\(\int \frac {\cos ^4(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [281]

Optimal result

Integrand size = 25, antiderivative size = 194 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {3 \left (a^2-2 a b+5 b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 a^{7/2} f}+\frac {(3 a-5 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(a-3 b) b (3 a+5 b) \tan (e+f x)}{8 a^3 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 14.33 (sec) , antiderivative size = 2046, normalized size of antiderivative = 10.55 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Result too large to show} \] Input:

Integrate[Cos[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

((a + b)*AppellF1[1/2, -3, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a 
 + b)]*Cos[e + f*x]^10*Sin[e + f*x])/(2*f*Sqrt[a + 2*b + a*Cos[2*(e + f*x) 
]]*(a + b*Sec[e + f*x]^2)^(3/2)*(a + b - a*Sin[e + f*x]^2)*((a + b)*Appell 
F1[1/2, -3, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*App 
ellF1[3/2, -3, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 2*( 
a + b)*AppellF1[3/2, -2, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + 
 b)])*Sin[e + f*x]^2)*((a*(a + b)*AppellF1[1/2, -3, 3/2, 3/2, Sin[e + f*x] 
^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^7*Sin[e + f*x]^2)/(Sqrt[a + 2 
*b + a*Cos[2*(e + f*x)]]*(a + b - a*Sin[e + f*x]^2)^2*((a + b)*AppellF1[1/ 
2, -3, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*AppellF1 
[3/2, -3, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 2*(a + b 
)*AppellF1[3/2, -2, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]) 
*Sin[e + f*x]^2)) + ((a + b)*AppellF1[1/2, -3, 3/2, 3/2, Sin[e + f*x]^2, ( 
a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^7)/(2*Sqrt[a + 2*b + a*Cos[2*(e + 
f*x)]]*(a + b - a*Sin[e + f*x]^2)*((a + b)*AppellF1[1/2, -3, 3/2, 3/2, Sin 
[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*AppellF1[3/2, -3, 5/2, 5/2, 
Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 2*(a + b)*AppellF1[3/2, -2, 
3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e + f*x]^2)) - 
(3*(a + b)*AppellF1[1/2, -3, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/ 
(a + b)]*Cos[e + f*x]^5*Sin[e + f*x]^2)/(Sqrt[a + 2*b + a*Cos[2*(e + f*...
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4634, 316, 25, 402, 25, 402, 27, 291, 216}

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.

\(\displaystyle \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sec (e+f x)^4 \left (a+b \sec (e+f x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

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

Input:

Int[Cos[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

(Tan[e + f*x]/(4*a*(1 + Tan[e + f*x]^2)^2*Sqrt[a + b + b*Tan[e + f*x]^2]) 
+ (((3*a - 5*b)*Tan[e + f*x])/(2*a*(1 + Tan[e + f*x]^2)*Sqrt[a + b + b*Tan 
[e + f*x]^2]) + ((3*(a^2 - 2*a*b + 5*b^2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sq 
rt[a + b + b*Tan[e + f*x]^2]])/a^(3/2) + ((a - 3*b)*b*(3*a + 5*b)*Tan[e + 
f*x])/(a*(a + b)*Sqrt[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 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

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 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(857\) vs. \(2(174)=348\).

Time = 11.56 (sec) , antiderivative size = 858, normalized size of antiderivative = 4.42

method	result	size
default	\(\text {Expression too large to display}\)	\(858\)

