\(\int \frac {\cos ^2(c+d x) (A+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [571]

Optimal result

Integrand size = 33, antiderivative size = 262 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {\left (2 A b^2+\left (6 a^2+b^2\right ) C\right ) x}{2 b^4}-\frac {2 a \left (a^2 A b^2-2 A b^4+3 a^4 C-4 a^2 b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^4 (a+b)^{3/2} d}-\frac {a \left (A b^2+3 a^2 C-2 b^2 C\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (2 A b^2+3 a^2 C-b^2 C\right ) \cos (c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 2.36 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.68 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {2 \left (2 A b^2+\left (6 a^2+b^2\right ) C\right ) (c+d x)-\frac {8 a \left (-2 A b^4+a^2 b^2 (A-4 C)+3 a^4 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}-8 a b C \sin (c+d x)-\frac {4 a^2 b \left (A b^2+a^2 C\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+b^2 C \sin (2 (c+d x))}{4 b^4 d} \] Input:

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

Output:

(2*(2*A*b^2 + (6*a^2 + b^2)*C)*(c + d*x) - (8*a*(-2*A*b^4 + a^2*b^2*(A - 4 
*C) + 3*a^4*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 
 + b^2)^(3/2) - 8*a*b*C*Sin[c + d*x] - (4*a^2*b*(A*b^2 + a^2*C)*Sin[c + d* 
x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])) + b^2*C*Sin[2*(c + d*x)])/(4*b^ 
4*d)
 

Rubi [A] (verified)

Time = 1.38 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3527, 3042, 3528, 25, 3042, 3502, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3527

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

-(((A*b^2 + a^2*C)*Cos[c + d*x]^2*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Co 
s[c + d*x]))) - (-1/2*((2*A*b^2 + 3*a^2*C - b^2*C)*Cos[c + d*x]*Sin[c + d* 
x])/(b*d) - ((((a^2 - b^2)*(2*A*b^2 + (6*a^2 + b^2)*C)*x)/b + (4*a*(2*A*b^ 
4 - a^2*b^2*(A - 4*C) - 3*a^4*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqr 
t[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d))/b - (2*a*(A*b^2 + 3*a^2*C - 2*b^ 
2*C)*Sin[c + d*x])/(b*d))/(2*b))/(b*(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 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 3527

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + 
 f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^ 
2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A* 
d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b 
*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( 
A*d^2*(m + n + 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - 
b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -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 = 0.88 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.99

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (\left (-2 a b C -\frac {1}{2} C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-2 a b C +\frac {1}{2} C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (2 A \,b^{2}+6 a^{2} C +C \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}-\frac {2 a \left (\frac {a b \left (A \,b^{2}+a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (A \,a^{2} b^{2}-2 A \,b^{4}+3 a^{4} C -4 C \,a^{2} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}}{d}\)	\(259\)
default	\(\frac {\frac {\frac {2 \left (\left (-2 a b C -\frac {1}{2} C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-2 a b C +\frac {1}{2} C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (2 A \,b^{2}+6 a^{2} C +C \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}-\frac {2 a \left (\frac {a b \left (A \,b^{2}+a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (A \,a^{2} b^{2}-2 A \,b^{4}+3 a^{4} C -4 C \,a^{2} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}}{d}\)	\(259\)
risch	\(\frac {x A}{b^{2}}+\frac {3 x \,a^{2} C}{b^{4}}+\frac {C x}{2 b^{2}}-\frac {i C \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{2} d}-\frac {i C a \,{\mathrm e}^{-i \left (d x +c \right )}}{b^{3} d}-\frac {2 i a^{2} \left (A \,b^{2}+a^{2} C \right ) \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{4} \left (a^{2}-b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}+\frac {i C a \,{\mathrm e}^{i \left (d x +c \right )}}{b^{3} d}+\frac {i C \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 b^{2} d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {3 a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{4}}+\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {3 a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{4}}-\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}\)	\(870\)

Input:

int(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x,method=_RETURNVER 
BOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 809, normalized size of antiderivative = 3.09 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm= 
"fricas")
 

