\(\int \frac {\cos ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx\) [13]

Optimal result

Integrand size = 19, antiderivative size = 326 \[ \int \frac {\cos ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\frac {x}{2 c}+\frac {\left (b^2-a c\right ) x}{c^3}-\frac {2 \left (b^3-2 a b c-\frac {b^4-4 a b^2 c+2 a^2 c^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {b-2 c-\sqrt {b^2-4 a c}} \tan \left (\frac {x}{2}\right )}{\sqrt {b+2 c-\sqrt {b^2-4 a c}}}\right )}{c^3 \sqrt {b-2 c-\sqrt {b^2-4 a c}} \sqrt {b+2 c-\sqrt {b^2-4 a c}}}-\frac {2 \left (b^3-2 a b c+\frac {b^4-4 a b^2 c+2 a^2 c^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {b-2 c+\sqrt {b^2-4 a c}} \tan \left (\frac {x}{2}\right )}{\sqrt {b+2 c+\sqrt {b^2-4 a c}}}\right )}{c^3 \sqrt {b-2 c+\sqrt {b^2-4 a c}} \sqrt {b+2 c+\sqrt {b^2-4 a c}}}-\frac {b \sin (x)}{c^2}+\frac {\cos (x) \sin (x)}{2 c} \] Output:

1/2*x/c+(-a*c+b^2)*x/c^3-2*(b^3-2*a*b*c-(2*a^2*c^2-4*a*b^2*c+b^4)/(-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))/c^3/(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^3-2*a*b*c+(2*a^2*c^2-4*a*b^2*c+b^4)/(-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))/c^3/(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*sin(x)/c^2+1/2*cos(x)*sin(x)/c
 

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\frac {4 b^2 x+2 c (-2 a+c) x+\frac {4 \sqrt {2} \left (b^4-4 a b^2 c+2 a^2 c^2+b^3 \sqrt {b^2-4 a c}-2 a b c \sqrt {b^2-4 a c}\right ) \text {arctanh}\left (\frac {\left (b-2 c+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 c (a+c)-2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {-b^2+2 c (a+c)-b \sqrt {b^2-4 a c}}}-\frac {4 \sqrt {2} \left (-b^4+4 a b^2 c-2 a^2 c^2+b^3 \sqrt {b^2-4 a c}-2 a b c \sqrt {b^2-4 a c}\right ) \text {arctanh}\left (\frac {\left (-b+2 c+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 c (a+c)+2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {-b^2+2 c (a+c)+b \sqrt {b^2-4 a c}}}-4 b c \sin (x)+c^2 \sin (2 x)}{4 c^3} \] Input:

Integrate[Cos[x]^4/(a + b*Cos[x] + c*Cos[x]^2),x]
 

Output:

(4*b^2*x + 2*c*(-2*a + c)*x + (4*Sqrt[2]*(b^4 - 4*a*b^2*c + 2*a^2*c^2 + b^ 
3*Sqrt[b^2 - 4*a*c] - 2*a*b*c*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]]) - (4 
*Sqrt[2]*(-b^4 + 4*a*b^2*c - 2*a^2*c^2 + b^3*Sqrt[b^2 - 4*a*c] - 2*a*b*c*S 
qrt[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]]) - 4*b*c*Sin[x] + c^2*Sin[2*x])/(4* 
c^3)
 

Rubi [A] (verified)

Time = 2.70 (sec) , antiderivative size = 326, 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.

\(\displaystyle \int \frac {\cos ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (x)^4}{a+b \cos (x)+c \cos (x)^2}dx\)

\(\Big \downarrow \) 3738

\(\displaystyle \int \left (\frac {b^2-a c}{c^3}+\frac {-a b^2 \left (1-\frac {a c}{b^2}\right )-\left (b^3 \cos (x) \left (1-\frac {2 a c}{b^2}\right )\right )}{c^3 \left (a+b \cos (x)+c \cos ^2(x)\right )}-\frac {b \cos (x)}{c^2}+\frac {\cos ^2(x)}{c}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \left (-\frac {2 a^2 c^2-4 a b^2 c+b^4}{\sqrt {b^2-4 a c}}-2 a b c+b^3\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \sqrt {-\sqrt {b^2-4 a c}+b-2 c}}{\sqrt {-\sqrt {b^2-4 a c}+b+2 c}}\right )}{c^3 \sqrt {-\sqrt {b^2-4 a c}+b-2 c} \sqrt {-\sqrt {b^2-4 a c}+b+2 c}}-\frac {2 \left (\frac {2 a^2 c^2-4 a b^2 c+b^4}{\sqrt {b^2-4 a c}}-2 a b c+b^3\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \sqrt {\sqrt {b^2-4 a c}+b-2 c}}{\sqrt {\sqrt {b^2-4 a c}+b+2 c}}\right )}{c^3 \sqrt {\sqrt {b^2-4 a c}+b-2 c} \sqrt {\sqrt {b^2-4 a c}+b+2 c}}+\frac {x \left (b^2-a c\right )}{c^3}-\frac {b \sin (x)}{c^2}+\frac {x}{2 c}+\frac {\sin (x) \cos (x)}{2 c}\)

