\(\int \frac {\cos ^2(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [216]

Optimal result

Integrand size = 23, antiderivative size = 201 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(a-6 b) x}{2 a^4}+\frac {b^{3/2} \left (35 a^2+56 a b+24 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^4 (a+b)^{5/2} f}+\frac {\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {b (2 a+3 b) \tan (e+f x)}{4 a^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {b (4 a+3 b) (a+4 b) \tan (e+f x)}{8 a^3 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )} \] Output:

1/2*(a-6*b)*x/a^4+1/8*b^(3/2)*(35*a^2+56*a*b+24*b^2)*arctan(b^(1/2)*tan(f* 
x+e)/(a+b)^(1/2))/a^4/(a+b)^(5/2)/f+1/2*cos(f*x+e)*sin(f*x+e)/a/f/(a+b+b*t 
an(f*x+e)^2)^2+1/4*b*(2*a+3*b)*tan(f*x+e)/a^2/(a+b)/f/(a+b+b*tan(f*x+e)^2) 
^2+1/8*b*(4*a+3*b)*(a+4*b)*tan(f*x+e)/a^3/(a+b)^2/f/(a+b+b*tan(f*x+e)^2)
 

Mathematica [A] (verified)

Time = 2.99 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {4 (a-6 b) (e+f x)+\frac {b^{3/2} \left (35 a^2+56 a b+24 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}+a \left (2+\frac {13 a b^2}{(a+b)^2 (a+2 b+a \cos (2 (e+f x)))}+\frac {2 b^3 (3 a+8 b+5 a \cos (2 (e+f x)))}{(a+b)^2 (a+2 b+a \cos (2 (e+f x)))^2}\right ) \sin (2 (e+f x))}{8 a^4 f} \] Input:

Integrate[Cos[e + f*x]^2/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

(4*(a - 6*b)*(e + f*x) + (b^(3/2)*(35*a^2 + 56*a*b + 24*b^2)*ArcTan[(Sqrt[ 
b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(5/2) + a*(2 + (13*a*b^2)/((a + b)^ 
2*(a + 2*b + a*Cos[2*(e + f*x)])) + (2*b^3*(3*a + 8*b + 5*a*Cos[2*(e + f*x 
)]))/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2))*Sin[2*(e + f*x)])/(8*a^ 
4*f)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.19, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 4634, 316, 25, 402, 27, 402, 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.

\(\displaystyle \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 316

\(\displaystyle \frac {\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\int -\frac {5 b \tan ^2(e+f x)+a-b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^3}d\tan (e+f x)}{2 a}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {5 b \tan ^2(e+f x)+a-b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^3}d\tan (e+f x)}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {\int \frac {2 \left (2 a^2-4 b a-3 b^2+3 b (2 a+3 b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{4 a (a+b)}+\frac {b (2 a+3 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \frac {2 a^2-4 b a-3 b^2+3 b (2 a+3 b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{2 a (a+b)}+\frac {b (2 a+3 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {\frac {\int \frac {4 a^3-12 b a^2-25 b^2 a-12 b^3+b (4 a+3 b) (a+4 b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{2 a (a+b)}+\frac {b (4 a+3 b) (a+4 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{2 a (a+b)}+\frac {b (2 a+3 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\frac {\frac {\frac {b^2 \left (35 a^2+56 a b+24 b^2\right ) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}+\frac {4 (a-6 b) (a+b)^2 \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}}{2 a (a+b)}+\frac {b (4 a+3 b) (a+4 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{2 a (a+b)}+\frac {b (2 a+3 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {\frac {\frac {b^2 \left (35 a^2+56 a b+24 b^2\right ) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}+\frac {4 (a-6 b) (a+b)^2 \arctan (\tan (e+f x))}{a}}{2 a (a+b)}+\frac {b (4 a+3 b) (a+4 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{2 a (a+b)}+\frac {b (2 a+3 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\frac {\frac {b^{3/2} \left (35 a^2+56 a b+24 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}+\frac {4 (a-6 b) (a+b)^2 \arctan (\tan (e+f x))}{a}}{2 a (a+b)}+\frac {b (4 a+3 b) (a+4 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{2 a (a+b)}+\frac {b (2 a+3 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

