\(\int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx\) [58]

Optimal result

Integrand size = 31, antiderivative size = 147 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {(A-3 B) x}{a^3}-\frac {(7 A-27 B) \sin (c+d x)}{15 a^3 d}+\frac {(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(A-3 B) \sin (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

(A-3*B)*x/a^3-1/15*(7*A-27*B)*sin(d*x+c)/a^3/d+1/5*(A-B)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^3+1/15*(4*
A-9*B)*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2-(A-3*B)*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3056, 3047, 3102, 12, 2814, 2727} \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=-\frac {(7 A-27 B) \sin (c+d x)}{15 a^3 d}-\frac {(A-3 B) \sin (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {x (A-3 B)}{a^3}+\frac {(A-B) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac {(4 A-9 B) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

((A - 3*B)*x)/a^3 - ((7*A - 27*B)*Sin[c + d*x])/(15*a^3*d) + ((A - B)*Cos[c + d*x]^3*Sin[c + d*x])/(5*d*(a + a
*Cos[c + d*x])^3) + ((4*A - 9*B)*Cos[c + d*x]^2*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((A - 3*B)*Sin
[c + d*x])/(d*(a^3 + a^3*Cos[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

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 3056

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^n/(a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[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*} \text {integral}& = \frac {(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {\cos ^2(c+d x) (3 a (A-B)-a (A-6 B) \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = \frac {(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\cos (c+d x) \left (2 a^2 (4 A-9 B)-a^2 (7 A-27 B) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4} \\ & = \frac {(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {2 a^2 (4 A-9 B) \cos (c+d x)-a^2 (7 A-27 B) \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4} \\ & = -\frac {(7 A-27 B) \sin (c+d x)}{15 a^3 d}+\frac {(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {15 a^3 (A-3 B) \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^5} \\ & = -\frac {(7 A-27 B) \sin (c+d x)}{15 a^3 d}+\frac {(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(A-3 B) \int \frac {\cos (c+d x)}{a+a \cos (c+d x)} \, dx}{a^2} \\ & = \frac {(A-3 B) x}{a^3}-\frac {(7 A-27 B) \sin (c+d x)}{15 a^3 d}+\frac {(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(A-3 B) \int \frac {1}{a+a \cos (c+d x)} \, dx}{a^2} \\ & = \frac {(A-3 B) x}{a^3}-\frac {(7 A-27 B) \sin (c+d x)}{15 a^3 d}+\frac {(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(A-3 B) \sin (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(361\) vs. \(2(147)=294\).

Time = 1.86 (sec) , antiderivative size = 361, normalized size of antiderivative = 2.46 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (300 (A-3 B) d x \cos \left (\frac {d x}{2}\right )+300 (A-3 B) d x \cos \left (c+\frac {d x}{2}\right )+150 A d x \cos \left (c+\frac {3 d x}{2}\right )-450 B d x \cos \left (c+\frac {3 d x}{2}\right )+150 A d x \cos \left (2 c+\frac {3 d x}{2}\right )-450 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+30 A d x \cos \left (2 c+\frac {5 d x}{2}\right )-90 B d x \cos \left (2 c+\frac {5 d x}{2}\right )+30 A d x \cos \left (3 c+\frac {5 d x}{2}\right )-90 B d x \cos \left (3 c+\frac {5 d x}{2}\right )-740 A \sin \left (\frac {d x}{2}\right )+1755 B \sin \left (\frac {d x}{2}\right )+540 A \sin \left (c+\frac {d x}{2}\right )-1125 B \sin \left (c+\frac {d x}{2}\right )-460 A \sin \left (c+\frac {3 d x}{2}\right )+1215 B \sin \left (c+\frac {3 d x}{2}\right )+180 A \sin \left (2 c+\frac {3 d x}{2}\right )-225 B \sin \left (2 c+\frac {3 d x}{2}\right )-128 A \sin \left (2 c+\frac {5 d x}{2}\right )+363 B \sin \left (2 c+\frac {5 d x}{2}\right )+75 B \sin \left (3 c+\frac {5 d x}{2}\right )+15 B \sin \left (3 c+\frac {7 d x}{2}\right )+15 B \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{120 a^3 d (1+\cos (c+d x))^3} \]

