\(\int \frac {\cos (c+d x) (A+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [572]

Optimal result

Integrand size = 31, antiderivative size = 144 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 a C x}{b^3}-\frac {2 \left (A b^4-2 a^4 C+3 a^2 b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^3 (a+b)^{3/2} d}+\frac {C \sin (c+d x)}{b^2 d}+\frac {a \left (A b^2+a^2 C\right ) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 1.95 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.94 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {-2 a C (c+d x)-\frac {2 \left (A b^4-2 a^4 C+3 a^2 b^2 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+b C \sin (c+d x)+\frac {a b \left (A b^2+a^2 C\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}}{b^3 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.22, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 3511, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3511

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Defintions of rubi rules used

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 3511

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[ 
(-(b*c - a*d))*(A*b^2 + a^2*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/( 
b^2*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[(a 
 + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*(a*C*(b*c - a*d) + A*b*(a*c - b*d 
)) - ((b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] + b 
*C*d*(m + 1)*(a^2 - b^2)*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] && LtQ[m, -1]
 

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.38

method	result	size
derivativedivides	\(\frac {-\frac {2 C \left (-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{3}}-\frac {2 \left (-\frac {a b \left (A \,b^{2}+a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (A \,b^{4}-2 a^{4} C +3 C \,a^{2} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3}}}{d}\)	\(198\)
default	\(\frac {-\frac {2 C \left (-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{3}}-\frac {2 \left (-\frac {a b \left (A \,b^{2}+a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (A \,b^{4}-2 a^{4} C +3 C \,a^{2} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3}}}{d}\)	\(198\)
risch	\(-\frac {2 a C x}{b^{3}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 b^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 b^{2} d}-\frac {2 i a \left (A \,b^{2}+a^{2} C \right ) \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{3} \left (-a^{2}+b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {2 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C \,a^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {2 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C \,a^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}\)	\(637\)

Input:

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

Output:

1/d*(-2*C/b^3*(-b*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)+2*a*arctan(t 
an(1/2*d*x+1/2*c)))-2/b^3*(-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*b^4-2*C*a^4+3*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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (136) = 272\).

Time = 0.15 (sec) , antiderivative size = 632, normalized size of antiderivative = 4.39 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\left [-\frac {4 \, {\left (C a^{5} b - 2 \, C a^{3} b^{3} + C a b^{5}\right )} d x \cos \left (d x + c\right ) + 4 \, {\left (C a^{6} - 2 \, C a^{4} b^{2} + C a^{2} b^{4}\right )} d x + {\left (2 \, C a^{5} - 3 \, C a^{3} b^{2} - A a b^{4} + {\left (2 \, C a^{4} b - 3 \, C a^{2} b^{3} - A b^{5}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (2 \, C a^{5} b + {\left (A - 3 \, C\right )} a^{3} b^{3} - {\left (A - C\right )} a b^{5} + {\left (C a^{4} b^{2} - 2 \, C a^{2} b^{4} + C b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7}\right )} d\right )}}, -\frac {2 \, {\left (C a^{5} b - 2 \, C a^{3} b^{3} + C a b^{5}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (C a^{6} - 2 \, C a^{4} b^{2} + C a^{2} b^{4}\right )} d x - {\left (2 \, C a^{5} - 3 \, C a^{3} b^{2} - A a b^{4} + {\left (2 \, C a^{4} b - 3 \, C a^{2} b^{3} - A b^{5}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (2 \, C a^{5} b + {\left (A - 3 \, C\right )} a^{3} b^{3} - {\left (A - C\right )} a b^{5} + {\left (C a^{4} b^{2} - 2 \, C a^{2} b^{4} + C b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7}\right )} d}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,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. 998 vs. \(2 (136) = 272\).

Time = 0.23 (sec) , antiderivative size = 998, normalized size of antiderivative = 6.93 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((4*C*a^6*b^2 - 2*C*a^5*b^3 - 9*C*a^4*b^4 + 4*C*a^3*b^5 - A*a^2*b^6 + 5*C* 
a^2*b^6 - 2*C*a*b^7 + A*b^8 + 2*C*a^3*abs(-a^2*b^3 + b^5) - C*a^2*b*abs(-a 
^2*b^3 + b^5) - 2*C*a*b^2*abs(-a^2*b^3 + b^5) - A*b^3*abs(-a^2*b^3 + b^5)) 
*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(2*sqrt(1/2)*tan(1/2*d*x + 1/2* 
c)/sqrt((2*a^3*b^2 - 2*a*b^4 + sqrt(-4*(a^3*b^2 + a^2*b^3 - a*b^4 - b^5)*( 
a^3*b^2 - a^2*b^3 - a*b^4 + b^5) + 4*(a^3*b^2 - a*b^4)^2))/(a^3*b^2 - a^2* 
b^3 - a*b^4 + b^5))))/(a^3*b^2*abs(-a^2*b^3 + b^5) - a*b^4*abs(-a^2*b^3 + 
b^5) + (a^2*b^3 - b^5)^2) + (sqrt(a^2 - b^2)*A*b^3*abs(-a^2*b^3 + b^5)*abs 
(-a + b) - (2*a^3 - a^2*b - 2*a*b^2)*sqrt(a^2 - b^2)*C*abs(-a^2*b^3 + b^5) 
*abs(-a + b) - (a^2*b^6 - b^8)*sqrt(a^2 - b^2)*A*abs(-a + b) + (4*a^6*b^2 
- 2*a^5*b^3 - 9*a^4*b^4 + 4*a^3*b^5 + 5*a^2*b^6 - 2*a*b^7)*sqrt(a^2 - b^2) 
*C*abs(-a + b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(2*sqrt(1/2)*tan 
(1/2*d*x + 1/2*c)/sqrt((2*a^3*b^2 - 2*a*b^4 - sqrt(-4*(a^3*b^2 + a^2*b^3 - 
 a*b^4 - b^5)*(a^3*b^2 - a^2*b^3 - a*b^4 + b^5) + 4*(a^3*b^2 - a*b^4)^2))/ 
(a^3*b^2 - a^2*b^3 - a*b^4 + b^5))))/((a^2*b^3 - b^5)^2*(a^2 - 2*a*b + b^2 
) - (a^5*b^2 - 2*a^4*b^3 + 2*a^2*b^5 - a*b^6)*abs(-a^2*b^3 + b^5)) + 2*(2* 
C*a^3*tan(1/2*d*x + 1/2*c)^3 - C*a^2*b*tan(1/2*d*x + 1/2*c)^3 + A*a*b^2*ta 
n(1/2*d*x + 1/2*c)^3 - C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + C*b^3*tan(1/2*d*x 
+ 1/2*c)^3 + 2*C*a^3*tan(1/2*d*x + 1/2*c) + C*a^2*b*tan(1/2*d*x + 1/2*c) + 
 A*a*b^2*tan(1/2*d*x + 1/2*c) - C*a*b^2*tan(1/2*d*x + 1/2*c) - C*b^3*ta...
 

