\(\int \frac {\cos ^2(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\) [203]

Optimal result

Integrand size = 23, antiderivative size = 142 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a-4 b) x}{2 a^3}+\frac {b^{3/2} (5 a+4 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^3 (a+b)^{3/2} f}+\frac {\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {b (a+2 b) \tan (e+f x)}{2 a^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )} \] Output:

1/2*(a-4*b)*x/a^3+1/2*b^(3/2)*(5*a+4*b)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1 
/2))/a^3/(a+b)^(3/2)/f+1/2*cos(f*x+e)*sin(f*x+e)/a/f/(a+b+b*tan(f*x+e)^2)+ 
1/2*b*(a+2*b)*tan(f*x+e)/a^2/(a+b)/f/(a+b+b*tan(f*x+e)^2)
 

Mathematica [A] (verified)

Time = 1.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {2 (a-4 b) (e+f x)+\frac {2 b^{3/2} (5 a+4 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}+\left (a+\frac {2 a b^2}{(a+b) (a+2 b+a \cos (2 (e+f x)))}\right ) \sin (2 (e+f x))}{4 a^3 f} \] Input:

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

Output:

(2*(a - 4*b)*(e + f*x) + (2*b^(3/2)*(5*a + 4*b)*ArcTan[(Sqrt[b]*Tan[e + f* 
x])/Sqrt[a + b]])/(a + b)^(3/2) + (a + (2*a*b^2)/((a + b)*(a + 2*b + a*Cos 
[2*(e + f*x)])))*Sin[2*(e + f*x)])/(4*a^3*f)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4634, 316, 25, 402, 27, 397, 216, 218}

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.

\(\displaystyle \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sec (e+f x)^2 \left (a+b \sec (e+f x)^2\right )^2}dx\)

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 316

\(\displaystyle \frac {\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\int -\frac {3 b \tan ^2(e+f x)+a-b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{2 a}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {3 b \tan ^2(e+f x)+a-b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\frac {\frac {b^2 (5 a+4 b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}+\frac {(a-4 b) (a+b) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}}{a (a+b)}+\frac {b (a+2 b) \tan (e+f x)}{a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {\frac {b^2 (5 a+4 b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}+\frac {(a-4 b) (a+b) \arctan (\tan (e+f x))}{a}}{a (a+b)}+\frac {b (a+2 b) \tan (e+f x)}{a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\frac {b^{3/2} (5 a+4 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}+\frac {(a-4 b) (a+b) \arctan (\tan (e+f x))}{a}}{a (a+b)}+\frac {b (a+2 b) \tan (e+f x)}{a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

Input:

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

Output:

(Tan[e + f*x]/(2*a*(1 + Tan[e + f*x]^2)*(a + b + b*Tan[e + f*x]^2)) + (((( 
a - 4*b)*(a + b)*ArcTan[Tan[e + f*x]])/a + (b^(3/2)*(5*a + 4*b)*ArcTan[(Sq 
rt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/(a*(a + b)) + (b*(a + 2 
*b)*Tan[e + f*x])/(a*(a + b)*(a + b + b*Tan[e + f*x]^2)))/(2*a))/f
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

rule 397

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

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

rule 3042

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

rule 4634

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) 
)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f 
Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), 
x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ 
[m/2] && IntegerQ[n/2]
 

Maple [A] (verified)

