\(\int \frac {\sec ^2(x)}{a+c \sec (x)+b \tan (x)} \, dx\) [374]

Optimal result

Integrand size = 17, antiderivative size = 142 \[ \int \frac {\sec ^2(x)}{a+c \sec (x)+b \tan (x)} \, dx=-\frac {2 a c \text {arctanh}\left (\frac {b-(a-c) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2-c^2}}\right )}{\left (b^2-c^2\right ) \sqrt {a^2+b^2-c^2}}-\frac {\log \left (1-\tan \left (\frac {x}{2}\right )\right )}{b+c}-\frac {\log \left (1+\tan \left (\frac {x}{2}\right )\right )}{b-c}+\frac {b \log \left (a+c+2 b \tan \left (\frac {x}{2}\right )-(a-c) \tan ^2\left (\frac {x}{2}\right )\right )}{b^2-c^2} \] Output:

-2*a*c*arctanh((b-(a-c)*tan(1/2*x))/(a^2+b^2-c^2)^(1/2))/(b^2-c^2)/(a^2+b^ 
2-c^2)^(1/2)-ln(1-tan(1/2*x))/(b+c)-ln(1+tan(1/2*x))/(b-c)+b*ln(a+c+2*b*ta 
n(1/2*x)-(a-c)*tan(1/2*x)^2)/(b^2-c^2)
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^2(x)}{a+c \sec (x)+b \tan (x)} \, dx=\frac {\frac {2 a c \text {arctanh}\left (\frac {b+(-a+c) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2-c^2}}\right )}{\sqrt {a^2+b^2-c^2}}+(b-c) \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+(b+c) \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )-b \log (c+a \cos (x)+b \sin (x))}{(-b+c) (b+c)} \] Input:

Integrate[Sec[x]^2/(a + c*Sec[x] + b*Tan[x]),x]
 

Output:

((2*a*c*ArcTanh[(b + (-a + c)*Tan[x/2])/Sqrt[a^2 + b^2 - c^2]])/Sqrt[a^2 + 
 b^2 - c^2] + (b - c)*Log[Cos[x/2] - Sin[x/2]] + (b + c)*Log[Cos[x/2] + Si 
n[x/2]] - b*Log[c + a*Cos[x] + b*Sin[x]])/((-b + c)*(b + c))
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.824, Rules used = {3042, 4897, 3042, 4902, 2142, 27, 452, 219, 240, 1142, 27, 1083, 219, 1103}

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.

\(\displaystyle \int \frac {\sec ^2(x)}{a+b \tan (x)+c \sec (x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (x)^2}{a+b \tan (x)+c \sec (x)}dx\)

\(\Big \downarrow \) 4897

\(\displaystyle \int \frac {\sec (x)}{a \cos (x)+b \sin (x)+c}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (x)}{a \cos (x)+b \sin (x)+c}dx\)