Output:

[1/2*((6*C*a^6*b + (2*A - 11*C)*a^4*b^3 - 4*(A - C)*a^2*b^5 + (2*A + C)*b^ 
7)*d*x*cos(d*x + c) + (6*C*a^7 + (2*A - 11*C)*a^5*b^2 - 4*(A - C)*a^3*b^4 
+ (2*A + C)*a*b^6)*d*x - (3*C*a^6 + (A - 4*C)*a^4*b^2 - 2*A*a^2*b^4 + (3*C 
*a^5*b + (A - 4*C)*a^3*b^3 - 2*A*a*b^5)*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)) - (6*C*a^6*b + 2*(A - 5*C)*a^4*b^3 - 2*(A - 2*C)*a^ 
2*b^5 - (C*a^4*b^3 - 2*C*a^2*b^5 + C*b^7)*cos(d*x + c)^2 + 3*(C*a^5*b^2 - 
2*C*a^3*b^4 + C*a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^4*b^5 - 2*a^2*b^7 + 
 b^9)*d*cos(d*x + c) + (a^5*b^4 - 2*a^3*b^6 + a*b^8)*d), 1/2*((6*C*a^6*b + 
 (2*A - 11*C)*a^4*b^3 - 4*(A - C)*a^2*b^5 + (2*A + C)*b^7)*d*x*cos(d*x + c 
) + (6*C*a^7 + (2*A - 11*C)*a^5*b^2 - 4*(A - C)*a^3*b^4 + (2*A + C)*a*b^6) 
*d*x - 2*(3*C*a^6 + (A - 4*C)*a^4*b^2 - 2*A*a^2*b^4 + (3*C*a^5*b + (A - 4* 
C)*a^3*b^3 - 2*A*a*b^5)*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*C*a^6*b + 2*(A - 5*C)*a^4*b^ 
3 - 2*(A - 2*C)*a^2*b^5 - (C*a^4*b^3 - 2*C*a^2*b^5 + C*b^7)*cos(d*x + c)^2 
 + 3*(C*a^5*b^2 - 2*C*a^3*b^4 + C*a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^4 
*b^5 - 2*a^2*b^7 + b^9)*d*cos(d*x + c) + (a^5*b^4 - 2*a^3*b^6 + a*b^8)*d)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm= 
"maxima")
 

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 [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.19 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {4 \, {\left (3 \, C a^{5} + A a^{3} b^{2} - 4 \, C a^{3} b^{2} - 2 \, A a b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} - \frac {4 \, {\left (C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}} + \frac {{\left (6 \, C a^{2} + 2 \, A b^{2} + C b^{2}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {2 \, {\left (4 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{3}}}{2 \, d} \] Input:

