3.3.41 \(\int \cos (c+d x) (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \, dx\) [241]

Optimal. Leaf size=325 \[ \frac {1}{16} \left (32 a^3 A b+24 a A b^3+8 a^4 B+36 a^2 b^2 B+5 b^4 B\right ) x+\frac {\left (24 a^4 A b+224 a^2 A b^3+32 A b^5-4 a^5 B+121 a^3 b^2 B+128 a b^4 B\right ) \sin (c+d x)}{60 b d}+\frac {\left (48 a^3 A b+232 a A b^3-8 a^4 B+178 a^2 b^2 B+75 b^4 B\right ) \cos (c+d x) \sin (c+d x)}{240 d}+\frac {\left (24 a^2 A b+32 A b^3-4 a^3 B+53 a b^2 B\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{120 b d}+\frac {\left (24 a A b-4 a^2 B+25 b^2 B\right ) (a+b \cos (c+d x))^3 \sin (c+d x)}{120 b d}+\frac {(6 A b-a B) (a+b \cos (c+d x))^4 \sin (c+d x)}{30 b d}+\frac {B (a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d} \]

[Out]

1/16*(32*A*a^3*b+24*A*a*b^3+8*B*a^4+36*B*a^2*b^2+5*B*b^4)*x+1/60*(24*A*a^4*b+224*A*a^2*b^3+32*A*b^5-4*B*a^5+12
1*B*a^3*b^2+128*B*a*b^4)*sin(d*x+c)/b/d+1/240*(48*A*a^3*b+232*A*a*b^3-8*B*a^4+178*B*a^2*b^2+75*B*b^4)*cos(d*x+
c)*sin(d*x+c)/d+1/120*(24*A*a^2*b+32*A*b^3-4*B*a^3+53*B*a*b^2)*(a+b*cos(d*x+c))^2*sin(d*x+c)/b/d+1/120*(24*A*a
*b-4*B*a^2+25*B*b^2)*(a+b*cos(d*x+c))^3*sin(d*x+c)/b/d+1/30*(6*A*b-B*a)*(a+b*cos(d*x+c))^4*sin(d*x+c)/b/d+1/6*
B*(a+b*cos(d*x+c))^5*sin(d*x+c)/b/d

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Rubi [A]
time = 0.33, antiderivative size = 325, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3047, 3102, 2832, 2813} \begin {gather*} \frac {\left (-4 a^2 B+24 a A b+25 b^2 B\right ) \sin (c+d x) (a+b \cos (c+d x))^3}{120 b d}+\frac {\left (-4 a^3 B+24 a^2 A b+53 a b^2 B+32 A b^3\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{120 b d}+\frac {\left (-8 a^4 B+48 a^3 A b+178 a^2 b^2 B+232 a A b^3+75 b^4 B\right ) \sin (c+d x) \cos (c+d x)}{240 d}+\frac {1}{16} x \left (8 a^4 B+32 a^3 A b+36 a^2 b^2 B+24 a A b^3+5 b^4 B\right )+\frac {\left (-4 a^5 B+24 a^4 A b+121 a^3 b^2 B+224 a^2 A b^3+128 a b^4 B+32 A b^5\right ) \sin (c+d x)}{60 b d}+\frac {(6 A b-a B) \sin (c+d x) (a+b \cos (c+d x))^4}{30 b d}+\frac {B \sin (c+d x) (a+b \cos (c+d x))^5}{6 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x]),x]

[Out]

((32*a^3*A*b + 24*a*A*b^3 + 8*a^4*B + 36*a^2*b^2*B + 5*b^4*B)*x)/16 + ((24*a^4*A*b + 224*a^2*A*b^3 + 32*A*b^5
- 4*a^5*B + 121*a^3*b^2*B + 128*a*b^4*B)*Sin[c + d*x])/(60*b*d) + ((48*a^3*A*b + 232*a*A*b^3 - 8*a^4*B + 178*a
^2*b^2*B + 75*b^4*B)*Cos[c + d*x]*Sin[c + d*x])/(240*d) + ((24*a^2*A*b + 32*A*b^3 - 4*a^3*B + 53*a*b^2*B)*(a +
b*Cos[c + d*x])^2*Sin[c + d*x])/(120*b*d) + ((24*a*A*b - 4*a^2*B + 25*b^2*B)*(a + b*Cos[c + d*x])^3*Sin[c + d
*x])/(120*b*d) + ((6*A*b - a*B)*(a + b*Cos[c + d*x])^4*Sin[c + d*x])/(30*b*d) + (B*(a + b*Cos[c + d*x])^5*Sin[
c + d*x])/(6*b*d)