\(\Big \downarrow \) 4902

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

\(\Big \downarrow \) 2142

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 452

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 240

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1083

\(\displaystyle 2 \left (\frac {b \int \frac {b-(a-c) \tan \left (\frac {x}{2}\right )}{-\left ((a-c) \tan ^2\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a+c}d\tan \left (\frac {x}{2}\right )-2 a c \int \frac {1}{4 \left (a^2+b^2-c^2\right )-\left (2 b-2 (a-c) \tan \left (\frac {x}{2}\right )\right )^2}d\left (2 b-2 (a-c) \tan \left (\frac {x}{2}\right )\right )}{b^2-c^2}-\frac {c \text {arctanh}\left (\tan \left (\frac {x}{2}\right )\right )+\frac {1}{2} b \log \left (1-\tan ^2\left (\frac {x}{2}\right )\right )}{b^2-c^2}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle 2 \left (\frac {b \int \frac {b-(a-c) \tan \left (\frac {x}{2}\right )}{-\left ((a-c) \tan ^2\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a+c}d\tan \left (\frac {x}{2}\right )-\frac {a c \text {arctanh}\left (\frac {2 b-2 (a-c) \tan \left (\frac {x}{2}\right )}{2 \sqrt {a^2+b^2-c^2}}\right )}{\sqrt {a^2+b^2-c^2}}}{b^2-c^2}-\frac {c \text {arctanh}\left (\tan \left (\frac {x}{2}\right )\right )+\frac {1}{2} b \log \left (1-\tan ^2\left (\frac {x}{2}\right )\right )}{b^2-c^2}\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle 2 \left (\frac {\frac {1}{2} b \log \left (-\left ((a-c) \tan ^2\left (\frac {x}{2}\right )\right )+a+2 b \tan \left (\frac {x}{2}\right )+c\right )-\frac {a c \text {arctanh}\left (\frac {2 b-2 (a-c) \tan \left (\frac {x}{2}\right )}{2 \sqrt {a^2+b^2-c^2}}\right )}{\sqrt {a^2+b^2-c^2}}}{b^2-c^2}-\frac {c \text {arctanh}\left (\tan \left (\frac {x}{2}\right )\right )+\frac {1}{2} b \log \left (1-\tan ^2\left (\frac {x}{2}\right )\right )}{b^2-c^2}\right )\)

Input:

Int[Sec[x]^2/(a + c*Sec[x] + b*Tan[x]),x]
 

Output:

2*(-((c*ArcTanh[Tan[x/2]] + (b*Log[1 - Tan[x/2]^2])/2)/(b^2 - c^2)) + (-(( 
a*c*ArcTanh[(2*b - 2*(a - c)*Tan[x/2])/(2*Sqrt[a^2 + b^2 - c^2])])/Sqrt[a^ 
2 + b^2 - c^2]) + (b*Log[a + c + 2*b*Tan[x/2] - (a - c)*Tan[x/2]^2])/2)/(b 
^2 - c^2))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 240

Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x 
^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
 

rule 452

Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c   Int[1/ 
(a + b*x^2), x], x] + Simp[d   Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, 
 d}, x] && NeQ[b*c^2 + a*d^2, 0]
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 2142

Int[(Px_)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_) + (f_.)*(x_)^2)), x_Sym 
bol] :> With[{A = Coeff[Px, x, 0], B = Coeff[Px, x, 1], C = Coeff[Px, x, 2] 
, q = c^2*d^2 + b^2*d*f - 2*a*c*d*f + a^2*f^2}, Simp[1/q   Int[(A*c^2*d - a 
*c*C*d + A*b^2*f - a*b*B*f - a*A*c*f + a^2*C*f + c*(B*c*d - b*C*d + A*b*f - 
 a*B*f)*x)/(a + b*x + c*x^2), x], x] + Simp[1/q   Int[(c*C*d^2 + b*B*d*f - 
A*c*d*f - a*C*d*f + a*A*f^2 - f*(B*c*d - b*C*d + A*b*f - a*B*f)*x)/(d + f*x 
^2), x], x] /; NeQ[q, 0]] /; FreeQ[{a, b, c, d, f}, x] && PolyQ[Px, x, 2]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4897

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]
 

rule 4902

Int[u_, x_Symbol] :> With[{w = Block[{$ShowSteps = False, $StepCounter = Nu 
ll}, Int[SubstFor[1/(1 + FreeFactors[Tan[FunctionOfTrig[u, x]/2], x]^2*x^2) 
, Tan[FunctionOfTrig[u, x]/2]/FreeFactors[Tan[FunctionOfTrig[u, x]/2], x], 
u, x], x]]}, Module[{v = FunctionOfTrig[u, x], d}, Simp[d = FreeFactors[Tan 
[v/2], x]; 2*(d/Coefficient[v, x, 1])   Subst[Int[SubstFor[1/(1 + d^2*x^2), 
 Tan[v/2]/d, u, x], x], x, Tan[v/2]/d], x]] /; CalculusFreeQ[w, x]] /; Inve 
rseFunctionFreeQ[u, x] &&  !FalseQ[FunctionOfTrig[u, x]]
 

