\(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx\) [275]

Optimal result

Integrand size = 31, antiderivative size = 274 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\frac {\left (a^3 A+4 a A b^2-3 a^2 b B-2 b^3 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {a^2 (A b-a B) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {a \left (a^2 A b-6 A b^3-4 a^3 B+9 a b^2 B\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4 A b-10 a^2 A b^3-6 A b^5+2 a^5 B-5 a^3 b^2 B+18 a b^4 B\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))} \] Output:

(A*a^3+4*A*a*b^2-3*B*a^2*b-2*B*b^3)*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^2*(A*b-B*a)*sin(d*x+c)/b^2/(a 
^2-b^2)/d/(a+b*cos(d*x+c))^3+1/6*a*(A*a^2*b-6*A*b^3-4*B*a^3+9*B*a*b^2)*sin 
(d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^2+1/6*(A*a^4*b-10*A*a^2*b^3-6*A 
*b^5+2*B*a^5-5*B*a^3*b^2+18*B*a*b^4)*sin(d*x+c)/b^2/(a^2-b^2)^3/d/(a+b*cos 
(d*x+c))
 

Mathematica [A] (verified)

Time = 2.83 (sec) , antiderivative size = 251, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\frac {-\frac {24 \left (a^3 A+4 a A b^2-3 a^2 b B-2 b^3 B\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+\frac {2 \left (-25 a^4 A b-14 a^2 A b^3-6 A b^5+10 a^5 B+17 a^3 b^2 B+18 a b^4 B+6 a \left (a^4 A-9 a^2 A b^2-2 A b^4+a^3 b B+9 a b^3 B\right ) \cos (c+d x)+\left (a^4 A b-10 a^2 A b^3-6 A b^5+2 a^5 B-5 a^3 b^2 B+18 a b^4 B\right ) \cos (2 (c+d x))\right ) \sin (c+d x)}{(a+b \cos (c+d x))^3}}{24 \left (a^2-b^2\right )^3 d} \] Input:

Integrate[(Cos[c + d*x]^2*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^4,x]
 

Output:

((-24*(a^3*A + 4*a*A*b^2 - 3*a^2*b*B - 2*b^3*B)*ArcTanh[((a - b)*Tan[(c + 
d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + (2*(-25*a^4*A*b - 14*a^2*A* 
b^3 - 6*A*b^5 + 10*a^5*B + 17*a^3*b^2*B + 18*a*b^4*B + 6*a*(a^4*A - 9*a^2* 
A*b^2 - 2*A*b^4 + a^3*b*B + 9*a*b^3*B)*Cos[c + d*x] + (a^4*A*b - 10*a^2*A* 
b^3 - 6*A*b^5 + 2*a^5*B - 5*a^3*b^2*B + 18*a*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 = 1.08 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.16, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 3467, 3042, 3500, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3467

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

\(\displaystyle \frac {\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\int -\frac {2 \left (B a^3+2 A b a^2-6 b^2 B a+3 A b^3\right ) b^2+\left (2 B a^4+A b a^3-3 b^2 B a^2-6 A b^3 a+6 b^4 B\right ) \cos (c+d x) b}{(a+b \cos (c+d x))^2}dx}{2 b \left (a^2-b^2\right )}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {2 \left (B a^3+2 A b a^2-6 b^2 B a+3 A b^3\right ) b^2+\left (2 B a^4+A b a^3-3 b^2 B a^2-6 A b^3 a+6 b^4 B\right ) \cos (c+d x) b}{(a+b \cos (c+d x))^2}dx}{2 b \left (a^2-b^2\right )}+\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {2 \left (B a^3+2 A b a^2-6 b^2 B a+3 A b^3\right ) b^2+\left (2 B a^4+A b a^3-3 b^2 B a^2-6 A b^3 a+6 b^4 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 b \left (a^2-b^2\right )}+\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3233

\(\displaystyle \frac {\frac {\frac {b \left (2 a^5 B+a^4 A b-5 a^3 b^2 B-10 a^2 A b^3+18 a b^4 B-6 A b^5\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int -\frac {3 b^3 \left (A a^3-3 b B a^2+4 A b^2 a-2 b^3 B\right )}{a+b \cos (c+d x)}dx}{a^2-b^2}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 b^3 \left (a^3 A-3 a^2 b B+4 a A b^2-2 b^3 B\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a^2-b^2}+\frac {b \left (2 a^5 B+a^4 A b-5 a^3 b^2 B-10 a^2 A b^3+18 a b^4 B-6 A b^5\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 b^3 \left (a^3 A-3 a^2 b B+4 a A b^2-2 b^3 B\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}+\frac {b \left (2 a^5 B+a^4 A b-5 a^3 b^2 B-10 a^2 A b^3+18 a b^4 B-6 A b^5\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\frac {\frac {6 b^3 \left (a^3 A-3 a^2 b B+4 a A b^2-2 b^3 B\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}+\frac {b \left (2 a^5 B+a^4 A b-5 a^3 b^2 B-10 a^2 A b^3+18 a b^4 B-6 A b^5\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {\frac {6 b^3 \left (a^3 A-3 a^2 b B+4 a A b^2-2 b^3 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}+\frac {b \left (2 a^5 B+a^4 A b-5 a^3 b^2 B-10 a^2 A b^3+18 a b^4 B-6 A b^5\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 b \left (a^2-b^2\right )}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