Input:

Int[Cos[x]^4/(a + b*Cos[x] + c*Cos[x]^2),x]
 

Output:

x/(2*c) + ((b^2 - a*c)*x)/c^3 - (2*(b^3 - 2*a*b*c - (b^4 - 4*a*b^2*c + 2*a 
^2*c^2)/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]]])/(c^3*Sqrt[b - 2*c - Sqrt[b^2 - 4* 
a*c]]*Sqrt[b + 2*c - Sqrt[b^2 - 4*a*c]]) - (2*(b^3 - 2*a*b*c + (b^4 - 4*a* 
b^2*c + 2*a^2*c^2)/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]]])/(c^3*Sqrt[b - 2*c + Sq 
rt[b^2 - 4*a*c]]*Sqrt[b + 2*c + Sqrt[b^2 - 4*a*c]]) - (b*Sin[x])/c^2 + (Co 
s[x]*Sin[x])/(2*c)
 

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 = 3.16 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.32

method	result	size
default	\(\frac {2 \left (a -b +c \right ) \left (\frac {\left (a^{2} c \sqrt {-4 a c +b^{2}}-a \,b^{2} \sqrt {-4 a c +b^{2}}-2 c a b \sqrt {-4 a c +b^{2}}+b^{3} \sqrt {-4 a c +b^{2}}+3 c \,a^{2} b +2 a^{2} c^{2}-a \,b^{3}-4 a \,b^{2} c +b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b -c \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (\sqrt {-4 a c +b^{2}}-a +c \right ) \left (a -b +c \right )}}\right )}{2 \sqrt {-4 a c +b^{2}}\, \left (a -b +c \right ) \sqrt {\left (\sqrt {-4 a c +b^{2}}-a +c \right ) \left (a -b +c \right )}}+\frac {\left (a^{2} c \sqrt {-4 a c +b^{2}}-a \,b^{2} \sqrt {-4 a c +b^{2}}-2 c a b \sqrt {-4 a c +b^{2}}+b^{3} \sqrt {-4 a c +b^{2}}-3 c \,a^{2} b -2 a^{2} c^{2}+a \,b^{3}+4 a \,b^{2} c -b^{4}\right ) \arctan \left (\frac {\left (a -b +c \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (\sqrt {-4 a c +b^{2}}+a -c \right ) \left (a -b +c \right )}}\right )}{2 \sqrt {-4 a c +b^{2}}\, \left (a -b +c \right ) \sqrt {\left (\sqrt {-4 a c +b^{2}}+a -c \right ) \left (a -b +c \right )}}\right )}{c^{3}}-\frac {2 \left (\frac {\left (c b +\frac {1}{2} c^{2}\right ) \tan \left (\frac {x}{2}\right )^{3}+\left (c b -\frac {1}{2} c^{2}\right ) \tan \left (\frac {x}{2}\right )}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{2}}+\frac {\left (2 a c -2 b^{2}-c^{2}\right ) \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{2}\right )}{c^{3}}\)	\(431\)
risch	\(\text {Expression too large to display}\)	\(3886\)

Input:

int(cos(x)^4/(a+b*cos(x)+c*cos(x)^2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 8167 vs. \(2 (282) = 564\).

Time = 5.08 (sec) , antiderivative size = 8167, normalized size of antiderivative = 25.05 \[ \int \frac {\cos ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Too large to display} \] Input:

integrate(cos(x)^4/(a+b*cos(x)+c*cos(x)^2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Timed out} \] Input:

integrate(cos(x)**4/(a+b*cos(x)+c*cos(x)**2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\cos ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\int { \frac {\cos \left (x\right )^{4}}{c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a} \,d x } \] Input:

integrate(cos(x)^4/(a+b*cos(x)+c*cos(x)^2),x, algorithm="maxima")
 

Output:

1/4*(4*c^3*integrate(-2*(2*(b^4 - 2*a*b^2*c)*cos(3*x)^2 + 4*(2*a^2*b^2 - a 
^2*c^2 - (2*a^3 - a*b^2)*c)*cos(2*x)^2 + 2*(b^4 - 2*a*b^2*c)*cos(x)^2 + 2* 
(b^4 - 2*a*b^2*c)*sin(3*x)^2 + 4*(2*a^2*b^2 - a^2*c^2 - (2*a^3 - a*b^2)*c) 
*sin(2*x)^2 + 2*(4*a*b^3 - 2*a*b*c^2 - (6*a^2*b - b^3)*c)*sin(2*x)*sin(x) 
+ 2*(b^4 - 2*a*b^2*c)*sin(x)^2 + ((b^3*c - 2*a*b*c^2)*cos(3*x) + 2*(a*b^2* 
c - a^2*c^2)*cos(2*x) + (b^3*c - 2*a*b*c^2)*cos(x))*cos(4*x) + (b^3*c - 2* 
a*b*c^2 + 2*(4*a*b^3 - 2*a*b*c^2 - (6*a^2*b - b^3)*c)*cos(2*x) + 4*(b^4 - 
2*a*b^2*c)*cos(x))*cos(3*x) + 2*(a*b^2*c - a^2*c^2 + (4*a*b^3 - 2*a*b*c^2 
- (6*a^2*b - b^3)*c)*cos(x))*cos(2*x) + (b^3*c - 2*a*b*c^2)*cos(x) + ((b^3 
*c - 2*a*b*c^2)*sin(3*x) + 2*(a*b^2*c - a^2*c^2)*sin(2*x) + (b^3*c - 2*a*b 
*c^2)*sin(x))*sin(4*x) + 2*((4*a*b^3 - 2*a*b*c^2 - (6*a^2*b - b^3)*c)*sin( 
2*x) + 2*(b^4 - 2*a*b^2*c)*sin(x))*sin(3*x))/(c^5*cos(4*x)^2 + 4*b^2*c^3*c 
os(3*x)^2 + 4*b^2*c^3*cos(x)^2 + c^5*sin(4*x)^2 + 4*b^2*c^3*sin(3*x)^2 + 4 
*b^2*c^3*sin(x)^2 + 4*b*c^4*cos(x) + c^5 + 4*(4*a^2*c^3 + 4*a*c^4 + c^5)*c 
os(2*x)^2 + 4*(4*a^2*c^3 + 4*a*c^4 + c^5)*sin(2*x)^2 + 8*(2*a*b*c^3 + b*c^ 
4)*sin(2*x)*sin(x) + 2*(2*b*c^4*cos(3*x) + 2*b*c^4*cos(x) + c^5 + 2*(2*a*c 
^4 + c^5)*cos(2*x))*cos(4*x) + 4*(2*b^2*c^3*cos(x) + b*c^4 + 2*(2*a*b*c^3 
+ b*c^4)*cos(2*x))*cos(3*x) + 4*(2*a*c^4 + c^5 + 2*(2*a*b*c^3 + b*c^4)*cos 
(x))*cos(2*x) + 4*(b*c^4*sin(3*x) + b*c^4*sin(x) + (2*a*c^4 + c^5)*sin(2*x 
))*sin(4*x) + 8*(b^2*c^3*sin(x) + (2*a*b*c^3 + b*c^4)*sin(2*x))*sin(3*x...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 12585 vs. \(2 (282) = 564\).

Time = 2.09 (sec) , antiderivative size = 12585, normalized size of antiderivative = 38.60 \[ \int \frac {\cos ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Too large to display} \] Input:

integrate(cos(x)^4/(a+b*cos(x)+c*cos(x)^2),x, algorithm="giac")
 

Output:

((2*a^3*b^6 - 6*a^2*b^7 + 6*a*b^8 - 2*b^9 - 18*a^4*b^4*c + 56*a^3*b^5*c - 
54*a^2*b^6*c + 12*a*b^7*c + 4*b^8*c + 48*a^5*b^2*c^2 - 160*a^4*b^3*c^2 + 1 
40*a^3*b^4*c^2 + 12*a^2*b^5*c^2 - 38*a*b^6*c^2 - 2*b^7*c^2 - 32*a^6*c^3 + 
128*a^5*b*c^3 - 64*a^4*b^2*c^3 - 160*a^3*b^3*c^3 + 110*a^2*b^4*c^3 + 20*a* 
b^5*c^3 - 64*a^5*c^4 + 192*a^4*b*c^4 - 80*a^3*b^2*c^4 - 64*a^2*b^3*c^4 - 3 
2*a^4*c^5 + 64*a^3*b*c^5 + 3*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c 
)*(a - b + c))*sqrt(b^2 - 4*a*c)*a^3*b^4 - 2*(b^2 - 4*a*c)*a^3*b^4 - 5*sqr 
t(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c) 
*a^2*b^5 + 6*(b^2 - 4*a*c)*a^2*b^5 - 3*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b 
^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c)*a*b^6 - 6*(b^2 - 4*a*c)*a*b^6 + 
 5*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 
4*a*c)*b^7 + 2*(b^2 - 4*a*c)*b^7 - 15*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^ 
2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c)*a^4*b^2*c + 10*(b^2 - 4*a*c)*a^4 
*b^2*c + 28*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sq 
rt(b^2 - 4*a*c)*a^3*b^3*c - 32*(b^2 - 4*a*c)*a^3*b^3*c + 27*sqrt(a^2 - a*b 
 + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c)*a^2*b^4*c 
+ 30*(b^2 - 4*a*c)*a^2*b^4*c - 38*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 
4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c)*a*b^5*c - 4*(b^2 - 4*a*c)*a*b^5*c - 
6*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4 
*a*c)*b^6*c - 4*(b^2 - 4*a*c)*b^6*c + 12*sqrt(a^2 - a*b + b*c - c^2 + s...
 

