\(\int \cot ^6(e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\) [318]

Optimal result

Integrand size = 25, antiderivative size = 165 \[ \int \cot ^6(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=-\frac {(a-b)^{3/2} \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a f}+\frac {(5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 f}-\frac {a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f} \] Output:

-(a-b)^(3/2)*arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/f-1/1 
5*(15*a^2-20*a*b+3*b^2)*cot(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/a/f+1/15*(5*a- 
6*b)*cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2)/f-1/5*a*cot(f*x+e)^5*(a+b*tan(f 
*x+e)^2)^(1/2)/f
 

Mathematica [C] (warning: unable to verify)

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

Time = 4.25 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.85 \[ \int \cot ^6(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=-\frac {\cos (e+f x) \left (b+a \cot ^2(e+f x)\right )^2 \left (a \left (-2 b+3 a \cot ^2(e+f x)\right ) \operatorname {Hypergeometric2F1}\left (1,1,-\frac {1}{2},\frac {(a-b) \sin ^2(e+f x)}{a}\right )+2 (a-b) (a+b+(a-b) \cos (2 (e+f x))) \operatorname {Hypergeometric2F1}\left (2,2,\frac {1}{2},\frac {(a-b) \sin ^2(e+f x)}{a}\right )\right ) \sin (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^3 f} \] Input:

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

Output:

-1/15*(Cos[e + f*x]*(b + a*Cot[e + f*x]^2)^2*(a*(-2*b + 3*a*Cot[e + f*x]^2 
)*Hypergeometric2F1[1, 1, -1/2, ((a - b)*Sin[e + f*x]^2)/a] + 2*(a - b)*(a 
 + b + (a - b)*Cos[2*(e + f*x)])*Hypergeometric2F1[2, 2, 1/2, ((a - b)*Sin 
[e + f*x]^2)/a])*Sin[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(a^3*f)
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4153, 376, 25, 445, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 376

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {1}{5} \left (\frac {\int \frac {a \cot ^2(e+f x) \left (15 a^2-20 b a+3 b^2+2 (5 a-6 b) b \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{3 a}+\frac {1}{3} (5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}\right )-\frac {1}{5} a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\cot ^2(e+f x) \left (15 a^2-20 b a+3 b^2+2 (5 a-6 b) b \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)+\frac {1}{3} (5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}\right )-\frac {1}{5} a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (-\frac {\int \frac {15 a (a-b)^2}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{a}-\frac {\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}\right )+\frac {1}{3} (5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}\right )-\frac {1}{5} a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (-15 (a-b)^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)-\frac {\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}\right )+\frac {1}{3} (5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}\right )-\frac {1}{5} a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (-15 (a-b)^2 \int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}-\frac {\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}\right )+\frac {1}{3} (5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}\right )-\frac {1}{5} a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (-\frac {\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}-15 (a-b)^{3/2} \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )\right )+\frac {1}{3} (5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}\right )-\frac {1}{5} a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

Input:

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

Output:

(-1/5*(a*Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2]) + (((5*a - 6*b)*Cot[e 
+ f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/3 + (-15*(a - b)^(3/2)*ArcTan[(Sqrt[a 
 - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]] - ((15*a^2 - 20*a*b + 3*b^ 
2)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/a)/3)/5)/f
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 376

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

Leaf count of result is larger than twice the leaf count of optimal. \(533\) vs. \(2(147)=294\).

Time = 23.85 (sec) , antiderivative size = 534, normalized size of antiderivative = 3.24

method	result	size
default	\(-\frac {\left (\sin \left (f x +e \right )^{3} \left (-15 \cos \left (f x +e \right )+15\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 ) a^{3}+\sin \left (f x +e \right )^{3} \left (30 \cos \left (f x +e \right )-30\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 ) a^{2} b +\sin \left (f x +e \right )^{3} \left (-15 \cos \left (f x +e \right )+15\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 ) a \,b^{2}+\left (23 \cos \left (f x +e \right )^{4}-35 \cos \left (f x +e \right )^{2}+15\right ) \sqrt {a -b}\, \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}+\left (26 \cos \left (f x +e \right )^{2}-20\right ) \sin \left (f x +e \right )^{2} \sqrt {a -b}\, \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 b +3 \sin \left (f x +e \right )^{4} \sqrt {a -b}\, \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}}}\, b^{2}\right ) \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \cot \left (f x +e \right )^{3} \csc \left (f x +e \right )^{2}}{15 f a \sqrt {a -b}\, \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}}}\, \left (a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}\right )}\)	\(534\)

Input:

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

Output:

-1/15/f/a/(a-b)^(1/2)*(sin(f*x+e)^3*(-15*cos(f*x+e)+15)*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)/(co 
s(f*x+e)-1))*a^3+sin(f*x+e)^3*(30*cos(f*x+e)-30)*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^2*b+sin(f*x+e)^3*(-15*cos(f*x+e)+15)*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*b^2+(23*cos(f*x+e)^4-35*cos(f*x+e)^2+15)*(a-b)^(1/2)*((a*cos(f*x+e)^2+ 
b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^2+(26*cos(f*x+e)^2-20)*sin(f*x+e 
)^2*(a-b)^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a 
*b+3*sin(f*x+e)^4*(a-b)^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e) 
+1)^2)^(1/2)*b^2)*(a+b*tan(f*x+e)^2)^(3/2)/((a*cos(f*x+e)^2+b*sin(f*x+e)^2 
)/(cos(f*x+e)+1)^2)^(1/2)/(a*cos(f*x+e)^2+b*sin(f*x+e)^2)*cot(f*x+e)^3*csc 
(f*x+e)^2
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 385, normalized size of antiderivative = 2.33 \[ \int \cot ^6(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\left [-\frac {15 \, {\left (a^{2} - a b\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{5} + 4 \, {\left ({\left (15 \, a^{2} - 20 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - {\left (5 \, a^{2} - 6 \, a b\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{60 \, a f \tan \left (f x + e\right )^{5}}, -\frac {15 \, {\left (a^{2} - a b\right )} \sqrt {a - b} \arctan \left (-\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) \tan \left (f x + e\right )^{5} + 2 \, {\left ({\left (15 \, a^{2} - 20 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - {\left (5 \, a^{2} - 6 \, a b\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{30 \, a f \tan \left (f x + e\right )^{5}}\right ] \] Input:

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

Output:

[-1/60*(15*(a^2 - a*b)*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))*tan(f*x + e)^5 + 4*((15*a^2 - 20*a*b + 3*b^2)* 
tan(f*x + e)^4 - (5*a^2 - 6*a*b)*tan(f*x + e)^2 + 3*a^2)*sqrt(b*tan(f*x + 
e)^2 + a))/(a*f*tan(f*x + e)^5), -1/30*(15*(a^2 - a*b)*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))*tan(f*x + e)^5 + 2*((15*a^2 - 20*a*b + 3*b^2)*tan(f*x + e)^4 
- (5*a^2 - 6*a*b)*tan(f*x + e)^2 + 3*a^2)*sqrt(b*tan(f*x + e)^2 + a))/(a*f 
*tan(f*x + e)^5)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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