Rule 2813

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 2832

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)*Sim
p[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 3047

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 3102

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 \cos (c+d x) (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \, dx &=\int (a+b \cos (c+d x))^4 \left (A \cos (c+d x)+B \cos ^2(c+d x)\right ) \, dx\\ &=\frac {B (a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d}+\frac {\int (a+b \cos (c+d x))^4 (5 b B+(6 A b-a B) \cos (c+d x)) \, dx}{6 b}\\ &=\frac {(6 A b-a B) (a+b \cos (c+d x))^4 \sin (c+d x)}{30 b d}+\frac {B (a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d}+\frac {\int (a+b \cos (c+d x))^3 \left (3 b (8 A b+7 a B)+\left (24 a A b-4 a^2 B+25 b^2 B\right ) \cos (c+d x)\right ) \, dx}{30 b}\\ &=\frac {\left (24 a A b-4 a^2 B+25 b^2 B\right ) (a+b \cos (c+d x))^3 \sin (c+d x)}{120 b d}+\frac {(6 A b-a B) (a+b \cos (c+d x))^4 \sin (c+d x)}{30 b d}+\frac {B (a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d}+\frac {\int (a+b \cos (c+d x))^2 \left (3 b \left (56 a A b+24 a^2 B+25 b^2 B\right )+3 \left (24 a^2 A b+32 A b^3-4 a^3 B+53 a b^2 B\right ) \cos (c+d x)\right ) \, dx}{120 b}\\ &=\frac {\left (24 a^2 A b+32 A b^3-4 a^3 B+53 a b^2 B\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{120 b d}+\frac {\left (24 a A b-4 a^2 B+25 b^2 B\right ) (a+b \cos (c+d x))^3 \sin (c+d x)}{120 b d}+\frac {(6 A b-a B) (a+b \cos (c+d x))^4 \sin (c+d x)}{30 b d}+\frac {B (a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d}+\frac {\int (a+b \cos (c+d x)) \left (3 b \left (216 a^2 A b+64 A b^3+64 a^3 B+181 a b^2 B\right )+3 \left (48 a^3 A b+232 a A b^3-8 a^4 B+178 a^2 b^2 B+75 b^4 B\right ) \cos (c+d x)\right ) \, dx}{360 b}\\ &=\frac {1}{16} \left (32 a^3 A b+24 a A b^3+8 a^4 B+36 a^2 b^2 B+5 b^4 B\right ) x+\frac {\left (24 a^4 A b+224 a^2 A b^3+32 A b^5-4 a^5 B+121 a^3 b^2 B+128 a b^4 B\right ) \sin (c+d x)}{60 b d}+\frac {\left (48 a^3 A b+232 a A b^3-8 a^4 B+178 a^2 b^2 B+75 b^4 B\right ) \cos (c+d x) \sin (c+d x)}{240 d}+\frac {\left (24 a^2 A b+32 A b^3-4 a^3 B+53 a b^2 B\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{120 b d}+\frac {\left (24 a A b-4 a^2 B+25 b^2 B\right ) (a+b \cos (c+d x))^3 \sin (c+d x)}{120 b d}+\frac {(6 A b-a B) (a+b \cos (c+d x))^4 \sin (c+d x)}{30 b d}+\frac {B (a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d}\\ \end {align*}

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Mathematica [A]
time = 0.75, size = 333, normalized size = 1.02 \begin {gather*} \frac {1920 a^3 A b c+1440 a A b^3 c+480 a^4 B c+2160 a^2 b^2 B c+300 b^4 B c+1920 a^3 A b d x+1440 a A b^3 d x+480 a^4 B d x+2160 a^2 b^2 B d x+300 b^4 B d x+120 \left (8 a^4 A+36 a^2 A b^2+5 A b^4+24 a^3 b B+20 a b^3 B\right ) \sin (c+d x)+15 \left (64 a^3 A b+64 a A b^3+16 a^4 B+96 a^2 b^2 B+15 b^4 B\right ) \sin (2 (c+d x))+480 a^2 A b^2 \sin (3 (c+d x))+100 A b^4 \sin (3 (c+d x))+320 a^3 b B \sin (3 (c+d x))+400 a b^3 B \sin (3 (c+d x))+120 a A b^3 \sin (4 (c+d x))+180 a^2 b^2 B \sin (4 (c+d x))+45 b^4 B \sin (4 (c+d x))+12 A b^4 \sin (5 (c+d x))+48 a b^3 B \sin (5 (c+d x))+5 b^4 B \sin (6 (c+d x))}{960 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x]),x]

[Out]

