\(\int \frac {\sin ^2(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [127]

Optimal result

Integrand size = 25, antiderivative size = 167 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {(a+5 b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 a^{7/2} f}-\frac {\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {5 b \tan (e+f x)}{6 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (13 a+15 b) \tan (e+f x)}{6 a^3 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}} \] 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-1/2*cos(f*x+e)*sin(f*x+e)/a/f/(a+b+b*tan(f*x+e)^2)^(3/2)-5/6*b*tan(f*x+e 
)/a^2/f/(a+b+b*tan(f*x+e)^2)^(3/2)-1/6*b*(13*a+15*b)*tan(f*x+e)/a^3/(a+b)/ 
f/(a+b+b*tan(f*x+e)^2)^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(983\) vs. \(2(167)=334\).

Time = 8.39 (sec) , antiderivative size = 983, normalized size of antiderivative = 5.89 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/256*((a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Csc[e + f*x]*Sec[e + f*x]^5*( 
Sin[e + f*x]^2/(a + b) + ((a + 2*b + a*Cos[2*(e + f*x)])*Sin[e + f*x]^2)/( 
a + b)^2 - (12*Sin[e + f*x]^4)/(a + b) + (16*(a + b - a*Sin[e + f*x]^2)*(1 
 - (a*Sin[e + f*x]^2)/(a + b))*((-6*a*(a + b)*Sin[e + f*x]^2)/(a + 2*b + a 
*Cos[2*(e + f*x)]) + (a^2*(a + b)*Sin[e + f*x]^4)/(a + b - a*Sin[e + f*x]^ 
2)^2 + (3*Sqrt[a]*Sqrt[a + b]*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*S 
in[e + f*x])/Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b)]))/a^3))/(Sqrt[2]*f*( 
a + b*Sec[e + f*x]^2)^(5/2)*(a + b - a*Sin[e + f*x]^2)^(3/2)) - ((a + 2*b 
+ a*Cos[2*e + 2*f*x])^(5/2)*Csc[e + f*x]*Sec[e + f*x]^5*(Sin[e + f*x]^2/(a 
 + b) + ((a + 2*b + a*Cos[2*(e + f*x)])*Sin[e + f*x]^2)/(a + b)^2 - (24*Si 
n[e + f*x]^4)/(a + b) + (96*Sin[e + f*x]^6)/a + (80*(a + b - a*Sin[e + f*x 
]^2)*(1 - (a*Sin[e + f*x]^2)/(a + b))*((-6*a*(a + b)*Sin[e + f*x]^2)/(a + 
2*b + a*Cos[2*(e + f*x)]) + (a^2*(a + b)*Sin[e + f*x]^4)/(a + b - a*Sin[e 
+ f*x]^2)^2 + (3*Sqrt[a]*Sqrt[a + b]*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a 
+ b]]*Sin[e + f*x])/Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b)]))/a^3 - (160* 
(a + b - a*Sin[e + f*x]^2)*(1 - (a*Sin[e + f*x]^2)/(a + b))*((-6*a*(a + b) 
^2*Sin[e + f*x]^2)/(a + 2*b + a*Cos[2*(e + f*x)]) + (3*Sqrt[a]*(a + b)^(3/ 
2)*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*Sin[e + f*x])/Sqrt[(a + b - 
a*Sin[e + f*x]^2)/(a + b)] + (a^2*Sin[e + f*x]^4)/(-1 + (a*Sin[e + f*x]^2) 
/(a + b))^2))/a^4))/(768*Sqrt[2]*f*(a + b*Sec[e + f*x]^2)^(5/2)*(a + b ...
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4620, 373, 402, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4620

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

\(\Big \downarrow \) 373

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

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {\int \frac {(a+b) \left (-10 b \tan ^2(e+f x)+3 a+5 b\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 a (a+b)}-\frac {5 b \tan (e+f x)}{3 a \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \frac {-10 b \tan ^2(e+f x)+3 a+5 b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 a}-\frac {5 b \tan (e+f x)}{3 a \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {\frac {\int \frac {3 (a+b) (a+5 b)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a (a+b)}-\frac {b (13 a+15 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a}-\frac {5 b \tan (e+f x)}{3 a \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\frac {3 (a+5 b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a}-\frac {b (13 a+15 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a}-\frac {5 b \tan (e+f x)}{3 a \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\frac {\frac {3 (a+5 b) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{a}-\frac {b (13 a+15 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a}-\frac {5 b \tan (e+f x)}{3 a \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {\frac {3 (a+5 b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2}}-\frac {b (13 a+15 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a}-\frac {5 b \tan (e+f x)}{3 a \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

