\(\int \frac {1}{(a+b \cos (e+f x)) (c+d \sec (e+f x))} \, dx\) [13]

Optimal result

Integrand size = 25, antiderivative size = 121 \[ \int \frac {1}{(a+b \cos (e+f x)) (c+d \sec (e+f x))} \, dx=\frac {2 a \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} (a c-b d) f}-\frac {2 d \text {arctanh}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{\sqrt {c-d} \sqrt {c+d} (a c-b d) f} \] Output:

2*a*arctan((a-b)^(1/2)*tan(1/2*f*x+1/2*e)/(a+b)^(1/2))/(a-b)^(1/2)/(a+b)^( 
1/2)/(a*c-b*d)/f-2*d*arctanh((c-d)^(1/2)*tan(1/2*f*x+1/2*e)/(c+d)^(1/2))/( 
c-d)^(1/2)/(c+d)^(1/2)/(a*c-b*d)/f
 

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(a+b \cos (e+f x)) (c+d \sec (e+f x))} \, dx=\frac {-\frac {2 a \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+\frac {2 d \text {arctanh}\left (\frac {(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}}{a c f-b d f} \] Input:

Integrate[1/((a + b*Cos[e + f*x])*(c + d*Sec[e + f*x])),x]
 

Output:

((-2*a*ArcTanh[((a - b)*Tan[(e + f*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b 
^2] + (2*d*ArcTanh[((-c + d)*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/Sqrt[c^2 
- d^2])/(a*c*f - b*d*f)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3307, 3042, 3480, 3042, 3138, 218, 221}

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.

\(\displaystyle \int \frac {1}{(a+b \cos (e+f x)) (c+d \sec (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a+b \sin \left (e+f x+\frac {\pi }{2}\right )\right ) \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 3307

\(\displaystyle \int \frac {\cos (e+f x)}{(a+b \cos (e+f x)) (c \cos (e+f x)+d)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

\(\displaystyle \frac {a \int \frac {1}{a+b \cos (e+f x)}dx}{a c-b d}-\frac {d \int \frac {1}{d+c \cos (e+f x)}dx}{a c-b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \int \frac {1}{a+b \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{a c-b d}-\frac {d \int \frac {1}{d+c \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{a c-b d}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {2 a \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (e+f x)\right )+a+b}d\tan \left (\frac {1}{2} (e+f x)\right )}{f (a c-b d)}-\frac {2 d \int \frac {1}{-\left ((c-d) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+c+d}d\tan \left (\frac {1}{2} (e+f x)\right )}{f (a c-b d)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {2 a \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{f \sqrt {a-b} \sqrt {a+b} (a c-b d)}-\frac {2 d \int \frac {1}{-\left ((c-d) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+c+d}d\tan \left (\frac {1}{2} (e+f x)\right )}{f (a c-b d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 a \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{f \sqrt {a-b} \sqrt {a+b} (a c-b d)}-\frac {2 d \text {arctanh}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{f \sqrt {c-d} \sqrt {c+d} (a c-b d)}\)

Input:

Int[1/((a + b*Cos[e + f*x])*(c + d*Sec[e + f*x])),x]
 

Output:

(2*a*ArcTan[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt 
[a + b]*(a*c - b*d)*f) - (2*d*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[ 
c + d]])/(Sqrt[c - d]*Sqrt[c + d]*(a*c - b*d)*f)
 

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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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 3307

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + 
(f_.)*(x_)])^(m_.), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*((d + c*Sin[e + 
 f*x])^n/Sin[e + f*x]^n), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IntegerQ 
[n]
 

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]
 

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {\frac {2 a \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a c -b d \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 d \,\operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (a c -b d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}}{f}\)	\(108\)
default	\(\frac {\frac {2 a \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a c -b d \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 d \,\operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (a c -b d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}}{f}\)	\(108\)
risch	\(-\frac {a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a c -b d \right ) f}+\frac {a \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a c -b d \right ) f}+\frac {d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {-i c^{2}+i d^{2}+d \sqrt {c^{2}-d^{2}}}{\sqrt {c^{2}-d^{2}}\, c}\right )}{\sqrt {c^{2}-d^{2}}\, \left (a c -b d \right ) f}-\frac {d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c^{2}-i d^{2}+d \sqrt {c^{2}-d^{2}}}{\sqrt {c^{2}-d^{2}}\, c}\right )}{\sqrt {c^{2}-d^{2}}\, \left (a c -b d \right ) f}\)	\(318\)