(1920*a^3*A*b*c + 1440*a*A*b^3*c + 480*a^4*B*c + 2160*a^2*b^2*B*c + 300*b^4*B*c + 1920*a^3*A*b*d*x + 1440*a*A*
b^3*d*x + 480*a^4*B*d*x + 2160*a^2*b^2*B*d*x + 300*b^4*B*d*x + 120*(8*a^4*A + 36*a^2*A*b^2 + 5*A*b^4 + 24*a^3*
b*B + 20*a*b^3*B)*Sin[c + d*x] + 15*(64*a^3*A*b + 64*a*A*b^3 + 16*a^4*B + 96*a^2*b^2*B + 15*b^4*B)*Sin[2*(c +
d*x)] + 480*a^2*A*b^2*Sin[3*(c + d*x)] + 100*A*b^4*Sin[3*(c + d*x)] + 320*a^3*b*B*Sin[3*(c + d*x)] + 400*a*b^3
*B*Sin[3*(c + d*x)] + 120*a*A*b^3*Sin[4*(c + d*x)] + 180*a^2*b^2*B*Sin[4*(c + d*x)] + 45*b^4*B*Sin[4*(c + d*x)
] + 12*A*b^4*Sin[5*(c + d*x)] + 48*a*b^3*B*Sin[5*(c + d*x)] + 5*b^4*B*Sin[6*(c + d*x)])/(960*d)

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Maple [A]
time = 0.22, size = 316, normalized size = 0.97 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 307, normalized size = 0.94 \begin {gather*} \frac {240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 960 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} b - 1280 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} b - 1920 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} b^{2} + 180 \, {\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} b^{2} + 120 \, {\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 b^{3} + 256 \, {\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 b^{3} + 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 )} A b^{4} - 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 )} B b^{4} + 960 \, A a^{4} \sin \left (d x + c\right )}{960 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

1/960*(240*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 + 960*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3*b - 1280*(sin(d
*x + c)^3 - 3*sin(d*x + c))*B*a^3*b - 1920*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2*b^2 + 180*(12*d*x + 12*c +
sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^2*b^2 + 120*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))
*A*a*b^3 + 256*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*B*a*b^3 + 64*(3*sin(d*x + c)^5 - 10*si
n(d*x + c)^3 + 15*sin(d*x + c))*A*b^4 - 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))*B*b^4 + 960*A*a^4*sin(d*x + c))/d

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Fricas [A]
time = 0.39, size = 243, normalized size = 0.75 \begin {gather*} \frac {15 \, {\left (8 \, B a^{4} + 32 \, A a^{3} b + 36 \, B a^{2} b^{2} + 24 \, A a b^{3} + 5 \, B b^{4}\right )} d x + {\left (40 \, B b^{4} \cos \left (d x + c\right )^{5} + 240 \, A a^{4} + 640 \, B a^{3} b + 960 \, A a^{2} b^{2} + 512 \, B a b^{3} + 128 \, A b^{4} + 48 \, {\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{4} + 10 \, {\left (36 \, B a^{2} b^{2} + 24 \, A a b^{3} + 5 \, B b^{4}\right )} \cos \left (d x + c\right )^{3} + 32 \, {\left (10 \, B a^{3} b + 15 \, A a^{2} b^{2} + 8 \, B a b^{3} + 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (8 \, B a^{4} + 32 \, A a^{3} b + 36 \, B a^{2} b^{2} + 24 \, A a b^{3} + 5 \, B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

1/240*(15*(8*B*a^4 + 32*A*a^3*b + 36*B*a^2*b^2 + 24*A*a*b^3 + 5*B*b^4)*d*x + (40*B*b^4*cos(d*x + c)^5 + 240*A*
a^4 + 640*B*a^3*b + 960*A*a^2*b^2 + 512*B*a*b^3 + 128*A*b^4 + 48*(4*B*a*b^3 + A*b^4)*cos(d*x + c)^4 + 10*(36*B
*a^2*b^2 + 24*A*a*b^3 + 5*B*b^4)*cos(d*x + c)^3 + 32*(10*B*a^3*b + 15*A*a^2*b^2 + 8*B*a*b^3 + 2*A*b^4)*cos(d*x
+ c)^2 + 15*(8*B*a^4 + 32*A*a^3*b + 36*B*a^2*b^2 + 24*A*a*b^3 + 5*B*b^4)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 811 vs. \(2 (335) = 670\).
time = 0.56, size = 811, normalized size = 2.50 \begin {gather*} \begin {cases} \frac {A a^{4} \sin {\left (c + d x \right )}}{d} + 2 A a^{3} b x \sin ^{2}{\left (c + d x \right )} + 2 A a^{3} b x \cos ^{2}{\left (c + d x \right )} + \frac {2 A a^{3} b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {4 A a^{2} b^{2} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {6 A a^{2} b^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 A a b^{3} x \sin ^{4}{\left (c + d x \right )}}{2} + 3 A a b^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} + \frac {3 A a b^{3} x \cos ^{4}{\left (c + d x \right )}}{2} + \frac {3 A a b^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {5 A a b^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} + \frac {8 A b^{4} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 A b^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {A b^{4} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {B a^{4} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B a^{4} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B a^{4} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {8 B a^{3} b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {4 B a^{3} b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {9 B a^{2} b^{2} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {9 B a^{2} b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {9 B a^{2} b^{2} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {9 B a^{2} b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {15 B a^{2} b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {32 B a b^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {16 B a b^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {4 B a b^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 B b^{4} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 B b^{4} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 B b^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 B b^{4} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 B b^{4} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 B b^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 B b^{4} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )}\right ) \left (a + b \cos {\left (c \right )}\right )^{4} \cos {\left (c \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+b*cos(d*x+c))**4*(A+B*cos(d*x+c)),x)