Maple [A] (verified)

Time = 1.42 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.27

method	result	size
default	\(-\frac {2 \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{-2 c +2 b}-\frac {2 \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{2 b +2 c}+\frac {\frac {2 \left (b a -c b \right ) \ln \left (\tan \left (\frac {x}{2}\right )^{2} a -c \tan \left (\frac {x}{2}\right )^{2}-2 b \tan \left (\frac {x}{2}\right )-a -c \right )}{2 a -2 c}+\frac {2 \left (-a c -b^{2}+\frac {\left (b a -c b \right ) b}{a -c}\right ) \arctan \left (\frac {2 \left (a -c \right ) \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {-a^{2}-b^{2}+c^{2}}}\right )}{\sqrt {-a^{2}-b^{2}+c^{2}}}}{\left (b -c \right ) \left (b +c \right )}\)	\(180\)
risch	\(\text {Expression too large to display}\)	\(1344\)

Input:

int(sec(x)^2/(a+c*sec(x)+b*tan(x)),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 1.46 (sec) , antiderivative size = 663, normalized size of antiderivative = 4.67 \[ \int \frac {\sec ^2(x)}{a+c \sec (x)+b \tan (x)} \, dx =\text {Too large to display} \] Input:

integrate(sec(x)^2/(a+c*sec(x)+b*tan(x)),x, algorithm="fricas")
 

Output:

[-1/2*(sqrt(a^2 + b^2 - c^2)*a*c*log((2*a^4 + 3*a^2*b^2 + b^4 - (a^2 - b^2 
)*c^2 + 2*(a^3 + a*b^2)*c*cos(x) - (a^4 - b^4 - 2*(a^2 - b^2)*c^2)*cos(x)^ 
2 + 2*((a^2*b + b^3)*c - (a^3*b + a*b^3 - 2*a*b*c^2)*cos(x))*sin(x) + 2*(2 
*a*b*c*cos(x)^2 - a*b*c + (a^2*b + b^3)*cos(x) - (a^3 + a*b^2 + (a^2 - b^2 
)*c*cos(x))*sin(x))*sqrt(a^2 + b^2 - c^2))/(2*a*c*cos(x) + (a^2 - b^2)*cos 
(x)^2 + b^2 + c^2 + 2*(a*b*cos(x) + b*c)*sin(x))) - (a^2*b + b^3 - b*c^2)* 
log(2*a*c*cos(x) + (a^2 - b^2)*cos(x)^2 + b^2 + c^2 + 2*(a*b*cos(x) + b*c) 
*sin(x)) + (a^2*b + b^3 - b*c^2 - c^3 + (a^2 + b^2)*c)*log(sin(x) + 1) + ( 
a^2*b + b^3 - b*c^2 + c^3 - (a^2 + b^2)*c)*log(-sin(x) + 1))/(a^2*b^2 + b^ 
4 + c^4 - (a^2 + 2*b^2)*c^2), 1/2*(2*sqrt(-a^2 - b^2 + c^2)*a*c*arctan((a* 
c*cos(x) + b*c*sin(x) + a^2 + b^2)*sqrt(-a^2 - b^2 + c^2)/((a^2*b + b^3 - 
b*c^2)*cos(x) - (a^3 + a*b^2 - a*c^2)*sin(x))) + (a^2*b + b^3 - b*c^2)*log 
(2*a*c*cos(x) + (a^2 - b^2)*cos(x)^2 + b^2 + c^2 + 2*(a*b*cos(x) + b*c)*si 
n(x)) - (a^2*b + b^3 - b*c^2 - c^3 + (a^2 + b^2)*c)*log(sin(x) + 1) - (a^2 
*b + b^3 - b*c^2 + c^3 - (a^2 + b^2)*c)*log(-sin(x) + 1))/(a^2*b^2 + b^4 + 
 c^4 - (a^2 + 2*b^2)*c^2)]
 

Sympy [F]

