\(\int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx\) [276]

Optimal result

Integrand size = 29, antiderivative size = 263 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=-\frac {\left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac {a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))} \] Output:

-(4*A*a^2*b+A*b^3-B*a^3-4*B*a*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/( 
a+b)^(1/2))/(a-b)^(7/2)/(a+b)^(7/2)/d+1/3*a*(A*b-B*a)*sin(d*x+c)/b/(a^2-b^ 
2)/d/(a+b*cos(d*x+c))^3+1/6*(2*A*a^2*b+3*A*b^3+B*a^3-6*B*a*b^2)*sin(d*x+c) 
/b/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^2+1/6*(2*A*a^3*b+13*A*a*b^3+B*a^4-10*B*a 
^2*b^2-6*B*b^4)*sin(d*x+c)/b/(a^2-b^2)^3/d/(a+b*cos(d*x+c))
 

Mathematica [A] (verified)

Time = 2.21 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.96 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\frac {-\frac {24 \left (-4 a^2 A b-A b^3+a^3 B+4 a b^2 B\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+\frac {2 \left (12 a^5 A+22 a^3 A b^2+11 a A b^4-25 a^4 b B-14 a^2 b^3 B-6 b^5 B+6 \left (2 a^4 A b+9 a^2 A b^3-A b^5+a^5 B-9 a^3 b^2 B-2 a b^4 B\right ) \cos (c+d x)+b \left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \cos (2 (c+d x))\right ) \sin (c+d x)}{(a+b \cos (c+d x))^3}}{24 \left (a^2-b^2\right )^3 d} \] Input:

Integrate[(Cos[c + d*x]*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^4,x]
 

Output:

((-24*(-4*a^2*A*b - A*b^3 + a^3*B + 4*a*b^2*B)*ArcTanh[((a - b)*Tan[(c + d 
*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + (2*(12*a^5*A + 22*a^3*A*b^2 
+ 11*a*A*b^4 - 25*a^4*b*B - 14*a^2*b^3*B - 6*b^5*B + 6*(2*a^4*A*b + 9*a^2* 
A*b^3 - A*b^5 + a^5*B - 9*a^3*b^2*B - 2*a*b^4*B)*Cos[c + d*x] + b*(2*a^3*A 
*b + 13*a*A*b^3 + a^4*B - 10*a^2*b^2*B - 6*b^4*B)*Cos[2*(c + d*x)])*Sin[c 
+ d*x])/(a + b*Cos[c + d*x])^3)/(24*(a^2 - b^2)^3*d)
 

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.15, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.448, Rules used = {3042, 3447, 3042, 3500, 3042, 3233, 25, 3042, 3233, 27, 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.

\(\displaystyle \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4}dx\)

\(\Big \downarrow \) 3447

\(\displaystyle \int \frac {A \cos (c+d x)+B \cos ^2(c+d x)}{(a+b \cos (c+d x))^4}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )+B \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4}dx\)

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

\(\displaystyle \frac {a (A b-a B) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {-\frac {\int -\frac {2 b \left (-2 B a^2+5 A b a-3 b^2 B\right )-\left (B a^3+2 A b a^2-6 b^2 B a+3 A b^3\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {\left (a^3 B+2 a^2 A b-6 a b^2 B+3 A b^3\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {a (A b-a B) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {\frac {\int \frac {2 b \left (-2 B a^2+5 A b a-3 b^2 B\right )-\left (B a^3+2 A b a^2-6 b^2 B a+3 A b^3\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {\left (a^3 B+2 a^2 A b-6 a b^2 B+3 A b^3\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (A b-a B) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {\frac {\int \frac {2 b \left (-2 B a^2+5 A b a-3 b^2 B\right )+\left (-B a^3-2 A b a^2+6 b^2 B a-3 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 \left (a^2-b^2\right )}-\frac {\left (a^3 B+2 a^2 A b-6 a b^2 B+3 A b^3\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3233