Input:

Int[Cos[e + f*x]^2/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

(Tan[e + f*x]/(2*a*(1 + Tan[e + f*x]^2)*(a + b + b*Tan[e + f*x]^2)^2) + (( 
b*(2*a + 3*b)*Tan[e + f*x])/(2*a*(a + b)*(a + b + b*Tan[e + f*x]^2)^2) + ( 
((4*(a - 6*b)*(a + b)^2*ArcTan[Tan[e + f*x]])/a + (b^(3/2)*(35*a^2 + 56*a* 
b + 24*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/( 
2*a*(a + b)) + (b*(4*a + 3*b)*(a + 4*b)*Tan[e + f*x])/(2*a*(a + b)*(a + b 
+ b*Tan[e + f*x]^2)))/(2*a*(a + b)))/(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 = 3.14 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\frac {b^{2} \left (\frac {\frac {a b \left (11 a +8 b \right ) \tan \left (f x +e \right )^{3}}{8 a^{2}+16 a b +8 b^{2}}+\frac {\left (13 a +8 b \right ) a \tan \left (f x +e \right )}{8 a +8 b}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (35 a^{2}+56 a b +24 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{4}}+\frac {\frac {a \tan \left (f x +e \right )}{2+2 \tan \left (f x +e \right )^{2}}+\frac {\left (a -6 b \right ) \arctan \left (\tan \left (f x +e \right )\right )}{2}}{a^{4}}}{f}\)	\(177\)
default	\(\frac {\frac {b^{2} \left (\frac {\frac {a b \left (11 a +8 b \right ) \tan \left (f x +e \right )^{3}}{8 a^{2}+16 a b +8 b^{2}}+\frac {\left (13 a +8 b \right ) a \tan \left (f x +e \right )}{8 a +8 b}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (35 a^{2}+56 a b +24 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{4}}+\frac {\frac {a \tan \left (f x +e \right )}{2+2 \tan \left (f x +e \right )^{2}}+\frac {\left (a -6 b \right ) \arctan \left (\tan \left (f x +e \right )\right )}{2}}{a^{4}}}{f}\)	\(177\)
risch	\(\frac {x}{2 a^{3}}-\frac {3 x b}{a^{4}}-\frac {i {\mathrm e}^{2 i \left (f x +e \right )}}{8 a^{3} f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a^{3} f}+\frac {i b^{2} \left (13 a^{3} {\mathrm e}^{6 i \left (f x +e \right )}+40 a^{2} b \,{\mathrm e}^{6 i \left (f x +e \right )}+24 a \,b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+39 a^{3} {\mathrm e}^{4 i \left (f x +e \right )}+134 a^{2} b \,{\mathrm e}^{4 i \left (f x +e \right )}+184 a \,b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+80 b^{3} {\mathrm e}^{4 i \left (f x +e \right )}+39 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}+104 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{2} b +56 \,{\mathrm e}^{2 i \left (f x +e \right )} a \,b^{2}+13 a^{3}+10 a^{2} b \right )}{4 a^{4} \left (a +b \right )^{2} 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 )^{2}}+\frac {35 \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}{16 \left (a +b \right )^{3} f \,a^{2}}+\frac {7 \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^{2}}{2 \left (a +b \right )^{3} f \,a^{3}}+\frac {3 \sqrt {-\left (a +b \right ) b}\, b^{3} \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 )}{2 \left (a +b \right )^{3} f \,a^{4}}-\frac {35 \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}{16 \left (a +b \right )^{3} f \,a^{2}}-\frac {7 \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^{2}}{2 \left (a +b \right )^{3} f \,a^{3}}-\frac {3 \sqrt {-\left (a +b \right ) b}\, b^{3} \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 )}{2 \left (a +b \right )^{3} f \,a^{4}}\)	\(600\)

Input:

int(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)
 

Output:

1/f*(b^2/a^4*((1/8*a*b*(11*a+8*b)/(a^2+2*a*b+b^2)*tan(f*x+e)^3+1/8*(13*a+8 
*b)*a/(a+b)*tan(f*x+e))/(a+b+b*tan(f*x+e)^2)^2+1/8*(35*a^2+56*a*b+24*b^2)/ 
(a^2+2*a*b+b^2)/((a+b)*b)^(1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2)))+1/a^ 
4*(1/2*a*tan(f*x+e)/(1+tan(f*x+e)^2)+1/2*(a-6*b)*arctan(tan(f*x+e))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 442 vs. \(2 (183) = 366\).