Input:

int(1/(a+b*cos(f*x+e))/(c+d*sec(f*x+e)),x,method=_RETURNVERBOSE)
 

Output:

1/f*(2*a/(a*c-b*d)/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*f*x+1/2*e)/((a 
-b)*(a+b))^(1/2))-2*d/(a*c-b*d)/((c+d)*(c-d))^(1/2)*arctanh((c-d)*tan(1/2* 
f*x+1/2*e)/((c+d)*(c-d))^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 1.65 (sec) , antiderivative size = 1022, normalized size of antiderivative = 8.45 \[ \int \frac {1}{(a+b \cos (e+f x)) (c+d \sec (e+f x))} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*cos(f*x+e))/(c+d*sec(f*x+e)),x, algorithm="fricas")
 

Output:

[-1/2*((a^2 - b^2)*sqrt(c^2 - d^2)*d*log((2*c*d*cos(f*x + e) - (c^2 - 2*d^ 
2)*cos(f*x + e)^2 + 2*sqrt(c^2 - d^2)*(d*cos(f*x + e) + c)*sin(f*x + e) + 
2*c^2 - d^2)/(c^2*cos(f*x + e)^2 + 2*c*d*cos(f*x + e) + d^2)) - (a*c^2 - a 
*d^2)*sqrt(-a^2 + b^2)*log((2*a*b*cos(f*x + e) + (2*a^2 - b^2)*cos(f*x + e 
)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(f*x + e) + b)*sin(f*x + e) - a^2 + 2*b^2)/ 
(b^2*cos(f*x + e)^2 + 2*a*b*cos(f*x + e) + a^2)))/(((a^3 - a*b^2)*c^3 - (a 
^2*b - b^3)*c^2*d - (a^3 - a*b^2)*c*d^2 + (a^2*b - b^3)*d^3)*f), -1/2*(2*( 
a^2 - b^2)*sqrt(-c^2 + d^2)*d*arctan(-sqrt(-c^2 + d^2)*(d*cos(f*x + e) + c 
)/((c^2 - d^2)*sin(f*x + e))) - (a*c^2 - a*d^2)*sqrt(-a^2 + b^2)*log((2*a* 
b*cos(f*x + e) + (2*a^2 - b^2)*cos(f*x + e)^2 - 2*sqrt(-a^2 + b^2)*(a*cos( 
f*x + e) + b)*sin(f*x + e) - a^2 + 2*b^2)/(b^2*cos(f*x + e)^2 + 2*a*b*cos( 
f*x + e) + a^2)))/(((a^3 - a*b^2)*c^3 - (a^2*b - b^3)*c^2*d - (a^3 - a*b^2 
)*c*d^2 + (a^2*b - b^3)*d^3)*f), -1/2*((a^2 - b^2)*sqrt(c^2 - d^2)*d*log(( 
2*c*d*cos(f*x + e) - (c^2 - 2*d^2)*cos(f*x + e)^2 + 2*sqrt(c^2 - d^2)*(d*c 
os(f*x + e) + c)*sin(f*x + e) + 2*c^2 - d^2)/(c^2*cos(f*x + e)^2 + 2*c*d*c 
os(f*x + e) + d^2)) - 2*(a*c^2 - a*d^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(f*x 
 + e) + b)/(sqrt(a^2 - b^2)*sin(f*x + e))))/(((a^3 - a*b^2)*c^3 - (a^2*b - 
 b^3)*c^2*d - (a^3 - a*b^2)*c*d^2 + (a^2*b - b^3)*d^3)*f), -((a^2 - b^2)*s 
qrt(-c^2 + d^2)*d*arctan(-sqrt(-c^2 + d^2)*(d*cos(f*x + e) + c)/((c^2 - d^ 
2)*sin(f*x + e))) - (a*c^2 - a*d^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(f*x ...
 

