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

3.4.45.1 Optimal result
3.4.45.2 Mathematica [C] (verified)
3.4.45.3 Rubi [A] (verified)
3.4.45.4 Maple [C] (warning: unable to verify)
3.4.45.5 Fricas [A] (verification not implemented)
3.4.45.6 Sympy [F]
3.4.45.7 Maxima [F(-1)]
3.4.45.8 Giac [F(-1)]
3.4.45.9 Mupad [F(-1)]

3.4.45.1 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
 
3.4.45.2 Mathematica [C] (verified)

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

Time = 17.56 (sec) , antiderivative size = 850, normalized size of antiderivative = 3.37 \[ \int \frac {\cot ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {-\frac {b \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{a (a+b+(a-b) \cos (2 (e+f x)))}-\frac {4 b \sqrt {1+\cos (2 (e+f x))} \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{4 a \sqrt {1+\cos (2 (e+f x))} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}-\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticPi}\left (-\frac {b}{a-b},\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{2 (a-b) \sqrt {1+\cos (2 (e+f x))} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}\right )}{\sqrt {a+b+(a-b) \cos (2 (e+f x))}}}{(a-b) f}+\frac {\sqrt {\frac {a+b+a \cos (2 (e+f x))-b \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (\frac {\left (-23 a^2 \cos (e+f x)-34 a b \cos (e+f x)-33 b^2 \cos (e+f x)\right ) \csc (e+f x)}{15 a^4}+\frac {(11 a \cos (e+f x)+9 b \cos (e+f x)) \csc ^3(e+f x)}{15 a^3}-\frac {\cot (e+f x) \csc ^4(e+f x)}{5 a^2}+\frac {b^4 \sin (2 (e+f x))}{a^4 (a-b) (a+b+a \cos (2 (e+f x))-b \cos (2 (e+f x)))}\right )}{f} \]

input
Integrate[Cot[e + f*x]^6/(a + b*Tan[e + f*x]^2)^(3/2),x]
 
output
-((-((b*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*Sq 
rt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^ 
2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*Csc[2*( 
e + f*x)]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e 
+ f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*x]^4)/(a*(a + b + (a - b)*Cos[2*(e + f 
*x)]))) - (4*b*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[(a + b + (a - b)*Cos[2*(e + 
 f*x)])/(1 + Cos[2*(e + f*x)])]*((Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a* 
(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e 
 + f*x)])*Csc[e + f*x]^2)/b]*Csc[2*(e + f*x)]*EllipticF[ArcSin[Sqrt[((a + 
b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*x] 
^4)/(4*a*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]] 
) - (Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x)])*Csc[e 
+ f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]* 
Csc[2*(e + f*x)]*EllipticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Co 
s[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*x]^4)/(2*(a - b) 
*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])))/Sqrt 
[a + b + (a - b)*Cos[2*(e + f*x)]])/((a - b)*f)) + (Sqrt[(a + b + a*Cos[2* 
(e + f*x)] - b*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(((-23*a^2*Cos[e 
+ f*x] - 34*a*b*Cos[e + f*x] - 33*b^2*Cos[e + f*x])*Csc[e + f*x])/(15*a^4) 
 + ((11*a*Cos[e + f*x] + 9*b*Cos[e + f*x])*Csc[e + f*x]^3)/(15*a^3) - (...
 
3.4.45.3 Rubi [A] (verified)

Time = 0.48 (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
 

3.4.45.3.1 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 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]))
 
