3.428 \(\int \cos ^3(c+d x) (a+b \cos (c+d x))^3 \, dx\)

Optimal. Leaf size=170 \[ -\frac{a \left (a^2+6 b^2\right ) \sin ^3(c+d x)}{3 d}+\frac{a \left (a^2+3 b^2\right ) \sin (c+d x)}{d}+\frac{b \left (18 a^2+5 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{b \left (18 a^2+5 b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{9}{8} a^2 b x+\frac{3 a b^2 \sin ^5(c+d x)}{5 d}+\frac{b^3 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{5 b^3 x}{16} \]

[Out]

(9*a^2*b*x)/8 + (5*b^3*x)/16 + (a*(a^2 + 3*b^2)*Sin[c + d*x])/d + (b*(18*a^2 + 5*b^2)*Cos[c + d*x]*Sin[c + d*x
])/(16*d) + (b*(18*a^2 + 5*b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) + (b^3*Cos[c + d*x]^5*Sin[c + d*x])/(6*d)
- (a*(a^2 + 6*b^2)*Sin[c + d*x]^3)/(3*d) + (3*a*b^2*Sin[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.214977, antiderivative size = 193, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2793, 3023, 2748, 2633, 2635, 8} \[ -\frac{a \left (5 a^2+12 b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{a \left (5 a^2+12 b^2\right ) \sin (c+d x)}{5 d}+\frac{b \left (18 a^2+5 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{b \left (18 a^2+5 b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} b x \left (18 a^2+5 b^2\right )+\frac{b^2 \sin (c+d x) \cos ^4(c+d x) (a+b \cos (c+d x))}{6 d}+\frac{13 a b^2 \sin (c+d x) \cos ^4(c+d x)}{30 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(18*a^2 + 5*b^2)*x)/16 + (a*(5*a^2 + 12*b^2)*Sin[c + d*x])/(5*d) + (b*(18*a^2 + 5*b^2)*Cos[c + d*x]*Sin[c +
d*x])/(16*d) + (b*(18*a^2 + 5*b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) + (13*a*b^2*Cos[c + d*x]^4*Sin[c + d*x
])/(30*d) + (b^2*Cos[c + d*x]^4*(a + b*Cos[c + d*x])*Sin[c + d*x])/(6*d) - (a*(5*a^2 + 12*b^2)*Sin[c + d*x]^3)
/(15*d)

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[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 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 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (a+b \cos (c+d x))^3 \, dx &=\frac{b^2 \cos ^4(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{6 d}+\frac{1}{6} \int \cos ^3(c+d x) \left (2 a \left (3 a^2+2 b^2\right )+b \left (18 a^2+5 b^2\right ) \cos (c+d x)+13 a b^2 \cos ^2(c+d x)\right ) \, dx\\ &=\frac{13 a b^2 \cos ^4(c+d x) \sin (c+d x)}{30 d}+\frac{b^2 \cos ^4(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{6 d}+\frac{1}{30} \int \cos ^3(c+d x) \left (6 a \left (5 a^2+12 b^2\right )+5 b \left (18 a^2+5 b^2\right ) \cos (c+d x)\right ) \, dx\\ &=\frac{13 a b^2 \cos ^4(c+d x) \sin (c+d x)}{30 d}+\frac{b^2 \cos ^4(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{6 d}+\frac{1}{6} \left (b \left (18 a^2+5 b^2\right )\right ) \int \cos ^4(c+d x) \, dx+\frac{1}{5} \left (a \left (5 a^2+12 b^2\right )\right ) \int \cos ^3(c+d x) \, dx\\ &=\frac{b \left (18 a^2+5 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{13 a b^2 \cos ^4(c+d x) \sin (c+d x)}{30 d}+\frac{b^2 \cos ^4(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{6 d}+\frac{1}{8} \left (b \left (18 a^2+5 b^2\right )\right ) \int \cos ^2(c+d x) \, dx-\frac{\left (a \left (5 a^2+12 b^2\right )\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac{a \left (5 a^2+12 b^2\right ) \sin (c+d x)}{5 d}+\frac{b \left (18 a^2+5 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{b \left (18 a^2+5 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{13 a b^2 \cos ^4(c+d x) \sin (c+d x)}{30 d}+\frac{b^2 \cos ^4(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{6 d}-\frac{a \left (5 a^2+12 b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{1}{16} \left (b \left (18 a^2+5 b^2\right )\right ) \int 1 \, dx\\ &=\frac{1}{16} b \left (18 a^2+5 b^2\right ) x+\frac{a \left (5 a^2+12 b^2\right ) \sin (c+d x)}{5 d}+\frac{b \left (18 a^2+5 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{b \left (18 a^2+5 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{13 a b^2 \cos ^4(c+d x) \sin (c+d x)}{30 d}+\frac{b^2 \cos ^4(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{6 d}-\frac{a \left (5 a^2+12 b^2\right ) \sin ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.328951, size = 159, normalized size = 0.94 \[ \frac{360 a \left (2 a^2+5 b^2\right ) \sin (c+d x)+45 \left (16 a^2 b+5 b^3\right ) \sin (2 (c+d x))+90 a^2 b \sin (4 (c+d x))+1080 a^2 b c+1080 a^2 b d x+80 a^3 \sin (3 (c+d x))+300 a b^2 \sin (3 (c+d x))+36 a b^2 \sin (5 (c+d x))+45 b^3 \sin (4 (c+d x))+5 b^3 \sin (6 (c+d x))+300 b^3 c+300 b^3 d x}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1080*a^2*b*c + 300*b^3*c + 1080*a^2*b*d*x + 300*b^3*d*x + 360*a*(2*a^2 + 5*b^2)*Sin[c + d*x] + 45*(16*a^2*b +
5*b^3)*Sin[2*(c + d*x)] + 80*a^3*Sin[3*(c + d*x)] + 300*a*b^2*Sin[3*(c + d*x)] + 90*a^2*b*Sin[4*(c + d*x)] +
45*b^3*Sin[4*(c + d*x)] + 36*a*b^2*Sin[5*(c + d*x)] + 5*b^3*Sin[6*(c + d*x)])/(960*d)

