\(\int \frac {\sec ^2(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx\) [19]

Optimal result

Integrand size = 19, antiderivative size = 275 \[ \int \frac {\sec ^2(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\frac {2 b c \left (1+\frac {b^2-2 a c}{b \sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {b-2 c-\sqrt {b^2-4 a c}} \tan \left (\frac {x}{2}\right )}{\sqrt {b+2 c-\sqrt {b^2-4 a c}}}\right )}{a^2 \sqrt {b-2 c-\sqrt {b^2-4 a c}} \sqrt {b+2 c-\sqrt {b^2-4 a c}}}+\frac {2 b c \left (1-\frac {b^2-2 a c}{b \sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {b-2 c+\sqrt {b^2-4 a c}} \tan \left (\frac {x}{2}\right )}{\sqrt {b+2 c+\sqrt {b^2-4 a c}}}\right )}{a^2 \sqrt {b-2 c+\sqrt {b^2-4 a c}} \sqrt {b+2 c+\sqrt {b^2-4 a c}}}-\frac {b \text {arctanh}(\sin (x))}{a^2}+\frac {\tan (x)}{a} \] Output:

2*b*c*(1+(-2*a*c+b^2)/b/(-4*a*c+b^2)^(1/2))*arctan((b-2*c-(-4*a*c+b^2)^(1/ 
2))^(1/2)*tan(1/2*x)/(b+2*c-(-4*a*c+b^2)^(1/2))^(1/2))/a^2/(b-2*c-(-4*a*c+ 
b^2)^(1/2))^(1/2)/(b+2*c-(-4*a*c+b^2)^(1/2))^(1/2)+2*b*c*(1-(-2*a*c+b^2)/b 
/(-4*a*c+b^2)^(1/2))*arctan((b-2*c+(-4*a*c+b^2)^(1/2))^(1/2)*tan(1/2*x)/(b 
+2*c+(-4*a*c+b^2)^(1/2))^(1/2))/a^2/(b-2*c+(-4*a*c+b^2)^(1/2))^(1/2)/(b+2* 
c+(-4*a*c+b^2)^(1/2))^(1/2)-b*arctanh(sin(x))/a^2+tan(x)/a
 

Mathematica [A] (verified)

Time = 1.88 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.27 \[ \int \frac {\sec ^2(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\frac {-\frac {\sqrt {2} c \left (-b^2+2 a c+b \sqrt {b^2-4 a c}\right ) \text {arctanh}\left (\frac {\left (b-2 c+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 c (a+c)-2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {-b^2+2 c (a+c)-b \sqrt {b^2-4 a c}}}+\frac {\sqrt {2} c \left (b^2-2 a c+b \sqrt {b^2-4 a c}\right ) \text {arctanh}\left (\frac {\left (-b+2 c+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 c (a+c)+2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {-b^2+2 c (a+c)+b \sqrt {b^2-4 a c}}}+b \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-b \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )+\frac {a \sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )}+\frac {a \sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )}}{a^2} \] Input:

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

Output:

(-((Sqrt[2]*c*(-b^2 + 2*a*c + b*Sqrt[b^2 - 4*a*c])*ArcTanh[((b - 2*c + Sqr 
t[b^2 - 4*a*c])*Tan[x/2])/Sqrt[-2*b^2 + 4*c*(a + c) - 2*b*Sqrt[b^2 - 4*a*c 
]]])/(Sqrt[b^2 - 4*a*c]*Sqrt[-b^2 + 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c]])) + 
 (Sqrt[2]*c*(b^2 - 2*a*c + b*Sqrt[b^2 - 4*a*c])*ArcTanh[((-b + 2*c + Sqrt[ 
b^2 - 4*a*c])*Tan[x/2])/Sqrt[-2*b^2 + 4*c*(a + c) + 2*b*Sqrt[b^2 - 4*a*c]] 
])/(Sqrt[b^2 - 4*a*c]*Sqrt[-b^2 + 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]]) + b* 
Log[Cos[x/2] - Sin[x/2]] - b*Log[Cos[x/2] + Sin[x/2]] + (a*Sin[x/2])/(Cos[ 
x/2] - Sin[x/2]) + (a*Sin[x/2])/(Cos[x/2] + Sin[x/2]))/a^2
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 3738, 2009}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos (x)^2 \left (a+b \cos (x)+c \cos (x)^2\right )}dx\)

