

[Out]

1/16*a^3*(30*A+23*C)*x+1/10*a^3*(30*A+23*C)*sin(d*x+c)/d+3/80*a^3*(30*A+23*C)*cos(d*x+c)*sin(d*x+c)/d+1/120*(3
0*A+7*C)*(a+a*cos(d*x+c))^3*sin(d*x+c)/d+1/6*C*cos(d*x+c)^2*(a+a*cos(d*x+c))^3*sin(d*x+c)/d+1/10*C*(a+a*cos(d*
x+c))^4*sin(d*x+c)/a/d-1/120*a^3*(30*A+23*C)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.33, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3046, 2968, 3023, 2751, 2645, 2637, 2635, 8, 2633} \[ -\frac {a^3 (30 A+23 C) \sin ^3(c+d x)}{120 d}+\frac {a^3 (30 A+23 C) \sin (c+d x)}{10 d}+\frac {3 a^3 (30 A+23 C) \sin (c+d x) \cos (c+d x)}{80 d}+\frac {1}{16} a^3 x (30 A+23 C)+\frac {(30 A+7 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{120 d}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^3}{6 d}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{10 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(30*A + 23*C)*x)/16 + (a^3*(30*A + 23*C)*Sin[c + d*x])/(10*d) + (3*a^3*(30*A + 23*C)*Cos[c + d*x]*Sin[c +
d*x])/(80*d) + ((30*A + 7*C)*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(120*d) + (C*Cos[c + d*x]^2*(a + a*Cos[c +
d*x])^3*Sin[c + d*x])/(6*d) + (C*(a + a*Cos[c + d*x])^4*Sin[c + d*x])/(10*a*d) - (a^3*(30*A + 23*C)*Sin[c + d*
x]^3)/(120*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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2645

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTrig[(a + b*sin[c + d*x])^n, x], x] /;
FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0]

Rule 2751

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

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 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 3046

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/(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))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b
, c, d, e, f, A, 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 (c+d x) (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac {C \cos ^2(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{6 d}+\frac {\int \cos (c+d x) (a+a \cos (c+d x))^3 (2 a (3 A+C)+3 a C \cos (c+d x)) \, dx}{6 a}\\ &=\frac {C \cos ^2(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{6 d}+\frac {\int (a+a \cos (c+d x))^3 \left (2 a (3 A+C) \cos (c+d x)+3 a C \cos ^2(c+d x)\right ) \, dx}{6 a}\\ &=\frac {C \cos ^2(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{6 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{10 a d}+\frac {\int (a+a \cos (c+d x))^3 \left (12 a^2 C+a^2 (30 A+7 C) \cos (c+d x)\right ) \, dx}{30 a^2}\\ &=\frac {(30 A+7 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{120 d}+\frac {C \cos ^2(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{6 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{10 a d}+\frac {1}{40} (30 A+23 C) \int (a+a \cos (c+d x))^3 \, dx\\ &=\frac {(30 A+7 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{120 d}+\frac {C \cos ^2(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{6 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{10 a d}+\frac {1}{40} (30 A+23 C) \int \left (a^3+3 a^3 \cos (c+d x)+3 a^3 \cos ^2(c+d x)+a^3 \cos ^3(c+d x)\right ) \, dx\\ &=\frac {1}{40} a^3 (30 A+23 C) x+\frac {(30 A+7 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{120 d}+\frac {C \cos ^2(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{6 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{10 a d}+\frac {1}{40} \left (a^3 (30 A+23 C)\right ) \int \cos ^3(c+d x) \, dx+\frac {1}{40} \left (3 a^3 (30 A+23 C)\right ) \int \cos (c+d x) \, dx+\frac {1}{40} \left (3 a^3 (30 A+23 C)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {1}{40} a^3 (30 A+23 C) x+\frac {3 a^3 (30 A+23 C) \sin (c+d x)}{40 d}+\frac {3 a^3 (30 A+23 C) \cos (c+d x) \sin (c+d x)}{80 d}+\frac {(30 A+7 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{120 d}+\frac {C \cos ^2(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{6 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{10 a d}+\frac {1}{80} \left (3 a^3 (30 A+23 C)\right ) \int 1 \, dx-\frac {\left (a^3 (30 A+23 C)\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{40 d}\\ &=\frac {1}{16} a^3 (30 A+23 C) x+\frac {a^3 (30 A+23 C) \sin (c+d x)}{10 d}+\frac {3 a^3 (30 A+23 C) \cos (c+d x) \sin (c+d x)}{80 d}+\frac {(30 A+7 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{120 d}+\frac {C \cos ^2(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{6 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{10 a d}-\frac {a^3 (30 A+23 C) \sin ^3(c+d x)}{120 d}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 123, normalized size = 0.65 \[ \frac {a^3 (120 (26 A+21 C) \sin (c+d x)+15 (64 A+63 C) \sin (2 (c+d x))+240 A \sin (3 (c+d x))+30 A \sin (4 (c+d x))+1800 A d x+380 C \sin (3 (c+d x))+135 C \sin (4 (c+d x))+36 C \sin (5 (c+d x))+5 C \sin (6 (c+d x))+900 c C+1380 C d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(900*c*C + 1800*A*d*x + 1380*C*d*x + 120*(26*A + 21*C)*Sin[c + d*x] + 15*(64*A + 63*C)*Sin[2*(c + d*x)] +
240*A*Sin[3*(c + d*x)] + 380*C*Sin[3*(c + d*x)] + 30*A*Sin[4*(c + d*x)] + 135*C*Sin[4*(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 = 1.06, size = 126, normalized size = 0.67 \[ \frac {15 \, {\left (30 \, A + 23 \, C\right )} a^{3} d x + {\left (40 \, C a^{3} \cos \left (d x + c\right )^{5} + 144 \, C a^{3} \cos \left (d x + c\right )^{4} + 10 \, {\left (6 \, A + 23 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 16 \, {\left (15 \, A + 17 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 15 \, {\left (30 \, A + 23 \, C\right )} a^{3} \cos \left (d x + c\right ) + 16 \, {\left (45 \, A + 34 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(30*A + 23*C)*a^3*d*x + (40*C*a^3*cos(d*x + c)^5 + 144*C*a^3*cos(d*x + c)^4 + 10*(6*A + 23*C)*a^3*co
s(d*x + c)^3 + 16*(15*A + 17*C)*a^3*cos(d*x + c)^2 + 15*(30*A + 23*C)*a^3*cos(d*x + c) + 16*(45*A + 34*C)*a^3)
*sin(d*x + c))/d