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Maple [A]  time = 0.038, size = 145, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({b}^{3} \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) +{\frac{3\,a{b}^{2}\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 ) }+3\,{a}^{2}b \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{{a}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+b*cos(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 0.99068, size = 196, normalized size = 1.15 \begin{align*} -\frac{320 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} 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 )} a^{2} b - 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 )} a b^{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 )} b^{3}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.0297, size = 324, normalized size = 1.91 \begin{align*} \frac{15 \,{\left (18 \, a^{2} b + 5 \, b^{3}\right )} d x +{\left (40 \, b^{3} \cos \left (d x + c\right )^{5} + 144 \, a b^{2} \cos \left (d x + c\right )^{4} + 10 \,{\left (18 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 160 \, a^{3} + 384 \, a b^{2} + 16 \,{\left (5 \, a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (18 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(18*a^2*b + 5*b^3)*d*x + (40*b^3*cos(d*x + c)^5 + 144*a*b^2*cos(d*x + c)^4 + 10*(18*a^2*b + 5*b^3)*c
os(d*x + c)^3 + 160*a^3 + 384*a*b^2 + 16*(5*a^3 + 12*a*b^2)*cos(d*x + c)^2 + 15*(18*a^2*b + 5*b^3)*cos(d*x + c
))*sin(d*x + c))/d

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Sympy [A]  time = 4.89508, size = 393, normalized size = 2.31 \begin{align*} \begin{cases} \frac{2 a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{a^{3} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{9 a^{2} b x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{9 a^{2} b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{9 a^{2} b x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{9 a^{2} b \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{15 a^{2} b \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{8 a b^{2} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac{4 a b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 a b^{2} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac{5 b^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{15 b^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{15 b^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{5 b^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{5 b^{3} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{5 b^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac{11 b^{3} \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 )^{3} \cos ^{3}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+b*cos(d*x+c))**3,x)

[Out]

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

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Giac [A]  time = 1.3278, size = 203, normalized size = 1.19 \begin{align*} \frac{b^{3} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{3 \, a b^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{1}{16} \,{\left (18 \, a^{2} b + 5 \, b^{3}\right )} x + \frac{3 \,{\left (2 \, a^{2} b + b^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{{\left (4 \, a^{3} + 15 \, a b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{3 \,{\left (16 \, a^{2} b + 5 \, b^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac{3 \,{\left (2 \, a^{3} + 5 \, a b^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*b^3*sin(6*d*x + 6*c)/d + 3/80*a*b^2*sin(5*d*x + 5*c)/d + 1/16*(18*a^2*b + 5*b^3)*x + 3/64*(2*a^2*b + b^3
)*sin(4*d*x + 4*c)/d + 1/48*(4*a^3 + 15*a*b^2)*sin(3*d*x + 3*c)/d + 3/64*(16*a^2*b + 5*b^3)*sin(2*d*x + 2*c)/d
+ 3/8*(2*a^3 + 5*a*b^2)*sin(d*x + c)/d