\(\int \cos (c+d x) (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx\) [20]

Optimal result

Integrand size = 29, antiderivative size = 154 \[ \int \cos (c+d x) (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx=\frac {1}{8} a^3 (15 A+13 B) x+\frac {a^3 (15 A+13 B) \sin (c+d x)}{5 d}+\frac {3 a^3 (15 A+13 B) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {(5 A-B) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}-\frac {a^3 (15 A+13 B) \sin ^3(c+d x)}{60 d} \]

[Out]

1/8*a^3*(15*A+13*B)*x+1/5*a^3*(15*A+13*B)*sin(d*x+c)/d+3/40*a^3*(15*A+13*B)*cos(d*x+c)*sin(d*x+c)/d+1/20*(5*A-
B)*(a+a*cos(d*x+c))^3*sin(d*x+c)/d+1/5*B*(a+a*cos(d*x+c))^4*sin(d*x+c)/a/d-1/60*a^3*(15*A+13*B)*sin(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3047, 3102, 2830, 2724, 2717, 2715, 8, 2713} \[ \int \cos (c+d x) (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx=-\frac {a^3 (15 A+13 B) \sin ^3(c+d x)}{60 d}+\frac {a^3 (15 A+13 B) \sin (c+d x)}{5 d}+\frac {3 a^3 (15 A+13 B) \sin (c+d x) \cos (c+d x)}{40 d}+\frac {1}{8} a^3 x (15 A+13 B)+\frac {(5 A-B) \sin (c+d x) (a \cos (c+d x)+a)^3}{20 d}+\frac {B \sin (c+d x) (a \cos (c+d x)+a)^4}{5 a d} \]

[In]

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

[Out]

(a^3*(15*A + 13*B)*x)/8 + (a^3*(15*A + 13*B)*Sin[c + d*x])/(5*d) + (3*a^3*(15*A + 13*B)*Cos[c + d*x]*Sin[c + d
*x])/(40*d) + ((5*A - B)*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(20*d) + (B*(a + a*Cos[c + d*x])^4*Sin[c + d*x])
/(5*a*d) - (a^3*(15*A + 13*B)*Sin[c + d*x]^3)/(60*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 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2724

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTrig[(a + b*sin[c + d*x])^n, x], x] /;
FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0]

Rule 2830

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[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*S
in[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m
, -2^(-1)]

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*} \text {integral}& = \int (a+a \cos (c+d x))^3 \left (A \cos (c+d x)+B \cos ^2(c+d x)\right ) \, dx \\ & = \frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}+\frac {\int (a+a \cos (c+d x))^3 (4 a B+a (5 A-B) \cos (c+d x)) \, dx}{5 a} \\ & = \frac {(5 A-B) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}+\frac {1}{20} (15 A+13 B) \int (a+a \cos (c+d x))^3 \, dx \\ & = \frac {(5 A-B) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}+\frac {1}{20} (15 A+13 B) \int \left (a^3+3 a^3 \cos (c+d x)+3 a^3 \cos ^2(c+d x)+a^3 \cos ^3(c+d x)\right ) \, dx \\ & = \frac {1}{20} a^3 (15 A+13 B) x+\frac {(5 A-B) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}+\frac {1}{20} \left (a^3 (15 A+13 B)\right ) \int \cos ^3(c+d x) \, dx+\frac {1}{20} \left (3 a^3 (15 A+13 B)\right ) \int \cos (c+d x) \, dx+\frac {1}{20} \left (3 a^3 (15 A+13 B)\right ) \int \cos ^2(c+d x) \, dx \\ & = \frac {1}{20} a^3 (15 A+13 B) x+\frac {3 a^3 (15 A+13 B) \sin (c+d x)}{20 d}+\frac {3 a^3 (15 A+13 B) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {(5 A-B) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}+\frac {1}{40} \left (3 a^3 (15 A+13 B)\right ) \int 1 \, dx-\frac {\left (a^3 (15 A+13 B)\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{20 d} \\ & = \frac {1}{8} a^3 (15 A+13 B) x+\frac {a^3 (15 A+13 B) \sin (c+d x)}{5 d}+\frac {3 a^3 (15 A+13 B) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {(5 A-B) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}-\frac {a^3 (15 A+13 B) \sin ^3(c+d x)}{60 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.70 \[ \int \cos (c+d x) (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx=\frac {a^3 (780 B c+900 A d x+780 B d x+60 (26 A+23 B) \sin (c+d x)+480 (A+B) \sin (2 (c+d x))+120 A \sin (3 (c+d x))+170 B \sin (3 (c+d x))+15 A \sin (4 (c+d x))+45 B \sin (4 (c+d x))+6 B \sin (5 (c+d x)))}{480 d} \]

