\(\int \frac {\cot ^6(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\) [345]

Optimal result

Integrand size = 25, antiderivative size = 252 \[ \int \frac {\cot ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{3/2} f}-\frac {b \cot ^5(e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}}-\frac {\left (15 a^3+10 a^2 b+8 a b^2-48 b^3\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^4 (a-b) f}+\frac {\left (5 a^2+4 a b-24 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^3 (a-b) f}-\frac {(a-6 b) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a^2 (a-b) f} \] 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)^5/a/(a-b)/f/(a+b*tan(f*x+e)^2)^(1/2)-1/15*(15*a^3+10*a^2*b+8*a*b 
^2-48*b^3)*cot(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/a^4/(a-b)/f+1/15*(5*a^2+4*a 
*b-24*b^2)*cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2)/a^3/(a-b)/f-1/5*(a-6*b)*c 
ot(f*x+e)^5*(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 = 15.66 (sec) , antiderivative size = 1994, normalized size of antiderivative = 7.91 \[ \int \frac {\cot ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

Integrate[Cot[e + f*x]^6/(a + b*Tan[e + f*x]^2)^(3/2),x]
 

Output:

-1/5*(Cos[e + f*x]^2*Cot[e + f*x]^5*((3*a*Csc[e + f*x]^2)/(a - b) - (8*b*S 
ec[e + f*x]^2)/(a - b) - (16*(a - b)*HypergeometricPFQ[{2, 2, 2}, {1, 7/2} 
, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2)/(9*a) + (32*(a - b)*Hypergeo 
metricPFQ[{2, 2, 2, 2, 2}, {1, 1, 1, 7/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin 
[e + f*x]^2)/(45*a) + (48*b^2*Sec[e + f*x]^2*Tan[e + f*x]^2)/(a*(a - b)) - 
 (16*(a - b)*b*Hypergeometric2F1[2, 2, 7/2, ((a - b)*Sin[e + f*x]^2)/a]*Si 
n[e + f*x]^2*Tan[e + f*x]^2)/(9*a^2) + (32*(a - b)*b*HypergeometricPFQ[{2, 
 2, 2}, {1, 7/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^ 
2)/(9*a^2) + (64*(a - b)*b*HypergeometricPFQ[{2, 2, 2, 2}, {1, 1, 7/2}, (( 
a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^2)/(9*a^2) + (128*(a 
 - b)*b*HypergeometricPFQ[{2, 2, 2, 2, 2}, {1, 1, 1, 7/2}, ((a - b)*Sin[e 
+ f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^2)/(45*a^2) + (192*b^3*Sec[e + f* 
x]^2*Tan[e + f*x]^4)/(a^2*(a - b)) + (80*(a - b)*b^2*Hypergeometric2F1[2, 
2, 7/2, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^4)/(3*a^3) 
 + (112*(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)/(3*a^3) + (64*(a - b)*b^2*Hype 
rgeometricPFQ[{2, 2, 2, 2}, {1, 1, 7/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e 
 + f*x]^2*Tan[e + f*x]^4)/(3*a^3) + (64*(a - b)*b^2*HypergeometricPFQ[{2, 
2, 2, 2, 2}, {1, 1, 1, 7/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Ta 
n[e + f*x]^4)/(15*a^3) + (128*b^4*Sec[e + f*x]^2*Tan[e + f*x]^6)/(a^3*(...
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4153, 374, 445, 445, 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.

\(\displaystyle \int \frac {\cot ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 374

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

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {-\frac {\int \frac {\cot ^4(e+f x) \left (5 a^2+4 b a-24 b^2+4 (a-6 b) b \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{5 a}-\frac {(a-6 b) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a}}{a (a-b)}-\frac {b \cot ^5(e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{f}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {-\frac {-\frac {\int \frac {\cot ^2(e+f x) \left (15 a^3+10 b a^2+8 b^2 a-48 b^3+2 b \left (5 a^2+4 b a-24 b^2\right ) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{3 a}-\frac {\left (5 a^2+4 a b-24 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}}{5 a}-\frac {(a-6 b) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a}}{a (a-b)}-\frac {b \cot ^5(e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{f}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {-\frac {-\frac {-\frac {\int \frac {15 a^4}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{a}-\frac {\left (15 a^3+10 a^2 b+8 a b^2-48 b^3\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{3 a}-\frac {\left (5 a^2+4 a b-24 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}}{5 a}-\frac {(a-6 b) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a}}{a (a-b)}-\frac {b \cot ^5(e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {-\frac {-15 a^3 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)-\frac {\left (15 a^3+10 a^2 b+8 a b^2-48 b^3\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{3 a}-\frac {\left (5 a^2+4 a b-24 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}}{5 a}-\frac {(a-6 b) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a}}{a (a-b)}-\frac {b \cot ^5(e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {-\frac {-\frac {-15 a^3 \int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}-\frac {\left (15 a^3+10 a^2 b+8 a b^2-48 b^3\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{3 a}-\frac {\left (5 a^2+4 a b-24 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}}{5 a}-\frac {(a-6 b) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a}}{a (a-b)}-\frac {b \cot ^5(e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {-\frac {-\frac {\left (5 a^2+4 a b-24 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}-\frac {-\frac {15 a^3 \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b}}-\frac {\left (15 a^3+10 a^2 b+8 a b^2-48 b^3\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{3 a}}{5 a}-\frac {(a-6 b) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a}}{a (a-b)}-\frac {b \cot ^5(e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{f}\)

Input:

Int[Cot[e + f*x]^6/(a + b*Tan[e + f*x]^2)^(3/2),x]
 

Output:

(-((b*Cot[e + f*x]^5)/(a*(a - b)*Sqrt[a + b*Tan[e + f*x]^2])) + (-1/5*((a 
- 6*b)*Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/a - (-1/3*((5*a^2 + 4*a* 
b - 24*b^2)*Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/a - ((-15*a^3*ArcTa 
n[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/Sqrt[a - b] - (( 
15*a^3 + 10*a^2*b + 8*a*b^2 - 48*b^3)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x] 
^2])/a)/(3*a))/(5*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 [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 16.82 (sec) , antiderivative size = 1474, normalized size of antiderivative = 5.85

method	result	size
default	\(\text {Expression too large to display}\)	\(1474\)

Input:

int(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/f*(1/15*sin(f*x+e)^5*cos(f*x+e)^2*(60*cos(f*x+e)+60)*EllipticPi(((2*I*b^ 
(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*(cot(f*x+e)-csc(f*x+e)),-1/(2*I*b^(1/2)* 
(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1 
/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*(1/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I* 
b^(1/2)*(a-b)^(1/2)+a*cos(f*x+e)-cos(f*x+e)*b+b)/(cos(f*x+e)+1))^(1/2)*(-1 
/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-a*cos(f*x+e)+co 
s(f*x+e)*b-b)/(cos(f*x+e)+1))^(1/2)*a^5+1/15*sin(f*x+e)^7*(60*cos(f*x+e)+6 
0)*EllipticPi(((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*(cot(f*x+e)-csc(f* 
x+e)),-1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2* 
b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*(1/a*(I*cos(f*x+e)* 
b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+a*cos(f*x+e)-cos(f*x+e)*b+b)/(co 
s(f*x+e)+1))^(1/2)*(-1/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b) 
^(1/2)-a*cos(f*x+e)+cos(f*x+e)*b-b)/(cos(f*x+e)+1))^(1/2)*a^4*b+1/15*sin(f 
*x+e)^5*cos(f*x+e)^2*(-30*cos(f*x+e)-30)*EllipticF(((2*I*b^(1/2)*(a-b)^(1/ 
2)+a-2*b)/a)^(1/2)*(cot(f*x+e)-csc(f*x+e)),((8*I*b^(3/2)*(a-b)^(1/2)-4*I*b 
^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*(1/a*(I*cos(f*x+e)*b^(1/ 
2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+a*cos(f*x+e)-cos(f*x+e)*b+b)/(cos(f*x 
+e)+1))^(1/2)*(-1/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2 
)-a*cos(f*x+e)+cos(f*x+e)*b-b)/(cos(f*x+e)+1))^(1/2)*a^5+1/15*sin(f*x+e)^7 
*(-30*cos(f*x+e)-30)*EllipticF(((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2...
 

Fricas [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 687, normalized size of antiderivative = 2.73 \[ \int \frac {\cot ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[1/60*(15*(a^4*b*tan(f*x + e)^7 + a^5*tan(f*x + e)^5)*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)*sqrt(-a + b))/(tan(f*x + e)^4 + 2*tan(f*x + e)^2 + 1)) - 4*((15*a^4*b 
- 5*a^3*b^2 - 2*a^2*b^3 - 56*a*b^4 + 48*b^5)*tan(f*x + e)^6 + 3*a^5 - 6*a^ 
4*b + 3*a^3*b^2 + (15*a^5 - 10*a^4*b - a^3*b^2 - 28*a^2*b^3 + 24*a*b^4)*ta 
n(f*x + e)^4 - (5*a^5 - 4*a^4*b - 7*a^3*b^2 + 6*a^2*b^3)*tan(f*x + e)^2)*s 
qrt(b*tan(f*x + e)^2 + a))/((a^6*b - 2*a^5*b^2 + a^4*b^3)*f*tan(f*x + e)^7 
 + (a^7 - 2*a^6*b + a^5*b^2)*f*tan(f*x + e)^5), -1/30*(15*(a^4*b*tan(f*x + 
 e)^7 + a^5*tan(f*x + e)^5)*sqrt(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*((15*a^4*b 
 - 5*a^3*b^2 - 2*a^2*b^3 - 56*a*b^4 + 48*b^5)*tan(f*x + e)^6 + 3*a^5 - 6*a 
^4*b + 3*a^3*b^2 + (15*a^5 - 10*a^4*b - a^3*b^2 - 28*a^2*b^3 + 24*a*b^4)*t 
an(f*x + e)^4 - (5*a^5 - 4*a^4*b - 7*a^3*b^2 + 6*a^2*b^3)*tan(f*x + e)^2)* 
sqrt(b*tan(f*x + e)^2 + a))/((a^6*b - 2*a^5*b^2 + a^4*b^3)*f*tan(f*x + e)^ 
7 + (a^7 - 2*a^6*b + a^5*b^2)*f*tan(f*x + e)^5)]
 

Sympy [F]

\[ \int \frac {\cot ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\cot ^{6}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(cot(f*x+e)**6/(a+b*tan(f*x+e)**2)**(3/2),x)
 

Output:

Integral(cot(e + f*x)**6/(a + b*tan(e + f*x)**2)**(3/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\cot ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F(-1)]

Timed out. \[ \int \frac {\cot ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Hanged} \] Input:

int(cot(e + f*x)^6/(a + b*tan(e + f*x)^2)^(3/2),x)
 

Output:

\text{Hanged}
 

Reduce [F]

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

int(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x)
 

Output:

int(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x)