Time = 0.16 (sec) , antiderivative size = 970, normalized size of antiderivative = 4.83 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
 

Output:

[1/32*(16*(a^5 - 4*a^4*b - 11*a^3*b^2 - 6*a^2*b^3)*f*x*cos(f*x + e)^4 + 32 
*(a^4*b - 4*a^3*b^2 - 11*a^2*b^3 - 6*a*b^4)*f*x*cos(f*x + e)^2 + 16*(a^3*b 
^2 - 4*a^2*b^3 - 11*a*b^4 - 6*b^5)*f*x + (35*a^2*b^3 + 56*a*b^4 + 24*b^5 + 
 (35*a^4*b + 56*a^3*b^2 + 24*a^2*b^3)*cos(f*x + e)^4 + 2*(35*a^3*b^2 + 56* 
a^2*b^3 + 24*a*b^4)*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)*cos(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*(4*( 
a^5 + 2*a^4*b + a^3*b^2)*cos(f*x + e)^5 + (8*a^4*b + 29*a^3*b^2 + 18*a^2*b 
^3)*cos(f*x + e)^3 + (4*a^3*b^2 + 19*a^2*b^3 + 12*a*b^4)*cos(f*x + e))*sin 
(f*x + e))/((a^8 + 2*a^7*b + a^6*b^2)*f*cos(f*x + e)^4 + 2*(a^7*b + 2*a^6* 
b^2 + a^5*b^3)*f*cos(f*x + e)^2 + (a^6*b^2 + 2*a^5*b^3 + a^4*b^4)*f), 1/16 
*(8*(a^5 - 4*a^4*b - 11*a^3*b^2 - 6*a^2*b^3)*f*x*cos(f*x + e)^4 + 16*(a^4* 
b - 4*a^3*b^2 - 11*a^2*b^3 - 6*a*b^4)*f*x*cos(f*x + e)^2 + 8*(a^3*b^2 - 4* 
a^2*b^3 - 11*a*b^4 - 6*b^5)*f*x - (35*a^2*b^3 + 56*a*b^4 + 24*b^5 + (35*a^ 
4*b + 56*a^3*b^2 + 24*a^2*b^3)*cos(f*x + e)^4 + 2*(35*a^3*b^2 + 56*a^2*b^3 
 + 24*a*b^4)*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*(4*(a^5 + 
2*a^4*b + a^3*b^2)*cos(f*x + e)^5 + (8*a^4*b + 29*a^3*b^2 + 18*a^2*b^3)*co 
s(f*x + e)^3 + (4*a^3*b^2 + 19*a^2*b^3 + 12*a*b^4)*cos(f*x + e))*sin(f*...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \] Input:

integrate(cos(f*x+e)**2/(a+b*sec(f*x+e)**2)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.68 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {{\left (35 \, a^{2} b^{2} + 56 \, a b^{3} + 24 \, b^{4}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {{\left (4 \, a^{2} b^{2} + 19 \, a b^{3} + 12 \, b^{4}\right )} \tan \left (f x + e\right )^{5} + {\left (8 \, a^{3} b + 37 \, a^{2} b^{2} + 56 \, a b^{3} + 24 \, b^{4}\right )} \tan \left (f x + e\right )^{3} + {\left (4 \, a^{4} + 16 \, a^{3} b + 37 \, a^{2} b^{2} + 37 \, a b^{3} + 12 \, b^{4}\right )} \tan \left (f x + e\right )}{a^{7} + 4 \, a^{6} b + 6 \, a^{5} b^{2} + 4 \, a^{4} b^{3} + a^{3} b^{4} + {\left (a^{5} b^{2} + 2 \, a^{4} b^{3} + a^{3} b^{4}\right )} \tan \left (f x + e\right )^{6} + {\left (2 \, a^{6} b + 7 \, a^{5} b^{2} + 8 \, a^{4} b^{3} + 3 \, a^{3} b^{4}\right )} \tan \left (f x + e\right )^{4} + {\left (a^{7} + 6 \, a^{6} b + 12 \, a^{5} b^{2} + 10 \, a^{4} b^{3} + 3 \, a^{3} b^{4}\right )} \tan \left (f x + e\right )^{2}} + \frac {4 \, {\left (f x + e\right )} {\left (a - 6 \, b\right )}}{a^{4}}}{8 \, f} \] Input:

integrate(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
 

Output:

1/8*((35*a^2*b^2 + 56*a*b^3 + 24*b^4)*arctan(b*tan(f*x + e)/sqrt((a + b)*b 
))/((a^6 + 2*a^5*b + a^4*b^2)*sqrt((a + b)*b)) + ((4*a^2*b^2 + 19*a*b^3 + 
12*b^4)*tan(f*x + e)^5 + (8*a^3*b + 37*a^2*b^2 + 56*a*b^3 + 24*b^4)*tan(f* 
x + e)^3 + (4*a^4 + 16*a^3*b + 37*a^2*b^2 + 37*a*b^3 + 12*b^4)*tan(f*x + e 
))/(a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4 + (a^5*b^2 + 2*a^4*b^3 
 + a^3*b^4)*tan(f*x + e)^6 + (2*a^6*b + 7*a^5*b^2 + 8*a^4*b^3 + 3*a^3*b^4) 
*tan(f*x + e)^4 + (a^7 + 6*a^6*b + 12*a^5*b^2 + 10*a^4*b^3 + 3*a^3*b^4)*ta 
n(f*x + e)^2) + 4*(f*x + e)*(a - 6*b)/a^4)/f
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.13 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {{\left (35 \, a^{2} b^{2} + 56 \, a b^{3} + 24 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} \sqrt {a b + b^{2}}} + \frac {11 \, a b^{3} \tan \left (f x + e\right )^{3} + 8 \, b^{4} \tan \left (f x + e\right )^{3} + 13 \, a^{2} b^{2} \tan \left (f x + e\right ) + 21 \, a b^{3} \tan \left (f x + e\right ) + 8 \, b^{4} \tan \left (f x + e\right )}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} + \frac {4 \, {\left (f x + e\right )} {\left (a - 6 \, b\right )}}{a^{4}} + \frac {4 \, \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )} a^{3}}}{8 \, f} \] Input:

integrate(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")
 

Output:

1/8*((35*a^2*b^2 + 56*a*b^3 + 24*b^4)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) 
 + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/((a^6 + 2*a^5*b + a^4*b^2)*sqrt 
(a*b + b^2)) + (11*a*b^3*tan(f*x + e)^3 + 8*b^4*tan(f*x + e)^3 + 13*a^2*b^ 
2*tan(f*x + e) + 21*a*b^3*tan(f*x + e) + 8*b^4*tan(f*x + e))/((a^5 + 2*a^4 
*b + a^3*b^2)*(b*tan(f*x + e)^2 + a + b)^2) + 4*(f*x + e)*(a - 6*b)/a^4 + 
4*tan(f*x + e)/((tan(f*x + e)^2 + 1)*a^3))/f
 

Mupad [B] (verification not implemented)

Time = 19.59 (sec) , antiderivative size = 3708, normalized size of antiderivative = 18.45 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \] Input:

int(cos(e + f*x)^2/(a + b/cos(e + f*x)^2)^3,x)
 

Output:

((tan(e + f*x)^5*(19*a*b^3 + 12*b^4 + 4*a^2*b^2))/(8*a^3*(a + b)^2) + (tan 
(e + f*x)*(25*a*b^2 + 12*a^2*b + 4*a^3 + 12*b^3))/(8*a^3*(a + b)) + (b*tan 
(e + f*x)^3*(56*a*b^2 + 37*a^2*b + 8*a^3 + 24*b^3))/(8*a^3*(a + b)^2))/(f* 
(2*a*b + tan(e + f*x)^2*(4*a*b + a^2 + 3*b^2) + a^2 + b^2 + tan(e + f*x)^4 
*(2*a*b + 3*b^2) + b^2*tan(e + f*x)^6)) + (atan(((((((6*a^8*b^7 + (49*a^9* 
b^6)/2 + 37*a^10*b^5 + (45*a^11*b^4)/2 + 2*a^12*b^3 - 2*a^13*b^2)/(4*a^12* 
b + a^13 + a^9*b^4 + 4*a^10*b^3 + 6*a^11*b^2) - (tan(e + f*x)*(a*1i - b*6i 
)*(512*a^8*b^7 + 2304*a^9*b^6 + 4096*a^10*b^5 + 3584*a^11*b^4 + 1536*a^12* 
b^3 + 256*a^13*b^2))/(128*a^4*(4*a^9*b + a^10 + a^6*b^4 + 4*a^7*b^3 + 6*a^ 
8*b^2)))*(a*1i - b*6i))/(4*a^4) - (tan(e + f*x)*(4800*a*b^8 + 1152*b^9 + 7 
520*a^2*b^7 + 5136*a^3*b^6 + 1129*a^4*b^5 - 128*a^5*b^4 + 16*a^6*b^3))/(32 
*(4*a^9*b + a^10 + a^6*b^4 + 4*a^7*b^3 + 6*a^8*b^2)))*(a*1i - b*6i)*1i)/(4 
*a^4) - (((((6*a^8*b^7 + (49*a^9*b^6)/2 + 37*a^10*b^5 + (45*a^11*b^4)/2 + 
2*a^12*b^3 - 2*a^13*b^2)/(4*a^12*b + a^13 + a^9*b^4 + 4*a^10*b^3 + 6*a^11* 
b^2) + (tan(e + f*x)*(a*1i - b*6i)*(512*a^8*b^7 + 2304*a^9*b^6 + 4096*a^10 
*b^5 + 3584*a^11*b^4 + 1536*a^12*b^3 + 256*a^13*b^2))/(128*a^4*(4*a^9*b + 
a^10 + a^6*b^4 + 4*a^7*b^3 + 6*a^8*b^2)))*(a*1i - b*6i))/(4*a^4) + (tan(e 
+ f*x)*(4800*a*b^8 + 1152*b^9 + 7520*a^2*b^7 + 5136*a^3*b^6 + 1129*a^4*b^5 
 - 128*a^5*b^4 + 16*a^6*b^3))/(32*(4*a^9*b + a^10 + a^6*b^4 + 4*a^7*b^3 + 
6*a^8*b^2)))*(a*1i - b*6i)*1i)/(4*a^4))/(((405*a*b^8)/4 + 27*b^9 + (261...
 

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 1949, normalized size of antiderivative = 9.70 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

int(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x)
 

Output:

(35*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt 
(b))*sin(e + f*x)**4*a**4*b + 56*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan 
((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**3*b**2 + 24*sqrt(b)*s 
qrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + 
f*x)**4*a**2*b**3 - 70*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x) 
/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**4*b - 182*sqrt(b)*sqrt(a + b)*a 
tan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**3 
*b**2 - 160*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**3 - 48*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**4 + 35*s 
qrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))* 
a**4*b + 126*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt 
(a))/sqrt(b))*a**3*b**2 + 171*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e 
 + f*x)/2) - sqrt(a))/sqrt(b))*a**2*b**3 + 104*sqrt(b)*sqrt(a + b)*atan((s 
qrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a*b**4 + 24*sqrt(b)*sqrt(a 
 + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*b**5 + 35*sqr 
t(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*si 
n(e + f*x)**4*a**4*b + 56*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f 
*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**3*b**2 + 24*sqrt(b)*sqrt(a + 
 b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)...