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

integrate(sec(x)**2/(a+c*sec(x)+b*tan(x)),x)
 

Output:

Integral(sec(x)**2/(a + b*tan(x) + c*sec(x)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec ^2(x)}{a+c \sec (x)+b \tan (x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(sec(x)^2/(a+c*sec(x)+b*tan(x)),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(c^2-b^2-a^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^2(x)}{a+c \sec (x)+b \tan (x)} \, dx=\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, c\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - c \tan \left (\frac {1}{2} \, x\right ) - b}{\sqrt {-a^{2} - b^{2} + c^{2}}}\right )\right )} a c}{\sqrt {-a^{2} - b^{2} + c^{2}} {\left (b^{2} - c^{2}\right )}} + \frac {b \log \left (-a \tan \left (\frac {1}{2} \, x\right )^{2} + c \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a + c\right )}{b^{2} - c^{2}} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{b - c} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{b + c} \] Input:

integrate(sec(x)^2/(a+c*sec(x)+b*tan(x)),x, algorithm="giac")
 

Output:

2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*c) + arctan(-(a*tan(1/2*x) - c*ta 
n(1/2*x) - b)/sqrt(-a^2 - b^2 + c^2)))*a*c/(sqrt(-a^2 - b^2 + c^2)*(b^2 - 
c^2)) + b*log(-a*tan(1/2*x)^2 + c*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a + c)/( 
b^2 - c^2) - log(abs(tan(1/2*x) + 1))/(b - c) - log(abs(tan(1/2*x) - 1))/( 
b + c)
 

Mupad [B] (verification not implemented)

Time = 25.37 (sec) , antiderivative size = 977, normalized size of antiderivative = 6.88 \[ \int \frac {\sec ^2(x)}{a+c \sec (x)+b \tan (x)} \, dx =\text {Too large to display} \] Input:

int(1/(cos(x)^2*(a + b*tan(x) + c/cos(x))),x)
 

Output:

(log(32*a*c - 32*a^2 - 32*b*tan(x/2)*(a - c) - ((a^2*b - b*c^2 + b^3 + a*c 
*(a^2 + b^2 - c^2)^(1/2))*(32*a^2*b - 32*b*c^2 + 32*tan(x/2)*(a - c)*(2*a^ 
2 - 2*a*c + 3*b^2 + c^2) - ((a^2*b - b*c^2 + b^3 + a*c*(a^2 + b^2 - c^2)^( 
1/2))*(32*c^4 - 64*a*c^3 + 32*a^2*b^2 + 32*a^2*c^2 + 64*b^2*c^2 - 96*a*b^2 
*c + 32*b*tan(x/2)*(a - c)*(4*a*c + b^2 - 4*c^2) + (32*(a - c)*(a^2*b - b* 
c^2 + b^3 + a*c*(a^2 + b^2 - c^2)^(1/2))*(3*b^4*tan(x/2) + a*b^3 - 3*b*c^3 
 + 3*b^3*c + 2*a^2*b^2*tan(x/2) + 2*a^2*c^2*tan(x/2) - 3*b^2*c^2*tan(x/2) 
- 2*a*c^3*tan(x/2) - a*b*c^2 + 4*a^2*b*c + 2*a*b^2*c*tan(x/2)))/((b^2 - c^ 
2)*(a^2 + b^2 - c^2))))/((b^2 - c^2)*(a^2 + b^2 - c^2))))/((b^2 - c^2)*(a^ 
2 + b^2 - c^2)))*(b*(a^2 - c^2) + b^3 + a*c*(a^2 + b^2 - c^2)^(1/2)))/((b^ 
2 - c^2)*(a^2 + b^2 - c^2)) - log(tan(x/2) - 1)/(b + c) - log(tan(x/2) + 1 
)/(b - c) + (log(32*a*c - 32*a^2 - 32*b*tan(x/2)*(a - c) - ((a^2*b - b*c^2 
 + b^3 - a*c*(a^2 + b^2 - c^2)^(1/2))*(32*a^2*b - 32*b*c^2 + 32*tan(x/2)*( 
a - c)*(2*a^2 - 2*a*c + 3*b^2 + c^2) - ((a^2*b - b*c^2 + b^3 - a*c*(a^2 + 
b^2 - c^2)^(1/2))*(32*c^4 - 64*a*c^3 + 32*a^2*b^2 + 32*a^2*c^2 + 64*b^2*c^ 
2 - 96*a*b^2*c + 32*b*tan(x/2)*(a - c)*(4*a*c + b^2 - 4*c^2) + (32*(a - c) 
*(a^2*b - b*c^2 + b^3 - a*c*(a^2 + b^2 - c^2)^(1/2))*(3*b^4*tan(x/2) + a*b 
^3 - 3*b*c^3 + 3*b^3*c + 2*a^2*b^2*tan(x/2) + 2*a^2*c^2*tan(x/2) - 3*b^2*c 
^2*tan(x/2) - 2*a*c^3*tan(x/2) - a*b*c^2 + 4*a^2*b*c + 2*a*b^2*c*tan(x/2)) 
)/((b^2 - c^2)*(a^2 + b^2 - c^2))))/((b^2 - c^2)*(a^2 + b^2 - c^2))))/(...
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 347, normalized size of antiderivative = 2.44 \[ \int \frac {\sec ^2(x)}{a+c \sec (x)+b \tan (x)} \, dx=\frac {2 \sqrt {-a^{2}-b^{2}+c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) c -b}{\sqrt {-a^{2}-b^{2}+c^{2}}}\right ) a c -\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) a^{2} b +\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) a^{2} c -\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) b^{3}+\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) b^{2} c +\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) b \,c^{2}-\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) c^{3}-\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) a^{2} b -\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) a^{2} c -\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) b^{3}-\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) b^{2} c +\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) b \,c^{2}+\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) c^{3}+\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2} a -\tan \left (\frac {x}{2}\right )^{2} c -2 \tan \left (\frac {x}{2}\right ) b -a -c \right ) a^{2} b +\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2} a -\tan \left (\frac {x}{2}\right )^{2} c -2 \tan \left (\frac {x}{2}\right ) b -a -c \right ) b^{3}-\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2} a -\tan \left (\frac {x}{2}\right )^{2} c -2 \tan \left (\frac {x}{2}\right ) b -a -c \right ) b \,c^{2}}{a^{2} b^{2}-a^{2} c^{2}+b^{4}-2 b^{2} c^{2}+c^{4}} \] Input:

int(sec(x)^2/(a+c*sec(x)+b*tan(x)),x)
 

Output:

(2*sqrt( - a**2 - b**2 + c**2)*atan((tan(x/2)*a - tan(x/2)*c - b)/sqrt( - 
a**2 - b**2 + c**2))*a*c - log(tan(x/2) - 1)*a**2*b + log(tan(x/2) - 1)*a* 
*2*c - log(tan(x/2) - 1)*b**3 + log(tan(x/2) - 1)*b**2*c + log(tan(x/2) - 
1)*b*c**2 - log(tan(x/2) - 1)*c**3 - log(tan(x/2) + 1)*a**2*b - log(tan(x/ 
2) + 1)*a**2*c - log(tan(x/2) + 1)*b**3 - log(tan(x/2) + 1)*b**2*c + log(t 
an(x/2) + 1)*b*c**2 + log(tan(x/2) + 1)*c**3 + log(tan(x/2)**2*a - tan(x/2 
)**2*c - 2*tan(x/2)*b - a - c)*a**2*b + log(tan(x/2)**2*a - tan(x/2)**2*c 
- 2*tan(x/2)*b - a - c)*b**3 - log(tan(x/2)**2*a - tan(x/2)**2*c - 2*tan(x 
/2)*b - a - c)*b*c**2)/(a**2*b**2 - a**2*c**2 + b**4 - 2*b**2*c**2 + c**4)