Input:

int(cos(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/8/f/(a+b)/a^3/(-a)^(1/2)*(cos(f*x+e)^2*(3*cos(f*x+e)+3)*((b+a*cos(f*x+e) 
^2)/(1+cos(f*x+e))^2)^(1/2)*a^4*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos 
(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e 
))^2)^(1/2)-4*sin(f*x+e)*a)+(3*cos(f*x+e)+3)*sin(f*x+e)^2*((b+a*cos(f*x+e) 
^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+c 
os(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x 
+e))^2)^(1/2)-4*sin(f*x+e)*a)+(9*cos(f*x+e)^3+9*cos(f*x+e)^2-3*cos(f*x+e)- 
3)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b^2*ln(4*(-a)^(1/2)*((b 
+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos 
(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)+(15*cos(f*x+e)^3+15*cos 
(f*x+e)^2+9*cos(f*x+e)+9)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b^ 
3*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4 
*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)+(1 
5*cos(f*x+e)+15)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^4*ln(4*(-a) 
^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2) 
*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)+sin(f*x+e)*co 
s(f*x+e)^4*(2*cos(f*x+e)^2+3)*(-a)^(1/2)*a^4+sin(f*x+e)*cos(f*x+e)^2*(2*co 
s(f*x+e)^4+6)*(-a)^(1/2)*a^3*b+(-3*cos(f*x+e)^4-6*cos(f*x+e)^2+3)*sin(f*x+ 
e)*(-a)^(1/2)*a^2*b^2+(-20*cos(f*x+e)^2-4)*sin(f*x+e)*(-a)^(1/2)*a*b^3-15* 
(-a)^(1/2)*b^4*sin(f*x+e))/(a+b*sec(f*x+e)^2)^(3/2)*sec(f*x+e)^3
 

Fricas [A] (verification not implemented)

Time = 1.25 (sec) , antiderivative size = 811, normalized size of antiderivative = 4.18 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(cos(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[-1/64*(3*(a^3*b - a^2*b^2 + 3*a*b^3 + 5*b^4 + (a^4 - a^3*b + 3*a^2*b^2 + 
5*a*b^3)*cos(f*x + e)^2)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - 
a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + 
 a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2* 
b^2 - a*b^3)*cos(f*x + e)^2 + 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)* 
cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7* 
a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/ 
cos(f*x + e)^2)*sin(f*x + e)) - 8*(2*(a^4 + a^3*b)*cos(f*x + e)^5 + (3*a^4 
 - 2*a^3*b - 5*a^2*b^2)*cos(f*x + e)^3 + (3*a^3*b - 4*a^2*b^2 - 15*a*b^3)* 
cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(( 
a^6 + a^5*b)*f*cos(f*x + e)^2 + (a^5*b + a^4*b^2)*f), -1/32*(3*(a^3*b - a^ 
2*b^2 + 3*a*b^3 + 5*b^4 + (a^4 - a^3*b + 3*a^2*b^2 + 5*a*b^3)*cos(f*x + e) 
^2)*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^ 
3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/ 
cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*c 
os(f*x + e)^2)*sin(f*x + e))) - 4*(2*(a^4 + a^3*b)*cos(f*x + e)^5 + (3*a^4 
 - 2*a^3*b - 5*a^2*b^2)*cos(f*x + e)^3 + (3*a^3*b - 4*a^2*b^2 - 15*a*b^3)* 
cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*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]

\[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\cos ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(cos(f*x+e)**4/(a+b*sec(f*x+e)**2)**(3/2),x)
 

Output:

Integral(cos(e + f*x)**4/(a + b*sec(e + f*x)**2)**(3/2), x)
 

Maxima [F]

\[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{4}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(cos(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate(cos(f*x + e)^4/(b*sec(f*x + e)^2 + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{4}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(cos(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

integrate(cos(f*x + e)^4/(b*sec(f*x + e)^2 + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^4}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x \] Input:

int(cos(e + f*x)^4/(a + b/cos(e + f*x)^2)^(3/2),x)
                                                                                    
                                                                                    
 

Output:

int(cos(e + f*x)^4/(a + b/cos(e + f*x)^2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \cos \left (f x +e \right )^{4}}{\sec \left (f x +e \right )^{4} b^{2}+2 \sec \left (f x +e \right )^{2} a b +a^{2}}d x \] Input:

int(cos(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x)
 

Output:

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