\(\displaystyle \frac {a (A b-a B) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {\frac {-\frac {\int -\frac {3 b \left (-B a^3+4 A b a^2-4 b^2 B a+A b^3\right )}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {\left (a^4 B+2 a^3 A b-10 a^2 b^2 B+13 a A b^3-6 b^4 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (a^3 B+2 a^2 A b-6 a b^2 B+3 A b^3\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a (A b-a B) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {\frac {\frac {3 b \left (a^3 (-B)+4 a^2 A b-4 a b^2 B+A b^3\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {\left (a^4 B+2 a^3 A b-10 a^2 b^2 B+13 a A b^3-6 b^4 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (a^3 B+2 a^2 A b-6 a b^2 B+3 A b^3\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (A b-a B) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {\frac {\frac {3 b \left (a^3 (-B)+4 a^2 A b-4 a b^2 B+A b^3\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {\left (a^4 B+2 a^3 A b-10 a^2 b^2 B+13 a A b^3-6 b^4 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (a^3 B+2 a^2 A b-6 a b^2 B+3 A b^3\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {a (A b-a B) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {\frac {\frac {6 b \left (a^3 (-B)+4 a^2 A b-4 a b^2 B+A b^3\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}-\frac {\left (a^4 B+2 a^3 A b-10 a^2 b^2 B+13 a A b^3-6 b^4 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (a^3 B+2 a^2 A b-6 a b^2 B+3 A b^3\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {a (A b-a B) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {\frac {\frac {6 b \left (a^3 (-B)+4 a^2 A b-4 a b^2 B+A b^3\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {\left (a^4 B+2 a^3 A b-10 a^2 b^2 B+13 a A b^3-6 b^4 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (a^3 B+2 a^2 A b-6 a b^2 B+3 A b^3\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}\)

Input:

Int[(Cos[c + d*x]*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^4,x]
 

Output:

(a*(A*b - a*B)*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^3) - 
(-1/2*((2*a^2*A*b + 3*A*b^3 + a^3*B - 6*a*b^2*B)*Sin[c + d*x])/((a^2 - b^2 
)*d*(a + b*Cos[c + d*x])^2) + ((6*b*(4*a^2*A*b + A*b^3 - a^3*B - 4*a*b^2*B 
)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a 
+ b]*(a^2 - b^2)*d) - ((2*a^3*A*b + 13*a*A*b^3 + a^4*B - 10*a^2*b^2*B - 6* 
b^4*B)*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x])))/(2*(a^2 - b^2)) 
)/(3*b*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3233

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + 
 f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( 
m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
 

rule 3447

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 3500

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 1.77 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.46

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {\left (2 a^{3} A +2 A \,a^{2} b +6 A a \,b^{2}+A \,b^{3}-a^{3} B -6 B \,a^{2} b -2 B a \,b^{2}-2 b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}-\frac {2 \left (3 a^{3} A +7 A a \,b^{2}-7 B \,a^{2} b -3 b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (2 a^{3} A -2 A \,a^{2} b +6 A a \,b^{2}-A \,b^{3}+a^{3} B -6 B \,a^{2} b +2 B a \,b^{2}-2 b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{3}}-\frac {\left (4 A \,a^{2} b +A \,b^{3}-a^{3} B -4 B a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(384\)
default	\(\frac {-\frac {2 \left (-\frac {\left (2 a^{3} A +2 A \,a^{2} b +6 A a \,b^{2}+A \,b^{3}-a^{3} B -6 B \,a^{2} b -2 B a \,b^{2}-2 b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}-\frac {2 \left (3 a^{3} A +7 A a \,b^{2}-7 B \,a^{2} b -3 b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (2 a^{3} A -2 A \,a^{2} b +6 A a \,b^{2}-A \,b^{3}+a^{3} B -6 B \,a^{2} b +2 B a \,b^{2}-2 b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{3}}-\frac {\left (4 A \,a^{2} b +A \,b^{3}-a^{3} B -4 B a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(384\)
risch	\(\text {Expression too large to display}\)	\(1282\)

Input:

int(cos(d*x+c)*(A+B*cos(d*x+c))/(a+cos(d*x+c)*b)^4,x,method=_RETURNVERBOSE 
)
 

Output:

1/d*(-2*(-1/2*(2*A*a^3+2*A*a^2*b+6*A*a*b^2+A*b^3-B*a^3-6*B*a^2*b-2*B*a*b^2 
-2*B*b^3)/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5-2/3*(3*A*a^ 
3+7*A*a*b^2-7*B*a^2*b-3*B*b^3)/(a^2-2*a*b+b^2)/(a^2+2*a*b+b^2)*tan(1/2*d*x 
+1/2*c)^3-1/2*(2*A*a^3-2*A*a^2*b+6*A*a*b^2-A*b^3+B*a^3-6*B*a^2*b+2*B*a*b^2 
-2*B*b^3)/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x 
+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3-(4*A*a^2*b+A*b^3-B*a^3-4*B*a*b^2 
)/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d 
*x+1/2*c)/((a-b)*(a+b))^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 581 vs. \(2 (247) = 494\).

