\(\int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx\) [59]

Optimal result

Integrand size = 28, antiderivative size = 216 \[ \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {5 a^3 x}{16}+\frac {3}{16} a b^2 x-\frac {a^2 b \cos ^6(c+d x)}{2 d}+\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {3 a b^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a b^2 \cos ^3(c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {a b^2 \cos ^5(c+d x) \sin (c+d x)}{2 d}+\frac {b^3 \sin ^4(c+d x)}{4 d}-\frac {b^3 \sin ^6(c+d x)}{6 d} \]

[Out]

5/16*a^3*x+3/16*a*b^2*x-1/2*a^2*b*cos(d*x+c)^6/d+5/16*a^3*cos(d*x+c)*sin(d*x+c)/d+3/16*a*b^2*cos(d*x+c)*sin(d*
x+c)/d+5/24*a^3*cos(d*x+c)^3*sin(d*x+c)/d+1/8*a*b^2*cos(d*x+c)^3*sin(d*x+c)/d+1/6*a^3*cos(d*x+c)^5*sin(d*x+c)/
d-1/2*a*b^2*cos(d*x+c)^5*sin(d*x+c)/d+1/4*b^3*sin(d*x+c)^4/d-1/6*b^3*sin(d*x+c)^6/d

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3169, 2715, 8, 2645, 30, 2648, 2644, 14} \[ \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {a^3 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5 a^3 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {5 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a^3 x}{16}-\frac {a^2 b \cos ^6(c+d x)}{2 d}-\frac {a b^2 \sin (c+d x) \cos ^5(c+d x)}{2 d}+\frac {a b^2 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac {3 a b^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {3}{16} a b^2 x-\frac {b^3 \sin ^6(c+d x)}{6 d}+\frac {b^3 \sin ^4(c+d x)}{4 d} \]

[In]

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

[Out]

(5*a^3*x)/16 + (3*a*b^2*x)/16 - (a^2*b*Cos[c + d*x]^6)/(2*d) + (5*a^3*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (3*a
*b^2*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (5*a^3*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) + (a*b^2*Cos[c + d*x]^3*Si
n[c + d*x])/(8*d) + (a^3*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (a*b^2*Cos[c + d*x]^5*Sin[c + d*x])/(2*d) + (b^3
*Sin[c + d*x]^4)/(4*d) - (b^3*Sin[c + d*x]^6)/(6*d)

Rule 8

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2648

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-a)*(b*Cos[e
+ f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Dist[a^2*((m - 1)/(m + n)), Int[(b*Cos[e + f*x
])^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[
2*m, 2*n]

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 3169

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (a^3 \cos ^6(c+d x)+3 a^2 b \cos ^5(c+d x) \sin (c+d x)+3 a b^2 \cos ^4(c+d x) \sin ^2(c+d x)+b^3 \cos ^3(c+d x) \sin ^3(c+d x)\right ) \, dx \\ & = a^3 \int \cos ^6(c+d x) \, dx+\left (3 a^2 b\right ) \int \cos ^5(c+d x) \sin (c+d x) \, dx+\left (3 a b^2\right ) \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx+b^3 \int \cos ^3(c+d x) \sin ^3(c+d x) \, dx \\ & = \frac {a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {a b^2 \cos ^5(c+d x) \sin (c+d x)}{2 d}+\frac {1}{6} \left (5 a^3\right ) \int \cos ^4(c+d x) \, dx+\frac {1}{2} \left (a b^2\right ) \int \cos ^4(c+d x) \, dx-\frac {\left (3 a^2 b\right ) \text {Subst}\left (\int x^5 \, dx,x,\cos (c+d x)\right )}{d}+\frac {b^3 \text {Subst}\left (\int x^3 \left (1-x^2\right ) \, dx,x,\sin (c+d x)\right )}{d} \\ & = -\frac {a^2 b \cos ^6(c+d x)}{2 d}+\frac {5 a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a b^2 \cos ^3(c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {a b^2 \cos ^5(c+d x) \sin (c+d x)}{2 d}+\frac {1}{8} \left (5 a^3\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{8} \left (3 a b^2\right ) \int \cos ^2(c+d x) \, dx+\frac {b^3 \text {Subst}\left (\int \left (x^3-x^5\right ) \, dx,x,\sin (c+d x)\right )}{d} \\ & = -\frac {a^2 b \cos ^6(c+d x)}{2 d}+\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {3 a b^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a b^2 \cos ^3(c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {a b^2 \cos ^5(c+d x) \sin (c+d x)}{2 d}+\frac {b^3 \sin ^4(c+d x)}{4 d}-\frac {b^3 \sin ^6(c+d x)}{6 d}+\frac {1}{16} \left (5 a^3\right ) \int 1 \, dx+\frac {1}{16} \left (3 a b^2\right ) \int 1 \, dx \\ & = \frac {5 a^3 x}{16}+\frac {3}{16} a b^2 x-\frac {a^2 b \cos ^6(c+d x)}{2 d}+\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {3 a b^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a b^2 \cos ^3(c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {a b^2 \cos ^5(c+d x) \sin (c+d x)}{2 d}+\frac {b^3 \sin ^4(c+d x)}{4 d}-\frac {b^3 \sin ^6(c+d x)}{6 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.90 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.79 \[ \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {a \left (5 a^2+3 b^2\right ) (c+d x)}{16 d}-\frac {3 b \left (5 a^2+b^2\right ) \cos (2 (c+d x))}{64 d}-\frac {3 a^2 b \cos (4 (c+d x))}{32 d}-\frac {b \left (3 a^2-b^2\right ) \cos (6 (c+d x))}{192 d}+\frac {3 a \left (5 a^2+b^2\right ) \sin (2 (c+d x))}{64 d}+\frac {3 a \left (a^2-b^2\right ) \sin (4 (c+d x))}{64 d}+\frac {a \left (a^2-3 b^2\right ) \sin (6 (c+d x))}{192 d} \]

