\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x)}{a+b \cos (c+d x)} \, dx\) [981]

Optimal result

Integrand size = 39, antiderivative size = 94 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {C x}{b}-\frac {2 \left (A b^2-a (b B-a C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} b \sqrt {a+b} d}+\frac {A \text {arctanh}(\sin (c+d x))}{a d} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.75 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.72 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \left (\left (a C d x-A b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+A b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2\right )}+2 \left (A b^2+a (-b B+a C)\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (b \sin (c)+(-a+b \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2\right )}}\right ) (i \cos (c)+\sin (c))\right )}{a b d (2 A+C+2 B \cos (c+d x)+C \cos (2 (c+d x))) \sqrt {\left (-a^2+b^2\right ) (\cos (2 c)-i \sin (2 c))}} \] Input:

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

Output:

(2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*((a*C*d*x - A*b*Log[Cos[(c + d* 
x)/2] - Sin[(c + d*x)/2]] + A*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])* 
Sqrt[-((a^2 - b^2)*(Cos[c] - I*Sin[c])^2)] + 2*(A*b^2 + a*(-(b*B) + a*C))* 
ArcTan[((I*Cos[c] + Sin[c])*(b*Sin[c] + (-a + b*Cos[c])*Tan[(d*x)/2]))/Sqr 
t[-((a^2 - b^2)*(Cos[c] - I*Sin[c])^2)]]*(I*Cos[c] + Sin[c])))/(a*b*d*(2*A 
 + C + 2*B*Cos[c + d*x] + C*Cos[2*(c + d*x)])*Sqrt[(-a^2 + b^2)*(Cos[2*c] 
- I*Sin[2*c])])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3042, 3536, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3536

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

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

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 3536

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) 
*(x_)])), x_Symbol] :> Simp[C*(x/(b*d)), x] + (Simp[(A*b^2 - a*b*B + a^2*C) 
/(b*(b*c - a*d))   Int[1/(a + b*Sin[e + f*x]), x], x] - Simp[(c^2*C - B*c*d 
 + A*d^2)/(d*(b*c - a*d))   Int[1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a 
, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 
NeQ[c^2 - d^2, 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 = 0.70 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.28

method	result	size
derivativedivides	\(\frac {\frac {2 \left (-A \,b^{2}+B a b -a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a b \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a}}{d}\)	\(120\)
default	\(\frac {\frac {2 \left (-A \,b^{2}+B a b -a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a b \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a}}{d}\)	\(120\)
risch	\(\frac {C x}{b}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d b}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d b}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a d}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a d}\)	\(484\)

Input:

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

Output:

1/d*(2*(-A*b^2+B*a*b-C*a^2)/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))+2*C/b*arctan(tan(1/2*d*x+1/2*c))-A/a*ln(tan 
(1/2*d*x+1/2*c)-1)+A/a*ln(tan(1/2*d*x+1/2*c)+1))
 

Fricas [A] (verification not implemented)

Time = 1.45 (sec) , antiderivative size = 363, normalized size of antiderivative = 3.86 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\left [\frac {2 \, {\left (C a^{3} - C a b^{2}\right )} d x - {\left (C a^{2} - B a b + A b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + {\left (A a^{2} b - A b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} b - A b^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{3} b - a b^{3}\right )} d}, \frac {2 \, {\left (C a^{3} - C a b^{2}\right )} d x - 2 \, {\left (C a^{2} - B a b + A b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) + {\left (A a^{2} b - A b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} b - A b^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{3} b - a b^{3}\right )} d}\right ] \] Input:

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

Output:

[1/2*(2*(C*a^3 - C*a*b^2)*d*x - (C*a^2 - B*a*b + A*b^2)*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)) + (A*a^2*b - A*b^3)*log(sin(d*x + c) + 1) - (A*a^ 
2*b - A*b^3)*log(-sin(d*x + c) + 1))/((a^3*b - a*b^3)*d), 1/2*(2*(C*a^3 - 
C*a*b^2)*d*x - 2*(C*a^2 - B*a*b + A*b^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(d* 
x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) + (A*a^2*b - A*b^3)*log(sin(d* 
x + c) + 1) - (A*a^2*b - A*b^3)*log(-sin(d*x + c) + 1))/((a^3*b - a*b^3)*d 
)]
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c)),x, a 
lgorithm="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 = 0.15 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.57 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {{\left (d x + c\right )} C}{b} + \frac {A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {2 \, {\left (C a^{2} - B a b + A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a b}}{d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 6.83 (sec) , antiderivative size = 18201, normalized size of antiderivative = 193.63 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

(2*C*atan((16384*C^5*a^5*tan(c/2 + (d*x)/2))/(16384*C^5*a^5 + 32768*A*C^4* 
a^5 + 32768*B*C^4*a^5 - 16384*A^4*C*b^5 - 16384*C^5*a^4*b + 16384*B^2*C^3* 
a^5 - 32768*A^2*C^3*a^2*b^3 + 32768*A^2*C^3*a^3*b^2 - 32768*A^3*C^2*a^2*b^ 
3 + 32768*A^3*C^2*a^3*b^2 - 32768*A*C^4*a^4*b + 16384*A^4*C*a*b^4 - 16384* 
B^2*C^3*a^4*b - (32768*B*C^4*a^6)/b + 32768*A*B*C^3*a^3*b^2 - 32768*A^2*B* 
C^2*a^4*b - 32768*A^3*B*C*a^2*b^3 + 32768*A^2*B*C^2*a^3*b^2 - 16384*A^2*B^ 
2*C*a^2*b^3 + 16384*A^2*B^2*C*a^3*b^2 - 32768*A*B*C^3*a^4*b + 32768*A^3*B* 
C*a*b^4) + (16384*C^5*a^4*tan(c/2 + (d*x)/2))/(16384*C^5*a^4 + 32768*A*C^4 
*a^4 + 16384*A^4*C*b^4 + 16384*B^2*C^3*a^4 - (16384*C^5*a^5)/b + 32768*A^2 
*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - (16384*B^2*C^3*a^5)/b + 32768*A*B*C 
^3*a^4 - 16384*A^4*C*a*b^3 + 32768*A^2*B*C^2*a^4 - 32768*A^2*C^3*a^3*b - 3 
2768*A^3*C^2*a^3*b - (32768*A*C^4*a^5)/b - (32768*B*C^4*a^5)/b + (32768*B* 
C^4*a^6)/b^2 - 32768*A^2*B*C^2*a^3*b - 16384*A^2*B^2*C*a^3*b + 32768*A^3*B 
*C*a^2*b^2 + 16384*A^2*B^2*C*a^2*b^2 - 32768*A*B*C^3*a^3*b - 32768*A^3*B*C 
*a*b^3) + (16384*B^2*C^3*a^4*tan(c/2 + (d*x)/2))/(16384*C^5*a^4 + 32768*A* 
C^4*a^4 + 16384*A^4*C*b^4 + 16384*B^2*C^3*a^4 - (16384*C^5*a^5)/b + 32768* 
A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - (16384*B^2*C^3*a^5)/b + 32768*A* 
B*C^3*a^4 - 16384*A^4*C*a*b^3 + 32768*A^2*B*C^2*a^4 - 32768*A^2*C^3*a^3*b 
- 32768*A^3*C^2*a^3*b - (32768*A*C^4*a^5)/b - (32768*B*C^4*a^5)/b + (32768 
*B*C^4*a^6)/b^2 - 32768*A^2*B*C^2*a^3*b - 16384*A^2*B^2*C*a^3*b + 32768...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.64 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a c -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{3}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{3}+a^{2} c d x -b^{2} c d x}{b d \left (a^{2}-b^{2}\right )} \] Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c)),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))*a*c - log(tan((c + d*x)/2) - 1)*a**2*b + log(tan((c + d*x) 
/2) - 1)*b**3 + log(tan((c + d*x)/2) + 1)*a**2*b - log(tan((c + d*x)/2) + 
1)*b**3 + a**2*c*d*x - b**2*c*d*x)/(b*d*(a**2 - b**2))