\(\Big \downarrow \) 3738

\(\displaystyle \int \left (\frac {b^2 \left (1-\frac {a c}{b^2}\right )+b c \cos (x)}{a^2 \left (a+b \cos (x)+c \cos ^2(x)\right )}-\frac {b \sec (x)}{a^2}+\frac {\sec ^2(x)}{a}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(2*b*c*(1 + (b^2 - 2*a*c)/(b*Sqrt[b^2 - 4*a*c]))*ArcTan[(Sqrt[b - 2*c - Sq 
rt[b^2 - 4*a*c]]*Tan[x/2])/Sqrt[b + 2*c - Sqrt[b^2 - 4*a*c]]])/(a^2*Sqrt[b 
 - 2*c - Sqrt[b^2 - 4*a*c]]*Sqrt[b + 2*c - Sqrt[b^2 - 4*a*c]]) + (2*b*c*(1 
 - (b^2 - 2*a*c)/(b*Sqrt[b^2 - 4*a*c]))*ArcTan[(Sqrt[b - 2*c + Sqrt[b^2 - 
4*a*c]]*Tan[x/2])/Sqrt[b + 2*c + Sqrt[b^2 - 4*a*c]]])/(a^2*Sqrt[b - 2*c + 
Sqrt[b^2 - 4*a*c]]*Sqrt[b + 2*c + Sqrt[b^2 - 4*a*c]]) - (b*ArcTanh[Sin[x]] 
)/a^2 + Tan[x]/a
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3738

Int[cos[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + cos[(d_.) + (e_.)*(x_)]^(n_.)*(b 
_.) + cos[(d_.) + (e_.)*(x_)]^(n2_.)*(c_.))^(p_), x_Symbol] :> Int[ExpandTr 
ig[cos[d + e*x]^m*(a + b*cos[d + e*x]^n + c*cos[d + e*x]^(2*n))^p, x], x] / 
; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && Integ 
ersQ[m, n, p]
 

Maple [A] (verified)

Time = 2.75 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.29

method	result	size
default	\(\frac {2 \left (a -b +c \right ) \left (\frac {\left (-\sqrt {-4 a c +b^{2}}\, a c +\sqrt {-4 a c +b^{2}}\, b^{2}-\sqrt {-4 a c +b^{2}}\, b c +3 c a b -2 a \,c^{2}-b^{3}+b^{2} c \right ) \operatorname {arctanh}\left (\frac {\left (-a +b -c \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (\sqrt {-4 a c +b^{2}}-a +c \right ) \left (a -b +c \right )}}\right )}{2 \sqrt {-4 a c +b^{2}}\, \left (a -b +c \right ) \sqrt {\left (\sqrt {-4 a c +b^{2}}-a +c \right ) \left (a -b +c \right )}}+\frac {\left (-\sqrt {-4 a c +b^{2}}\, a c +\sqrt {-4 a c +b^{2}}\, b^{2}-\sqrt {-4 a c +b^{2}}\, b c -3 c a b +2 a \,c^{2}+b^{3}-b^{2} c \right ) \arctan \left (\frac {\left (a -b +c \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (\sqrt {-4 a c +b^{2}}+a -c \right ) \left (a -b +c \right )}}\right )}{2 \sqrt {-4 a c +b^{2}}\, \left (a -b +c \right ) \sqrt {\left (\sqrt {-4 a c +b^{2}}+a -c \right ) \left (a -b +c \right )}}\right )}{a^{2}}-\frac {1}{a \left (\tan \left (\frac {x}{2}\right )-1\right )}+\frac {b \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{a^{2}}-\frac {1}{a \left (\tan \left (\frac {x}{2}\right )+1\right )}-\frac {b \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{a^{2}}\)	\(354\)
risch	\(\text {Expression too large to display}\)	\(2641\)

