3.3.93 \(\int \frac {\cos ^2(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [293]

3.3.93.1 Optimal result

Integrand size = 25, antiderivative size = 187 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {(a-5 b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 a^{7/2} f}+\frac {\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {b (3 a+5 b) \tan (e+f x)}{6 a^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {b \left (3 a^2+22 a b+15 b^2\right ) \tan (e+f x)}{6 a^3 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}} \]

output

1/2*(a-5*b)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(7/2)/ 
f+1/6*b*(3*a^2+22*a*b+15*b^2)*tan(f*x+e)/a^3/(a+b)^2/f/(a+b+b*tan(f*x+e)^2 
)^(1/2)+1/2*cos(f*x+e)*sin(f*x+e)/a/f/(a+b+b*tan(f*x+e)^2)^(3/2)+1/6*b*(3* 
a+5*b)*tan(f*x+e)/a^2/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(3/2)
 

3.3.93.2 Mathematica [C] (warning: unable to verify)

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

Time = 16.74 (sec) , antiderivative size = 1775, normalized size of antiderivative = 9.49 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \]

input

Integrate[Cos[e + f*x]^2/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

output

(3*(a + b)*AppellF1[1/2, -3, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/ 
(a + b)]*Cos[e + f*x]^8*Sin[e + f*x])/(4*Sqrt[2]*f*(a + b*Sec[e + f*x]^2)^ 
(5/2)*(a + b - a*Sin[e + f*x]^2)^(5/2)*(3*(a + b)*AppellF1[1/2, -3, 5/2, 3 
/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (5*a*AppellF1[3/2, -3, 7 
/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 6*(a + b)*AppellF1[ 
3/2, -2, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e + f* 
x]^2)*((15*a*(a + b)*AppellF1[1/2, -3, 5/2, 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)/(4*Sqrt[2]*(a + b - a*Si 
n[e + f*x]^2)^(7/2)*(3*(a + b)*AppellF1[1/2, -3, 5/2, 3/2, Sin[e + f*x]^2, 
 (a*Sin[e + f*x]^2)/(a + b)] + (5*a*AppellF1[3/2, -3, 7/2, 5/2, Sin[e + f* 
x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 6*(a + b)*AppellF1[3/2, -2, 5/2, 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, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*C 
os[e + f*x]^7)/(4*Sqrt[2]*(a + b - a*Sin[e + f*x]^2)^(5/2)*(3*(a + b)*Appe 
llF1[1/2, -3, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (5*a 
*AppellF1[3/2, -3, 7/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 
 6*(a + b)*AppellF1[3/2, -2, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/ 
(a + b)])*Sin[e + f*x]^2)) - (9*(a + b)*AppellF1[1/2, -3, 5/2, 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)/(2*Sq 
rt[2]*(a + b - a*Sin[e + f*x]^2)^(5/2)*(3*(a + b)*AppellF1[1/2, -3, 5/2...
 

3.3.93.3 Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4634, 316, 25, 402, 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.

\(\displaystyle \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sec (e+f x)^2 \left (a+b \sec (e+f x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {\int \frac {3 a^2-6 b a-5 b^2+2 b (3 a+5 b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 a (a+b)}+\frac {b (3 a+5 b) \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {\frac {\int \frac {3 (a-5 b) (a+b)^2}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a (a+b)}+\frac {b \left (3 a^2+22 a b+15 b^2\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}+\frac {b (3 a+5 b) \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\frac {3 (a-5 b) (a+b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a}+\frac {b \left (3 a^2+22 a b+15 b^2\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}+\frac {b (3 a+5 b) \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\frac {\frac {3 (a-5 b) (a+b) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{a}+\frac {b \left (3 a^2+22 a b+15 b^2\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}+\frac {b (3 a+5 b) \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {\frac {3 (a-5 b) (a+b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2}}+\frac {b \left (3 a^2+22 a b+15 b^2\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}+\frac {b (3 a+5 b) \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}+\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

input

Int[Cos[e + f*x]^2/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

output

(Tan[e + f*x]/(2*a*(1 + Tan[e + f*x]^2)*(a + b + b*Tan[e + f*x]^2)^(3/2)) 
+ ((b*(3*a + 5*b)*Tan[e + f*x])/(3*a*(a + b)*(a + b + b*Tan[e + f*x]^2)^(3 
/2)) + ((3*(a - 5*b)*(a + b)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b* 
Tan[e + f*x]^2]])/a^(3/2) + (b*(3*a^2 + 22*a*b + 15*b^2)*Tan[e + f*x])/(a* 
(a + b)*Sqrt[a + b + b*Tan[e + f*x]^2]))/(3*a*(a + b)))/(2*a))/f
 

3.3.93.3.1 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]
 

3.3.93.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2142\) vs. \(2(167)=334\).