[In]

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

[Out]

(a^3*(780*B*c + 900*A*d*x + 780*B*d*x + 60*(26*A + 23*B)*Sin[c + d*x] + 480*(A + B)*Sin[2*(c + d*x)] + 120*A*S
in[3*(c + d*x)] + 170*B*Sin[3*(c + d*x)] + 15*A*Sin[4*(c + d*x)] + 45*B*Sin[4*(c + d*x)] + 6*B*Sin[5*(c + d*x)
]))/(480*d)

Maple [A] (verified)

Time = 3.36 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.60

method	result	size
parallelrisch	\(\frac {\left (32 \left (A +B \right ) \sin \left (2 d x +2 c \right )+2 \left (4 A +\frac {17 B}{3}\right ) \sin \left (3 d x +3 c \right )+\left (A +3 B \right ) \sin \left (4 d x +4 c \right )+\frac {2 B \sin \left (5 d x +5 c \right )}{5}+4 \left (26 A +23 B \right ) \sin \left (d x +c \right )+60 \left (A +\frac {13 B}{15}\right ) x d \right ) a^{3}}{32 d}\)	\(93\)
risch	\(\frac {15 a^{3} A x}{8}+\frac {13 a^{3} B x}{8}+\frac {13 a^{3} A \sin \left (d x +c \right )}{4 d}+\frac {23 a^{3} B \sin \left (d x +c \right )}{8 d}+\frac {\sin \left (5 d x +5 c \right ) B \,a^{3}}{80 d}+\frac {\sin \left (4 d x +4 c \right ) A \,a^{3}}{32 d}+\frac {3 \sin \left (4 d x +4 c \right ) B \,a^{3}}{32 d}+\frac {\sin \left (3 d x +3 c \right ) A \,a^{3}}{4 d}+\frac {17 \sin \left (3 d x +3 c \right ) B \,a^{3}}{48 d}+\frac {\sin \left (2 d x +2 c \right ) A \,a^{3}}{d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{3}}{d}\)	\(170\)
parts	\(\frac {\left (A \,a^{3}+3 B \,a^{3}\right ) \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {\left (3 A \,a^{3}+B \,a^{3}\right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (3 A \,a^{3}+3 B \,a^{3}\right ) \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {a^{3} A \sin \left (d x +c \right )}{d}+\frac {B \,a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5 d}\)	\(172\)
derivativedivides	\(\frac {A \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+A \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 B \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+A \,a^{3} \sin \left (d x +c \right )+B \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(223\)
default	\(\frac {A \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+A \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 B \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+A \,a^{3} \sin \left (d x +c \right )+B \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(223\)
norman	\(\frac {\frac {a^{3} \left (15 A +13 B \right ) x}{8}+\frac {32 a^{3} \left (15 A +13 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {7 a^{3} \left (15 A +13 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {a^{3} \left (15 A +13 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {5 a^{3} \left (15 A +13 B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {5 a^{3} \left (15 A +13 B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 a^{3} \left (15 A +13 B \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 a^{3} \left (15 A +13 B \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {a^{3} \left (15 A +13 B \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {a^{3} \left (49 A +51 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{3} \left (183 A +133 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\)	\(279\)

[In]

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

[Out]

1/32*(32*(A+B)*sin(2*d*x+2*c)+2*(4*A+17/3*B)*sin(3*d*x+3*c)+(A+3*B)*sin(4*d*x+4*c)+2/5*B*sin(5*d*x+5*c)+4*(26*
A+23*B)*sin(d*x+c)+60*(A+13/15*B)*x*d)*a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.71 \[ \int \cos (c+d x) (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx=\frac {15 \, {\left (15 \, A + 13 \, B\right )} a^{3} d x + {\left (24 \, B a^{3} \cos \left (d x + c\right )^{4} + 30 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (15 \, A + 19 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 15 \, {\left (15 \, A + 13 \, B\right )} a^{3} \cos \left (d x + c\right ) + 8 \, {\left (45 \, A + 38 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{120 \, d} \]