Mupad [B] (verification not implemented)

Time = 4.98 (sec) , antiderivative size = 4124, normalized size of antiderivative = 28.64 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((2*tan(c/2 + (d*x)/2)^3*(2*C*a^3 + C*b^3 + A*a*b^2 - C*a*b^2 - C*a^2*b))/ 
(b^2*(a + b)*(a - b)) + (2*tan(c/2 + (d*x)/2)*(2*C*a^3 - C*b^3 + A*a*b^2 - 
 C*a*b^2 + C*a^2*b))/(b^2*(a + b)*(a - b)))/(d*(a + b + tan(c/2 + (d*x)/2) 
^4*(a - b) + 2*a*tan(c/2 + (d*x)/2)^2)) - (4*C*a*atan(((2*C*a*((32*tan(c/2 
 + (d*x)/2)*(A^2*b^8 + 8*C^2*a^8 - 8*C^2*a^7*b + 4*C^2*a^2*b^6 - 8*C^2*a^3 
*b^5 + 5*C^2*a^4*b^4 + 16*C^2*a^5*b^3 - 16*C^2*a^6*b^2 + 6*A*C*a^2*b^6 - 4 
*A*C*a^4*b^4))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) + (C*a*((32*(A*b^12 - A*a 
^2*b^10 + A*a^3*b^9 + 3*C*a^2*b^10 + 3*C*a^3*b^9 - 5*C*a^4*b^8 - C*a^5*b^7 
 + 2*C*a^6*b^6 - A*a*b^11 - 2*C*a*b^11))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) 
 - (C*a*tan(c/2 + (d*x)/2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 
+ 2*a^5*b^7 - 2*a^6*b^6)*64i)/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4)))*2i) 
/b^3))/b^3 + (2*C*a*((32*tan(c/2 + (d*x)/2)*(A^2*b^8 + 8*C^2*a^8 - 8*C^2*a 
^7*b + 4*C^2*a^2*b^6 - 8*C^2*a^3*b^5 + 5*C^2*a^4*b^4 + 16*C^2*a^5*b^3 - 16 
*C^2*a^6*b^2 + 6*A*C*a^2*b^6 - 4*A*C*a^4*b^4))/(a*b^6 + b^7 - a^2*b^5 - a^ 
3*b^4) - (C*a*((32*(A*b^12 - A*a^2*b^10 + A*a^3*b^9 + 3*C*a^2*b^10 + 3*C*a 
^3*b^9 - 5*C*a^4*b^8 - C*a^5*b^7 + 2*C*a^6*b^6 - A*a*b^11 - 2*C*a*b^11))/( 
a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (C*a*tan(c/2 + (d*x)/2)*(2*a*b^11 - 2*a 
^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6)*64i)/(b^3*(a*b^6 
+ b^7 - a^2*b^5 - a^3*b^4)))*2i)/b^3))/b^3)/((64*(8*C^3*a^8 - 4*C^3*a^7*b 
+ 12*C^3*a^4*b^4 + 6*C^3*a^5*b^3 - 20*C^3*a^6*b^2 + 2*A^2*C*a*b^7 + 4*A...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 692, normalized size of antiderivative = 4.81 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:

int(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,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**4*b*c - 6*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**3*c 
- 2*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**5 + 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*c - 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**3* 
b**2*c - 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**2*b**4 + cos(c + d*x)*sin(c + d*x)*a**4*b**2*c - 2 
*cos(c + d*x)*sin(c + d*x)*a**2*b**4*c + cos(c + d*x)*sin(c + d*x)*b**6*c 
- 2*cos(c + d*x)*a**5*b*c**2 - 2*cos(c + d*x)*a**5*b*c*d*x + 4*cos(c + d*x 
)*a**3*b**3*c**2 + 4*cos(c + d*x)*a**3*b**3*c*d*x - 2*cos(c + d*x)*a*b**5* 
c**2 - 2*cos(c + d*x)*a*b**5*c*d*x + 2*sin(c + d*x)*a**5*b*c + sin(c + d*x 
)*a**4*b**3 - 3*sin(c + d*x)*a**3*b**3*c - sin(c + d*x)*a**2*b**5 + sin(c 
+ d*x)*a*b**5*c - 2*a**6*c**2 - 2*a**6*c*d*x + 4*a**4*b**2*c**2 + 4*a**4*b 
**2*c*d*x - 2*a**2*b**4*c**2 - 2*a**2*b**4*c*d*x)/(b**3*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))