\(\int \frac {\cot ^2(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\) [343]

Optimal result

Integrand size = 25, antiderivative size = 128 \[ \int \frac {\cot ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{3/2} f}-\frac {b \cot (e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}}-\frac {(a-2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a^2 (a-b) f} \] Output:

-arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(3/2)/f-b*c 
ot(f*x+e)/a/(a-b)/f/(a+b*tan(f*x+e)^2)^(1/2)-(a-2*b)*cot(f*x+e)*(a+b*tan(f 
*x+e)^2)^(1/2)/a^2/(a-b)/f
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 9.65 (sec) , antiderivative size = 882, normalized size of antiderivative = 6.89 \[ \int \frac {\cot ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

Integrate[Cot[e + f*x]^2/(a + b*Tan[e + f*x]^2)^(3/2),x]
 

Output:

-((Cos[e + f*x]^2*Cot[e + f*x]*((3*a*Csc[e + f*x]^2)/(a - b) + (12*b*Sec[e 
 + f*x]^2)/(a - b) + (16*(a - b)*Hypergeometric2F1[2, 2, 7/2, ((a - b)*Sin 
[e + f*x]^2)/a]*Sin[e + f*x]^2)/(15*a) + (8*(a - b)*HypergeometricPFQ[{2, 
2, 2}, {1, 7/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2)/(15*a) + (8*b 
^2*Sec[e + f*x]^2*Tan[e + f*x]^2)/(a*(a - b)) + (8*(a - b)*b*Hypergeometri 
c2F1[2, 2, 7/2, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^2) 
/(3*a^2) + (16*(a - b)*b*HypergeometricPFQ[{2, 2, 2}, {1, 7/2}, ((a - b)*S 
in[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^2)/(15*a^2) + (8*(a - b)*b^2 
*Hypergeometric2F1[2, 2, 7/2, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*T 
an[e + f*x]^4)/(5*a^3) + (8*(a - b)*b^2*HypergeometricPFQ[{2, 2, 2}, {1, 7 
/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^4)/(15*a^3) - 
 (3*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]])/((((a - b)*Sin[e + f*x]^2)/a 
)^(3/2)*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a]) - (12*b*ArcSin[Sq 
rt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^2)/(a*(((a - b)*Sin[e + f*x]^ 
2)/a)^(3/2)*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a]) - (8*b^2*ArcS 
in[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^4)/(a^2*(((a - b)*Sin[e 
+ f*x]^2)/a)^(3/2)*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a]) + (3*A 
rcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]])/Sqrt[((a - b)*Cos[e + f*x]^2*Sin[ 
e + f*x]^2*(a + b*Tan[e + f*x]^2))/a^2] + (12*b*ArcSin[Sqrt[((a - b)*Sin[e 
 + f*x]^2)/a]]*Tan[e + f*x]^2)/(a*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + ...
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4153, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 374

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

\(\Big \downarrow \) 445

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

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

Input:

Int[Cot[e + f*x]^2/(a + b*Tan[e + f*x]^2)^(3/2),x]
 

Output:

(-((b*Cot[e + f*x])/(a*(a - b)*Sqrt[a + b*Tan[e + f*x]^2])) + (-((a*ArcTan 
[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/Sqrt[a - b]) - (( 
a - 2*b)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/a)/(a*(a - b)))/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 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 3042

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

rule 4153

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], 
 x]}, Simp[c*(ff/f)   Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f 
f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, 
n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio 
nalQ[n]))
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(498\) vs. \(2(118)=236\).

Time = 25.71 (sec) , antiderivative size = 499, normalized size of antiderivative = 3.90