[In]

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

[Out]

(a*(5*a^2 + 3*b^2)*(c + d*x))/(16*d) - (3*b*(5*a^2 + b^2)*Cos[2*(c + d*x)])/(64*d) - (3*a^2*b*Cos[4*(c + d*x)]
)/(32*d) - (b*(3*a^2 - b^2)*Cos[6*(c + d*x)])/(192*d) + (3*a*(5*a^2 + b^2)*Sin[2*(c + d*x)])/(64*d) + (3*a*(a^
2 - b^2)*Sin[4*(c + d*x)])/(64*d) + (a*(a^2 - 3*b^2)*Sin[6*(c + d*x)])/(192*d)

Maple [A] (verified)

Time = 1.10 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.71

method	result	size
parallelrisch	\(\frac {\left (-45 a^{2} b -9 b^{3}\right ) \cos \left (2 d x +2 c \right )+\left (-3 a^{2} b +b^{3}\right ) \cos \left (6 d x +6 c \right )+\left (45 a^{3}+9 a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+\left (9 a^{3}-9 a \,b^{2}\right ) \sin \left (4 d x +4 c \right )+\left (a^{3}-3 a \,b^{2}\right ) \sin \left (6 d x +6 c \right )+60 a^{3} x d +36 a \,b^{2} d x -18 \cos \left (4 d x +4 c \right ) a^{2} b +66 a^{2} b +8 b^{3}}{192 d}\)	\(154\)
derivativedivides	\(\frac {a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )-\frac {a^{2} b \cos \left (d x +c \right )^{6}}{2}+3 a \,b^{2} \left (-\frac {\cos \left (d x +c \right )^{5} \sin \left (d x +c \right )}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )+b^{3} \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}{6}-\frac {\cos \left (d x +c \right )^{4}}{12}\right )}{d}\)	\(155\)
default	\(\frac {a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )-\frac {a^{2} b \cos \left (d x +c \right )^{6}}{2}+3 a \,b^{2} \left (-\frac {\cos \left (d x +c \right )^{5} \sin \left (d x +c \right )}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )+b^{3} \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}{6}-\frac {\cos \left (d x +c \right )^{4}}{12}\right )}{d}\)	\(155\)
parts	\(\frac {a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}+\frac {b^{3} \left (-\frac {\sin \left (d x +c \right )^{6}}{6}+\frac {\sin \left (d x +c \right )^{4}}{4}\right )}{d}+\frac {3 a \,b^{2} \left (-\frac {\cos \left (d x +c \right )^{5} \sin \left (d x +c \right )}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )}{d}-\frac {a^{2} b \cos \left (d x +c \right )^{6}}{2 d}\)	\(155\)
risch	\(\frac {5 a^{3} x}{16}+\frac {3 a \,b^{2} x}{16}-\frac {b \cos \left (6 d x +6 c \right ) a^{2}}{64 d}+\frac {b^{3} \cos \left (6 d x +6 c \right )}{192 d}+\frac {a^{3} \sin \left (6 d x +6 c \right )}{192 d}-\frac {a \sin \left (6 d x +6 c \right ) b^{2}}{64 d}-\frac {3 b \cos \left (4 d x +4 c \right ) a^{2}}{32 d}+\frac {3 a^{3} \sin \left (4 d x +4 c \right )}{64 d}-\frac {3 a \sin \left (4 d x +4 c \right ) b^{2}}{64 d}-\frac {15 b \cos \left (2 d x +2 c \right ) a^{2}}{64 d}-\frac {3 b^{3} \cos \left (2 d x +2 c \right )}{64 d}+\frac {15 a^{3} \sin \left (2 d x +2 c \right )}{64 d}+\frac {3 a \sin \left (2 d x +2 c \right ) b^{2}}{64 d}\)	\(208\)
norman	\(\frac {\left (\frac {5}{16} a^{3}+\frac {3}{16} a \,b^{2}\right ) x +\left (\frac {5}{16} a^{3}+\frac {3}{16} a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (\frac {15}{8} a^{3}+\frac {9}{8} a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {15}{8} a^{3}+\frac {9}{8} a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {25}{4} a^{3}+\frac {15}{4} a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {75}{16} a^{3}+\frac {45}{16} a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {75}{16} a^{3}+\frac {45}{16} a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {4 b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {4 b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {6 a^{2} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {6 a^{2} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {4 \left (15 a^{2} b -2 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}-\frac {a \left (5 a^{2}-141 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d}+\frac {a \left (5 a^{2}-141 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 d}+\frac {3 a \left (5 a^{2}-13 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}-\frac {3 a \left (5 a^{2}-13 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {a \left (11 a^{2}-3 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}-\frac {a \left (11 a^{2}-3 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}\)	\(454\)