Input:

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

Output:

2/a^2*(a-b+c)*(1/2*(-(-4*a*c+b^2)^(1/2)*a*c+(-4*a*c+b^2)^(1/2)*b^2-(-4*a*c 
+b^2)^(1/2)*b*c+3*c*a*b-2*a*c^2-b^3+b^2*c)/(-4*a*c+b^2)^(1/2)/(a-b+c)/(((- 
4*a*c+b^2)^(1/2)-a+c)*(a-b+c))^(1/2)*arctanh((-a+b-c)*tan(1/2*x)/(((-4*a*c 
+b^2)^(1/2)-a+c)*(a-b+c))^(1/2))+1/2*(-(-4*a*c+b^2)^(1/2)*a*c+(-4*a*c+b^2) 
^(1/2)*b^2-(-4*a*c+b^2)^(1/2)*b*c-3*c*a*b+2*a*c^2+b^3-b^2*c)/(-4*a*c+b^2)^ 
(1/2)/(a-b+c)/(((-4*a*c+b^2)^(1/2)+a-c)*(a-b+c))^(1/2)*arctan((a-b+c)*tan( 
1/2*x)/(((-4*a*c+b^2)^(1/2)+a-c)*(a-b+c))^(1/2)))-1/a/(tan(1/2*x)-1)+b/a^2 
*ln(tan(1/2*x)-1)-1/a/(tan(1/2*x)+1)-b/a^2*ln(tan(1/2*x)+1)
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/2*(2*(a^2*cos(2*x)^2 + a^2*sin(2*x)^2 + 2*a^2*cos(2*x) + a^2)*integrate( 
2*(2*b^2*c*cos(3*x)^2 + 2*b^2*c*cos(x)^2 + 2*b^2*c*sin(3*x)^2 + 2*b^2*c*si 
n(x)^2 + b*c^2*cos(x) + 4*(2*a*b^2 - a*c^2 - (2*a^2 - b^2)*c)*cos(2*x)^2 + 
 4*(2*a*b^2 - a*c^2 - (2*a^2 - b^2)*c)*sin(2*x)^2 + 2*(2*b^3 + b*c^2)*sin( 
2*x)*sin(x) + (b*c^2*cos(3*x) + b*c^2*cos(x) + 2*(b^2*c - a*c^2)*cos(2*x)) 
*cos(4*x) + (4*b^2*c*cos(x) + b*c^2 + 2*(2*b^3 + b*c^2)*cos(2*x))*cos(3*x) 
 + 2*(b^2*c - a*c^2 + (2*b^3 + b*c^2)*cos(x))*cos(2*x) + (b*c^2*sin(3*x) + 
 b*c^2*sin(x) + 2*(b^2*c - a*c^2)*sin(2*x))*sin(4*x) + 2*(2*b^2*c*sin(x) + 
 (2*b^3 + b*c^2)*sin(2*x))*sin(3*x))/(a^2*c^2*cos(4*x)^2 + 4*a^2*b^2*cos(3 
*x)^2 + 4*a^2*b^2*cos(x)^2 + a^2*c^2*sin(4*x)^2 + 4*a^2*b^2*sin(3*x)^2 + 4 
*a^2*b^2*sin(x)^2 + 4*a^2*b*c*cos(x) + a^2*c^2 + 4*(4*a^4 + 4*a^3*c + a^2* 
c^2)*cos(2*x)^2 + 4*(4*a^4 + 4*a^3*c + a^2*c^2)*sin(2*x)^2 + 8*(2*a^3*b + 
a^2*b*c)*sin(2*x)*sin(x) + 2*(2*a^2*b*c*cos(3*x) + 2*a^2*b*c*cos(x) + a^2* 
c^2 + 2*(2*a^3*c + a^2*c^2)*cos(2*x))*cos(4*x) + 4*(2*a^2*b^2*cos(x) + a^2 
*b*c + 2*(2*a^3*b + a^2*b*c)*cos(2*x))*cos(3*x) + 4*(2*a^3*c + a^2*c^2 + 2 
*(2*a^3*b + a^2*b*c)*cos(x))*cos(2*x) + 4*(a^2*b*c*sin(3*x) + a^2*b*c*sin( 
x) + (2*a^3*c + a^2*c^2)*sin(2*x))*sin(4*x) + 8*(a^2*b^2*sin(x) + (2*a^3*b 
 + a^2*b*c)*sin(2*x))*sin(3*x)), x) - (b*cos(2*x)^2 + b*sin(2*x)^2 + 2*b*c 
os(2*x) + b)*log(cos(x)^2 + sin(x)^2 + 2*sin(x) + 1) + (b*cos(2*x)^2 + b*s 
in(2*x)^2 + 2*b*cos(2*x) + b)*log(cos(x)^2 + sin(x)^2 - 2*sin(x) + 1) +...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4784 vs. \(2 (234) = 468\).