3.4.45.4 Maple [C] (warning: unable to verify)

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

Time = 5.33 (sec) , antiderivative size = 1971, normalized size of antiderivative = 7.82

method result size
default \(\text {Expression too large to display}\) \(1971\)

input
int(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 
output
-1/480/f/a^4/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/(a-b)*(-656*((2*I*b 
^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2*b^2*(-cos(f*x+e)+1)^6*csc(f*x+e)^6- 
1280*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b^3*(-cos(f*x+e)+1)^6*csc 
(f*x+e)^6-51*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^3*b*(-cos(f*x+e)+ 
1)^4*csc(f*x+e)^4+32*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2*b^2*(-c 
os(f*x+e)+1)^4*csc(f*x+e)^4+384*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)* 
a*b^3*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-14*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a 
)^(1/2)*a^3*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2-24*((2*I*b^(1/2)*(a-b)^(1/2)+ 
a-2*b)/a)^(1/2)*a^2*b^2*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+3*((2*I*b^(1/2)*(a- 
b)^(1/2)+a-2*b)/a)^(1/2)*a^3*b*(-cos(f*x+e)+1)^12*csc(f*x+e)^12-14*((2*I*b 
^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^3*b*(-cos(f*x+e)+1)^10*csc(f*x+e)^10+ 
960*a^4*(-(2*I*(-cos(f*x+e)+1)^2*b^(1/2)*(a-b)^(1/2)*csc(f*x+e)^2+a*(-cos( 
f*x+e)+1)^2*csc(f*x+e)^2-2*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2-a)/a)^(1/2)*(( 
2*I*(-cos(f*x+e)+1)^2*b^(1/2)*(a-b)^(1/2)*csc(f*x+e)^2-a*(-cos(f*x+e)+1)^2 
*csc(f*x+e)^2+2*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)/a)^(1/2)*EllipticF(((2 
*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*(csc(f*x+e)-cot(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))*(-co 
s(f*x+e)+1)^5*csc(f*x+e)^5-24*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^ 
2*b^2*(-cos(f*x+e)+1)^10*csc(f*x+e)^10-51*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b) 
/a)^(1/2)*a^3*b*(-cos(f*x+e)+1)^8*csc(f*x+e)^8+32*((2*I*b^(1/2)*(a-b)^(...
 
3.4.45.5 Fricas [A] (verification not implemented)

Time = 0.37 (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=\left [\frac {15 \, {\left (a^{4} b \tan \left (f x + e\right )^{7} + a^{5} \tan \left (f x + e\right )^{5}\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 ({\left (15 \, a^{4} b - 5 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 56 \, a b^{4} + 48 \, b^{5}\right )} \tan \left (f x + e\right )^{6} + 3 \, a^{5} - 6 \, a^{4} b + 3 \, a^{3} b^{2} + {\left (15 \, a^{5} - 10 \, a^{4} b - a^{3} b^{2} - 28 \, a^{2} b^{3} + 24 \, a b^{4}\right )} \tan \left (f x + e\right )^{4} - {\left (5 \, a^{5} - 4 \, a^{4} b - 7 \, a^{3} b^{2} + 6 \, a^{2} b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{60 \, {\left ({\left (a^{6} b - 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \tan \left (f x + e\right )^{7} + {\left (a^{7} - 2 \, a^{6} b + a^{5} b^{2}\right )} f \tan \left (f x + e\right )^{5}\right )}}, -\frac {15 \, {\left (a^{4} b \tan \left (f x + e\right )^{7} + a^{5} \tan \left (f x + e\right )^{5}\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 ({\left (15 \, a^{4} b - 5 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 56 \, a b^{4} + 48 \, b^{5}\right )} \tan \left (f x + e\right )^{6} + 3 \, a^{5} - 6 \, a^{4} b + 3 \, a^{3} b^{2} + {\left (15 \, a^{5} - 10 \, a^{4} b - a^{3} b^{2} - 28 \, a^{2} b^{3} + 24 \, a b^{4}\right )} \tan \left (f x + e\right )^{4} - {\left (5 \, a^{5} - 4 \, a^{4} b - 7 \, a^{3} b^{2} + 6 \, a^{2} b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{30 \, {\left ({\left (a^{6} b - 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \tan \left (f x + e\right )^{7} + {\left (a^{7} - 2 \, a^{6} b + a^{5} b^{2}\right )} f \tan \left (f x + e\right )^{5}\right )}}\right ] \]

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)]
 
3.4.45.6 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)
 
3.4.45.7 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
 
3.4.45.8 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
 
3.4.45.9 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}