[Out]

Piecewise((A*a**4*sin(c + d*x)/d + 2*A*a**3*b*x*sin(c + d*x)**2 + 2*A*a**3*b*x*cos(c + d*x)**2 + 2*A*a**3*b*si
n(c + d*x)*cos(c + d*x)/d + 4*A*a**2*b**2*sin(c + d*x)**3/d + 6*A*a**2*b**2*sin(c + d*x)*cos(c + d*x)**2/d + 3
*A*a*b**3*x*sin(c + d*x)**4/2 + 3*A*a*b**3*x*sin(c + d*x)**2*cos(c + d*x)**2 + 3*A*a*b**3*x*cos(c + d*x)**4/2
+ 3*A*a*b**3*sin(c + d*x)**3*cos(c + d*x)/(2*d) + 5*A*a*b**3*sin(c + d*x)*cos(c + d*x)**3/(2*d) + 8*A*b**4*sin
(c + d*x)**5/(15*d) + 4*A*b**4*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + A*b**4*sin(c + d*x)*cos(c + d*x)**4/d +
B*a**4*x*sin(c + d*x)**2/2 + B*a**4*x*cos(c + d*x)**2/2 + B*a**4*sin(c + d*x)*cos(c + d*x)/(2*d) + 8*B*a**3*b
*sin(c + d*x)**3/(3*d) + 4*B*a**3*b*sin(c + d*x)*cos(c + d*x)**2/d + 9*B*a**2*b**2*x*sin(c + d*x)**4/4 + 9*B*a
**2*b**2*x*sin(c + d*x)**2*cos(c + d*x)**2/2 + 9*B*a**2*b**2*x*cos(c + d*x)**4/4 + 9*B*a**2*b**2*sin(c + d*x)*
*3*cos(c + d*x)/(4*d) + 15*B*a**2*b**2*sin(c + d*x)*cos(c + d*x)**3/(4*d) + 32*B*a*b**3*sin(c + d*x)**5/(15*d)
+ 16*B*a*b**3*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 4*B*a*b**3*sin(c + d*x)*cos(c + d*x)**4/d + 5*B*b**4*x*
sin(c + d*x)**6/16 + 15*B*b**4*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*B*b**4*x*sin(c + d*x)**2*cos(c + d*x)
**4/16 + 5*B*b**4*x*cos(c + d*x)**6/16 + 5*B*b**4*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*B*b**4*sin(c + d*x)*
*3*cos(c + d*x)**3/(6*d) + 11*B*b**4*sin(c + d*x)*cos(c + d*x)**5/(16*d), Ne(d, 0)), (x*(A + B*cos(c))*(a + b*
cos(c))**4*cos(c), True))

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Giac [A]
time = 0.47, size = 263, normalized size = 0.81 \begin {gather*} \frac {B b^{4} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {1}{16} \, {\left (8 \, B a^{4} + 32 \, A a^{3} b + 36 \, B a^{2} b^{2} + 24 \, A a b^{3} + 5 \, B b^{4}\right )} x + \frac {{\left (4 \, B a b^{3} + A b^{4}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (12 \, B a^{2} b^{2} + 8 \, A a b^{3} + 3 \, B b^{4}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (16 \, B a^{3} b + 24 \, A a^{2} b^{2} + 20 \, B a b^{3} + 5 \, A b^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (16 \, B a^{4} + 64 \, A a^{3} b + 96 \, B a^{2} b^{2} + 64 \, A a b^{3} + 15 \, B b^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (8 \, A a^{4} + 24 \, B a^{3} b + 36 \, A a^{2} b^{2} + 20 \, B a b^{3} + 5 \, A b^{4}\right )} \sin \left (d x + c\right )}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