Time = 5.35 (sec) , antiderivative size = 2143, normalized size of antiderivative = 11.46

method	result	size
default	\(\text {Expression too large to display}\)	\(2143\)

input

int(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
 

output

1/6/f/(a+b)^2/a^3/(-a)^(1/2)*(b+a*cos(f*x+e)^2)*(-15*((b+a*cos(f*x+e)^2)/( 
1+cos(f*x+e))^2)^(1/2)*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)*a*b^3*cos(f*x+e)^2+6*(-a)^(1/2)*a^3*b*cos(f*x+e)^2*sin( 
f*x+e)+30*(-a)^(1/2)*a^2*b^2*cos(f*x+e)^2*sin(f*x+e)+20*(-a)^(1/2)*a*b^3*c 
os(f*x+e)^2*sin(f*x+e)+3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*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)*a^3*b*cos 
(f*x+e)-9*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*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)*a^2*b^2*cos(f*x+e)-27*(( 
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*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)*a*b^3*cos(f*x+e)+6*(-a)^(1/2)*a^3*b* 
cos(f*x+e)^4*sin(f*x+e)+3*(-a)^(1/2)*a^2*b^2*cos(f*x+e)^4*sin(f*x+e)-9*((b 
+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*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)*a^3*b*cos(f*x+e)^3-27*((b+a*cos(f*x+e 
)^2)/(1+cos(f*x+e))^2)^(1/2)*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...
 

3.3.93.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 451 vs. \(2 (167) = 334\).

Time = 1.78 (sec) , antiderivative size = 1023, normalized size of antiderivative = 5.47 \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]

input

integrate(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")
 

output

[1/48*(3*(a^3*b^2 - 3*a^2*b^3 - 9*a*b^4 - 5*b^5 + (a^5 - 3*a^4*b - 9*a^3*b 
^2 - 5*a^2*b^3)*cos(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 - 9*a^2*b^3 - 5*a*b^ 
4)*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*(3*(a^5 + 2*a^4*b + a^3*b^2)*cos(f*x + e)^5 + 
2*(3*a^4*b + 15*a^3*b^2 + 10*a^2*b^3)*cos(f*x + e)^3 + (3*a^3*b^2 + 22*a^2 
*b^3 + 15*a*b^4)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2) 
*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), 
-1/24*(3*(a^3*b^2 - 3*a^2*b^3 - 9*a*b^4 - 5*b^5 + (a^5 - 3*a^4*b - 9*a^3*b 
^2 - 5*a^2*b^3)*cos(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 - 9*a^2*b^3 - 5*a*b^ 
4)*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)*cos(f*x + e)^2)*sin(f*x + e))) - 4*(3*(a^5 + 2*a^4*b + a^3*b 
^2)*cos(f*x + e)^5 + 2*(3*a^4*b + 15*a^3*b^2 + 10*a^2*b^3)*cos(f*x + e)...
 

3.3.93.6 Sympy [F]

\[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\cos ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

input

integrate(cos(f*x+e)**2/(a+b*sec(f*x+e)**2)**(5/2),x)
 

output

Integral(cos(e + f*x)**2/(a + b*sec(e + f*x)**2)**(5/2), x)
 

3.3.93.7 Maxima [F]

\[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{2}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")
 

output

integrate(cos(f*x + e)^2/(b*sec(f*x + e)^2 + a)^(5/2), x)
 

3.3.93.8 Giac [F]

\[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{2}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")
 

output

sage0*x
 

3.3.93.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^2}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \]

input

int(cos(e + f*x)^2/(a + b/cos(e + f*x)^2)^(5/2),x)
 

output

int(cos(e + f*x)^2/(a + b/cos(e + f*x)^2)^(5/2), x)