\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) [991]

Optimal result

Integrand size = 41, antiderivative size = 211 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 \left (3 a^2 A b^2-2 A b^4-2 a^3 b B+a b^3 B+a^4 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 (a-b)^{3/2} (a+b)^{3/2} d}-\frac {(2 A b-a B) \text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {\left (2 A b^2-a b B-a^2 (A-C)\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 4.74 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.57 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 \cos ^2(c+d x) \left (C+B \sec (c+d x)+A \sec ^2(c+d x)\right ) \left (\frac {2 \left (3 a^2 A b^2-2 A b^4-2 a^3 b B+a b^3 B+a^4 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+(2 A b-a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+(-2 A b+a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a A \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {a A \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {a b \left (A b^2+a (-b B+a C)\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}\right )}{a^3 d (2 A+C+2 B \cos (c+d x)+C \cos (2 (c+d x)))} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.31 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.268, Rules used = {3042, 3534, 25, 3042, 3534, 3042, 3480, 3042, 3138, 218, 4257}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

((A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x 
])) - (((-2*(3*a^2*A*b^2 - 2*A*b^4 - 2*a^3*b*B + a*b^3*B + a^4*C)*ArcTan[( 
Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) 
+ ((a^2 - b^2)*(2*A*b - a*B)*ArcTanh[Sin[c + d*x]])/(a*d))/a + ((2*A*b^2 - 
 a*b*B - a^2*(A - C))*Tan[c + d*x])/(a*d))/(a*(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 3480

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b 
- a*B)/(b*c - a*d)   Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ 
(b*c - a*d)   Int[1/(c + d*Sin[e + f*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]
 

rule 3534

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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* 
c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int 
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 0])))
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 1.24 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.21

method	result	size
derivativedivides	\(\frac {\frac {-\frac {2 a \left (A \,b^{2}-B a b +a^{2} C \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {2 \left (3 A \,a^{2} b^{2}-2 A \,b^{4}-2 B \,a^{3} b +B a \,b^{3}+a^{4} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{a^{3}}-\frac {A}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (2 A b -B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}-\frac {A}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-2 A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}}{d}\)	\(256\)
default	\(\frac {\frac {-\frac {2 a \left (A \,b^{2}-B a b +a^{2} C \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {2 \left (3 A \,a^{2} b^{2}-2 A \,b^{4}-2 B \,a^{3} b +B a \,b^{3}+a^{4} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{a^{3}}-\frac {A}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (2 A b -B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}-\frac {A}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-2 A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}}{d}\)	\(256\)
risch	\(\text {Expression too large to display}\)	\(1204\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x,meth 
od=_RETURNVERBOSE)
 

Output:

1/d*(2/a^3*(-a*(A*b^2-B*a*b+C*a^2)*b/(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)+(3*A*a^2*b^2-2*A*b^4-2*B*a^3*b 
+B*a*b^3+C*a^4)/(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)))-A/a^2/(tan(1/2*d*x+1/2*c)-1)+(2*A*b-B*a)/a^3*l 
n(tan(1/2*d*x+1/2*c)-1)-A/a^2/(tan(1/2*d*x+1/2*c)+1)+1/a^3*(-2*A*b+B*a)*ln 
(tan(1/2*d*x+1/2*c)+1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 529 vs. \(2 (196) = 392\).

Time = 15.23 (sec) , antiderivative size = 1126, normalized size of antiderivative = 5.34 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/2*(((C*a^4*b - 2*B*a^3*b^2 + 3*A*a^2*b^3 + B*a*b^4 - 2*A*b^5)*cos(d*x 
+ c)^2 + (C*a^5 - 2*B*a^4*b + 3*A*a^3*b^2 + B*a^2*b^3 - 2*A*a*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)) - ((B*a^5*b - 2*A*a^4*b^ 
2 - 2*B*a^3*b^3 + 4*A*a^2*b^4 + B*a*b^5 - 2*A*b^6)*cos(d*x + c)^2 + (B*a^6 
 - 2*A*a^5*b - 2*B*a^4*b^2 + 4*A*a^3*b^3 + B*a^2*b^4 - 2*A*a*b^5)*cos(d*x 
+ c))*log(sin(d*x + c) + 1) + ((B*a^5*b - 2*A*a^4*b^2 - 2*B*a^3*b^3 + 4*A* 
a^2*b^4 + B*a*b^5 - 2*A*b^6)*cos(d*x + c)^2 + (B*a^6 - 2*A*a^5*b - 2*B*a^4 
*b^2 + 4*A*a^3*b^3 + B*a^2*b^4 - 2*A*a*b^5)*cos(d*x + c))*log(-sin(d*x + c 
) + 1) - 2*(A*a^6 - 2*A*a^4*b^2 + A*a^2*b^4 + ((A - C)*a^5*b + B*a^4*b^2 - 
 (3*A - C)*a^3*b^3 - B*a^2*b^4 + 2*A*a*b^5)*cos(d*x + c))*sin(d*x + c))/(( 
a^7*b - 2*a^5*b^3 + a^3*b^5)*d*cos(d*x + c)^2 + (a^8 - 2*a^6*b^2 + a^4*b^4 
)*d*cos(d*x + c)), 1/2*(2*((C*a^4*b - 2*B*a^3*b^2 + 3*A*a^2*b^3 + B*a*b^4 
- 2*A*b^5)*cos(d*x + c)^2 + (C*a^5 - 2*B*a^4*b + 3*A*a^3*b^2 + B*a^2*b^3 - 
 2*A*a*b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sq 
rt(a^2 - b^2)*sin(d*x + c))) + ((B*a^5*b - 2*A*a^4*b^2 - 2*B*a^3*b^3 + 4*A 
*a^2*b^4 + B*a*b^5 - 2*A*b^6)*cos(d*x + c)^2 + (B*a^6 - 2*A*a^5*b - 2*B*a^ 
4*b^2 + 4*A*a^3*b^3 + B*a^2*b^4 - 2*A*a*b^5)*cos(d*x + c))*log(sin(d*x + c 
) + 1) - ((B*a^5*b - 2*A*a^4*b^2 - 2*B*a^3*b^3 + 4*A*a^2*b^4 + B*a*b^5 ...
 