1/192*B*b^4*sin(6*d*x + 6*c)/d + 1/16*(8*B*a^4 + 32*A*a^3*b + 36*B*a^2*b^2 + 24*A*a*b^3 + 5*B*b^4)*x + 1/80*(4
*B*a*b^3 + A*b^4)*sin(5*d*x + 5*c)/d + 1/64*(12*B*a^2*b^2 + 8*A*a*b^3 + 3*B*b^4)*sin(4*d*x + 4*c)/d + 1/48*(16
*B*a^3*b + 24*A*a^2*b^2 + 20*B*a*b^3 + 5*A*b^4)*sin(3*d*x + 3*c)/d + 1/64*(16*B*a^4 + 64*A*a^3*b + 96*B*a^2*b^
2 + 64*A*a*b^3 + 15*B*b^4)*sin(2*d*x + 2*c)/d + 1/8*(8*A*a^4 + 24*B*a^3*b + 36*A*a^2*b^2 + 20*B*a*b^3 + 5*A*b^
4)*sin(d*x + c)/d

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Mupad [B]
time = 1.37, size = 403, normalized size = 1.24 \begin {gather*} \frac {B\,a^4\,x}{2}+\frac {5\,B\,b^4\,x}{16}+\frac {3\,A\,a\,b^3\,x}{2}+2\,A\,a^3\,b\,x+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {5\,A\,b^4\,\sin \left (c+d\,x\right )}{8\,d}+\frac {9\,B\,a^2\,b^2\,x}{4}+\frac {B\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {5\,A\,b^4\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {A\,b^4\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {15\,B\,b^4\,\sin \left (2\,c+2\,d\,x\right )}{64\,d}+\frac {3\,B\,b^4\,\sin \left (4\,c+4\,d\,x\right )}{64\,d}+\frac {B\,b^4\,\sin \left (6\,c+6\,d\,x\right )}{192\,d}+\frac {A\,a\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {A\,a^3\,b\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {A\,a\,b^3\,\sin \left (4\,c+4\,d\,x\right )}{8\,d}+\frac {9\,A\,a^2\,b^2\,\sin \left (c+d\,x\right )}{2\,d}+\frac {5\,B\,a\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,a^3\,b\,\sin \left (3\,c+3\,d\,x\right )}{3\,d}+\frac {B\,a\,b^3\,\sin \left (5\,c+5\,d\,x\right )}{20\,d}+\frac {A\,a^2\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{2\,d}+\frac {3\,B\,a^2\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {3\,B\,a^2\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{16\,d}+\frac {5\,B\,a\,b^3\,\sin \left (c+d\,x\right )}{2\,d}+\frac {3\,B\,a^3\,b\,\sin \left (c+d\,x\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)*(A + B*cos(c + d*x))*(a + b*cos(c + d*x))^4,x)

[Out]

(B*a^4*x)/2 + (5*B*b^4*x)/16 + (3*A*a*b^3*x)/2 + 2*A*a^3*b*x + (A*a^4*sin(c + d*x))/d + (5*A*b^4*sin(c + d*x))
/(8*d) + (9*B*a^2*b^2*x)/4 + (B*a^4*sin(2*c + 2*d*x))/(4*d) + (5*A*b^4*sin(3*c + 3*d*x))/(48*d) + (A*b^4*sin(5
*c + 5*d*x))/(80*d) + (15*B*b^4*sin(2*c + 2*d*x))/(64*d) + (3*B*b^4*sin(4*c + 4*d*x))/(64*d) + (B*b^4*sin(6*c
+ 6*d*x))/(192*d) + (A*a*b^3*sin(2*c + 2*d*x))/d + (A*a^3*b*sin(2*c + 2*d*x))/d + (A*a*b^3*sin(4*c + 4*d*x))/(
8*d) + (9*A*a^2*b^2*sin(c + d*x))/(2*d) + (5*B*a*b^3*sin(3*c + 3*d*x))/(12*d) + (B*a^3*b*sin(3*c + 3*d*x))/(3*
d) + (B*a*b^3*sin(5*c + 5*d*x))/(20*d) + (A*a^2*b^2*sin(3*c + 3*d*x))/(2*d) + (3*B*a^2*b^2*sin(2*c + 2*d*x))/(
2*d) + (3*B*a^2*b^2*sin(4*c + 4*d*x))/(16*d) + (5*B*a*b^3*sin(c + d*x))/(2*d) + (3*B*a^3*b*sin(c + d*x))/d

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