Mupad [B] (verification not implemented)

Time = 8.47 (sec) , antiderivative size = 45364, normalized size of antiderivative = 139.15 \[ \int \frac {\cos ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Too large to display} \] Input:

int(cos(x)^4/(a + b*cos(x) + c*cos(x)^2),x)
 

Output:

atan(((((2048*(12*a^3*c^11 - 28*a^4*c^10 - 44*a^5*c^9 + 72*a^6*c^8 + 88*a^ 
7*c^7 + 12*a^8*c^6 + b^5*c^9 - 4*b^6*c^8 + 10*b^7*c^7 - 20*b^8*c^6 + 29*b^ 
9*c^5 - 30*b^10*c^4 + 26*b^11*c^3 - 12*b^12*c^2 - 6*a*b^3*c^10 + 27*a*b^4* 
c^9 - 72*a*b^5*c^8 + 154*a*b^6*c^7 - 238*a*b^7*c^6 + 251*a*b^8*c^5 - 228*a 
*b^9*c^4 + 98*a*b^10*c^3 + 20*a*b^11*c^2 + 8*a^2*b*c^11 - 68*a^3*b*c^10 + 
112*a^4*b*c^9 + 100*a^5*b*c^8 - 200*a^6*b*c^7 - 96*a^7*b*c^6 - 47*a^2*b^2* 
c^10 + 145*a^2*b^3*c^9 - 354*a^2*b^4*c^8 + 612*a^2*b^5*c^7 - 655*a^2*b^6*c 
^6 + 635*a^2*b^7*c^5 - 202*a^2*b^8*c^4 - 222*a^2*b^9*c^3 + 4*a^2*b^10*c^2 
+ 239*a^3*b^2*c^9 - 524*a^3*b^3*c^8 + 536*a^3*b^4*c^7 - 564*a^3*b^5*c^6 - 
115*a^3*b^6*c^5 + 856*a^3*b^7*c^4 + 2*a^3*b^8*c^3 - 20*a^3*b^9*c^2 - 37*a^ 
4*b^2*c^8 + 9*a^4*b^3*c^7 + 583*a^4*b^4*c^6 - 1362*a^4*b^5*c^5 - 152*a^4*b 
^6*c^4 + 156*a^4*b^7*c^3 + 8*a^4*b^8*c^2 - 399*a^5*b^2*c^7 + 904*a^5*b^3*c 
^6 + 394*a^5*b^4*c^5 - 388*a^5*b^5*c^4 - 60*a^5*b^6*c^3 - 340*a^6*b^2*c^6 
+ 364*a^6*b^3*c^5 + 136*a^6*b^4*c^4 - 100*a^7*b^2*c^5))/c^8 + (((2048*(16* 
a^3*c^13 - 32*a^2*c^14 + 176*a^4*c^12 + 176*a^5*c^11 + 48*a^6*c^10 - 2*b^4 
*c^12 + 6*b^5*c^11 - 18*b^6*c^10 + 26*b^7*c^9 - 12*b^8*c^8 + 16*a*b^2*c^13 
 - 40*a*b^3*c^12 + 122*a*b^4*c^11 - 192*a*b^5*c^10 + 74*a*b^6*c^9 + 20*a*b 
^7*c^8 + 64*a^2*b*c^13 - 144*a^3*b*c^12 - 352*a^4*b*c^11 - 144*a^5*b*c^10 
- 204*a^2*b^2*c^12 + 388*a^2*b^3*c^11 - 50*a^2*b^4*c^10 - 182*a^2*b^5*c^9 
+ 4*a^2*b^6*c^8 - 260*a^3*b^2*c^11 + 496*a^3*b^3*c^10 + 10*a^3*b^4*c^9 ...
 

Reduce [F]

\[ \int \frac {\cos ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\int \frac {\cos \left (x \right )^{4}}{a +\cos \left (x \right ) b +c \cos \left (x \right )^{2}}d x \] Input:

int(cos(x)^4/(a+b*cos(x)+c*cos(x)^2),x)
 

Output:

int(cos(x)^4/(a+b*cos(x)+c*cos(x)^2),x)