3.5.25 \(\int \frac {\cot ^2(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [425]

3.5.25.1 Optimal result
3.5.25.2 Mathematica [A] (verified)
3.5.25.3 Rubi [A] (verified)
3.5.25.4 Maple [B] (warning: unable to verify)
3.5.25.5 Fricas [B] (verification not implemented)
3.5.25.6 Sympy [F]
3.5.25.7 Maxima [F]
3.5.25.8 Giac [F]
3.5.25.9 Mupad [F(-1)]

3.5.25.1 Optimal result

Integrand size = 25, antiderivative size = 119 \[ \int \frac {\cot ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{3/2} f}-\frac {b \cot (e+f x)}{a (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {(a-b) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{a (a+b)^2 f} \]

output
-arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(3/2)/f-b*cot(f*x 
+e)/a/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(1/2)-(a-b)*cot(f*x+e)*(a+b+b*tan(f*x+e 
)^2)^(1/2)/a/(a+b)^2/f
 
3.5.25.2 Mathematica [A] (verified)

Time = 4.46 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.53 \[ \int \frac {\cot ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right ) (a+2 b+a \cos (2 e+2 f x))^{3/2} \sec ^3(e+f x)}{2 \sqrt {2} a^{3/2} f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(a+2 b+a \cos (2 (e+f x))) \left (a^2+2 a b-b^2+\left (a^2+b^2\right ) \cos (2 (e+f x))\right ) \csc (e+f x) \sec ^3(e+f x)}{4 a (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

input
Integrate[Cot[e + f*x]^2/(a + b*Sec[e + f*x]^2)^(3/2),x]
 
output
-1/2*(ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]]*(a + 2 
*b + a*Cos[2*e + 2*f*x])^(3/2)*Sec[e + f*x]^3)/(Sqrt[2]*a^(3/2)*f*(a + b*S 
ec[e + f*x]^2)^(3/2)) - ((a + 2*b + a*Cos[2*(e + f*x)])*(a^2 + 2*a*b - b^2 
 + (a^2 + b^2)*Cos[2*(e + f*x)])*Csc[e + f*x]*Sec[e + f*x]^3)/(4*a*(a + b) 
^2*f*(a + b*Sec[e + f*x]^2)^(3/2))
 
3.5.25.3 Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4629, 2075, 374, 445, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 374

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

\(\Big \downarrow \) 445

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {-(a+b) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-\frac {(a-b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{a (a+b)}-\frac {b \cot (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {-\frac {(a+b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {a}}-\frac {(a-b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{a (a+b)}-\frac {b \cot (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\)

input
Int[Cot[e + f*x]^2/(a + b*Sec[e + f*x]^2)^(3/2),x]
 
output
(-((b*Cot[e + f*x])/(a*(a + b)*Sqrt[a + b + b*Tan[e + f*x]^2])) + (-(((a + 
 b)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/Sqrt[a] 
) - ((a - b)*Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(a + b))/(a*(a + 
 b)))/f
 

3.5.25.3.1 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 374
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q 
 + 1)/(a*e*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(e*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[b*c*(m + 1) + 2*(b*c - 
a*d)*(p + 1) + d*b*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, 
c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IntBinomialQ[a, b, 
 c, d, e, m, 2, p, q, x]
 

rule 445
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ 
.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p 
+ 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) 
 Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c 
+ a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 
2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
 

rule 2075
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa 
ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi 
alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  ! 
BinomialMatchQ[{u, v}, x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4629
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f 
_.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim 
p[ff/f   Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 
)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte 
gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
 
3.5.25.4 Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1227\) vs. \(2(109)=218\).

