\(\int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx\) [250]

Optimal result

Integrand size = 31, antiderivative size = 178 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {\left (2 a^2+b^2\right ) (A b-a B) x}{2 b^4}-\frac {2 a^3 (A b-a B) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^4 \sqrt {a+b} d}-\frac {\left (3 a A b-3 a^2 B-2 b^2 B\right ) \sin (c+d x)}{3 b^3 d}+\frac {(A b-a B) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {B \cos ^2(c+d x) \sin (c+d x)}{3 b d} \] Output:

1/2*(2*a^2+b^2)*(A*b-B*a)*x/b^4-2*a^3*(A*b-B*a)*arctan((a-b)^(1/2)*tan(1/2 
*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(1/2)/b^4/(a+b)^(1/2)/d-1/3*(3*A*a*b-3*B*a^ 
2-2*B*b^2)*sin(d*x+c)/b^3/d+1/2*(A*b-B*a)*cos(d*x+c)*sin(d*x+c)/b^2/d+1/3* 
B*cos(d*x+c)^2*sin(d*x+c)/b/d
 

Mathematica [A] (verified)

Time = 1.69 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {6 \left (2 a^2+b^2\right ) (A b-a B) (c+d x)-\frac {24 a^3 (-A b+a B) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+3 b \left (-4 a A b+4 a^2 B+3 b^2 B\right ) \sin (c+d x)+3 b^2 (A b-a B) \sin (2 (c+d x))+b^3 B \sin (3 (c+d x))}{12 b^4 d} \] Input:

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

Output:

(6*(2*a^2 + b^2)*(A*b - a*B)*(c + d*x) - (24*a^3*(-(A*b) + a*B)*ArcTanh[(( 
a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + 3*b*(-4*a*A 
*b + 4*a^2*B + 3*b^2*B)*Sin[c + d*x] + 3*b^2*(A*b - a*B)*Sin[2*(c + d*x)] 
+ b^3*B*Sin[3*(c + d*x)])/(12*b^4*d)
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.10, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 3469, 3042, 3528, 3042, 3502, 27, 3042, 3214, 3042, 3138, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 3469

\(\displaystyle \frac {\int \frac {\cos (c+d x) \left (3 (A b-a B) \cos ^2(c+d x)+2 b B \cos (c+d x)+2 a B\right )}{a+b \cos (c+d x)}dx}{3 b}+\frac {B \sin (c+d x) \cos ^2(c+d x)}{3 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (3 (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b B \sin \left (c+d x+\frac {\pi }{2}\right )+2 a B\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b}+\frac {B \sin (c+d x) \cos ^2(c+d x)}{3 b d}\)

\(\Big \downarrow \) 3528

\(\displaystyle \frac {\frac {\int \frac {-2 \left (-3 B a^2+3 A b a-2 b^2 B\right ) \cos ^2(c+d x)+b (3 A b+a B) \cos (c+d x)+3 a (A b-a B)}{a+b \cos (c+d x)}dx}{2 b}+\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {B \sin (c+d x) \cos ^2(c+d x)}{3 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {-2 \left (-3 B a^2+3 A b a-2 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (3 A b+a B) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a (A b-a B)}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}+\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {B \sin (c+d x) \cos ^2(c+d x)}{3 b d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {\frac {\int \frac {3 \left (a b (A b-a B)+\left (2 a^2+b^2\right ) \cos (c+d x) (A b-a B)\right )}{a+b \cos (c+d x)}dx}{b}-\frac {2 \left (-3 a^2 B+3 a A b-2 b^2 B\right ) \sin (c+d x)}{b d}}{2 b}+\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {B \sin (c+d x) \cos ^2(c+d x)}{3 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 \int \frac {a b (A b-a B)+\left (2 a^2+b^2\right ) \cos (c+d x) (A b-a B)}{a+b \cos (c+d x)}dx}{b}-\frac {2 \left (-3 a^2 B+3 a A b-2 b^2 B\right ) \sin (c+d x)}{b d}}{2 b}+\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {B \sin (c+d x) \cos ^2(c+d x)}{3 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \int \frac {a b (A b-a B)+\left (2 a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) (A b-a B)}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 \left (-3 a^2 B+3 a A b-2 b^2 B\right ) \sin (c+d x)}{b d}}{2 b}+\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {B \sin (c+d x) \cos ^2(c+d x)}{3 b d}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {x \left (2 a^2+b^2\right ) (A b-a B)}{b}-\frac {2 a^3 (A b-a B) \int \frac {1}{a+b \cos (c+d x)}dx}{b}\right )}{b}-\frac {2 \left (-3 a^2 B+3 a A b-2 b^2 B\right ) \sin (c+d x)}{b d}}{2 b}+\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {B \sin (c+d x) \cos ^2(c+d x)}{3 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {x \left (2 a^2+b^2\right ) (A b-a B)}{b}-\frac {2 a^3 (A b-a B) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )}{b}-\frac {2 \left (-3 a^2 B+3 a A b-2 b^2 B\right ) \sin (c+d x)}{b d}}{2 b}+\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {B \sin (c+d x) \cos ^2(c+d x)}{3 b d}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {x \left (2 a^2+b^2\right ) (A b-a B)}{b}-\frac {4 a^3 (A b-a B) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}\right )}{b}-\frac {2 \left (-3 a^2 B+3 a A b-2 b^2 B\right ) \sin (c+d x)}{b d}}{2 b}+\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {B \sin (c+d x) \cos ^2(c+d x)}{3 b d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {3 \left (\frac {x \left (2 a^2+b^2\right ) (A b-a B)}{b}-\frac {4 a^3 (A b-a B) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}\right )}{b}-\frac {2 \left (-3 a^2 B+3 a A b-2 b^2 B\right ) \sin (c+d x)}{b d}}{2 b}+\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 b d}}{3 b}+\frac {B \sin (c+d x) \cos ^2(c+d x)}{3 b d}\)