Time = 0.17 (sec) , antiderivative size = 1232, normalized size of antiderivative = 4.68 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^4,x, algorithm="fri 
cas")
 

Output:

[-1/12*(3*(B*a^6 - 4*A*a^5*b + 4*B*a^4*b^2 - A*a^3*b^3 + (B*a^3*b^3 - 4*A* 
a^2*b^4 + 4*B*a*b^5 - A*b^6)*cos(d*x + c)^3 + 3*(B*a^4*b^2 - 4*A*a^3*b^3 + 
 4*B*a^2*b^4 - A*a*b^5)*cos(d*x + c)^2 + 3*(B*a^5*b - 4*A*a^4*b^2 + 4*B*a^ 
3*b^3 - A*a^2*b^4)*cos(d*x + c))*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)*s 
in(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2) 
) - 2*(6*A*a^7 - 13*B*a^6*b + 4*A*a^5*b^2 + 11*B*a^4*b^3 - 11*A*a^3*b^4 + 
2*B*a^2*b^5 + A*a*b^6 + (B*a^6*b + 2*A*a^5*b^2 - 11*B*a^4*b^3 + 11*A*a^3*b 
^4 + 4*B*a^2*b^5 - 13*A*a*b^6 + 6*B*b^7)*cos(d*x + c)^2 + 3*(B*a^7 + 2*A*a 
^6*b - 10*B*a^5*b^2 + 7*A*a^4*b^3 + 7*B*a^3*b^4 - 10*A*a^2*b^5 + 2*B*a*b^6 
 + A*b^7)*cos(d*x + c))*sin(d*x + c))/((a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 
4*a^2*b^9 + b^11)*d*cos(d*x + c)^3 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 
4*a^3*b^8 + a*b^10)*d*cos(d*x + c)^2 + 3*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 
 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c) + (a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a 
^5*b^6 + a^3*b^8)*d), 1/6*(3*(B*a^6 - 4*A*a^5*b + 4*B*a^4*b^2 - A*a^3*b^3 
+ (B*a^3*b^3 - 4*A*a^2*b^4 + 4*B*a*b^5 - A*b^6)*cos(d*x + c)^3 + 3*(B*a^4* 
b^2 - 4*A*a^3*b^3 + 4*B*a^2*b^4 - A*a*b^5)*cos(d*x + c)^2 + 3*(B*a^5*b - 4 
*A*a^4*b^2 + 4*B*a^3*b^3 - A*a^2*b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan 
(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) + (6*A*a^7 - 13*B*a 
^6*b + 4*A*a^5*b^2 + 11*B*a^4*b^3 - 11*A*a^3*b^4 + 2*B*a^2*b^5 + A*a*b^...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))**4,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^4,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. 722 vs. \(2 (247) = 494\).

Time = 0.20 (sec) , antiderivative size = 722, normalized size of antiderivative = 2.75 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^4,x, algorithm="gia 
c")
 

Output:

-1/3*(3*(B*a^3 - 4*A*a^2*b + 4*B*a*b^2 - A*b^3)*(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)))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(a^2 
- b^2)) - (6*A*a^5*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^5*tan(1/2*d*x + 1/2*c)^5 
 - 6*A*a^4*b*tan(1/2*d*x + 1/2*c)^5 - 12*B*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 
12*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 27*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 
- 27*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 12*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^ 
5 + 12*A*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 6*B*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 
 3*A*b^5*tan(1/2*d*x + 1/2*c)^5 - 6*B*b^5*tan(1/2*d*x + 1/2*c)^5 + 12*A*a^ 
5*tan(1/2*d*x + 1/2*c)^3 - 28*B*a^4*b*tan(1/2*d*x + 1/2*c)^3 + 16*A*a^3*b^ 
2*tan(1/2*d*x + 1/2*c)^3 + 16*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 28*A*a*b^ 
4*tan(1/2*d*x + 1/2*c)^3 + 12*B*b^5*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^5*tan(1 
/2*d*x + 1/2*c) + 3*B*a^5*tan(1/2*d*x + 1/2*c) + 6*A*a^4*b*tan(1/2*d*x + 1 
/2*c) - 12*B*a^4*b*tan(1/2*d*x + 1/2*c) + 12*A*a^3*b^2*tan(1/2*d*x + 1/2*c 
) - 27*B*a^3*b^2*tan(1/2*d*x + 1/2*c) + 27*A*a^2*b^3*tan(1/2*d*x + 1/2*c) 
- 12*B*a^2*b^3*tan(1/2*d*x + 1/2*c) + 12*A*a*b^4*tan(1/2*d*x + 1/2*c) - 6* 
B*a*b^4*tan(1/2*d*x + 1/2*c) - 3*A*b^5*tan(1/2*d*x + 1/2*c) - 6*B*b^5*tan( 
1/2*d*x + 1/2*c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(a*tan(1/2*d*x + 1/ 
2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^3))/d
 

Mupad [B] (verification not implemented)