Time = 5.74 (sec) , antiderivative size = 1228, normalized size of antiderivative = 10.32

method result size
default \(\text {Expression too large to display}\) \(1228\)

input
int(cot(f*x+e)^2/(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 
output
1/2/f/(a+b)^2/a/(-a)^(1/2)*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e 
))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*c 
sc(f*x+e)^2+a+b)*((-a)^(1/2)*a^2*(1-cos(f*x+e))^4*csc(f*x+e)^4+(-a)^(1/2)* 
a*b*(1-cos(f*x+e))^4*csc(f*x+e)^4+2*ln(4*((-a)^(1/2)*(a*(1-cos(f*x+e))^4*c 
sc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e 
)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)-2*a*(csc(f*x+e)-cot(f*x+e 
)))/((1-cos(f*x+e))^2*csc(f*x+e)^2+1))*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b* 
(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos 
(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)*a^2*(csc(f*x+e)-cot(f*x+e))+4*ln(4*((-a 
)^(1/2)*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2 
*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1 
/2)-2*a*(csc(f*x+e)-cot(f*x+e)))/((1-cos(f*x+e))^2*csc(f*x+e)^2+1))*(a*(1- 
cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+ 
e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)*a*b*(csc(f 
*x+e)-cot(f*x+e))+2*ln(4*((-a)^(1/2)*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1 
-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f 
*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)-2*a*(csc(f*x+e)-cot(f*x+e)))/((1-cos(f*x+ 
e))^2*csc(f*x+e)^2+1))*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4 
*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f 
*x+e)^2+a+b)^(1/2)*b^2*(csc(f*x+e)-cot(f*x+e))-2*a^2*(1-cos(f*x+e))^2*(...
 
3.5.25.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (109) = 218\).

Time = 0.73 (sec) , antiderivative size = 741, normalized size of antiderivative = 6.23 \[ \int \frac {\cot ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\left [-\frac {{\left (a^{2} b + 2 \, a b^{2} + b^{3} + {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a^{4} - a^{3} b\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{4} - 14 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 28 \, a^{3} b + 70 \, a^{2} b^{2} - 28 \, a b^{3} + b^{4} - 32 \, {\left (a^{4} - 7 \, a^{3} b + 7 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (f x + e\right )^{2} - 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a^{3} - a^{2} b\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 14 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 7 \, a^{2} b + 7 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right ) \sin \left (f x + e\right ) + 8 \, {\left ({\left (a^{3} + a b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} b - a b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{8 \, {\left ({\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f\right )} \sin \left (f x + e\right )}, \frac {{\left (a^{2} b + 2 \, a b^{2} + b^{3} + {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 8 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} - a^{2} b + a b^{2} - {\left (a^{3} - 3 \, a^{2} b\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (a^{3} + a b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} b - a b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left ({\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f\right )} \sin \left (f x + e\right )}\right ] \]

input
integrate(cot(f*x+e)^2/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
 
output
[-1/8*((a^2*b + 2*a*b^2 + b^3 + (a^3 + 2*a^2*b + a*b^2)*cos(f*x + e)^2)*sq 
rt(-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 
))*sin(f*x + e) + 8*((a^3 + a*b^2)*cos(f*x + e)^3 + (a^2*b - a*b^2)*cos(f* 
x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a^5 + 2*a^4*b + a^ 
3*b^2)*f*cos(f*x + e)^2 + (a^4*b + 2*a^3*b^2 + a^2*b^3)*f)*sin(f*x + e)), 
1/4*((a^2*b + 2*a*b^2 + b^3 + (a^3 + 2*a^2*b + a*b^2)*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)))*sin(f*x + e) - 4*((a^3 + a*b^2)*cos(f*x + e)^3 + (a^ 
2*b - a*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(( 
(a^5 + 2*a^4*b + a^3*b^2)*f*cos(f*x + e)^2 + (a^4*b + 2*a^3*b^2 + a^2*b^3) 
*f)*sin(f*x + e))]
 
3.5.25.6 Sympy [F]

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

input
integrate(cot(f*x+e)**2/(a+b*sec(f*x+e)**2)**(3/2),x)
 
output
Integral(cot(e + f*x)**2/(a + b*sec(e + f*x)**2)**(3/2), x)
 
3.5.25.7 Maxima [F]

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

input
integrate(cot(f*x+e)^2/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
 
output
integrate(cot(f*x + e)^2/(b*sec(f*x + e)^2 + a)^(3/2), x)
 
3.5.25.8 Giac [F]

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

input
integrate(cot(f*x+e)^2/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
 
output
sage0*x
 
3.5.25.9 Mupad [F(-1)]

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

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