Input:

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

Output:

(-1/2*Tan[e + f*x]/(a*(1 + Tan[e + f*x]^2)*(a + b + b*Tan[e + f*x]^2)^(3/2 
)) + ((-5*b*Tan[e + f*x])/(3*a*(a + b + b*Tan[e + f*x]^2)^(3/2)) + ((3*(a 
+ 5*b)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/a^(3 
/2) - (b*(13*a + 15*b)*Tan[e + f*x])/(a*(a + b)*Sqrt[a + b + b*Tan[e + f*x 
]^2]))/(3*a))/(2*a))/f
 

Defintions of rubi rules used

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 373

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 
1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1))   Int[(e 
*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 
 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, 
m, 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 4620

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ 
)]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m 
+ 1)/f   Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f 
f^2*x^2)^(m/2 + 1)), 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. \(887\) vs. \(2(147)=294\).

Time = 7.73 (sec) , antiderivative size = 888, normalized size of antiderivative = 5.32

method	result	size
default	\(\text {Expression too large to display}\)	\(888\)

Input:

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

Output:

-1/6/f/(a+b)/a^3/(-a)^(1/2)/(a+b*sec(f*x+e)^2)^(5/2)*(((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)*(-3-3*sec(f*x+e))+((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+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)*(-18-18*sec(f*x+e)-6*sec(f*x+e)^2-6*sec(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)*(-15-15*sec(f*x+e)-36*sec(f*x+e)^2-3 
6*sec(f*x+e)^3-3*sec(f*x+e)^4-3*sec(f*x+e)^5)+((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)*(-30*sec(f*x+e)^2-30*sec(f*x+e)^3-18*sec(f*x+e)^4-18*sec 
(f*x+e)^5)+((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)*(-15*sec(f*x+e)^4-1 
5*sec(f*x+e)^5)+3*(-a)^(1/2)*a^4*cos(f*x+e)*sin(f*x+e)+(3*cos(f*x+e)^2+21) 
*(-a)^(1/2)*a^3*b*tan(f*x+e)+(-a)^(1/2)*a^2*b^2*(23*tan(f*x+e)+31*tan(f*x+ 
e)*sec(f*x+e)^2)+(35*cos(f*x+e)^2+13)*(-a)^(1/2)*a*b^3*tan(f*x+e)*sec(f...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (147) = 294\).

Time = 1.80 (sec) , antiderivative size = 879, normalized size of antiderivative = 5.26 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/48*(3*((a^4 + 6*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^2*b^2 + 6*a*b^3 
+ 5*b^4 + 2*(a^3*b + 6*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*c 
os(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^4 
 + a^3*b)*cos(f*x + e)^5 + 2*(9*a^3*b + 10*a^2*b^2)*cos(f*x + e)^3 + (13*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^7 + a^6*b)*f*cos(f*x + e)^4 + 2*(a^6*b + a^5*b^2)*f*c 
os(f*x + e)^2 + (a^5*b^2 + a^4*b^3)*f), -1/24*(3*((a^4 + 6*a^3*b + 5*a^2*b 
^2)*cos(f*x + e)^4 + a^2*b^2 + 6*a*b^3 + 5*b^4 + 2*(a^3*b + 6*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*c 
os(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^4 + a^3*b)*cos( 
f*x + e)^5 + 2*(9*a^3*b + 10*a^2*b^2)*cos(f*x + e)^3 + (13*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^7 + a^6*b)*f*cos(f*x + e)^4 + 2*(a^6*b + a^5*b^2)*f*cos(f*x + e)...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sin \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{6} b^{3}+3 \sec \left (f x +e \right )^{4} a \,b^{2}+3 \sec \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:

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

Output:

int((sqrt(sec(e + f*x)**2*b + a)*sin(e + f*x)**2)/(sec(e + f*x)**6*b**3 + 
3*sec(e + f*x)**4*a*b**2 + 3*sec(e + f*x)**2*a**2*b + a**3),x)