[In]

int(cos(d*x+c)^3*(cos(d*x+c)*a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/192*((-45*a^2*b-9*b^3)*cos(2*d*x+2*c)+(-3*a^2*b+b^3)*cos(6*d*x+6*c)+(45*a^3+9*a*b^2)*sin(2*d*x+2*c)+(9*a^3-9
*a*b^2)*sin(4*d*x+4*c)+(a^3-3*a*b^2)*sin(6*d*x+6*c)+60*a^3*x*d+36*a*b^2*d*x-18*cos(4*d*x+4*c)*a^2*b+66*a^2*b+8
*b^3)/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.59 \[ \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=-\frac {12 \, b^{3} \cos \left (d x + c\right )^{4} + 8 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} d x - {\left (8 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d} \]

[In]

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

[Out]

-1/48*(12*b^3*cos(d*x + c)^4 + 8*(3*a^2*b - b^3)*cos(d*x + c)^6 - 3*(5*a^3 + 3*a*b^2)*d*x - (8*(a^3 - 3*a*b^2)
*cos(d*x + c)^5 + 2*(5*a^3 + 3*a*b^2)*cos(d*x + c)^3 + 3*(5*a^3 + 3*a*b^2)*cos(d*x + c))*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 400, normalized size of antiderivative = 1.85 \[ \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\begin {cases} \frac {5 a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 a^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 a^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {a^{2} b \cos ^{6}{\left (c + d x \right )}}{2 d} + \frac {3 a b^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {9 a b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {9 a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 a b^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 a b^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {a b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} - \frac {3 a b^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac {b^{3} \sin ^{6}{\left (c + d x \right )}}{12 d} + \frac {b^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + b \sin {\left (c \right )}\right )^{3} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((5*a**3*x*sin(c + d*x)**6/16 + 15*a**3*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*a**3*x*sin(c + d*x)
**2*cos(c + d*x)**4/16 + 5*a**3*x*cos(c + d*x)**6/16 + 5*a**3*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*a**3*sin
(c + d*x)**3*cos(c + d*x)**3/(6*d) + 11*a**3*sin(c + d*x)*cos(c + d*x)**5/(16*d) - a**2*b*cos(c + d*x)**6/(2*d
) + 3*a*b**2*x*sin(c + d*x)**6/16 + 9*a*b**2*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 9*a*b**2*x*sin(c + d*x)**2
*cos(c + d*x)**4/16 + 3*a*b**2*x*cos(c + d*x)**6/16 + 3*a*b**2*sin(c + d*x)**5*cos(c + d*x)/(16*d) + a*b**2*si
n(c + d*x)**3*cos(c + d*x)**3/(2*d) - 3*a*b**2*sin(c + d*x)*cos(c + d*x)**5/(16*d) + b**3*sin(c + d*x)**6/(12*
d) + b**3*sin(c + d*x)**4*cos(c + d*x)**2/(4*d), Ne(d, 0)), (x*(a*cos(c) + b*sin(c))**3*cos(c)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.61 \[ \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=-\frac {96 \, a^{2} b \cos \left (d x + c\right )^{6} + {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 3 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a b^{2} + 16 \, {\left (2 \, \sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4}\right )} b^{3}}{192 \, d} \]