method	result	size
default	\(\frac {\left (-2 \sec \left (f x +e \right ) \csc \left (f x +e \right )+3 \cot \left (f x +e \right )\right ) b \,a^{2} \left (a -b \right )^{\frac {3}{2}}+\left (-\tan \left (f x +e \right ) \sec \left (f x +e \right )^{2}+4 \tan \left (f x +e \right )\right ) b^{2} a \left (a -b \right )^{\frac {3}{2}}-\left (a -b \right )^{\frac {3}{2}} a^{3} \cot \left (f x +e \right )+2 \left (a -b \right )^{\frac {3}{2}} b^{3} \tan \left (f x +e \right )^{3}-\arctan \left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )}{\sqrt {a -b}\, \left (\cos \left (f x +e \right )-1\right )}\right ) \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a^{4} \left (\sec \left (f x +e \right )+1\right )-\arctan \left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )}{\sqrt {a -b}\, \left (\cos \left (f x +e \right )-1\right )}\right ) \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a^{3} b \left (-2-2 \sec \left (f x +e \right )+\sec \left (f x +e \right )^{2}+\sec \left (f x +e \right )^{3}\right )-\arctan \left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )}{\sqrt {a -b}\, \left (\cos \left (f x +e \right )-1\right )}\right ) \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a^{2} b^{2} \left (-\tan \left (f x +e \right )^{2}-\tan \left (f x +e \right )^{2} \sec \left (f x +e \right )\right )}{f \,a^{2} \left (a -b \right )^{\frac {5}{2}} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\)	\(499\)

Input:

int(cot(f*x+e)^2/(a+b*tan(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 471, normalized size of antiderivative = 3.68 \[ \int \frac {\cot ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\left [\frac {{\left (a^{2} b \tan \left (f x + e\right )^{3} + a^{3} \tan \left (f x + e\right )\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} - 4 \, {\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) - 4 \, {\left (a^{3} - 2 \, a^{2} b + a b^{2} + {\left (a^{2} b - 3 \, a b^{2} + 2 \, b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{4 \, {\left ({\left (a^{4} b - 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f \tan \left (f x + e\right )^{3} + {\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} f \tan \left (f x + e\right )\right )}}, -\frac {{\left (a^{2} b \tan \left (f x + e\right )^{3} + a^{3} \tan \left (f x + e\right )\right )} \sqrt {a - b} \arctan \left (-\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) + 2 \, {\left (a^{3} - 2 \, a^{2} b + a b^{2} + {\left (a^{2} b - 3 \, a b^{2} + 2 \, b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{2 \, {\left ({\left (a^{4} b - 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f \tan \left (f x + e\right )^{3} + {\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} f \tan \left (f x + e\right )\right )}}\right ] \] Input:

integrate(cot(f*x+e)^2/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[1/4*((a^2*b*tan(f*x + e)^3 + a^3*tan(f*x + e))*sqrt(-a + b)*log(-((a^2 - 
8*a*b + 8*b^2)*tan(f*x + e)^4 - 2*(3*a^2 - 4*a*b)*tan(f*x + e)^2 + a^2 - 4 
*((a - 2*b)*tan(f*x + e)^3 - a*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a)*sq 
rt(-a + b))/(tan(f*x + e)^4 + 2*tan(f*x + e)^2 + 1)) - 4*(a^3 - 2*a^2*b + 
a*b^2 + (a^2*b - 3*a*b^2 + 2*b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + 
a))/((a^4*b - 2*a^3*b^2 + a^2*b^3)*f*tan(f*x + e)^3 + (a^5 - 2*a^4*b + a^3 
*b^2)*f*tan(f*x + e)), -1/2*((a^2*b*tan(f*x + e)^3 + a^3*tan(f*x + e))*sqr 
t(a - b)*arctan(-2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b)*tan(f*x + e)/((a 
 - 2*b)*tan(f*x + e)^2 - a)) + 2*(a^3 - 2*a^2*b + a*b^2 + (a^2*b - 3*a*b^2 
 + 2*b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^4*b - 2*a^3*b^2 
+ a^2*b^3)*f*tan(f*x + e)^3 + (a^5 - 2*a^4*b + a^3*b^2)*f*tan(f*x + e))]
 

Sympy [F]

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

integrate(cot(f*x+e)**2/(a+b*tan(f*x+e)**2)**(3/2),x)
 

Output:

Integral(cot(e + f*x)**2/(a + b*tan(e + f*x)**2)**(3/2), x)
 

Maxima [F(-1)]

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

integrate(cot(f*x+e)^2/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F(-1)]

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

integrate(cot(f*x+e)^2/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(tan(e + f*x)**2*b + a)*cot(e + f*x)**2)/(tan(e + f*x)**4*b**2 + 
2*tan(e + f*x)**2*a*b + a**2),x)