\(\int \frac {\cos ^4(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\) [204]

Optimal result

Integrand size = 23, antiderivative size = 203 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\left (3 a^2-8 a b+24 b^2\right ) x}{8 a^4}-\frac {b^{5/2} (7 a+6 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^4 (a+b)^{3/2} f}+\frac {3 (a-2 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {(a-3 b) b (3 a+4 b) \tan (e+f x)}{8 a^3 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )} \] Output:

1/8*(3*a^2-8*a*b+24*b^2)*x/a^4-1/2*b^(5/2)*(7*a+6*b)*arctan(b^(1/2)*tan(f* 
x+e)/(a+b)^(1/2))/a^4/(a+b)^(3/2)/f+3/8*(a-2*b)*cos(f*x+e)*sin(f*x+e)/a^2/ 
f/(a+b+b*tan(f*x+e)^2)+1/4*cos(f*x+e)^3*sin(f*x+e)/a/f/(a+b+b*tan(f*x+e)^2 
)+1/8*(a-3*b)*b*(3*a+4*b)*tan(f*x+e)/a^3/(a+b)/f/(a+b+b*tan(f*x+e)^2)
 

Mathematica [A] (verified)

Time = 1.50 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.68 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {4 \left (3 a^2-8 a b+24 b^2\right ) (e+f x)-\frac {16 b^{5/2} (7 a+6 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}+8 a (a-2 b) \sin (2 (e+f x))-\frac {16 a b^3 \sin (2 (e+f x))}{(a+b) (a+2 b+a \cos (2 (e+f x)))}+a^2 \sin (4 (e+f x))}{32 a^4 f} \] Input:

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

Output:

(4*(3*a^2 - 8*a*b + 24*b^2)*(e + f*x) - (16*b^(5/2)*(7*a + 6*b)*ArcTan[(Sq 
rt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(3/2) + 8*a*(a - 2*b)*Sin[2*(e + 
 f*x)] - (16*a*b^3*Sin[2*(e + f*x)])/((a + b)*(a + 2*b + a*Cos[2*(e + f*x) 
])) + a^2*Sin[4*(e + f*x)])/(32*a^4*f)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.14, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 4634, 316, 25, 402, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {3 (a-2 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\int -\frac {3 a^2+b a+6 b^2+9 (a-2 b) b \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{2 a}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\int \frac {3 a^2+b a+6 b^2+9 (a-2 b) b \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{2 a}+\frac {3 (a-2 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {\frac {\int \frac {2 \left (3 a^3-2 b a^2+11 b^2 a+12 b^3+(a-3 b) b (3 a+4 b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{2 a (a+b)}+\frac {b (a-3 b) (3 a+4 b) \tan (e+f x)}{a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}+\frac {3 (a-2 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\frac {\int \frac {3 a^3-2 b a^2+11 b^2 a+12 b^3+(a-3 b) b (3 a+4 b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{a (a+b)}+\frac {b (a-3 b) (3 a+4 b) \tan (e+f x)}{a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}+\frac {3 (a-2 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\frac {\frac {\frac {(a+b) \left (3 a^2-8 a b+24 b^2\right ) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}-\frac {4 b^3 (7 a+6 b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{a (a+b)}+\frac {b (a-3 b) (3 a+4 b) \tan (e+f x)}{a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}+\frac {3 (a-2 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {\frac {\frac {(a+b) \left (3 a^2-8 a b+24 b^2\right ) \arctan (\tan (e+f x))}{a}-\frac {4 b^3 (7 a+6 b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{a (a+b)}+\frac {b (a-3 b) (3 a+4 b) \tan (e+f x)}{a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}+\frac {3 (a-2 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\frac {\frac {(a+b) \left (3 a^2-8 a b+24 b^2\right ) \arctan (\tan (e+f x))}{a}-\frac {4 b^{5/2} (7 a+6 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}}{a (a+b)}+\frac {b (a-3 b) (3 a+4 b) \tan (e+f x)}{a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}+\frac {3 (a-2 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

Input:

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

Output:

