3.5.28 \(\int \cos ^3(c+d x) (a+b \cos (c+d x))^3 \, dx\) [428]

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

[Out]

9/8*a^2*b*x+5/16*b^3*x+a*(a^2+3*b^2)*sin(d*x+c)/d+1/16*b*(18*a^2+5*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/24*b*(18*a^2
+5*b^2)*cos(d*x+c)^3*sin(d*x+c)/d+1/6*b^3*cos(d*x+c)^5*sin(d*x+c)/d-1/3*a*(a^2+6*b^2)*sin(d*x+c)^3/d+3/5*a*b^2
*sin(d*x+c)^5/d

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Rubi [A]
time = 0.15, 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 = {2872, 3102, 2827, 2713, 2715, 8} \begin {gather*} -\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} \end {gather*}

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 8

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

Rule 2713

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 2715

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] && Integ
erQ[2*n]

Rule 2827

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 2872

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-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 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 ^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 ) \text {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*}

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Mathematica [A]
time = 0.34, size = 159, normalized size = 0.94 \begin {gather*} \frac {1080 a^2 b c+300 b^3 c+1080 a^2 b d x+300 b^3 d x+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))+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} \end {gather*}

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 = 0.16, size = 145, normalized size = 0.85

method	result	size
derivativedivides	\(\frac {b^{3} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {3 b^{2} a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 a^{2} b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{3} \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(145\)
default	\(\frac {b^{3} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {3 b^{2} a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 a^{2} b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{3} \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(145\)
risch	\(\frac {9 a^{2} b x}{8}+\frac {5 b^{3} x}{16}+\frac {3 a^{3} \sin \left (d x +c \right )}{4 d}+\frac {15 a \,b^{2} \sin \left (d x +c \right )}{8 d}+\frac {b^{3} \sin \left (6 d x +6 c \right )}{192 d}+\frac {3 b^{2} a \sin \left (5 d x +5 c \right )}{80 d}+\frac {3 \sin \left (4 d x +4 c \right ) a^{2} b}{32 d}+\frac {3 \sin \left (4 d x +4 c \right ) b^{3}}{64 d}+\frac {a^{3} \sin \left (3 d x +3 c \right )}{12 d}+\frac {5 \sin \left (3 d x +3 c \right ) b^{2} a}{16 d}+\frac {3 \sin \left (2 d x +2 c \right ) a^{2} b}{4 d}+\frac {15 \sin \left (2 d x +2 c \right ) b^{3}}{64 d}\)	\(184\)
norman	\(\frac {\left (\frac {9}{8} a^{2} b +\frac {5}{16} b^{3}\right ) x +\left (\frac {9}{8} a^{2} b +\frac {5}{16} b^{3}\right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {27}{4} a^{2} b +\frac {15}{8} b^{3}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {27}{4} a^{2} b +\frac {15}{8} b^{3}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {45}{2} a^{2} b +\frac {25}{4} b^{3}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {135}{8} a^{2} b +\frac {75}{16} b^{3}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {135}{8} a^{2} b +\frac {75}{16} b^{3}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (16 a^{3}-30 a^{2} b +48 b^{2} a -11 b^{3}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {\left (16 a^{3}+30 a^{2} b +48 b^{2} a +11 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {3 \left (80 a^{3}-10 a^{2} b +208 b^{2} a -25 b^{3}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d}+\frac {3 \left (80 a^{3}+10 a^{2} b +208 b^{2} a +25 b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d}+\frac {\left (176 a^{3}-126 a^{2} b +336 b^{2} a +5 b^{3}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {\left (176 a^{3}+126 a^{2} b +336 b^{2} a -5 b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\)	\(414\)

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,method=_RETURNVERBOSE)

[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*b^2*a*(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*(cos(d*x+c)^2+2)*sin(d*x+c))

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Maxima [A]
time = 0.28, size = 145, normalized size = 0.85 \begin {gather*} -\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 {gather*}

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 = 0.41, size = 132, normalized size = 0.78 \begin {gather*} \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 {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 393 vs. \(2 (158) = 316\).
time = 0.45, size = 393, normalized size = 2.31 \begin {gather*} \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 {gather*}

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 = 0.41, size = 150, normalized size = 0.88 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 2.10, size = 380, normalized size = 2.24 \begin {gather*} \frac {\left (2\,a^3-\frac {15\,a^2\,b}{4}+6\,a\,b^2-\frac {11\,b^3}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {22\,a^3}{3}-\frac {21\,a^2\,b}{4}+14\,a\,b^2+\frac {5\,b^3}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (12\,a^3-\frac {3\,a^2\,b}{2}+\frac {156\,a\,b^2}{5}-\frac {15\,b^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (12\,a^3+\frac {3\,a^2\,b}{2}+\frac {156\,a\,b^2}{5}+\frac {15\,b^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {22\,a^3}{3}+\frac {21\,a^2\,b}{4}+14\,a\,b^2-\frac {5\,b^3}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^3+\frac {15\,a^2\,b}{4}+6\,a\,b^2+\frac {11\,b^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 {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (18\,a^2+5\,b^2\right )}{8\,\left (\frac {9\,a^2\,b}{4}+\frac {5\,b^3}{8}\right )}\right )\,\left (18\,a^2+5\,b^2\right )}{8\,d}-\frac {b\,\left (18\,a^2+5\,b^2\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{8\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^11*(6*a*b^2 - (15*a^2*b)/4 + 2*a^3 - (11*b^3)/8) + tan(c/2 + (d*x)/2)^3*(14*a*b^2 + (21*a^
2*b)/4 + (22*a^3)/3 - (5*b^3)/24) + tan(c/2 + (d*x)/2)^9*(14*a*b^2 - (21*a^2*b)/4 + (22*a^3)/3 + (5*b^3)/24) +
tan(c/2 + (d*x)/2)^5*((156*a*b^2)/5 + (3*a^2*b)/2 + 12*a^3 + (15*b^3)/4) + tan(c/2 + (d*x)/2)^7*((156*a*b^2)/
5 - (3*a^2*b)/2 + 12*a^3 - (15*b^3)/4) + tan(c/2 + (d*x)/2)*(6*a*b^2 + (15*a^2*b)/4 + 2*a^3 + (11*b^3)/8))/(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)) + (b*atan((b*tan(c/2 + (d*x)/2)*(18*a^2 + 5*b^2))/(8*((9*a^2*b
)/4 + (5*b^3)/8)))*(18*a^2 + 5*b^2))/(8*d) - (b*(18*a^2 + 5*b^2)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(8*d)

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