Input:

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

Output:

(B*Cos[c + d*x]^2*Sin[c + d*x])/(3*b*d) + ((3*(A*b - a*B)*Cos[c + d*x]*Sin 
[c + d*x])/(2*b*d) + ((3*(((2*a^2 + b^2)*(A*b - a*B)*x)/b - (4*a^3*(A*b - 
a*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sq 
rt[a + b]*d)))/b - (2*(3*a*A*b - 3*a^2*B - 2*b^2*B)*Sin[c + d*x])/(b*d))/( 
2*b))/(3*b)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(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 3469

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^( 
n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
 f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*( 
m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c - b*d*(m 
+ n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin 
[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !(IGt 
Q[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[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 3528

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((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*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + 
n + 2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* 
d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a 
*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + 
 n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} 
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ 
m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
 

Maple [A] (verified)

Time = 1.76 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.35

method	result	size
derivativedivides	\(\frac {-\frac {2 a^{3} \left (A b -B a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\frac {2 \left (\left (-A a \,b^{2}-\frac {1}{2} A \,b^{3}+B \,a^{2} b +\frac {1}{2} B a \,b^{2}+b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-2 A a \,b^{2}+2 B \,a^{2} b +\frac {2}{3} b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-A a \,b^{2}+B \,a^{2} b +b^{3} B +\frac {1}{2} A \,b^{3}-\frac {1}{2} B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\left (2 A \,a^{2} b +A \,b^{3}-2 a^{3} B -B a \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}}{d}\)	\(240\)
default	\(\frac {-\frac {2 a^{3} \left (A b -B a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\frac {2 \left (\left (-A a \,b^{2}-\frac {1}{2} A \,b^{3}+B \,a^{2} b +\frac {1}{2} B a \,b^{2}+b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-2 A a \,b^{2}+2 B \,a^{2} b +\frac {2}{3} b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-A a \,b^{2}+B \,a^{2} b +b^{3} B +\frac {1}{2} A \,b^{3}-\frac {1}{2} B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\left (2 A \,a^{2} b +A \,b^{3}-2 a^{3} B -B a \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}}{d}\)	\(240\)
risch	\(\frac {x A \,a^{2}}{b^{3}}+\frac {x A}{2 b}-\frac {x \,a^{3} B}{b^{4}}-\frac {a B x}{2 b^{2}}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B}{8 d b}+\frac {i {\mathrm e}^{i \left (d x +c \right )} A a}{2 d \,b^{2}}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2} B}{2 d \,b^{3}}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} A a}{2 d \,b^{2}}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B}{8 d b}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2} B}{2 d \,b^{3}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}+\frac {B \sin \left (3 d x +3 c \right )}{12 b d}+\frac {\sin \left (2 d x +2 c \right ) A}{4 b d}-\frac {\sin \left (2 d x +2 c \right ) B a}{4 b^{2} d}\)	\(515\)

