\(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx\) [267]

Optimal result

Integrand size = 31, antiderivative size = 211 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx=\frac {B x}{b^3}+\frac {\left (a^2 A b^3+2 A b^5-2 a^5 B+5 a^3 b^2 B-6 a b^4 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} b^3 (a+b)^{5/2} d}-\frac {a^2 (A b-a B) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {a \left (a^2 A b-4 A b^3-3 a^3 B+6 a b^2 B\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 2.78 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx=\frac {2 B (c+d x)+\frac {2 \left (-a^2 A b^3-2 A b^5+2 a^5 B-5 a^3 b^2 B+6 a b^4 B\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}+\frac {a^2 b (-A b+a B) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}+\frac {a b \left (a^2 A b-4 A b^3-3 a^3 B+6 a b^2 B\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}}{2 b^3 d} \] Input:

Integrate[(Cos[c + d*x]^2*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^3,x]
 

Output:

(2*B*(c + d*x) + (2*(-(a^2*A*b^3) - 2*A*b^5 + 2*a^5*B - 5*a^3*b^2*B + 6*a* 
b^4*B)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^ 
(5/2) + (a^2*b*(-(A*b) + a*B)*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c 
+ d*x])^2) + (a*b*(a^2*A*b - 4*A*b^3 - 3*a^3*B + 6*a*b^2*B)*Sin[c + d*x])/ 
((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])))/(2*b^3*d)
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.17, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 3467, 3042, 3500, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3467

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

Input:

Int[(Cos[c + d*x]^2*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^3,x]
 

Output:

-1/2*(a^2*(A*b - a*B)*Sin[c + d*x])/(b^2*(a^2 - b^2)*d*(a + b*Cos[c + d*x] 
)^2) + ((2*(a^2 - b^2)^2*B*x + (2*(a^2*A*b^3 + 2*A*b^5 - 2*a^5*B + 5*a^3*b 
^2*B - 6*a*b^4*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqr 
t[a - b]*Sqrt[a + b]*d))/(b*(a^2 - b^2)) + (a*(a^2*A*b - 4*A*b^3 - 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*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 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 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 = 1.72 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.34

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (-\frac {\left (A a \,b^{2}+4 A \,b^{3}+2 a^{3} B -B \,a^{2} b -6 B a \,b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {b a \left (A a \,b^{2}-4 A \,b^{3}-2 a^{3} B -B \,a^{2} b +6 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (A \,a^{2} b^{3}+2 A \,b^{5}-2 B \,a^{5}+5 B \,a^{3} b^{2}-6 B a \,b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{b^{3}}+\frac {2 B \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}}{d}\)	\(282\)
default	\(\frac {\frac {\frac {2 \left (-\frac {\left (A a \,b^{2}+4 A \,b^{3}+2 a^{3} B -B \,a^{2} b -6 B a \,b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {b a \left (A a \,b^{2}-4 A \,b^{3}-2 a^{3} B -B \,a^{2} b +6 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (A \,a^{2} b^{3}+2 A \,b^{5}-2 B \,a^{5}+5 B \,a^{3} b^{2}-6 B a \,b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{b^{3}}+\frac {2 B \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}}{d}\)	\(282\)
risch	\(\text {Expression too large to display}\)	\(1180\)

Input:

int(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+cos(d*x+c)*b)^3,x,method=_RETURNVERBO 
SE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 542 vs. \(2 (201) = 402\).

Time = 0.16 (sec) , antiderivative size = 1152, normalized size of antiderivative = 5.46 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^3,x, algorithm="f 
ricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^3,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. 455 vs. \(2 (201) = 402\).

Time = 0.20 (sec) , antiderivative size = 455, normalized size of antiderivative = 2.16 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx=-\frac {\frac {{\left (2 \, B a^{5} - 5 \, B a^{3} b^{2} - A a^{2} b^{3} + 6 \, B a b^{4} - 2 \, A b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \sqrt {a^{2} - b^{2}}} - \frac {{\left (d x + c\right )} B}{b^{3}} + \frac {2 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}}}{d} \] Input:

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^3,x, algorithm="g 
iac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 31.62 (sec) , antiderivative size = 6923, normalized size of antiderivative = 32.81 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

int((cos(c + d*x)^2*(A + B*cos(c + d*x)))/(a + b*cos(c + d*x))^3,x)
 

Output:

