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

Optimal. Leaf size=256 \[ \frac{b \left (32 a^2 b^2 C+95 a^3 b B-83 a^4 C+80 a b^3 B+16 b^4 C\right ) \sin (c+d x)}{30 d}+\frac{b \left (-23 a^2 C+35 a b B+16 b^2 C\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{60 d}+\frac{b^2 \left (130 a^2 b B-106 a^3 C+71 a b^2 C+45 b^3 B\right ) \sin (c+d x) \cos (c+d x)}{120 d}+\frac{1}{8} x \left (24 a^2 b^3 B-8 a^3 b^2 C+8 a^4 b B-8 a^5 C+9 a b^4 C+3 b^5 B\right )+\frac{b (5 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^3}{20 d}+\frac{b C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d} \]

[Out]

((8*a^4*b*B + 24*a^2*b^3*B + 3*b^5*B - 8*a^5*C - 8*a^3*b^2*C + 9*a*b^4*C)*x)/8 + (b*(95*a^3*b*B + 80*a*b^3*B -
83*a^4*C + 32*a^2*b^2*C + 16*b^4*C)*Sin[c + d*x])/(30*d) + (b^2*(130*a^2*b*B + 45*b^3*B - 106*a^3*C + 71*a*b^
2*C)*Cos[c + d*x]*Sin[c + d*x])/(120*d) + (b*(35*a*b*B - 23*a^2*C + 16*b^2*C)*(a + b*Cos[c + d*x])^2*Sin[c + d
*x])/(60*d) + (b*(5*b*B - a*C)*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(20*d) + (b*C*(a + b*Cos[c + d*x])^4*Sin[c
+ d*x])/(5*d)

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Rubi [A]  time = 0.551879, antiderivative size = 256, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 48, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {3015, 2753, 2734} \[ \frac{b \left (32 a^2 b^2 C+95 a^3 b B-83 a^4 C+80 a b^3 B+16 b^4 C\right ) \sin (c+d x)}{30 d}+\frac{b \left (-23 a^2 C+35 a b B+16 b^2 C\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{60 d}+\frac{b^2 \left (130 a^2 b B-106 a^3 C+71 a b^2 C+45 b^3 B\right ) \sin (c+d x) \cos (c+d x)}{120 d}+\frac{1}{8} x \left (24 a^2 b^3 B-8 a^3 b^2 C+8 a^4 b B-8 a^5 C+9 a b^4 C+3 b^5 B\right )+\frac{b (5 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^3}{20 d}+\frac{b C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^4*b*B + 24*a^2*b^3*B + 3*b^5*B - 8*a^5*C - 8*a^3*b^2*C + 9*a*b^4*C)*x)/8 + (b*(95*a^3*b*B + 80*a*b^3*B -
83*a^4*C + 32*a^2*b^2*C + 16*b^4*C)*Sin[c + d*x])/(30*d) + (b^2*(130*a^2*b*B + 45*b^3*B - 106*a^3*C + 71*a*b^
2*C)*Cos[c + d*x]*Sin[c + d*x])/(120*d) + (b*(35*a*b*B - 23*a^2*C + 16*b^2*C)*(a + b*Cos[c + d*x])^2*Sin[c + d
*x])/(60*d) + (b*(5*b*B - a*C)*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(20*d) + (b*C*(a + b*Cos[c + d*x])^4*Sin[c
+ d*x])/(5*d)

Rule 3015

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((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)*Simp[b*B - a*C + b*C*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 2753

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)*Simp[
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 2734

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]