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giac [A]  time = 0.56, size = 158, normalized size = 0.84 \[ \frac {C a^{3} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {3 \, C a^{3} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {1}{16} \, {\left (30 \, A a^{3} + 23 \, C a^{3}\right )} x + \frac {{\left (2 \, A a^{3} + 9 \, C a^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (12 \, A a^{3} + 19 \, C a^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (64 \, A a^{3} + 63 \, C a^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (26 \, A a^{3} + 21 \, C a^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*C*a^3*sin(6*d*x + 6*c)/d + 3/80*C*a^3*sin(5*d*x + 5*c)/d + 1/16*(30*A*a^3 + 23*C*a^3)*x + 1/64*(2*A*a^3
+ 9*C*a^3)*sin(4*d*x + 4*c)/d + 1/48*(12*A*a^3 + 19*C*a^3)*sin(3*d*x + 3*c)/d + 1/64*(64*A*a^3 + 63*C*a^3)*sin
(2*d*x + 2*c)/d + 1/8*(26*A*a^3 + 21*C*a^3)*sin(d*x + c)/d

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maple [A]  time = 0.31, size = 245, normalized size = 1.30 \[ \frac {A \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+C \,a^{3} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+A \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\frac {3 C \,a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 C \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+A \,a^{3} \sin \left (d x +c \right )+\frac {C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 239, normalized size = 1.27 \[ -\frac {960 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 720 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 192 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a^{3} + 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} - 90 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 960 \, A a^{3} \sin \left (d x + c\right )}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*(960*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3 - 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c
))*A*a^3 - 720*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*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 -
960*A*a^3*sin(d*x + c))/d

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mupad [B]  time = 2.23, size = 315, normalized size = 1.68 \[ \frac {\left (\frac {15\,A\,a^3}{4}+\frac {23\,C\,a^3}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {85\,A\,a^3}{4}+\frac {391\,C\,a^3}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {99\,A\,a^3}{2}+\frac {759\,C\,a^3}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {125\,A\,a^3}{2}+\frac {969\,C\,a^3}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {171\,A\,a^3}{4}+\frac {211\,C\,a^3}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {49\,A\,a^3}{4}+\frac {105\,C\,a^3}{8}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a^3\,\left (30\,A+23\,C\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{8\,d}+\frac {a^3\,\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (30\,A+23\,C\right )}{8\,\left (\frac {15\,A\,a^3}{4}+\frac {23\,C\,a^3}{8}\right )}\right )\,\left (30\,A+23\,C\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((49*A*a^3)/4 + (105*C*a^3)/8) + tan(c/2 + (d*x)/2)^11*((15*A*a^3)/4 + (23*C*a^3)/8) + tan
(c/2 + (d*x)/2)^3*((171*A*a^3)/4 + (211*C*a^3)/8) + tan(c/2 + (d*x)/2)^9*((85*A*a^3)/4 + (391*C*a^3)/24) + tan
(c/2 + (d*x)/2)^7*((99*A*a^3)/2 + (759*C*a^3)/20) + tan(c/2 + (d*x)/2)^5*((125*A*a^3)/2 + (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*(30*A + 23*C)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(8*
d) + (a^3*atan((a^3*tan(c/2 + (d*x)/2)*(30*A + 23*C))/(8*((15*A*a^3)/4 + (23*C*a^3)/8)))*(30*A + 23*C))/(8*d)

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sympy [A]  time = 4.68, size = 646, normalized size = 3.44 \[ \begin {cases} \frac {3 A a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 A a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 A a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {5 A a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {3 A a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 A a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {A a^{3} \sin {\left (c + d x \right )}}{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 (A + 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*(A+C*cos(d*x+c)**2),x)

[Out]

Piecewise((3*A*a**3*x*sin(c + d*x)**4/8 + 3*A*a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*A*a**3*x*sin(c + d*
x)**2/2 + 3*A*a**3*x*cos(c + d*x)**4/8 + 3*A*a**3*x*cos(c + d*x)**2/2 + 3*A*a**3*sin(c + d*x)**3*cos(c + d*x)/
(8*d) + 2*A*a**3*sin(c + d*x)**3/d + 5*A*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 3*A*a**3*sin(c + d*x)*cos(c
+ d*x)**2/d + 3*A*a**3*sin(c + d*x)*cos(c + d*x)/(2*d) + A*a**3*sin(c + d*x)/d + 5*C*a**3*x*sin(c + d*x)**6/1
6 + 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*(A + C*cos(c)**2)*(a*cos(c) + a)**3*cos(c), True))

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