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(300*(A - 3*B)*d*x*Cos[(d*x)/2] + 300*(A - 3*B)*d*x*Cos[c + (d*x)/2] + 150*A*d*x*Co
s[c + (3*d*x)/2] - 450*B*d*x*Cos[c + (3*d*x)/2] + 150*A*d*x*Cos[2*c + (3*d*x)/2] - 450*B*d*x*Cos[2*c + (3*d*x)
/2] + 30*A*d*x*Cos[2*c + (5*d*x)/2] - 90*B*d*x*Cos[2*c + (5*d*x)/2] + 30*A*d*x*Cos[3*c + (5*d*x)/2] - 90*B*d*x
*Cos[3*c + (5*d*x)/2] - 740*A*Sin[(d*x)/2] + 1755*B*Sin[(d*x)/2] + 540*A*Sin[c + (d*x)/2] - 1125*B*Sin[c + (d*
x)/2] - 460*A*Sin[c + (3*d*x)/2] + 1215*B*Sin[c + (3*d*x)/2] + 180*A*Sin[2*c + (3*d*x)/2] - 225*B*Sin[2*c + (3
*d*x)/2] - 128*A*Sin[2*c + (5*d*x)/2] + 363*B*Sin[2*c + (5*d*x)/2] + 75*B*Sin[3*c + (5*d*x)/2] + 15*B*Sin[3*c
+ (7*d*x)/2] + 15*B*Sin[4*c + (7*d*x)/2]))/(120*a^3*d*(1 + Cos[c + d*x])^3)

Maple [A] (verified)

Time = 1.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.61

method	result	size
parallelrisch	\(\frac {-204 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (\frac {16 A}{51}-\frac {39 B}{34}\right ) \cos \left (2 d x +2 c \right )-\frac {5 B \cos \left (3 d x +3 c \right )}{68}+\left (A -\frac {243 B}{68}\right ) \cos \left (d x +c \right )+\frac {38 A}{51}-\frac {87 B}{34}\right ) \left (\sec ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+240 d x \left (A -3 B \right )}{240 a^{3} d}\)	\(89\)
derivativedivides	\(\frac {-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B -7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+17 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {8 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+8 \left (A -3 B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(134\)
default	\(\frac {-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B -7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+17 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {8 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+8 \left (A -3 B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(134\)
risch	\(\frac {A x}{a^{3}}-\frac {3 B x}{a^{3}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{2 a^{3} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{2 a^{3} d}-\frac {2 i \left (45 A \,{\mathrm e}^{4 i \left (d x +c \right )}-90 B \,{\mathrm e}^{4 i \left (d x +c \right )}+135 A \,{\mathrm e}^{3 i \left (d x +c \right )}-300 B \,{\mathrm e}^{3 i \left (d x +c \right )}+185 A \,{\mathrm e}^{2 i \left (d x +c \right )}-420 B \,{\mathrm e}^{2 i \left (d x +c \right )}+115 A \,{\mathrm e}^{i \left (d x +c \right )}-270 B \,{\mathrm e}^{i \left (d x +c \right )}+32 A -72 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(178\)
norman	\(\frac {\frac {\left (A -3 B \right ) x}{a}+\frac {\left (A -3 B \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {4 \left (A -3 B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {6 \left (A -3 B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {4 \left (A -3 B \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {\left (A -B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}+\frac {\left (4 A -9 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 a d}-\frac {\left (7 A -25 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {5 \left (8 A -27 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {\left (26 A -81 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}-\frac {\left (43 A -153 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}-\frac {\left (553 A -1773 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} a^{2}}\)	\(296\)

[In]

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

[Out]