[In]

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

[Out]

1/120*(15*(15*A + 13*B)*a^3*d*x + (24*B*a^3*cos(d*x + c)^4 + 30*(A + 3*B)*a^3*cos(d*x + c)^3 + 8*(15*A + 19*B)
*a^3*cos(d*x + c)^2 + 15*(15*A + 13*B)*a^3*cos(d*x + c) + 8*(45*A + 38*B)*a^3)*sin(d*x + c))/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 530 vs. \(2 (136) = 272\).

Time = 0.30 (sec) , antiderivative size = 530, normalized size of antiderivative = 3.44 \[ \int \cos (c+d x) (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx=\begin {cases} \frac {3 A a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 A a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 A a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {5 A a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {3 A a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 A a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {A a^{3} \sin {\left (c + d x \right )}}{d} + \frac {9 B a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 B a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {B a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {9 B a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {B a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {8 B a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 B a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {9 B a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 B a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {B a^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {15 B a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {3 B a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )}\right ) \left (a \cos {\left (c \right )} + a\right )^{3} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.38 \[ \int \cos (c+d x) (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx=-\frac {480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{3} + 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 480 \, A a^{3} \sin \left (d x + c\right )}{480 \, d} \]

[In]

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

[Out]

-1/480*(480*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3 - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c
))*A*a^3 - 360*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 - 32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x
+ c))*B*a^3 + 480*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 - 45*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x
+ 2*c))*B*a^3 - 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 - 480*A*a^3*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.88 \[ \int \cos (c+d x) (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx=\frac {B a^{3} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {1}{8} \, {\left (15 \, A a^{3} + 13 \, B a^{3}\right )} x + \frac {{\left (A a^{3} + 3 \, B a^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (12 \, A a^{3} + 17 \, B a^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (A a^{3} + B a^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{d} + \frac {{\left (26 \, A a^{3} + 23 \, B a^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \]

[In]

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

[Out]

1/80*B*a^3*sin(5*d*x + 5*c)/d + 1/8*(15*A*a^3 + 13*B*a^3)*x + 1/32*(A*a^3 + 3*B*a^3)*sin(4*d*x + 4*c)/d + 1/48
*(12*A*a^3 + 17*B*a^3)*sin(3*d*x + 3*c)/d + (A*a^3 + B*a^3)*sin(2*d*x + 2*c)/d + 1/8*(26*A*a^3 + 23*B*a^3)*sin
(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 1.51 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.80 \[ \int \cos (c+d x) (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx=\frac {\left (\frac {15\,A\,a^3}{4}+\frac {13\,B\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {35\,A\,a^3}{2}+\frac {91\,B\,a^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (32\,A\,a^3+\frac {416\,B\,a^3}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {61\,A\,a^3}{2}+\frac {133\,B\,a^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {49\,A\,a^3}{4}+\frac {51\,B\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a^3\,\left (15\,A+13\,B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d}+\frac {a^3\,\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (15\,A+13\,B\right )}{4\,\left (\frac {15\,A\,a^3}{4}+\frac {13\,B\,a^3}{4}\right )}\right )\,\left (15\,A+13\,B\right )}{4\,d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((49*A*a^3)/4 + (51*B*a^3)/4) + tan(c/2 + (d*x)/2)^9*((15*A*a^3)/4 + (13*B*a^3)/4) + tan(c
/2 + (d*x)/2)^7*((35*A*a^3)/2 + (91*B*a^3)/6) + tan(c/2 + (d*x)/2)^3*((61*A*a^3)/2 + (133*B*a^3)/6) + tan(c/2
+ (d*x)/2)^5*(32*A*a^3 + (416*B*a^3)/15))/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 +
(d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) - (a^3*(15*A + 13*B)*(atan(tan(c/2 + (d*x)/2
)) - (d*x)/2))/(4*d) + (a^3*atan((a^3*tan(c/2 + (d*x)/2)*(15*A + 13*B))/(4*((15*A*a^3)/4 + (13*B*a^3)/4)))*(15
*A + 13*B))/(4*d)