[In]

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

[Out]

-1/192*(96*a^2*b*cos(d*x + c)^6 + (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))*a^3 - 3*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a*b^2 + 16*(2*sin(d*x + c)^6 - 3*sin
(d*x + c)^4)*b^3)/d

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.73 \[ \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=-\frac {3 \, a^{2} b \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {1}{16} \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} x - \frac {{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {3 \, {\left (5 \, a^{2} b + b^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (a^{3} - 3 \, a b^{2}\right )} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {3 \, {\left (a^{3} - a b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {3 \, {\left (5 \, a^{3} + a b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

[In]

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

[Out]

-3/32*a^2*b*cos(4*d*x + 4*c)/d + 1/16*(5*a^3 + 3*a*b^2)*x - 1/192*(3*a^2*b - b^3)*cos(6*d*x + 6*c)/d - 3/64*(5
*a^2*b + b^3)*cos(2*d*x + 2*c)/d + 1/192*(a^3 - 3*a*b^2)*sin(6*d*x + 6*c)/d + 3/64*(a^3 - a*b^2)*sin(4*d*x + 4
*c)/d + 3/64*(5*a^3 + a*b^2)*sin(2*d*x + 2*c)/d

Mupad [B] (verification not implemented)

Time = 24.12 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.88 \[ \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {4\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a\,b^2}{8}-\frac {11\,a^3}{8}\right )+4\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (\frac {3\,a\,b^2}{8}-\frac {11\,a^3}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {39\,a\,b^2}{4}-\frac {15\,a^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {39\,a\,b^2}{4}-\frac {15\,a^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {47\,a\,b^2}{8}-\frac {5\,a^3}{24}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {47\,a\,b^2}{8}-\frac {5\,a^3}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (20\,a^2\,b-\frac {8\,b^3}{3}\right )+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,a^2+3\,b^2\right )}{8\,\left (\frac {5\,a^3}{8}+\frac {3\,a\,b^2}{8}\right )}\right )\,\left (5\,a^2+3\,b^2\right )}{8\,d}-\frac {a\,\left (5\,a^2+3\,b^2\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{8\,d} \]

[In]

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

[Out]

(4*b^3*tan(c/2 + (d*x)/2)^4 - tan(c/2 + (d*x)/2)*((3*a*b^2)/8 - (11*a^3)/8) + 4*b^3*tan(c/2 + (d*x)/2)^8 + tan
(c/2 + (d*x)/2)^11*((3*a*b^2)/8 - (11*a^3)/8) - tan(c/2 + (d*x)/2)^5*((39*a*b^2)/4 - (15*a^3)/4) + tan(c/2 + (
d*x)/2)^7*((39*a*b^2)/4 - (15*a^3)/4) + tan(c/2 + (d*x)/2)^3*((47*a*b^2)/8 - (5*a^3)/24) - tan(c/2 + (d*x)/2)^
9*((47*a*b^2)/8 - (5*a^3)/24) + tan(c/2 + (d*x)/2)^6*(20*a^2*b - (8*b^3)/3) + 6*a^2*b*tan(c/2 + (d*x)/2)^2 + 6
*a^2*b*tan(c/2 + (d*x)/2)^10)/(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*atan((a*tan(c/2 + (d*x)/
2)*(5*a^2 + 3*b^2))/(8*((3*a*b^2)/8 + (5*a^3)/8)))*(5*a^2 + 3*b^2))/(8*d) - (a*(5*a^2 + 3*b^2)*(atan(tan(c/2 +
(d*x)/2)) - (d*x)/2))/(8*d)