1/240*(-204*tan(1/2*d*x+1/2*c)*((16/51*A-39/34*B)*cos(2*d*x+2*c)-5/68*B*cos(3*d*x+3*c)+(A-243/68*B)*cos(d*x+c)
+38/51*A-87/34*B)*sec(1/2*d*x+1/2*c)^4+240*d*x*(A-3*B))/a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.12 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, {\left (A - 3 \, B\right )} d x \cos \left (d x + c\right )^{3} + 45 \, {\left (A - 3 \, B\right )} d x \cos \left (d x + c\right )^{2} + 45 \, {\left (A - 3 \, B\right )} d x \cos \left (d x + c\right ) + 15 \, {\left (A - 3 \, B\right )} d x + {\left (15 \, B \cos \left (d x + c\right )^{3} - {\left (32 \, A - 117 \, B\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (17 \, A - 57 \, B\right )} \cos \left (d x + c\right ) - 22 \, A + 72 \, B\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

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

[Out]

1/15*(15*(A - 3*B)*d*x*cos(d*x + c)^3 + 45*(A - 3*B)*d*x*cos(d*x + c)^2 + 45*(A - 3*B)*d*x*cos(d*x + c) + 15*(
A - 3*B)*d*x + (15*B*cos(d*x + c)^3 - (32*A - 117*B)*cos(d*x + c)^2 - 3*(17*A - 57*B)*cos(d*x + c) - 22*A + 72
*B)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 496 vs. \(2 (134) = 268\).

Time = 2.37 (sec) , antiderivative size = 496, normalized size of antiderivative = 3.37 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\begin {cases} \frac {60 A d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {60 A d x}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {3 A \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {17 A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {85 A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {105 A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {180 B d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {180 B d x}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {3 B \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {27 B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {225 B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {375 B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\left (c \right )}\right ) \cos ^{3}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((60*A*d*x*tan(c/2 + d*x/2)**2/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 60*A*d*x/(60*a**3*d*tan(
c/2 + d*x/2)**2 + 60*a**3*d) - 3*A*tan(c/2 + d*x/2)**7/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 17*A*tan(
c/2 + d*x/2)**5/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 85*A*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c/2 + d*
x/2)**2 + 60*a**3*d) - 105*A*tan(c/2 + d*x/2)/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 180*B*d*x*tan(c/2
+ d*x/2)**2/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 180*B*d*x/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d
) + 3*B*tan(c/2 + d*x/2)**7/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 27*B*tan(c/2 + d*x/2)**5/(60*a**3*d*
tan(c/2 + d*x/2)**2 + 60*a**3*d) + 225*B*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 375
*B*tan(c/2 + d*x/2)/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d), Ne(d, 0)), (x*(A + B*cos(c))*cos(c)**3/(a*cos
(c) + a)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.57 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {3 \, B {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{60 \, d} \]

[In]

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

[Out]

1/60*(3*B*(40*sin(d*x + c)/((a^3 + a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x
+ c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 -
120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3) - A*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3
/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1
))/a^3))/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {60 \, {\left (d x + c\right )} {\left (A - 3 \, B\right )}}{a^{3}} + \frac {120 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 255 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

[In]

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

[Out]

1/60*(60*(d*x + c)*(A - 3*B)/a^3 + 120*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3) - (3*A*a^12*t
an(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 20*A*a^12*tan(1/2*d*x + 1/2*c)^3 + 30*B*a^12*tan(1/2
*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2*d*x + 1/2*c) - 255*B*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{6\,a^3}+\frac {2\,A-4\,B}{12\,a^3}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B\right )}{4\,a^3}-\frac {3\,B}{2\,a^3}+\frac {2\,A-4\,B}{2\,a^3}\right )}{d}+\frac {x\,\left (A-3\,B\right )}{a^3}+\frac {2\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^3*((A - B)/(6*a^3) + (2*A - 4*B)/(12*a^3)))/d - (tan(c/2 + (d*x)/2)*((3*(A - B))/(4*a^3) -
(3*B)/(2*a^3) + (2*A - 4*B)/(2*a^3)))/d + (x*(A - 3*B))/a^3 + (2*B*tan(c/2 + (d*x)/2))/(d*(a^3*tan(c/2 + (d*x
)/2)^2 + a^3)) - (tan(c/2 + (d*x)/2)^5*(A - B))/(20*a^3*d)