Time = 2.38 (sec) , antiderivative size = 4784, normalized size of antiderivative = 17.40 \[ \int \frac {\sec ^2(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Too large to display} \] Input:

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

Output:

(2*a^2*b^5 - 2*b^7 - 14*a^3*b^3*c - 6*a^2*b^4*c + 26*a*b^5*c - 2*b^6*c + 2 
4*a^4*b*c^2 + 36*a^3*b^2*c^2 - 108*a^2*b^3*c^2 + 16*a*b^4*c^2 + 2*b^5*c^2 
- 48*a^4*c^3 + 144*a^3*b*c^3 - 24*a^2*b^2*c^3 - 22*a*b^3*c^3 + 2*b^4*c^3 - 
 32*a^3*c^4 + 56*a^2*b*c^4 - 12*a*b^2*c^4 + 16*a^2*c^5 + 3*sqrt(a^2 - a*b 
+ b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*a^2*b^4 - 2*sqrt(a^2 - a*b + 
b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*a*b^5 - 5*sqrt(a^2 - a*b + b*c 
- c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*b^6 - 15*sqrt(a^2 - a*b + b*c - c^2 
 + sqrt(b^2 - 4*a*c)*(a - b + c))*a^3*b^2*c + 13*sqrt(a^2 - a*b + b*c - c^ 
2 + sqrt(b^2 - 4*a*c)*(a - b + c))*a^2*b^3*c + 37*sqrt(a^2 - a*b + b*c - c 
^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*a*b^4*c + sqrt(a^2 - a*b + b*c - c^2 + 
 sqrt(b^2 - 4*a*c)*(a - b + c))*b^5*c + 12*sqrt(a^2 - a*b + b*c - c^2 + sq 
rt(b^2 - 4*a*c)*(a - b + c))*a^4*c^2 - 20*sqrt(a^2 - a*b + b*c - c^2 + sqr 
t(b^2 - 4*a*c)*(a - b + c))*a^3*b*c^2 - 82*sqrt(a^2 - a*b + b*c - c^2 + sq 
rt(b^2 - 4*a*c)*(a - b + c))*a^2*b^2*c^2 + 4*sqrt(a^2 - a*b + b*c - c^2 + 
sqrt(b^2 - 4*a*c)*(a - b + c))*a*b^3*c^2 + sqrt(a^2 - a*b + b*c - c^2 + sq 
rt(b^2 - 4*a*c)*(a - b + c))*b^4*c^2 + 56*sqrt(a^2 - a*b + b*c - c^2 + sqr 
t(b^2 - 4*a*c)*(a - b + c))*a^3*c^3 - 32*sqrt(a^2 - a*b + b*c - c^2 + sqrt 
(b^2 - 4*a*c)*(a - b + c))*a^2*b*c^3 + sqrt(a^2 - a*b + b*c - c^2 + sqrt(b 
^2 - 4*a*c)*(a - b + c))*a*b^2*c^3 - 5*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b 
^2 - 4*a*c)*(a - b + c))*b^3*c^3 - 20*sqrt(a^2 - a*b + b*c - c^2 + sqrt...
 