(Tan[e + f*x]/(4*a*(1 + Tan[e + f*x]^2)^2*(a + b + b*Tan[e + f*x]^2)) + (( 
3*(a - 2*b)*Tan[e + f*x])/(2*a*(1 + Tan[e + f*x]^2)*(a + b + b*Tan[e + f*x 
]^2)) + ((((a + b)*(3*a^2 - 8*a*b + 24*b^2)*ArcTan[Tan[e + f*x]])/a - (4*b 
^(5/2)*(7*a + 6*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a*Sqrt[a + 
 b]))/(a*(a + b)) + ((a - 3*b)*b*(3*a + 4*b)*Tan[e + f*x])/(a*(a + b)*(a + 
 b + b*Tan[e + f*x]^2)))/(2*a))/(4*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 = 2.55 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (\frac {3}{8} a^{2}-a b \right ) \tan \left (f x +e \right )^{3}+\left (-a b +\frac {5}{8} a^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-8 a b +24 b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{a^{4}}-\frac {b^{3} \left (\frac {a \tan \left (f x +e \right )}{2 \left (a +b \right ) \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (7 a +6 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{4}}}{f}\)	\(160\)
default	\(\frac {\frac {\frac {\left (\frac {3}{8} a^{2}-a b \right ) \tan \left (f x +e \right )^{3}+\left (-a b +\frac {5}{8} a^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-8 a b +24 b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{a^{4}}-\frac {b^{3} \left (\frac {a \tan \left (f x +e \right )}{2 \left (a +b \right ) \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (7 a +6 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{4}}}{f}\)	\(160\)
risch	\(\frac {3 x}{8 a^{2}}-\frac {x b}{a^{3}}+\frac {3 x \,b^{2}}{a^{4}}-\frac {i {\mathrm e}^{4 i \left (f x +e \right )}}{64 a^{2} f}-\frac {i {\mathrm e}^{2 i \left (f x +e \right )}}{8 a^{2} f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} b}{4 a^{3} f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a^{2} f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} b}{4 a^{3} f}+\frac {i {\mathrm e}^{-4 i \left (f x +e \right )}}{64 a^{2} f}-\frac {i b^{3} \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}{a^{4} \left (a +b \right ) f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}+\frac {7 \sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{4 \left (a +b \right )^{2} f \,a^{3}}+\frac {3 \sqrt {-\left (a +b \right ) b}\, b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 \left (a +b \right )^{2} f \,a^{4}}-\frac {7 \sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{4 \left (a +b \right )^{2} f \,a^{3}}-\frac {3 \sqrt {-\left (a +b \right ) b}\, b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 \left (a +b \right )^{2} f \,a^{4}}\)	\(441\)

Input:

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

Output:

1/f*(1/a^4*(((3/8*a^2-a*b)*tan(f*x+e)^3+(-a*b+5/8*a^2)*tan(f*x+e))/(1+tan( 
f*x+e)^2)^2+1/8*(3*a^2-8*a*b+24*b^2)*arctan(tan(f*x+e)))-b^3/a^4*(1/2*a/(a 
+b)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)+1/2*(7*a+6*b)/(a+b)/((a+b)*b)^(1/2)*ar 
ctan(b*tan(f*x+e)/((a+b)*b)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 656, normalized size of antiderivative = 3.23 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/8*((3*a^4 - 5*a^3*b + 16*a^2*b^2 + 24*a*b^3)*f*x*cos(f*x + e)^2 + (3*a^ 
3*b - 5*a^2*b^2 + 16*a*b^3 + 24*b^4)*f*x + (7*a*b^3 + 6*b^4 + (7*a^2*b^2 + 
 6*a*b^3)*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)*c 
os(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)) + (2*(a^4 + a^3*b) 
*cos(f*x + e)^5 + 3*(a^4 - a^3*b - 2*a^2*b^2)*cos(f*x + e)^3 + (3*a^3*b - 
5*a^2*b^2 - 12*a*b^3)*cos(f*x + e))*sin(f*x + e))/((a^6 + a^5*b)*f*cos(f*x 
 + e)^2 + (a^5*b + a^4*b^2)*f), 1/8*((3*a^4 - 5*a^3*b + 16*a^2*b^2 + 24*a* 
b^3)*f*x*cos(f*x + e)^2 + (3*a^3*b - 5*a^2*b^2 + 16*a*b^3 + 24*b^4)*f*x + 
2*(7*a*b^3 + 6*b^4 + (7*a^2*b^2 + 6*a*b^3)*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^4 + a^3*b)*cos(f*x + e)^5 + 3*(a^4 - a^3*b - 2*a^2 
*b^2)*cos(f*x + e)^3 + (3*a^3*b - 5*a^2*b^2 - 12*a*b^3)*cos(f*x + e))*sin( 
f*x + e))/((a^6 + a^5*b)*f*cos(f*x + e)^2 + (a^5*b + a^4*b^2)*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.32 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {4 \, {\left (7 \, a b^{3} + 6 \, b^{4}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{5} + a^{4} b\right )} \sqrt {{\left (a + b\right )} b}} - \frac {{\left (3 \, a^{2} b - 5 \, a b^{2} - 12 \, b^{3}\right )} \tan \left (f x + e\right )^{5} + {\left (3 \, a^{3} + 3 \, a^{2} b - 16 \, a b^{2} - 24 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + {\left (5 \, a^{3} + 2 \, a^{2} b - 11 \, a b^{2} - 12 \, b^{3}\right )} \tan \left (f x + e\right )}{{\left (a^{4} b + a^{3} b^{2}\right )} \tan \left (f x + e\right )^{6} + a^{5} + 2 \, a^{4} b + a^{3} b^{2} + {\left (a^{5} + 4 \, a^{4} b + 3 \, a^{3} b^{2}\right )} \tan \left (f x + e\right )^{4} + {\left (2 \, a^{5} + 5 \, a^{4} b + 3 \, a^{3} b^{2}\right )} \tan \left (f x + e\right )^{2}} - \frac {{\left (3 \, a^{2} - 8 \, a b + 24 \, b^{2}\right )} {\left (f x + e\right )}}{a^{4}}}{8 \, f} \] Input:

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

Output:

-1/8*(4*(7*a*b^3 + 6*b^4)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^5 + a 
^4*b)*sqrt((a + b)*b)) - ((3*a^2*b - 5*a*b^2 - 12*b^3)*tan(f*x + e)^5 + (3 
*a^3 + 3*a^2*b - 16*a*b^2 - 24*b^3)*tan(f*x + e)^3 + (5*a^3 + 2*a^2*b - 11 
*a*b^2 - 12*b^3)*tan(f*x + e))/((a^4*b + a^3*b^2)*tan(f*x + e)^6 + a^5 + 2 
*a^4*b + a^3*b^2 + (a^5 + 4*a^4*b + 3*a^3*b^2)*tan(f*x + e)^4 + (2*a^5 + 5 
*a^4*b + 3*a^3*b^2)*tan(f*x + e)^2) - (3*a^2 - 8*a*b + 24*b^2)*(f*x + e)/a 
^4)/f
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.96 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {4 \, b^{3} \tan \left (f x + e\right )}{{\left (a^{4} + a^{3} b\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}} + \frac {4 \, {\left (7 \, a b^{3} + 6 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{5} + a^{4} b\right )} \sqrt {a b + b^{2}}} - \frac {{\left (3 \, a^{2} - 8 \, a b + 24 \, b^{2}\right )} {\left (f x + e\right )}}{a^{4}} - \frac {3 \, a \tan \left (f x + e\right )^{3} - 8 \, b \tan \left (f x + e\right )^{3} + 5 \, a \tan \left (f x + e\right ) - 8 \, b \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2} a^{3}}}{8 \, f} \] Input:

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

Output:

-1/8*(4*b^3*tan(f*x + e)/((a^4 + a^3*b)*(b*tan(f*x + e)^2 + a + b)) + 4*(7 
*a*b^3 + 6*b^4)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + 
e)/sqrt(a*b + b^2)))/((a^5 + a^4*b)*sqrt(a*b + b^2)) - (3*a^2 - 8*a*b + 24 
*b^2)*(f*x + e)/a^4 - (3*a*tan(f*x + e)^3 - 8*b*tan(f*x + e)^3 + 5*a*tan(f 
*x + e) - 8*b*tan(f*x + e))/((tan(f*x + e)^2 + 1)^2*a^3))/f
 

Mupad [B] (verification not implemented)