Input:

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

Output:

1/d*(-2*a^3*(A*b-B*a)/b^4/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2 
*c)/((a-b)*(a+b))^(1/2))+2/b^4*(((-A*a*b^2-1/2*A*b^3+B*a^2*b+1/2*B*a*b^2+b 
^3*B)*tan(1/2*d*x+1/2*c)^5+(-2*A*a*b^2+2*B*a^2*b+2/3*b^3*B)*tan(1/2*d*x+1/ 
2*c)^3+(-A*a*b^2+B*a^2*b+b^3*B+1/2*A*b^3-1/2*B*a*b^2)*tan(1/2*d*x+1/2*c))/ 
(1+tan(1/2*d*x+1/2*c)^2)^3+1/2*(2*A*a^2*b+A*b^3-2*B*a^3-B*a*b^2)*arctan(ta 
n(1/2*d*x+1/2*c))))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 541, normalized size of antiderivative = 3.04 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\left [-\frac {3 \, {\left (2 \, B a^{5} - 2 \, A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} d x - 3 \, {\left (B a^{4} - A a^{3} b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (6 \, B a^{4} b - 6 \, A a^{3} b^{2} - 2 \, B a^{2} b^{3} + 6 \, A a b^{4} - 4 \, B b^{5} + 2 \, {\left (B a^{2} b^{3} - B b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (B a^{3} b^{2} - A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} b^{4} - b^{6}\right )} d}, -\frac {3 \, {\left (2 \, B a^{5} - 2 \, A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} d x - 6 \, {\left (B a^{4} - A a^{3} b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (6 \, B a^{4} b - 6 \, A a^{3} b^{2} - 2 \, B a^{2} b^{3} + 6 \, A a b^{4} - 4 \, B b^{5} + 2 \, {\left (B a^{2} b^{3} - B b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (B a^{3} b^{2} - A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} b^{4} - b^{6}\right )} d}\right ] \] Input:

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

Output:

[-1/6*(3*(2*B*a^5 - 2*A*a^4*b - B*a^3*b^2 + A*a^2*b^3 - B*a*b^4 + A*b^5)*d 
*x - 3*(B*a^4 - A*a^3*b)*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 
 - b^2)*cos(d*x + c)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + 
 c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - (6*B 
*a^4*b - 6*A*a^3*b^2 - 2*B*a^2*b^3 + 6*A*a*b^4 - 4*B*b^5 + 2*(B*a^2*b^3 - 
B*b^5)*cos(d*x + c)^2 - 3*(B*a^3*b^2 - A*a^2*b^3 - B*a*b^4 + A*b^5)*cos(d* 
x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d), -1/6*(3*(2*B*a^5 - 2*A*a^4*b - 
B*a^3*b^2 + A*a^2*b^3 - B*a*b^4 + A*b^5)*d*x - 6*(B*a^4 - A*a^3*b)*sqrt(a^ 
2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (6 
*B*a^4*b - 6*A*a^3*b^2 - 2*B*a^2*b^3 + 6*A*a*b^4 - 4*B*b^5 + 2*(B*a^2*b^3 
- B*b^5)*cos(d*x + c)^2 - 3*(B*a^3*b^2 - A*a^2*b^3 - B*a*b^4 + A*b^5)*cos( 
d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (161) = 322\).

Time = 0.18 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.02 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=-\frac {\frac {3 \, {\left (2 \, B a^{3} - 2 \, A a^{2} b + B a b^{2} - A b^{3}\right )} {\left (d x + c\right )}}{b^{4}} + \frac {12 \, {\left (B a^{4} - A a^{3} b\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} - \frac {2 \, {\left (6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \] Input:

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

Output:

-1/6*(3*(2*B*a^3 - 2*A*a^2*b + B*a*b^2 - A*b^3)*(d*x + c)/b^4 + 12*(B*a^4 
- A*a^3*b)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a* 
tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/(sqrt(a^2 
 - b^2)*b^4) - 2*(6*B*a^2*tan(1/2*d*x + 1/2*c)^5 - 6*A*a*b*tan(1/2*d*x + 1 
/2*c)^5 + 3*B*a*b*tan(1/2*d*x + 1/2*c)^5 - 3*A*b^2*tan(1/2*d*x + 1/2*c)^5 
+ 6*B*b^2*tan(1/2*d*x + 1/2*c)^5 + 12*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 12*A* 
a*b*tan(1/2*d*x + 1/2*c)^3 + 4*B*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*B*a^2*tan( 
1/2*d*x + 1/2*c) - 6*A*a*b*tan(1/2*d*x + 1/2*c) - 3*B*a*b*tan(1/2*d*x + 1/ 
2*c) + 3*A*b^2*tan(1/2*d*x + 1/2*c) + 6*B*b^2*tan(1/2*d*x + 1/2*c))/((tan( 
1/2*d*x + 1/2*c)^2 + 1)^3*b^3))/d
 