Sympy [F]

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

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

Output:

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sec(c + d*x)**2/(a + b*c 
os(c + d*x))**2, x)
 

Maxima [F(-2)]

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

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 442 vs. \(2 (196) = 392\).

Time = 0.19 (sec) , antiderivative size = 442, normalized size of antiderivative = 2.09 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (C a^{4} - 2 \, B a^{3} b + 3 \, A a^{2} b^{2} + B a b^{3} - 2 \, 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^{5} - a^{3} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )} {\left (a^{4} - a^{2} b^{2}\right )}} - \frac {{\left (B a - 2 \, A b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} + \frac {{\left (B a - 2 \, A b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}}}{d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 6.05 (sec) , antiderivative size = 6450, normalized size of antiderivative = 30.57 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((2*tan(c/2 + (d*x)/2)*(A*a^3 - 2*A*b^3 - A*a*b^2 + A*a^2*b + B*a*b^2 - C* 
a^2*b))/(a^2*(a + b)*(a - b)) + (2*tan(c/2 + (d*x)/2)^3*(A*a^3 + 2*A*b^3 - 
 A*a*b^2 - A*a^2*b - B*a*b^2 + C*a^2*b))/(a^2*(a + b)*(a - b)))/(d*(a + b 
- tan(c/2 + (d*x)/2)^4*(a - b) - 2*b*tan(c/2 + (d*x)/2)^2)) + (atan((((2*A 
*b - B*a)*((32*tan(c/2 + (d*x)/2)*(8*A^2*b^8 + B^2*a^8 + C^2*a^8 - 8*A^2*a 
*b^7 - 2*B^2*a^7*b - 16*A^2*a^2*b^6 + 16*A^2*a^3*b^5 + 5*A^2*a^4*b^4 - 8*A 
^2*a^5*b^3 + 4*A^2*a^6*b^2 + 2*B^2*a^2*b^6 - 2*B^2*a^3*b^5 - 5*B^2*a^4*b^4 
 + 4*B^2*a^5*b^3 + 3*B^2*a^6*b^2 - 8*A*B*a*b^7 - 4*A*B*a^7*b - 4*B*C*a^7*b 
 + 8*A*B*a^2*b^6 + 18*A*B*a^3*b^5 - 16*A*B*a^4*b^4 - 8*A*B*a^5*b^3 + 8*A*B 
*a^6*b^2 - 4*A*C*a^4*b^4 + 6*A*C*a^6*b^2 + 2*B*C*a^5*b^3))/(a^6*b + a^7 - 
a^4*b^3 - a^5*b^2) + ((2*A*b - B*a)*((32*(A*a^7*b^5 - C*a^12 - 2*A*a^6*b^6 
 - B*a^12 + 5*A*a^8*b^4 - 3*A*a^9*b^3 - 3*A*a^10*b^2 + B*a^7*b^5 - 3*B*a^9 
*b^3 + B*a^10*b^2 - C*a^9*b^3 + C*a^10*b^2 + 2*A*a^11*b + 2*B*a^11*b + C*a 
^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (32*tan(c/2 + (d*x)/2)*(2*A*b 
- B*a)*(2*a^11*b - 2*a^6*b^6 + 2*a^7*b^5 + 4*a^8*b^4 - 4*a^9*b^3 - 2*a^10* 
b^2))/(a^3*(a^6*b + a^7 - a^4*b^3 - a^5*b^2))))/a^3)*1i)/a^3 + ((2*A*b - B 
*a)*((32*tan(c/2 + (d*x)/2)*(8*A^2*b^8 + B^2*a^8 + C^2*a^8 - 8*A^2*a*b^7 - 
 2*B^2*a^7*b - 16*A^2*a^2*b^6 + 16*A^2*a^3*b^5 + 5*A^2*a^4*b^4 - 8*A^2*a^5 
*b^3 + 4*A^2*a^6*b^2 + 2*B^2*a^2*b^6 - 2*B^2*a^3*b^5 - 5*B^2*a^4*b^4 + 4*B 
^2*a^5*b^3 + 3*B^2*a^6*b^2 - 8*A*B*a*b^7 - 4*A*B*a^7*b - 4*B*C*a^7*b + ...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 1187, normalized size of antiderivative = 5.63 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(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**4*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**3*b**2 - 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**4 - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2) 
*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**3*b*c - 2*s 
qrt(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**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))*sin(c + d*x)**2*b**5 + 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**3*b*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**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))*b**5 + cos(c + d* 
x)*log(tan((c + d*x)/2) - 1)*a**5*b - 2*cos(c + d*x)*log(tan((c + d*x)/2) 
- 1)*a**3*b**3 + cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b**5 - cos(c + d 
*x)*log(tan((c + d*x)/2) + 1)*a**5*b + 2*cos(c + d*x)*log(tan((c + d*x)/2) 
 + 1)*a**3*b**3 - cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b**5 + cos(c + 
d*x)*sin(c + d*x)*a**5*b - cos(c + d*x)*sin(c + d*x)*a**4*b*c - 2*cos(c + 
d*x)*sin(c + d*x)*a**3*b**3 + cos(c + d*x)*sin(c + d*x)*a**2*b**3*c + cos( 
c + d*x)*sin(c + d*x)*a*b**5 - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**...