Sympy [F]

\[ \int \frac {1}{(a+b \cos (e+f x)) (c+d \sec (e+f x))} \, dx=\int \frac {1}{\left (a + b \cos {\left (e + f x \right )}\right ) \left (c + d \sec {\left (e + f x \right )}\right )}\, dx \] Input:

integrate(1/(a+b*cos(f*x+e))/(c+d*sec(f*x+e)),x)
 

Output:

Integral(1/((a + b*cos(e + f*x))*(c + d*sec(e + f*x))), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \cos (e+f x)) (c+d \sec (e+f x))} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(a+b*cos(f*x+e))/(c+d*sec(f*x+e)),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*c^2-4*d^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. 507 vs. \(2 (103) = 206\).

Time = 0.19 (sec) , antiderivative size = 507, normalized size of antiderivative = 4.19 \[ \int \frac {1}{(a+b \cos (e+f x)) (c+d \sec (e+f x))} \, dx=\frac {\frac {{\left (\sqrt {a^{2} - b^{2}} a c {\left | a - b \right |} - \sqrt {a^{2} - b^{2}} {\left (2 \, a - b\right )} d {\left | a - b \right |} + \sqrt {a^{2} - b^{2}} {\left | a c - b d \right |} {\left | a - b \right |}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {\frac {b c - a d + \sqrt {{\left (a c + b c + a d + b d\right )} {\left (a c - b c - a d + b d\right )} + {\left (b c - a d\right )}^{2}}}{a c - b c - a d + b d}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} {\left (a c - b d\right )}^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} c {\left | a c - b d \right |} - {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} d {\left | a c - b d \right |}} + \frac {{\left (\sqrt {-c^{2} + d^{2}} a {\left (c - 2 \, d\right )} {\left | -c + d \right |} + \sqrt {-c^{2} + d^{2}} b d {\left | -c + d \right |} - \sqrt {-c^{2} + d^{2}} {\left | a c - b d \right |} {\left | -c + d \right |}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {\frac {b c - a d - \sqrt {{\left (a c + b c + a d + b d\right )} {\left (a c - b c - a d + b d\right )} + {\left (b c - a d\right )}^{2}}}{a c - b c - a d + b d}}}\right )\right )}}{{\left (a c - b d\right )}^{2} {\left (c^{2} - 2 \, c d + d^{2}\right )} + {\left (c^{2} d - 2 \, c d^{2} + d^{3}\right )} a {\left | a c - b d \right |} - {\left (c^{3} - 2 \, c^{2} d + c d^{2}\right )} b {\left | a c - b d \right |}}}{f} \] Input:

integrate(1/(a+b*cos(f*x+e))/(c+d*sec(f*x+e)),x, algorithm="giac")
 

Output:

((sqrt(a^2 - b^2)*a*c*abs(a - b) - sqrt(a^2 - b^2)*(2*a - b)*d*abs(a - b) 
+ sqrt(a^2 - b^2)*abs(a*c - b*d)*abs(a - b))*(pi*floor(1/2*(f*x + e)/pi + 
1/2) + arctan(tan(1/2*f*x + 1/2*e)/sqrt((b*c - a*d + sqrt((a*c + b*c + a*d 
 + b*d)*(a*c - b*c - a*d + b*d) + (b*c - a*d)^2))/(a*c - b*c - a*d + b*d)) 
))/((a^2 - 2*a*b + b^2)*(a*c - b*d)^2 + (a^2*b - 2*a*b^2 + b^3)*c*abs(a*c 
- b*d) - (a^3 - 2*a^2*b + a*b^2)*d*abs(a*c - b*d)) + (sqrt(-c^2 + d^2)*a*( 
c - 2*d)*abs(-c + d) + sqrt(-c^2 + d^2)*b*d*abs(-c + d) - sqrt(-c^2 + d^2) 
*abs(a*c - b*d)*abs(-c + d))*(pi*floor(1/2*(f*x + e)/pi + 1/2) + arctan(ta 
n(1/2*f*x + 1/2*e)/sqrt((b*c - a*d - sqrt((a*c + b*c + a*d + b*d)*(a*c - b 
*c - a*d + b*d) + (b*c - a*d)^2))/(a*c - b*c - a*d + b*d))))/((a*c - b*d)^ 
2*(c^2 - 2*c*d + d^2) + (c^2*d - 2*c*d^2 + d^3)*a*abs(a*c - b*d) - (c^3 - 
2*c^2*d + c*d^2)*b*abs(a*c - b*d)))/f
 

Mupad [B] (verification not implemented)