Input:

Int[(Cos[c + d*x]^2*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^4,x]
 

Output:

-1/3*(a^2*(A*b - a*B)*Sin[c + d*x])/(b^2*(a^2 - b^2)*d*(a + b*Cos[c + d*x] 
)^3) + ((a*(a^2*A*b - 6*A*b^3 - 4*a^3*B + 9*a*b^2*B)*Sin[c + d*x])/(2*(a^2 
 - b^2)*d*(a + b*Cos[c + d*x])^2) + ((6*b^3*(a^3*A + 4*a*A*b^2 - 3*a^2*b*B 
 - 2*b^3*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - 
b]*Sqrt[a + b]*(a^2 - b^2)*d) + (b*(a^4*A*b - 10*a^2*A*b^3 - 6*A*b^5 + 2*a 
^5*B - 5*a^3*b^2*B + 18*a*b^4*B)*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c 
 + d*x])))/(2*b*(a^2 - b^2)))/(3*b^2*(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 3467

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

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 = 1.93 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.35

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\left (a^{3} A +6 A \,a^{2} b +2 A a \,b^{2}+2 A \,b^{3}-2 a^{3} B -3 B \,a^{2} b -6 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}-\frac {4 \left (7 A \,a^{2} b +3 A \,b^{3}-a^{3} B -9 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (a^{3} A -6 A \,a^{2} b +2 A a \,b^{2}-2 A \,b^{3}+2 a^{3} B -3 B \,a^{2} b +6 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{3}}+\frac {\left (a^{3} A +4 A a \,b^{2}-3 B \,a^{2} b -2 b^{3} B \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(371\)
default	\(\frac {\frac {-\frac {\left (a^{3} A +6 A \,a^{2} b +2 A a \,b^{2}+2 A \,b^{3}-2 a^{3} B -3 B \,a^{2} b -6 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}-\frac {4 \left (7 A \,a^{2} b +3 A \,b^{3}-a^{3} B -9 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (a^{3} A -6 A \,a^{2} b +2 A a \,b^{2}-2 A \,b^{3}+2 a^{3} B -3 B \,a^{2} b +6 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{3}}+\frac {\left (a^{3} A +4 A a \,b^{2}-3 B \,a^{2} b -2 b^{3} B \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(371\)
risch	\(\text {Expression too large to display}\)	\(1386\)

Input:

int(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+cos(d*x+c)*b)^4,x,method=_RETURNVERBO 
SE)
 

Output:

1/d*(2*(-1/2*(A*a^3+6*A*a^2*b+2*A*a*b^2+2*A*b^3-2*B*a^3-3*B*a^2*b-6*B*a*b^ 
2)/(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*(7*A*a^2*b+3*A 
*b^3-B*a^3-9*B*a*b^2)/(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*(A*a^3-6*A*a^2*b+2*A*a*b^2-2*A*b^3+2*B*a^3-3*B*a^2*b+6*B*a*b^2)/(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+(A*a^3+4*A*a*b^2-3*B*a^2*b-2*B*b^3)/(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. 575 vs. \(2 (260) = 520\).

Time = 0.16 (sec) , antiderivative size = 1220, normalized size of antiderivative = 4.45 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^4,x, algorithm="f 
ricas")
 

Output:

[-1/12*(3*(A*a^6 - 3*B*a^5*b + 4*A*a^4*b^2 - 2*B*a^3*b^3 + (A*a^3*b^3 - 3* 
B*a^2*b^4 + 4*A*a*b^5 - 2*B*b^6)*cos(d*x + c)^3 + 3*(A*a^4*b^2 - 3*B*a^3*b 
^3 + 4*A*a^2*b^4 - 2*B*a*b^5)*cos(d*x + c)^2 + 3*(A*a^5*b - 3*B*a^4*b^2 + 
4*A*a^3*b^3 - 2*B*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)*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)) - 2*(4*B*a^7 - 13*A*a^6*b + 7*B*a^5*b^2 + 11*A*a^4*b^3 - 11*B*a^ 
3*b^4 + 2*A*a^2*b^5 + (2*B*a^7 + A*a^6*b - 7*B*a^5*b^2 - 11*A*a^4*b^3 + 23 
*B*a^3*b^4 + 4*A*a^2*b^5 - 18*B*a*b^6 + 6*A*b^7)*cos(d*x + c)^2 + 3*(A*a^7 
 + B*a^6*b - 10*A*a^5*b^2 + 8*B*a^4*b^3 + 7*A*a^3*b^4 - 9*B*a^2*b^5 + 2*A* 
a*b^6)*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*(A*a^6 - 3*B*a^5*b + 4*A*a^4*b^2 - 2*B*a^3*b^3 + 
 (A*a^3*b^3 - 3*B*a^2*b^4 + 4*A*a*b^5 - 2*B*b^6)*cos(d*x + c)^3 + 3*(A*a^4 
*b^2 - 3*B*a^3*b^3 + 4*A*a^2*b^4 - 2*B*a*b^5)*cos(d*x + c)^2 + 3*(A*a^5*b 
- 3*B*a^4*b^2 + 4*A*a^3*b^3 - 2*B*a^2*b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*a 
rctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) + (4*B*a^7 - 1 
3*A*a^6*b + 7*B*a^5*b^2 + 11*A*a^4*b^3 - 11*B*a^3*b^4 + 2*A*a^2*b^5 + (...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))**4,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^4,x, algorithm="m 
axima")
 