Rubi steps

\begin{align*} \int (a+b \cos (c+d x))^3 \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx &=\frac{\int (a+b \cos (c+d x))^4 \left (b^2 (b B-a C)+b^3 C \cos (c+d x)\right ) \, dx}{b^2}\\ &=\frac{b C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{\int (a+b \cos (c+d x))^3 \left (b^2 \left (4 b^2 C+5 a (b B-a C)\right )+b^3 (5 b B-a C) \cos (c+d x)\right ) \, dx}{5 b^2}\\ &=\frac{b (5 b B-a C) (a+b \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac{b C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{\int (a+b \cos (c+d x))^2 \left (b^2 \left (20 a^2 b B+15 b^3 B-20 a^3 C+13 a b^2 C\right )+b^3 \left (35 a b B-23 a^2 C+16 b^2 C\right ) \cos (c+d x)\right ) \, dx}{20 b^2}\\ &=\frac{b \left (35 a b B-23 a^2 C+16 b^2 C\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{60 d}+\frac{b (5 b B-a C) (a+b \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac{b C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{\int (a+b \cos (c+d x)) \left (b^2 \left (60 a^3 b B+115 a b^3 B-60 a^4 C-7 a^2 b^2 C+32 b^4 C\right )+b^3 \left (130 a^2 b B+45 b^3 B-106 a^3 C+71 a b^2 C\right ) \cos (c+d x)\right ) \, dx}{60 b^2}\\ &=\frac{1}{8} \left (8 a^4 b B+24 a^2 b^3 B+3 b^5 B-8 a^5 C-8 a^3 b^2 C+9 a b^4 C\right ) x+\frac{b \left (95 a^3 b B+80 a b^3 B-83 a^4 C+32 a^2 b^2 C+16 b^4 C\right ) \sin (c+d x)}{30 d}+\frac{b^2 \left (130 a^2 b B+45 b^3 B-106 a^3 C+71 a b^2 C\right ) \cos (c+d x) \sin (c+d x)}{120 d}+\frac{b \left (35 a b B-23 a^2 C+16 b^2 C\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{60 d}+\frac{b (5 b B-a C) (a+b \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac{b C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 1.10825, size = 287, normalized size = 1.12 \[ \frac{60 b \left (12 a^2 b^2 C+32 a^3 b B-24 a^4 C+24 a b^3 B+5 b^4 C\right ) \sin (c+d x)+120 b^2 \left (6 a^2 b B-2 a^3 C+3 a b^2 C+b^3 B\right ) \sin (2 (c+d x))+1440 a^2 b^3 B c+1440 a^2 b^3 B d x+80 a^2 b^3 C \sin (3 (c+d x))-480 a^3 b^2 c C-480 a^3 b^2 C d x+480 a^4 b B c+480 a^4 b B d x-480 a^5 c C-480 a^5 C d x+160 a b^4 B \sin (3 (c+d x))+45 a b^4 C \sin (4 (c+d x))+540 a b^4 c C+540 a b^4 C d x+15 b^5 B \sin (4 (c+d x))+180 b^5 B c+180 b^5 B d x+50 b^5 C \sin (3 (c+d x))+6 b^5 C \sin (5 (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(480*a^4*b*B*c + 1440*a^2*b^3*B*c + 180*b^5*B*c - 480*a^5*c*C - 480*a^3*b^2*c*C + 540*a*b^4*c*C + 480*a^4*b*B*
d*x + 1440*a^2*b^3*B*d*x + 180*b^5*B*d*x - 480*a^5*C*d*x - 480*a^3*b^2*C*d*x + 540*a*b^4*C*d*x + 60*b*(32*a^3*
b*B + 24*a*b^3*B - 24*a^4*C + 12*a^2*b^2*C + 5*b^4*C)*Sin[c + d*x] + 120*b^2*(6*a^2*b*B + b^3*B - 2*a^3*C + 3*
a*b^2*C)*Sin[2*(c + d*x)] + 160*a*b^4*B*Sin[3*(c + d*x)] + 80*a^2*b^3*C*Sin[3*(c + d*x)] + 50*b^5*C*Sin[3*(c +
d*x)] + 15*b^5*B*Sin[4*(c + d*x)] + 45*a*b^4*C*Sin[4*(c + d*x)] + 6*b^5*C*Sin[5*(c + d*x)])/(480*d)

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Maple [A]  time = 0.021, size = 276, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{C{b}^{5}\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+{b}^{5}B \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +3\,Ca{b}^{4} \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +{\frac{4\,a{b}^{4}B \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{2\,C{a}^{2}{b}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+6\,{a}^{2}{b}^{3}B \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) -2\,{a}^{3}{b}^{2}C \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +4\,{a}^{3}{b}^{2}B\sin \left ( dx+c \right ) -3\,C{a}^{4}b\sin \left ( dx+c \right ) +{a}^{4}bB \left ( dx+c \right ) -{a}^{5}C \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01086, size = 355, normalized size = 1.39 \begin{align*} -\frac{480 \,{\left (d x + c\right )} C a^{5} - 480 \,{\left (d x + c\right )} B a^{4} b + 240 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} b^{2} - 720 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} b^{3} + 320 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} b^{3} + 640 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b^{4} - 45 \,{\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 b^{4} - 15 \,{\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 b^{5} - 32 \,{\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 b^{5} + 1440 \, C a^{4} b \sin \left (d x + c\right ) - 1920 \, B a^{3} b^{2} \sin \left (d x + c\right )}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/480*(480*(d*x + c)*C*a^5 - 480*(d*x + c)*B*a^4*b + 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3*b^2 - 720*(2*
d*x + 2*c + sin(2*d*x + 2*c))*B*a^2*b^3 + 320*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^2*b^3 + 640*(sin(d*x + c)^
3 - 3*sin(d*x + c))*B*a*b^4 - 45*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a*b^4 - 15*(12*d*x
+ 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*b^5 - 32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x
+ c))*C*b^5 + 1440*C*a^4*b*sin(d*x + c) - 1920*B*a^3*b^2*sin(d*x + c))/d