Time = 2.56 (sec) , antiderivative size = 2665, normalized size of antiderivative = 22.02 \[ \int \frac {1}{(a+b \cos (e+f x)) (c+d \sec (e+f x))} \, dx=\text {Too large to display} \] Input:

int(1/((c + d/cos(e + f*x))*(a + b*cos(e + f*x))),x)
 

Output:

(a*c^2*atan((a^5*c^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)*1i + a^3*d^2*tan 
(e/2 + (f*x)/2)*(b^2 - a^2)^(3/2)*2i + a^5*d^2*tan(e/2 + (f*x)/2)*(b^2 - a 
^2)^(1/2)*2i - b^5*d^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)*1i - a^2*b^3*c 
^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)*1i - a^3*b^2*c^2*tan(e/2 + (f*x)/2 
)*(b^2 - a^2)^(1/2)*1i + a^2*b^3*d^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)* 
1i - a^3*b^2*d^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)*3i - a^3*c*d*tan(e/2 
 + (f*x)/2)*(b^2 - a^2)^(3/2)*2i - a^5*c*d*tan(e/2 + (f*x)/2)*(b^2 - a^2)^ 
(1/2)*2i + a^2*b*c^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(3/2)*2i + a^4*b*c^2*t 
an(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)*1i + a*b^4*d^2*tan(e/2 + (f*x)/2)*(b^2 
 - a^2)^(1/2)*1i + a^2*b^3*c*d*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)*2i + a 
^3*b^2*c*d*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)*2i - a^2*b*c*d*tan(e/2 + ( 
f*x)/2)*(b^2 - a^2)^(3/2)*2i - a^4*b*c*d*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1 
/2)*2i)/(a^6*c^2 - b^6*d^2 + a^2*b^4*c^2 - 2*a^4*b^2*c^2 + 2*a^2*b^4*d^2 - 
 a^4*b^2*d^2))*(b^2 - a^2)^(1/2)*2i)/(f*(a^3*c^3 - b^3*d^3 - a*b^2*c^3 + a 
^2*b*d^3 - a^3*c*d^2 + b^3*c^2*d + a*b^2*c*d^2 - a^2*b*c^2*d)) - (a*d^2*at 
an((a^5*c^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)*1i + a^3*d^2*tan(e/2 + (f 
*x)/2)*(b^2 - a^2)^(3/2)*2i + a^5*d^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2) 
*2i - b^5*d^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)*1i - a^2*b^3*c^2*tan(e/ 
2 + (f*x)/2)*(b^2 - a^2)^(1/2)*1i - a^3*b^2*c^2*tan(e/2 + (f*x)/2)*(b^2 - 
a^2)^(1/2)*1i + a^2*b^3*d^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)*1i - a...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.40 \[ \int \frac {1}{(a+b \cos (e+f x)) (c+d \sec (e+f x))} \, dx=\frac {2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a -\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a \,c^{2}-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a -\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a \,d^{2}-2 \sqrt {-c^{2}+d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c -\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d}{\sqrt {-c^{2}+d^{2}}}\right ) a^{2} d +2 \sqrt {-c^{2}+d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c -\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d}{\sqrt {-c^{2}+d^{2}}}\right ) b^{2} d}{f \left (a^{3} c^{3}-a^{3} c \,d^{2}-a^{2} b \,c^{2} d +a^{2} b \,d^{3}-a \,b^{2} c^{3}+a \,b^{2} c \,d^{2}+b^{3} c^{2} d -b^{3} d^{3}\right )} \] Input:

int(1/(a+b*cos(f*x+e))/(c+d*sec(f*x+e)),x)
 

Output:

(2*(sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt( 
a**2 - b**2))*a*c**2 - sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a - tan((e 
 + f*x)/2)*b)/sqrt(a**2 - b**2))*a*d**2 - sqrt( - c**2 + d**2)*atan((tan(( 
e + f*x)/2)*c - tan((e + f*x)/2)*d)/sqrt( - c**2 + d**2))*a**2*d + sqrt( - 
 c**2 + d**2)*atan((tan((e + f*x)/2)*c - tan((e + f*x)/2)*d)/sqrt( - c**2 
+ d**2))*b**2*d))/(f*(a**3*c**3 - a**3*c*d**2 - a**2*b*c**2*d + a**2*b*d** 
3 - a*b**2*c**3 + a*b**2*c*d**2 + b**3*c**2*d - b**3*d**3))