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. 689 vs. \(2 (260) = 520\).

Time = 0.23 (sec) , antiderivative size = 689, normalized size of antiderivative = 2.51 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^4,x, algorithm="g 
iac")
 

Output:

-1/3*(3*(A*a^3 - 3*B*a^2*b + 4*A*a*b^2 - 2*B*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)) + (3*A*a^5*tan(1/2*d*x + 1/2*c)^5 - 6*B*a^5*tan(1/2*d*x + 1/2*c) 
^5 + 12*A*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 3*B*a^4*b*tan(1/2*d*x + 1/2*c)^5 
- 27*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 
 + 12*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 27*B*a^2*b^3*tan(1/2*d*x + 1/2*c) 
^5 - 6*A*a*b^4*tan(1/2*d*x + 1/2*c)^5 - 18*B*a*b^4*tan(1/2*d*x + 1/2*c)^5 
+ 6*A*b^5*tan(1/2*d*x + 1/2*c)^5 - 4*B*a^5*tan(1/2*d*x + 1/2*c)^3 + 28*A*a 
^4*b*tan(1/2*d*x + 1/2*c)^3 - 32*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 16*A*a 
^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 36*B*a*b^4*tan(1/2*d*x + 1/2*c)^3 - 12*A*b 
^5*tan(1/2*d*x + 1/2*c)^3 - 3*A*a^5*tan(1/2*d*x + 1/2*c) - 6*B*a^5*tan(1/2 
*d*x + 1/2*c) + 12*A*a^4*b*tan(1/2*d*x + 1/2*c) - 3*B*a^4*b*tan(1/2*d*x + 
1/2*c) + 27*A*a^3*b^2*tan(1/2*d*x + 1/2*c) - 6*B*a^3*b^2*tan(1/2*d*x + 1/2 
*c) + 12*A*a^2*b^3*tan(1/2*d*x + 1/2*c) - 27*B*a^2*b^3*tan(1/2*d*x + 1/2*c 
) + 6*A*a*b^4*tan(1/2*d*x + 1/2*c) - 18*B*a*b^4*tan(1/2*d*x + 1/2*c) + 6*A 
*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 = 26.48 (sec) , antiderivative size = 440, normalized size of antiderivative = 1.61 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (A\,a^3-3\,B\,a^2\,b+4\,A\,a\,b^2-2\,B\,b^3\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}-\frac {\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (-B\,a^3+7\,A\,a^2\,b-9\,B\,a\,b^2+3\,A\,b^3\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A\,a^3+2\,A\,b^3-2\,B\,a^3+2\,A\,a\,b^2+6\,A\,a^2\,b-6\,B\,a\,b^2-3\,B\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,a^3-2\,A\,b^3+2\,B\,a^3+2\,A\,a\,b^2-6\,A\,a^2\,b+6\,B\,a\,b^2-3\,B\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )} \] Input:

int((cos(c + d*x)^2*(A + B*cos(c + d*x)))/(a + b*cos(c + d*x))^4,x)
 

Output:

(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*a^3 - 2*B*b^3 + 4*A*a*b^2 - 3*B*a^2*b))/( 
d*(a + b)^(7/2)*(a - b)^(7/2)) - ((4*tan(c/2 + (d*x)/2)^3*(3*A*b^3 - B*a^3 
 + 7*A*a^2*b - 9*B*a*b^2))/(3*(a + b)^2*(a^2 - 2*a*b + b^2)) + (tan(c/2 + 
(d*x)/2)^5*(A*a^3 + 2*A*b^3 - 2*B*a^3 + 2*A*a*b^2 + 6*A*a^2*b - 6*B*a*b^2 
- 3*B*a^2*b))/((a + b)^3*(a - b)) - (tan(c/2 + (d*x)/2)*(A*a^3 - 2*A*b^3 + 
 2*B*a^3 + 2*A*a*b^2 - 6*A*a^2*b + 6*B*a*b^2 - 3*B*a^2*b))/((a + b)*(3*a*b 
^2 - 3*a^2*b + a^3 - b^3)))/(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)))
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 632, normalized size of antiderivative = 2.31 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:

int(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^4,x)
 

Output:

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