Mupad [B] (verification not implemented)

Time = 7.72 (sec) , antiderivative size = 29417, normalized size of antiderivative = 106.97 \[ \int \frac {\sec ^2(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Too large to display} \] Input:

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

Output:

(b*atan(((b*((8192*tan(x/2)*(a*b^8 + 5*b^8*c - b^9 + a^2*c^7 + a^3*c^6 + b 
^4*c^5 - 5*b^5*c^4 + 10*b^6*c^3 - 10*b^7*c^2 - 2*a*b^2*c^6 + 14*a*b^3*c^5 
- 35*a*b^4*c^4 + 40*a*b^5*c^3 - 20*a*b^6*c^2 - a^2*b*c^6 - 6*a^2*b^6*c + 1 
0*a^2*b^2*c^5 - 20*a^2*b^3*c^4 + 5*a^2*b^4*c^3 + 11*a^2*b^5*c^2 + 10*a^3*b 
^2*c^4 - 18*a^3*b^3*c^3 + 9*a^3*b^4*c^2 - 2*a^4*b^2*c^3 + 2*a*b^7*c))/a^4 
+ (b*((8192*(6*a^2*b^8 - 3*a*b^9 - 4*a^3*b^7 + a^4*b^6 + 3*a^4*c^6 + 2*a^5 
*c^5 - a^6*c^4 + 2*a*b^5*c^4 - 5*a*b^6*c^3 + a*b^7*c^2 + 16*a^2*b^7*c + 8* 
a^3*b*c^6 - 38*a^3*b^6*c + 10*a^4*b*c^5 + 23*a^4*b^5*c + 6*a^5*b*c^4 - 5*a 
^5*b^4*c - 10*a^2*b^3*c^5 + 25*a^2*b^4*c^4 + 4*a^2*b^5*c^3 - 41*a^2*b^6*c^ 
2 - 20*a^3*b^2*c^5 - 36*a^3*b^3*c^4 + 91*a^3*b^4*c^3 - 3*a^3*b^5*c^2 - 24* 
a^4*b^2*c^4 - 55*a^4*b^3*c^3 + 57*a^4*b^4*c^2 - 3*a^5*b^2*c^3 - 28*a^5*b^3 
*c^2 + 4*a^6*b^2*c^2 + 5*a*b^8*c))/a^4 + (b*((b*((8192*(3*a^5*b^7 - 7*a^6* 
b^6 + 5*a^7*b^5 - a^8*b^4 + 12*a^7*c^5 + 20*a^8*c^4 + 4*a^9*c^3 - 4*a^10*c 
^2 - 5*a^5*b^6*c + 8*a^6*b*c^5 - 15*a^6*b^5*c + 28*a^7*b*c^4 + 46*a^7*b^4* 
c + 64*a^8*b*c^3 - 31*a^8*b^3*c + 44*a^9*b*c^2 + 5*a^9*b^2*c - 2*a^5*b^3*c 
^4 + 5*a^5*b^4*c^3 - a^5*b^5*c^2 - 23*a^6*b^2*c^4 - 3*a^6*b^3*c^3 + 40*a^6 
*b^4*c^2 - 85*a^7*b^2*c^3 - 4*a^7*b^3*c^2 - 73*a^8*b^2*c^2))/a^4 + (8192*b 
*tan(x/2)*(8*a^12*c + 2*a^6*b^7 - 6*a^7*b^6 + 8*a^8*b^5 - 8*a^9*b^4 + 6*a^ 
10*b^3 - 2*a^11*b^2 + 24*a^8*c^5 + 16*a^9*c^4 - 32*a^10*c^3 - 16*a^11*c^2 
- 2*a^6*b^6*c - 14*a^7*b^5*c - 8*a^8*b*c^4 + 46*a^8*b^4*c + 88*a^9*b*c^...
 

Reduce [F]

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

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

Output:

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