(2*B*atan(-((B*((B*((8*(4*A*b^15 + 4*B*b^15 - 6*A*a^2*b^13 + 6*A*a^3*b^12 
+ 2*A*a^6*b^9 - 2*A*a^7*b^8 - 8*B*a^2*b^13 + 34*B*a^3*b^12 + 6*B*a^4*b^11 
- 36*B*a^5*b^10 - 4*B*a^6*b^9 + 18*B*a^7*b^8 + 2*B*a^8*b^7 - 4*B*a^9*b^6 - 
 4*A*a*b^14 - 12*B*a*b^14))/(a*b^12 + b^13 - 3*a^2*b^11 - 3*a^3*b^10 + 3*a 
^4*b^9 + 3*a^5*b^8 - a^6*b^7 - a^7*b^6) - (B*tan(c/2 + (d*x)/2)*(8*a*b^15 
- 8*a^2*b^14 - 32*a^3*b^13 + 32*a^4*b^12 + 48*a^5*b^11 - 48*a^6*b^10 - 32* 
a^7*b^9 + 32*a^8*b^8 + 8*a^9*b^7 - 8*a^10*b^6)*8i)/(b^3*(a*b^10 + b^11 - 3 
*a^2*b^9 - 3*a^3*b^8 + 3*a^4*b^7 + 3*a^5*b^6 - a^6*b^5 - a^7*b^4)))*1i)/b^ 
3 + (8*tan(c/2 + (d*x)/2)*(4*A^2*b^10 + 8*B^2*a^10 + 4*B^2*b^10 - 8*B^2*a* 
b^9 - 8*B^2*a^9*b + 4*A^2*a^2*b^8 + A^2*a^4*b^6 + 24*B^2*a^2*b^8 + 32*B^2* 
a^3*b^7 - 52*B^2*a^4*b^6 - 48*B^2*a^5*b^5 + 57*B^2*a^6*b^4 + 32*B^2*a^7*b^ 
3 - 32*B^2*a^8*b^2 - 24*A*B*a*b^9 + 8*A*B*a^3*b^7 + 2*A*B*a^5*b^5 - 4*A*B* 
a^7*b^3))/(a*b^10 + b^11 - 3*a^2*b^9 - 3*a^3*b^8 + 3*a^4*b^7 + 3*a^5*b^6 - 
 a^6*b^5 - a^7*b^4)))/b^3 - (B*((B*((8*(4*A*b^15 + 4*B*b^15 - 6*A*a^2*b^13 
 + 6*A*a^3*b^12 + 2*A*a^6*b^9 - 2*A*a^7*b^8 - 8*B*a^2*b^13 + 34*B*a^3*b^12 
 + 6*B*a^4*b^11 - 36*B*a^5*b^10 - 4*B*a^6*b^9 + 18*B*a^7*b^8 + 2*B*a^8*b^7 
 - 4*B*a^9*b^6 - 4*A*a*b^14 - 12*B*a*b^14))/(a*b^12 + b^13 - 3*a^2*b^11 - 
3*a^3*b^10 + 3*a^4*b^9 + 3*a^5*b^8 - a^6*b^7 - a^7*b^6) + (B*tan(c/2 + (d* 
x)/2)*(8*a*b^15 - 8*a^2*b^14 - 32*a^3*b^13 + 32*a^4*b^12 + 48*a^5*b^11 - 4 
8*a^6*b^10 - 32*a^7*b^9 + 32*a^8*b^8 + 8*a^9*b^7 - 8*a^10*b^6)*8i)/(b^3...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.80 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) a^{3} b +4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) a \,b^{3}-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{4}+4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2} b^{2}+\cos \left (d x +c \right ) a^{4} b d x -2 \cos \left (d x +c \right ) a^{2} b^{3} d x +\cos \left (d x +c \right ) b^{5} d x -\sin \left (d x +c \right ) a^{4} b +\sin \left (d x +c \right ) a^{2} b^{3}+a^{5} d x -2 a^{3} b^{2} d x +a \,b^{4} d x}{b^{2} d \left (\cos \left (d x +c \right ) a^{4} b -2 \cos \left (d x +c \right ) a^{2} b^{3}+\cos \left (d x +c \right ) b^{5}+a^{5}-2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:

int(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^3,x)
 

Output:

( - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t(a**2 - b**2))*cos(c + d*x)*a**3*b + 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*b**3 - 2* 
sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 
 - b**2))*a**4 + 4*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 + cos(c + d*x)*a**4*b*d*x - 2*cos(c 
 + d*x)*a**2*b**3*d*x + cos(c + d*x)*b**5*d*x - sin(c + d*x)*a**4*b + sin( 
c + d*x)*a**2*b**3 + a**5*d*x - 2*a**3*b**2*d*x + a*b**4*d*x)/(b**2*d*(cos 
(c + d*x)*a**4*b - 2*cos(c + d*x)*a**2*b**3 + cos(c + d*x)*b**5 + a**5 - 2 
*a**3*b**2 + a*b**4))