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Fricas [A]  time = 1.87124, size = 500, normalized size = 1.95 \begin{align*} -\frac{15 \,{\left (8 \, C a^{5} - 8 \, B a^{4} b + 8 \, C a^{3} b^{2} - 24 \, B a^{2} b^{3} - 9 \, C a b^{4} - 3 \, B b^{5}\right )} d x -{\left (24 \, C b^{5} \cos \left (d x + c\right )^{4} - 360 \, C a^{4} b + 480 \, B a^{3} b^{2} + 160 \, C a^{2} b^{3} + 320 \, B a b^{4} + 64 \, C b^{5} + 30 \,{\left (3 \, C a b^{4} + B b^{5}\right )} \cos \left (d x + c\right )^{3} + 16 \,{\left (5 \, C a^{2} b^{3} + 10 \, B a b^{4} + 2 \, C b^{5}\right )} \cos \left (d x + c\right )^{2} - 15 \,{\left (8 \, C a^{3} b^{2} - 24 \, B a^{2} b^{3} - 9 \, C a b^{4} - 3 \, B b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(15*(8*C*a^5 - 8*B*a^4*b + 8*C*a^3*b^2 - 24*B*a^2*b^3 - 9*C*a*b^4 - 3*B*b^5)*d*x - (24*C*b^5*cos(d*x +
c)^4 - 360*C*a^4*b + 480*B*a^3*b^2 + 160*C*a^2*b^3 + 320*B*a*b^4 + 64*C*b^5 + 30*(3*C*a*b^4 + B*b^5)*cos(d*x +
c)^3 + 16*(5*C*a^2*b^3 + 10*B*a*b^4 + 2*C*b^5)*cos(d*x + c)^2 - 15*(8*C*a^3*b^2 - 24*B*a^2*b^3 - 9*C*a*b^4 -
3*B*b^5)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 3.848, size = 619, normalized size = 2.42 \begin{align*} \begin{cases} B a^{4} b x + \frac{4 B a^{3} b^{2} \sin{\left (c + d x \right )}}{d} + 3 B a^{2} b^{3} x \sin ^{2}{\left (c + d x \right )} + 3 B a^{2} b^{3} x \cos ^{2}{\left (c + d x \right )} + \frac{3 B a^{2} b^{3} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{8 B a b^{4} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{4 B a b^{4} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 B b^{5} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 B b^{5} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 B b^{5} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 B b^{5} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 B b^{5} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - C a^{5} x - \frac{3 C a^{4} b \sin{\left (c + d x \right )}}{d} - C a^{3} b^{2} x \sin ^{2}{\left (c + d x \right )} - C a^{3} b^{2} x \cos ^{2}{\left (c + d x \right )} - \frac{C a^{3} b^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{4 C a^{2} b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{2 C a^{2} b^{3} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{9 C a b^{4} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{9 C a b^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{9 C a b^{4} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{9 C a b^{4} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{15 C a b^{4} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{8 C b^{5} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 C b^{5} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{C b^{5} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \cos{\left (c \right )}\right )^{3} \left (B a b + B b^{2} \cos{\left (c \right )} - C a^{2} + C b^{2} \cos ^{2}{\left (c \right )}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18481, size = 306, normalized size = 1.2 \begin{align*} \frac{C b^{5} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac{1}{8} \,{\left (8 \, C a^{5} - 8 \, B a^{4} b + 8 \, C a^{3} b^{2} - 24 \, B a^{2} b^{3} - 9 \, C a b^{4} - 3 \, B b^{5}\right )} x + \frac{{\left (3 \, C a b^{4} + B b^{5}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{{\left (8 \, C a^{2} b^{3} + 16 \, B a b^{4} + 5 \, C b^{5}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac{{\left (2 \, C a^{3} b^{2} - 6 \, B a^{2} b^{3} - 3 \, C a b^{4} - B b^{5}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} - \frac{{\left (24 \, C a^{4} b - 32 \, B a^{3} b^{2} - 12 \, C a^{2} b^{3} - 24 \, B a b^{4} - 5 \, C b^{5}\right )} \sin \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*C*b^5*sin(5*d*x + 5*c)/d - 1/8*(8*C*a^5 - 8*B*a^4*b + 8*C*a^3*b^2 - 24*B*a^2*b^3 - 9*C*a*b^4 - 3*B*b^5)*x
+ 1/32*(3*C*a*b^4 + B*b^5)*sin(4*d*x + 4*c)/d + 1/48*(8*C*a^2*b^3 + 16*B*a*b^4 + 5*C*b^5)*sin(3*d*x + 3*c)/d
- 1/4*(2*C*a^3*b^2 - 6*B*a^2*b^3 - 3*C*a*b^4 - B*b^5)*sin(2*d*x + 2*c)/d - 1/8*(24*C*a^4*b - 32*B*a^3*b^2 - 12
*C*a^2*b^3 - 24*B*a*b^4 - 5*C*b^5)*sin(d*x + c)/d