Mupad [B] (verification not implemented)

Time = 45.84 (sec) , antiderivative size = 4568, normalized size of antiderivative = 25.66 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

((tan(c/2 + (d*x)/2)*(A*b^2 + 2*B*a^2 + 2*B*b^2 - 2*A*a*b - B*a*b))/b^3 + 
(tan(c/2 + (d*x)/2)^5*(2*B*a^2 - A*b^2 + 2*B*b^2 - 2*A*a*b + B*a*b))/b^3 + 
 (4*tan(c/2 + (d*x)/2)^3*(3*B*a^2 + B*b^2 - 3*A*a*b))/(3*b^3))/(d*(3*tan(c 
/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) + (a 
tan((((2*a^2 + b^2)*(A*b - B*a)*((8*tan(c/2 + (d*x)/2)*(A^2*b^9 - 8*B^2*a^ 
9 - 3*A^2*a*b^8 + 16*B^2*a^8*b + 7*A^2*a^2*b^7 - 13*A^2*a^3*b^6 + 16*A^2*a 
^4*b^5 - 16*A^2*a^5*b^4 + 16*A^2*a^6*b^3 - 8*A^2*a^7*b^2 + B^2*a^2*b^7 - 3 
*B^2*a^3*b^6 + 7*B^2*a^4*b^5 - 13*B^2*a^5*b^4 + 16*B^2*a^6*b^3 - 16*B^2*a^ 
7*b^2 - 2*A*B*a*b^8 + 16*A*B*a^8*b + 6*A*B*a^2*b^7 - 14*A*B*a^3*b^6 + 26*A 
*B*a^4*b^5 - 32*A*B*a^5*b^4 + 32*A*B*a^6*b^3 - 32*A*B*a^7*b^2))/b^6 + ((2* 
a^2 + b^2)*(A*b - B*a)*((8*(2*A*b^13 + 2*A*a^2*b^11 - 6*A*a^3*b^10 + 4*A*a 
^4*b^9 + 2*B*a^2*b^11 - 2*B*a^3*b^10 + 6*B*a^4*b^9 - 4*B*a^5*b^8 - 2*A*a*b 
^12 - 2*B*a*b^12))/b^9 - (tan(c/2 + (d*x)/2)*(2*a^2 + b^2)*(A*b - B*a)*(8* 
a*b^10 - 16*a^2*b^9 + 8*a^3*b^8)*4i)/b^10)*1i)/(2*b^4)))/(2*b^4) + ((2*a^2 
 + b^2)*(A*b - B*a)*((8*tan(c/2 + (d*x)/2)*(A^2*b^9 - 8*B^2*a^9 - 3*A^2*a* 
b^8 + 16*B^2*a^8*b + 7*A^2*a^2*b^7 - 13*A^2*a^3*b^6 + 16*A^2*a^4*b^5 - 16* 
A^2*a^5*b^4 + 16*A^2*a^6*b^3 - 8*A^2*a^7*b^2 + B^2*a^2*b^7 - 3*B^2*a^3*b^6 
 + 7*B^2*a^4*b^5 - 13*B^2*a^5*b^4 + 16*B^2*a^6*b^3 - 16*B^2*a^7*b^2 - 2*A* 
B*a*b^8 + 16*A*B*a^8*b + 6*A*B*a^2*b^7 - 14*A*B*a^3*b^6 + 26*A*B*a^4*b^5 - 
 32*A*B*a^5*b^4 + 32*A*B*a^6*b^3 - 32*A*B*a^7*b^2))/b^6 - ((2*a^2 + b^2...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.13 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {\sin \left (d x +c \right ) \left (-\sin \left (d x +c \right )^{2}+3\right )}{3 d} \] Input:

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

Output:

(sin(c + d*x)*( - sin(c + d*x)**2 + 3))/(3*d)