\(\int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) [191]

Optimal result

Integrand size = 23, antiderivative size = 117 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\left (3 a^2-4 a b+8 b^2\right ) x}{8 a^3}-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^3 \sqrt {a+b} f}+\frac {(3 a-4 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f} \] Output:

1/8*(3*a^2-4*a*b+8*b^2)*x/a^3-b^(5/2)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2 
))/a^3/(a+b)^(1/2)/f+1/8*(3*a-4*b)*cos(f*x+e)*sin(f*x+e)/a^2/f+1/4*cos(f*x 
+e)^3*sin(f*x+e)/a/f
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {4 \left (3 a^2-4 a b+8 b^2\right ) (e+f x)-\frac {32 b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+8 a (a-b) \sin (2 (e+f x))+a^2 \sin (4 (e+f x))}{32 a^3 f} \] Input:

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

Output:

(4*(3*a^2 - 4*a*b + 8*b^2)*(e + f*x) - (32*b^(5/2)*ArcTan[(Sqrt[b]*Tan[e + 
 f*x])/Sqrt[a + b]])/Sqrt[a + b] + 8*a*(a - b)*Sin[2*(e + f*x)] + a^2*Sin[ 
4*(e + f*x)])/(32*a^3*f)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.21, 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, 25, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sec (e+f x)^4 \left (a+b \sec (e+f x)^2\right )}dx\)

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\frac {\frac {\left (3 a^2-4 a b+8 b^2\right ) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}-\frac {8 b^3 \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{2 a}+\frac {(3 a-4 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right )}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {\frac {\left (3 a^2-4 a b+8 b^2\right ) \arctan (\tan (e+f x))}{a}-\frac {8 b^3 \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{2 a}+\frac {(3 a-4 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right )}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{f}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\frac {\left (3 a^2-4 a b+8 b^2\right ) \arctan (\tan (e+f x))}{a}-\frac {8 b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}}{2 a}+\frac {(3 a-4 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right )}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{f}\)

Input:

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

Output:

(Tan[e + f*x]/(4*a*(1 + Tan[e + f*x]^2)^2) + ((((3*a^2 - 4*a*b + 8*b^2)*Ar 
cTan[Tan[e + f*x]])/a - (8*b^(5/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + 
b]])/(a*Sqrt[a + b]))/(2*a) + ((3*a - 4*b)*Tan[e + f*x])/(2*a*(1 + Tan[e + 
 f*x]^2)))/(4*a))/f
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 = 1.22 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.99

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (\frac {3}{8} a^{2}-\frac {1}{2} a b \right ) \tan \left (f x +e \right )^{3}+\left (-\frac {1}{2} a b +\frac {5}{8} a^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-4 a b +8 b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{a^{3}}-\frac {b^{3} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{3} \sqrt {\left (a +b \right ) b}}}{f}\)	\(116\)
default	\(\frac {\frac {\frac {\left (\frac {3}{8} a^{2}-\frac {1}{2} a b \right ) \tan \left (f x +e \right )^{3}+\left (-\frac {1}{2} a b +\frac {5}{8} a^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-4 a b +8 b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{a^{3}}-\frac {b^{3} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{3} \sqrt {\left (a +b \right ) b}}}{f}\)	\(116\)
risch	\(\frac {3 x}{8 a}-\frac {x b}{2 a^{2}}+\frac {x \,b^{2}}{a^{3}}-\frac {i {\mathrm e}^{2 i \left (f x +e \right )}}{8 a f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} b}{8 a^{2} f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} b}{8 a^{2} f}+\frac {\sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 \left (a +b \right ) f \,a^{3}}-\frac {\sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 \left (a +b \right ) f \,a^{3}}+\frac {\sin \left (4 f x +4 e \right )}{32 a f}\)	\(227\)

Input:

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

Output:

1/f*(1/a^3*(((3/8*a^2-1/2*a*b)*tan(f*x+e)^3+(-1/2*a*b+5/8*a^2)*tan(f*x+e)) 
/(1+tan(f*x+e)^2)^2+1/8*(3*a^2-4*a*b+8*b^2)*arctan(tan(f*x+e)))-b^3/a^3/(( 
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 = 343, normalized size of antiderivative = 2.93 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [\frac {2 \, b^{2} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) + {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} f x + {\left (2 \, a^{2} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, a^{3} f}, \frac {4 \, b^{2} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} f x + {\left (2 \, a^{2} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, a^{3} f}\right ] \] Input:

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

Output:

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

Sympy [F]

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

integrate(cos(f*x+e)**4/(a+b*sec(f*x+e)**2),x)
 

Output:

Integral(cos(e + f*x)**4/(a + b*sec(e + f*x)**2), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {8 \, b^{3} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{3}} - \frac {{\left (3 \, a - 4 \, b\right )} \tan \left (f x + e\right )^{3} + {\left (5 \, a - 4 \, b\right )} \tan \left (f x + e\right )}{a^{2} \tan \left (f x + e\right )^{4} + 2 \, a^{2} \tan \left (f x + e\right )^{2} + a^{2}} - \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )}}{a^{3}}}{8 \, f} \] Input:

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

Output:

-1/8*(8*b^3*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt((a + b)*b)*a^3) - 
 ((3*a - 4*b)*tan(f*x + e)^3 + (5*a - 4*b)*tan(f*x + e))/(a^2*tan(f*x + e) 
^4 + 2*a^2*tan(f*x + e)^2 + a^2) - (3*a^2 - 4*a*b + 8*b^2)*(f*x + e)/a^3)/ 
f
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.21 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {8 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b^{3}}{\sqrt {a b + b^{2}} a^{3}} - \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )}}{a^{3}} - \frac {3 \, a \tan \left (f x + e\right )^{3} - 4 \, b \tan \left (f x + e\right )^{3} + 5 \, a \tan \left (f x + e\right ) - 4 \, b \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2} a^{2}}}{8 \, f} \] Input:

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

Output:

-1/8*(8*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt( 
a*b + b^2)))*b^3/(sqrt(a*b + b^2)*a^3) - (3*a^2 - 4*a*b + 8*b^2)*(f*x + e) 
/a^3 - (3*a*tan(f*x + e)^3 - 4*b*tan(f*x + e)^3 + 5*a*tan(f*x + e) - 4*b*t 
an(f*x + e))/((tan(f*x + e)^2 + 1)^2*a^2))/f
 

Mupad [B] (verification not implemented)

Time = 16.45 (sec) , antiderivative size = 1114, normalized size of antiderivative = 9.52 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

((tan(e + f*x)*(5*a - 4*b))/(8*a^2) + (tan(e + f*x)^3*(3*a - 4*b))/(8*a^2) 
)/(f*(2*tan(e + f*x)^2 + tan(e + f*x)^4 + 1)) - (atan((((-b^5*(a + b))^(1/ 
2)*((tan(e + f*x)*(128*b^7 - 64*a*b^6 + 64*a^2*b^5 - 24*a^3*b^4 + 9*a^4*b^ 
3))/(64*a^4) - ((-b^5*(a + b))^(1/2)*((2*a^6*b^4 - (a^7*b^3)/2 + (3*a^8*b^ 
2)/2)/(2*a^6) - (tan(e + f*x)*(512*a^6*b^3 + 256*a^7*b^2)*(-b^5*(a + b))^( 
1/2))/(128*a^4*(a^3*b + a^4))))/(2*(a^3*b + a^4)))*1i)/(a^3*b + a^4) + ((- 
b^5*(a + b))^(1/2)*((tan(e + f*x)*(128*b^7 - 64*a*b^6 + 64*a^2*b^5 - 24*a^ 
3*b^4 + 9*a^4*b^3))/(64*a^4) + ((-b^5*(a + b))^(1/2)*((2*a^6*b^4 - (a^7*b^ 
3)/2 + (3*a^8*b^2)/2)/(2*a^6) + (tan(e + f*x)*(512*a^6*b^3 + 256*a^7*b^2)* 
(-b^5*(a + b))^(1/2))/(128*a^4*(a^3*b + a^4))))/(2*(a^3*b + a^4)))*1i)/(a^ 
3*b + a^4))/(((5*a*b^7)/4 - b^8 - (3*a^2*b^6)/4 + (9*a^3*b^5)/32)/a^6 + (( 
-b^5*(a + b))^(1/2)*((tan(e + f*x)*(128*b^7 - 64*a*b^6 + 64*a^2*b^5 - 24*a 
^3*b^4 + 9*a^4*b^3))/(64*a^4) - ((-b^5*(a + b))^(1/2)*((2*a^6*b^4 - (a^7*b 
^3)/2 + (3*a^8*b^2)/2)/(2*a^6) - (tan(e + f*x)*(512*a^6*b^3 + 256*a^7*b^2) 
*(-b^5*(a + b))^(1/2))/(128*a^4*(a^3*b + a^4))))/(2*(a^3*b + a^4))))/(a^3* 
b + a^4) - ((-b^5*(a + b))^(1/2)*((tan(e + f*x)*(128*b^7 - 64*a*b^6 + 64*a 
^2*b^5 - 24*a^3*b^4 + 9*a^4*b^3))/(64*a^4) + ((-b^5*(a + b))^(1/2)*((2*a^6 
*b^4 - (a^7*b^3)/2 + (3*a^8*b^2)/2)/(2*a^6) + (tan(e + f*x)*(512*a^6*b^3 + 
 256*a^7*b^2)*(-b^5*(a + b))^(1/2))/(128*a^4*(a^3*b + a^4))))/(2*(a^3*b + 
a^4))))/(a^3*b + a^4)))*(-b^5*(a + b))^(1/2)*1i)/(f*(a^3*b + a^4)) - (a...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.97 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {-8 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) b^{2}-8 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) b^{2}-2 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} a^{3}-2 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} a^{2} b +5 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a^{3}+\cos \left (f x +e \right ) \sin \left (f x +e \right ) a^{2} b -4 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,b^{2}+3 a^{3} e +3 a^{3} f x -a^{2} b e -a^{2} b f x +4 a \,b^{2} e +4 a \,b^{2} f x +8 b^{3} e +8 b^{3} f x}{8 a^{3} f \left (a +b \right )} \] Input:

int(cos(f*x+e)^4/(a+b*sec(f*x+e)^2),x)
 

Output:

( - 8*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sq 
rt(b))*b**2 - 8*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + s 
qrt(a))/sqrt(b))*b**2 - 2*cos(e + f*x)*sin(e + f*x)**3*a**3 - 2*cos(e + f* 
x)*sin(e + f*x)**3*a**2*b + 5*cos(e + f*x)*sin(e + f*x)*a**3 + cos(e + f*x 
)*sin(e + f*x)*a**2*b - 4*cos(e + f*x)*sin(e + f*x)*a*b**2 + 3*a**3*e + 3* 
a**3*f*x - a**2*b*e - a**2*b*f*x + 4*a*b**2*e + 4*a*b**2*f*x + 8*b**3*e + 
8*b**3*f*x)/(8*a**3*f*(a + b))