Time = 1.45 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {\frac {\frac {a \tan \left (f x +e \right )}{2+2 \tan \left (f x +e \right )^{2}}+\frac {\left (a -4 b \right ) \arctan \left (\tan \left (f x +e \right )\right )}{2}}{a^{3}}+\frac {b^{2} \left (\frac {a \tan \left (f x +e \right )}{2 \left (a +b \right ) \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (5 a +4 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{3}}}{f}\)	\(120\)
default	\(\frac {\frac {\frac {a \tan \left (f x +e \right )}{2+2 \tan \left (f x +e \right )^{2}}+\frac {\left (a -4 b \right ) \arctan \left (\tan \left (f x +e \right )\right )}{2}}{a^{3}}+\frac {b^{2} \left (\frac {a \tan \left (f x +e \right )}{2 \left (a +b \right ) \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (5 a +4 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{3}}}{f}\)	\(120\)
risch	\(\frac {x}{2 a^{2}}-\frac {2 x b}{a^{3}}-\frac {i {\mathrm e}^{2 i \left (f x +e \right )}}{8 a^{2} f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a^{2} f}+\frac {i b^{2} \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}{a^{3} \left (a +b \right ) f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}+\frac {5 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right ) b}{4 \left (a +b \right )^{2} f \,a^{2}}+\frac {\sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{\left (a +b \right )^{2} f \,a^{3}}-\frac {5 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right ) b}{4 \left (a +b \right )^{2} f \,a^{2}}-\frac {\sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{\left (a +b \right )^{2} f \,a^{3}}\)	\(353\)

Input:

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

Output:

1/f*(1/a^3*(1/2*a*tan(f*x+e)/(1+tan(f*x+e)^2)+1/2*(a-4*b)*arctan(tan(f*x+e 
)))+b^2/a^3*(1/2*a/(a+b)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)+1/2*(5*a+4*b)/(a+ 
b)/((a+b)*b)^(1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 544, normalized size of antiderivative = 3.83 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [\frac {4 \, {\left (a^{3} - 3 \, a^{2} b - 4 \, a b^{2}\right )} f x \cos \left (f x + e\right )^{2} + 4 \, {\left (a^{2} b - 3 \, a b^{2} - 4 \, b^{3}\right )} f x + {\left (5 \, a b^{2} + 4 \, b^{3} + {\left (5 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) + 4 \, {\left ({\left (a^{3} + a^{2} b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} b + 2 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, {\left ({\left (a^{5} + a^{4} b\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b + a^{3} b^{2}\right )} f\right )}}, \frac {2 \, {\left (a^{3} - 3 \, a^{2} b - 4 \, a b^{2}\right )} f x \cos \left (f x + e\right )^{2} + 2 \, {\left (a^{2} b - 3 \, a b^{2} - 4 \, b^{3}\right )} f x - {\left (5 \, a b^{2} + 4 \, b^{3} + {\left (5 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + 2 \, {\left ({\left (a^{3} + a^{2} b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} b + 2 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, {\left ({\left (a^{5} + a^{4} b\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b + a^{3} b^{2}\right )} f\right )}}\right ] \] Input:

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

Output:

[1/8*(4*(a^3 - 3*a^2*b - 4*a*b^2)*f*x*cos(f*x + e)^2 + 4*(a^2*b - 3*a*b^2 
- 4*b^3)*f*x + (5*a*b^2 + 4*b^3 + (5*a^2*b + 4*a*b^2)*cos(f*x + e)^2)*sqrt 
(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2) 
*cos(f*x + e)^2 - 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*co 
s(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a 
*b*cos(f*x + e)^2 + b^2)) + 4*((a^3 + a^2*b)*cos(f*x + e)^3 + (a^2*b + 2*a 
*b^2)*cos(f*x + e))*sin(f*x + e))/((a^5 + a^4*b)*f*cos(f*x + e)^2 + (a^4*b 
 + a^3*b^2)*f), 1/4*(2*(a^3 - 3*a^2*b - 4*a*b^2)*f*x*cos(f*x + e)^2 + 2*(a 
^2*b - 3*a*b^2 - 4*b^3)*f*x - (5*a*b^2 + 4*b^3 + (5*a^2*b + 4*a*b^2)*cos(f 
*x + e)^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt( 
b/(a + b))/(b*cos(f*x + e)*sin(f*x + e))) + 2*((a^3 + a^2*b)*cos(f*x + e)^ 
3 + (a^2*b + 2*a*b^2)*cos(f*x + e))*sin(f*x + e))/((a^5 + a^4*b)*f*cos(f*x 
 + e)^2 + (a^4*b + a^3*b^2)*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.23 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {{\left (5 \, a b^{2} + 4 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{4} + a^{3} b\right )} \sqrt {{\left (a + b\right )} b}} + \frac {{\left (a b + 2 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + {\left (a^{2} + 2 \, a b + 2 \, b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{3} b + a^{2} b^{2}\right )} \tan \left (f x + e\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2} + {\left (a^{4} + 3 \, a^{3} b + 2 \, a^{2} b^{2}\right )} \tan \left (f x + e\right )^{2}} + \frac {{\left (f x + e\right )} {\left (a - 4 \, b\right )}}{a^{3}}}{2 \, f} \] Input:

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

Output:

1/2*((5*a*b^2 + 4*b^3)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^4 + a^3* 
b)*sqrt((a + b)*b)) + ((a*b + 2*b^2)*tan(f*x + e)^3 + (a^2 + 2*a*b + 2*b^2 
)*tan(f*x + e))/((a^3*b + a^2*b^2)*tan(f*x + e)^4 + a^4 + 2*a^3*b + a^2*b^ 
2 + (a^4 + 3*a^3*b + 2*a^2*b^2)*tan(f*x + e)^2) + (f*x + e)*(a - 4*b)/a^3) 
/f
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.35 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {{\left (5 \, a b^{2} + 4 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{4} + a^{3} b\right )} \sqrt {a b + b^{2}}} + \frac {a b \tan \left (f x + e\right )^{3} + 2 \, b^{2} \tan \left (f x + e\right )^{3} + a^{2} \tan \left (f x + e\right ) + 2 \, a b \tan \left (f x + e\right ) + 2 \, b^{2} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{4} + a \tan \left (f x + e\right )^{2} + 2 \, b \tan \left (f x + e\right )^{2} + a + b\right )} {\left (a^{3} + a^{2} b\right )}} + \frac {{\left (f x + e\right )} {\left (a - 4 \, b\right )}}{a^{3}}}{2 \, f} \] Input:

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

Output:

1/2*((5*a*b^2 + 4*b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan 
(f*x + e)/sqrt(a*b + b^2)))/((a^4 + a^3*b)*sqrt(a*b + b^2)) + (a*b*tan(f*x 
 + e)^3 + 2*b^2*tan(f*x + e)^3 + a^2*tan(f*x + e) + 2*a*b*tan(f*x + e) + 2 
*b^2*tan(f*x + e))/((b*tan(f*x + e)^4 + a*tan(f*x + e)^2 + 2*b*tan(f*x + e 
)^2 + a + b)*(a^3 + a^2*b)) + (f*x + e)*(a - 4*b)/a^3)/f
 

Mupad [B] (verification not implemented)

Time = 18.40 (sec) , antiderivative size = 2401, normalized size of antiderivative = 16.91 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\text {Too large to display} \] Input:

int(cos(e + f*x)^2/(a + b/cos(e + f*x)^2)^2,x)
 

Output:

((tan(e + f*x)*(2*a*b + a^2 + 2*b^2))/(2*a^2*(a + b)) + (b*tan(e + f*x)^3* 
(a + 2*b))/(2*a^2*(a + b)))/(f*(a + b + b*tan(e + f*x)^4 + tan(e + f*x)^2* 
(a + 2*b))) - (atan(((((tan(e + f*x)*(64*a*b^6 + 32*b^7 + 26*a^2*b^5 - 6*a 
^3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^6 + a^4*b^2)) - (((4*a^6*b^5 + 8*a^7*b^ 
4 + 2*a^8*b^3 - 2*a^9*b^2)/(2*a^7*b + a^8 + a^6*b^2) - (tan(e + f*x)*(a*1i 
 - b*4i)*(32*a^6*b^5 + 80*a^7*b^4 + 64*a^8*b^3 + 16*a^9*b^2))/(8*a^3*(2*a^ 
5*b + a^6 + a^4*b^2)))*(a*1i - b*4i))/(4*a^3))*(a*1i - b*4i)*1i)/(4*a^3) + 
 (((tan(e + f*x)*(64*a*b^6 + 32*b^7 + 26*a^2*b^5 - 6*a^3*b^4 + a^4*b^3))/( 
2*(2*a^5*b + a^6 + a^4*b^2)) + (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^8*b^3 - 2*a^ 
9*b^2)/(2*a^7*b + a^8 + a^6*b^2) + (tan(e + f*x)*(a*1i - b*4i)*(32*a^6*b^5 
 + 80*a^7*b^4 + 64*a^8*b^3 + 16*a^9*b^2))/(8*a^3*(2*a^5*b + a^6 + a^4*b^2) 
))*(a*1i - b*4i))/(4*a^3))*(a*1i - b*4i)*1i)/(4*a^3))/((12*a*b^6 + 8*b^7 + 
 (3*a^2*b^5)/2 - (5*a^3*b^4)/4)/(2*a^7*b + a^8 + a^6*b^2) - (((tan(e + f*x 
)*(64*a*b^6 + 32*b^7 + 26*a^2*b^5 - 6*a^3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^ 
6 + a^4*b^2)) - (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^8*b^3 - 2*a^9*b^2)/(2*a^7*b 
 + a^8 + a^6*b^2) - (tan(e + f*x)*(a*1i - b*4i)*(32*a^6*b^5 + 80*a^7*b^4 + 
 64*a^8*b^3 + 16*a^9*b^2))/(8*a^3*(2*a^5*b + a^6 + a^4*b^2)))*(a*1i - b*4i 
))/(4*a^3))*(a*1i - b*4i))/(4*a^3) + (((tan(e + f*x)*(64*a*b^6 + 32*b^7 + 
26*a^2*b^5 - 6*a^3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^6 + a^4*b^2)) + (((4*a^ 
6*b^5 + 8*a^7*b^4 + 2*a^8*b^3 - 2*a^9*b^2)/(2*a^7*b + a^8 + a^6*b^2) + ...
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 792, normalized size of antiderivative = 5.58 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(5*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt( 
b))*sin(e + f*x)**2*a**2*b + 4*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan(( 
e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a*b**2 - 5*sqrt(b)*sqrt(a 
+ b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**2*b - 9*sqr 
t(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a* 
b**2 - 4*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a)) 
/sqrt(b))*b**3 + 5*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) 
+ sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**2*b + 4*sqrt(b)*sqrt(a + b)*atan((s 
qrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**2*a*b**2 - 5 
*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b) 
)*a**2*b - 9*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt 
(a))/sqrt(b))*a*b**2 - 4*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f* 
x)/2) + sqrt(a))/sqrt(b))*b**3 + cos(e + f*x)*sin(e + f*x)**3*a**4 + 2*cos 
(e + f*x)*sin(e + f*x)**3*a**3*b + cos(e + f*x)*sin(e + f*x)**3*a**2*b**2 
- cos(e + f*x)*sin(e + f*x)*a**4 - 3*cos(e + f*x)*sin(e + f*x)*a**3*b - 4* 
cos(e + f*x)*sin(e + f*x)*a**2*b**2 - 2*cos(e + f*x)*sin(e + f*x)*a*b**3 + 
 sin(e + f*x)**2*a**4*e + sin(e + f*x)**2*a**4*f*x - 2*sin(e + f*x)**2*a** 
3*b*e - 2*sin(e + f*x)**2*a**3*b*f*x - 7*sin(e + f*x)**2*a**2*b**2*e - 7*s 
in(e + f*x)**2*a**2*b**2*f*x - 4*sin(e + f*x)**2*a*b**3*e - 4*sin(e + f*x) 
**2*a*b**3*f*x - a**4*e - a**4*f*x + a**3*b*e + a**3*b*f*x + 9*a**2*b**...