Time = 27.05 (sec) , antiderivative size = 451, normalized size of antiderivative = 1.71 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\frac {\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,A\,a^3-7\,B\,a^2\,b+7\,A\,a\,b^2-3\,B\,b^3\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,a^3-A\,b^3+B\,a^3-2\,B\,b^3+6\,A\,a\,b^2-2\,A\,a^2\,b+2\,B\,a\,b^2-6\,B\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,A\,a^3+A\,b^3-B\,a^3-2\,B\,b^3+6\,A\,a\,b^2+2\,A\,a^2\,b-2\,B\,a\,b^2-6\,B\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}}{d\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}-\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (-B\,a^3+4\,A\,a^2\,b-4\,B\,a\,b^2+A\,b^3\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \] Input:

int((cos(c + d*x)*(A + B*cos(c + d*x)))/(a + b*cos(c + d*x))^4,x)
 

Output:

((4*tan(c/2 + (d*x)/2)^3*(3*A*a^3 - 3*B*b^3 + 7*A*a*b^2 - 7*B*a^2*b))/(3*( 
a + b)^2*(a^2 - 2*a*b + b^2)) + (tan(c/2 + (d*x)/2)*(2*A*a^3 - A*b^3 + B*a 
^3 - 2*B*b^3 + 6*A*a*b^2 - 2*A*a^2*b + 2*B*a*b^2 - 6*B*a^2*b))/((a + b)*(3 
*a*b^2 - 3*a^2*b + a^3 - b^3)) + (tan(c/2 + (d*x)/2)^5*(2*A*a^3 + A*b^3 - 
B*a^3 - 2*B*b^3 + 6*A*a*b^2 + 2*A*a^2*b - 2*B*a*b^2 - 6*B*a^2*b))/((a + b) 
^3*(a - b)))/(d*(3*a*b^2 - tan(c/2 + (d*x)/2)^4*(3*a*b^2 + 3*a^2*b - 3*a^3 
 - 3*b^3) - tan(c/2 + (d*x)/2)^2*(3*a*b^2 - 3*a^2*b - 3*a^3 + 3*b^3) + 3*a 
^2*b + a^3 + b^3 + tan(c/2 + (d*x)/2)^6*(3*a*b^2 - 3*a^2*b + a^3 - b^3))) 
- (atan((tan(c/2 + (d*x)/2)*(2*a - 2*b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3))/( 
2*(a + b)^(1/2)*(a - b)^(7/2)))*(A*b^3 - B*a^3 + 4*A*a^2*b - 4*B*a*b^2))/( 
d*(a + b)^(7/2)*(a - b)^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 467, normalized size of antiderivative = 1.78 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\frac {-12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) a^{2} b^{2}+6 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right )^{2} a \,b^{3}-6 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{3} b -6 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a \,b^{3}+\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{4} b +\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{3}-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{5}+2 \sin \left (d x +c \right ) a^{5}-\sin \left (d x +c \right ) a^{3} b^{2}-\sin \left (d x +c \right ) a \,b^{4}}{2 d \left (2 \cos \left (d x +c \right ) a^{7} b -6 \cos \left (d x +c \right ) a^{5} b^{3}+6 \cos \left (d x +c \right ) a^{3} b^{5}-2 \cos \left (d x +c \right ) a \,b^{7}-\sin \left (d x +c \right )^{2} a^{6} b^{2}+3 \sin \left (d x +c \right )^{2} a^{4} b^{4}-3 \sin \left (d x +c \right )^{2} a^{2} b^{6}+\sin \left (d x +c \right )^{2} b^{8}+a^{8}-2 a^{6} b^{2}+2 a^{2} b^{6}-b^{8}\right )} \] Input:

int(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^4,x)
 

Output:

( - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq 
rt(a**2 - b**2))*cos(c + d*x)*a**2*b**2 + 6*sqrt(a**2 - b**2)*atan((tan((c 
 + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a*b* 
*3 - 6*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq 
rt(a**2 - b**2))*a**3*b - 6*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - t 
an((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a*b**3 + cos(c + d*x)*sin(c + d*x)*a 
**4*b + cos(c + d*x)*sin(c + d*x)*a**2*b**3 - 2*cos(c + d*x)*sin(c + d*x)* 
b**5 + 2*sin(c + d*x)*a**5 - sin(c + d*x)*a**3*b**2 - sin(c + d*x)*a*b**4) 
/(2*d*(2*cos(c + d*x)*a**7*b - 6*cos(c + d*x)*a**5*b**3 + 6*cos(c + d*x)*a 
**3*b**5 - 2*cos(c + d*x)*a*b**7 - sin(c + d*x)**2*a**6*b**2 + 3*sin(c + d 
*x)**2*a**4*b**4 - 3*sin(c + d*x)**2*a**2*b**6 + sin(c + d*x)**2*b**8 + a* 
*8 - 2*a**6*b**2 + 2*a**2*b**6 - b**8))