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm= 
"giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 5.75 (sec) , antiderivative size = 6546, normalized size of antiderivative = 24.98 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan((((C*a^2*3i + b^2*(A*1i + (C*1i)/2))*((8*tan(c/2 + (d*x)/2)*(4*A^2*b 
^10 + 72*C^2*a^10 + C^2*b^10 - 8*A^2*a*b^9 - 2*C^2*a*b^9 - 72*C^2*a^9*b + 
12*A^2*a^2*b^8 + 16*A^2*a^3*b^7 - 20*A^2*a^4*b^6 - 8*A^2*a^5*b^5 + 8*A^2*a 
^6*b^4 + 11*C^2*a^2*b^8 - 20*C^2*a^3*b^7 + 23*C^2*a^4*b^6 - 26*C^2*a^5*b^5 
 + 17*C^2*a^6*b^4 + 120*C^2*a^7*b^3 - 120*C^2*a^8*b^2 + 4*A*C*b^10 - 8*A*C 
*a*b^9 + 20*A*C*a^2*b^8 - 32*A*C*a^3*b^7 + 36*A*C*a^4*b^6 + 88*A*C*a^5*b^5 
 - 100*A*C*a^6*b^4 - 48*A*C*a^7*b^3 + 48*A*C*a^8*b^2))/(a*b^8 + b^9 - a^2* 
b^7 - a^3*b^6) + (((8*(4*A*b^15 + 2*C*b^15 - 4*A*a^2*b^13 + 12*A*a^3*b^12 
- 4*A*a^5*b^10 + 6*C*a^2*b^13 - 16*C*a^3*b^12 - 14*C*a^4*b^11 + 28*C*a^5*b 
^10 + 6*C*a^6*b^9 - 12*C*a^7*b^8 - 8*A*a*b^14))/(a*b^11 + b^12 - a^2*b^10 
- a^3*b^9) - (8*tan(c/2 + (d*x)/2)*(C*a^2*3i + b^2*(A*1i + (C*1i)/2))*(8*a 
*b^13 - 8*a^2*b^12 - 16*a^3*b^11 + 16*a^4*b^10 + 8*a^5*b^9 - 8*a^6*b^8))/( 
b^4*(a*b^8 + b^9 - a^2*b^7 - a^3*b^6)))*(C*a^2*3i + b^2*(A*1i + (C*1i)/2)) 
)/b^4)*1i)/b^4 + ((C*a^2*3i + b^2*(A*1i + (C*1i)/2))*((8*tan(c/2 + (d*x)/2 
)*(4*A^2*b^10 + 72*C^2*a^10 + C^2*b^10 - 8*A^2*a*b^9 - 2*C^2*a*b^9 - 72*C^ 
2*a^9*b + 12*A^2*a^2*b^8 + 16*A^2*a^3*b^7 - 20*A^2*a^4*b^6 - 8*A^2*a^5*b^5 
 + 8*A^2*a^6*b^4 + 11*C^2*a^2*b^8 - 20*C^2*a^3*b^7 + 23*C^2*a^4*b^6 - 26*C 
^2*a^5*b^5 + 17*C^2*a^6*b^4 + 120*C^2*a^7*b^3 - 120*C^2*a^8*b^2 + 4*A*C*b^ 
10 - 8*A*C*a*b^9 + 20*A*C*a^2*b^8 - 32*A*C*a^3*b^7 + 36*A*C*a^4*b^6 + 88*A 
*C*a^5*b^5 - 100*A*C*a^6*b^4 - 48*A*C*a^7*b^3 + 48*A*C*a^8*b^2))/(a*b^8...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 1071, normalized size of antiderivative = 4.09 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:

int(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,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**5*b*c - 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**4*b** 
3 + 16*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**3*b**3*c + 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**2* 
b**5 - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b) 
/sqrt(a**2 - b**2))*a**6*c - 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**5*b**2 + 16*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*b** 
2*c + 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/s 
qrt(a**2 - b**2))*a**3*b**4 - 3*cos(c + d*x)*sin(c + d*x)*a**5*b**2*c + 6* 
cos(c + d*x)*sin(c + d*x)*a**3*b**4*c - 3*cos(c + d*x)*sin(c + d*x)*a*b**6 
*c + 6*cos(c + d*x)*a**6*b*c**2 + 6*cos(c + d*x)*a**6*b*c*d*x + 2*cos(c + 
d*x)*a**5*b**3*c + 2*cos(c + d*x)*a**5*b**3*d*x - 11*cos(c + d*x)*a**4*b** 
3*c**2 - 11*cos(c + d*x)*a**4*b**3*c*d*x - 4*cos(c + d*x)*a**3*b**5*c - 4* 
cos(c + d*x)*a**3*b**5*d*x + 4*cos(c + d*x)*a**2*b**5*c**2 + 4*cos(c + d*x 
)*a**2*b**5*c*d*x + 2*cos(c + d*x)*a*b**7*c + 2*cos(c + d*x)*a*b**7*d*x + 
cos(c + d*x)*b**7*c**2 + cos(c + d*x)*b**7*c*d*x - sin(c + d*x)**3*a**4*b* 
*3*c + 2*sin(c + d*x)**3*a**2*b**5*c - sin(c + d*x)**3*b**7*c - 6*sin(c...