Time = 19.49 (sec) , antiderivative size = 2880, normalized size of antiderivative = 14.19 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan(((((tan(e + f*x)*(2112*a*b^8 + 1152*b^9 + 800*a^2*b^7 - 16*a^3*b^6 + 
 121*a^4*b^5 - 30*a^5*b^4 + 9*a^6*b^3))/(32*(2*a^7*b + a^8 + a^6*b^2)) - ( 
((6*a^8*b^6 + (23*a^9*b^5)/2 + (9*a^10*b^4)/2 + (a^11*b^3)/2 + (3*a^12*b^2 
)/2)/(2*a^10*b + a^11 + a^9*b^2) - (tan(e + f*x)*(a^2*3i - a*b*8i + b^2*24 
i)*(512*a^8*b^5 + 1280*a^9*b^4 + 1024*a^10*b^3 + 256*a^11*b^2))/(512*a^4*( 
2*a^7*b + a^8 + a^6*b^2)))*(a^2*3i - a*b*8i + b^2*24i))/(16*a^4))*(a^2*3i 
- a*b*8i + b^2*24i)*1i)/(16*a^4) + (((tan(e + f*x)*(2112*a*b^8 + 1152*b^9 
+ 800*a^2*b^7 - 16*a^3*b^6 + 121*a^4*b^5 - 30*a^5*b^4 + 9*a^6*b^3))/(32*(2 
*a^7*b + a^8 + a^6*b^2)) + (((6*a^8*b^6 + (23*a^9*b^5)/2 + (9*a^10*b^4)/2 
+ (a^11*b^3)/2 + (3*a^12*b^2)/2)/(2*a^10*b + a^11 + a^9*b^2) + (tan(e + f* 
x)*(a^2*3i - a*b*8i + b^2*24i)*(512*a^8*b^5 + 1280*a^9*b^4 + 1024*a^10*b^3 
 + 256*a^11*b^2))/(512*a^4*(2*a^7*b + a^8 + a^6*b^2)))*(a^2*3i - a*b*8i + 
b^2*24i))/(16*a^4))*(a^2*3i - a*b*8i + b^2*24i)*1i)/(16*a^4))/(((135*a*b^9 
)/4 + 27*b^10 - (9*a^2*b^8)/2 - (149*a^3*b^7)/32 + (219*a^4*b^6)/64 - (63* 
a^5*b^5)/64)/(2*a^10*b + a^11 + a^9*b^2) - (((tan(e + f*x)*(2112*a*b^8 + 1 
152*b^9 + 800*a^2*b^7 - 16*a^3*b^6 + 121*a^4*b^5 - 30*a^5*b^4 + 9*a^6*b^3) 
)/(32*(2*a^7*b + a^8 + a^6*b^2)) - (((6*a^8*b^6 + (23*a^9*b^5)/2 + (9*a^10 
*b^4)/2 + (a^11*b^3)/2 + (3*a^12*b^2)/2)/(2*a^10*b + a^11 + a^9*b^2) - (ta 
n(e + f*x)*(a^2*3i - a*b*8i + b^2*24i)*(512*a^8*b^5 + 1280*a^9*b^4 + 1024* 
a^10*b^3 + 256*a^11*b^2))/(512*a^4*(2*a^7*b + a^8 + a^6*b^2)))*(a^2*3i ...
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 963, normalized size of antiderivative = 4.74 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 28*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/s 
qrt(b))*sin(e + f*x)**2*a**2*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)**2*a*b**3 + 28*sqrt(b 
)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**2* 
b**2 + 52*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a) 
)/sqrt(b))*a*b**3 + 24*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x) 
/2) - sqrt(a))/sqrt(b))*b**4 - 28*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*ta 
n((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**2*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)**2*a*b**3 + 28*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2 
) + sqrt(a))/sqrt(b))*a**2*b**2 + 52*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b) 
*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*a*b**3 + 24*sqrt(b)*sqrt(a + b)*atan 
((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*b**4 - 2*cos(e + f*x)*s 
in(e + f*x)**5*a**5 - 4*cos(e + f*x)*sin(e + f*x)**5*a**4*b - 2*cos(e + f* 
x)*sin(e + f*x)**5*a**3*b**2 + 7*cos(e + f*x)*sin(e + f*x)**3*a**5 + 8*cos 
(e + f*x)*sin(e + f*x)**3*a**4*b - 5*cos(e + f*x)*sin(e + f*x)**3*a**3*b** 
2 - 6*cos(e + f*x)*sin(e + f*x)**3*a**2*b**3 - 5*cos(e + f*x)*sin(e + f*x) 
*a**5 - 7*cos(e + f*x)*sin(e + f*x)*a**4*b + 9*cos(e + f*x)*sin(e + f*x)*a 
**3*b**2 + 23*cos(e + f*x)*sin(e + f*x)*a**2*b**3 + 12*cos(e + f*x)*sin(e 
+ f*x)*a*b**4 + 3*sin(e + f*x)**2*a**5*e + 3*sin(e + f*x)**2*a**5*f*x -...