\(\int \frac {\cot ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) [332]

Optimal result

Integrand size = 25, antiderivative size = 170 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b} f}-\frac {\left (15 a^2+10 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^3 f}+\frac {(5 a+4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a f} \] Output:

-arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(1/2)/f-1/1 
5*(15*a^2+10*a*b+8*b^2)*cot(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/a^3/f+1/15*(5* 
a+4*b)*cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2)/a^2/f-1/5*cot(f*x+e)^5*(a+b*t 
an(f*x+e)^2)^(1/2)/a/f
 

Mathematica [C] (warning: unable to verify)

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

Time = 11.65 (sec) , antiderivative size = 1214, normalized size of antiderivative = 7.14 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/45*(Cos[e + f*x]^4*Cot[e + f*x]^5*(1 + (b*Tan[e + f*x]^2)/a)*(9*a^4*Arc 
Sin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]] + 9*a^4*ArcSin[Sqrt[((a - b)*Sin[e + 
 f*x]^2)/a]]*Tan[e + f*x]^2 - 18*a^3*b*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2 
)/a]]*Tan[e + f*x]^2 - 18*a^3*b*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*T 
an[e + f*x]^4 + 72*a^2*b^2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e 
+ f*x]^4 + 72*a^2*b^2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x 
]^6 + 144*a*b^3*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^6 + 
144*a*b^3*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^8 - 4*a^4* 
Hypergeometric2F1[2, 2, 5/2, ((a - b)*Sin[e + f*x]^2)/a]*Tan[e + f*x]^2*Sq 
rt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a^2] + 4 
*a^3*b*Hypergeometric2F1[2, 2, 5/2, ((a - b)*Sin[e + f*x]^2)/a]*Tan[e + f* 
x]^2*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a 
^2] - 12*a^3*b*Hypergeometric2F1[2, 2, 5/2, ((a - b)*Sin[e + f*x]^2)/a]*Ta 
n[e + f*x]^4*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2*(a + b*Tan[e + f* 
x]^2))/a^2] + 12*a^2*b^2*Hypergeometric2F1[2, 2, 5/2, ((a - b)*Sin[e + f*x 
]^2)/a]*Tan[e + f*x]^4*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2*(a + b* 
Tan[e + f*x]^2))/a^2] + 168*a^2*b^2*Hypergeometric2F1[2, 2, 5/2, ((a - b)* 
Sin[e + f*x]^2)/a]*Tan[e + f*x]^6*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x 
]^2*(a + b*Tan[e + f*x]^2))/a^2] - 168*a*b^3*Hypergeometric2F1[2, 2, 5/2, 
((a - b)*Sin[e + f*x]^2)/a]*Tan[e + f*x]^6*Sqrt[((a - b)*Cos[e + f*x]^2...
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4153, 382, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 382

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 445

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

\(\Big \downarrow \) 445

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

(-1/5*(Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/a - (-1/3*((5*a + 4*b)*C 
ot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/a - ((-15*a^2*ArcTan[(Sqrt[a - b 
]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/Sqrt[a - b] - ((15*a^2 + 10*a 
*b + 8*b^2)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/a)/(3*a))/(5*a))/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 382

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) 
, x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/ 
(a*c*e*(m + 1))), x] - Simp[1/(a*c*e^2*(m + 1))   Int[(e*x)^(m + 2)*(a + b* 
x^2)^p*(c + d*x^2)^q*Simp[(b*c + a*d)*(m + 3) + 2*(b*c*p + a*d*q) + b*d*(m 
+ 2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[ 
b*c - a*d, 0] && 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 [C] (verified)

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

Time = 13.85 (sec) , antiderivative size = 773, normalized size of antiderivative = 4.55

method	result	size
default	\(\frac {-\frac {\sqrt {\frac {i \cos \left (f x +e \right ) \sqrt {b}\, \sqrt {a -b}-i \sqrt {b}\, \sqrt {a -b}+a \cos \left (f x +e \right )-\cos \left (f x +e \right ) b +b}{a \left (\cos \left (f x +e \right )+1\right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right ) \sqrt {b}\, \sqrt {a -b}-i \sqrt {b}\, \sqrt {a -b}-a \cos \left (f x +e \right )+\cos \left (f x +e \right ) b -b}{a \left (\cos \left (f x +e \right )+1\right )}}\, \operatorname {EllipticPi}\left (\sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), -\frac {a}{2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}, \frac {\sqrt {-\frac {2 i \sqrt {b}\, \sqrt {a -b}-a +2 b}{a}}}{\sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}}\right ) a^{3} \left (-60-60 \sec \left (f x +e \right )\right )}{15}-\frac {\sqrt {\frac {i \cos \left (f x +e \right ) \sqrt {b}\, \sqrt {a -b}-i \sqrt {b}\, \sqrt {a -b}+a \cos \left (f x +e \right )-\cos \left (f x +e \right ) b +b}{a \left (\cos \left (f x +e \right )+1\right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right ) \sqrt {b}\, \sqrt {a -b}-i \sqrt {b}\, \sqrt {a -b}-a \cos \left (f x +e \right )+\cos \left (f x +e \right ) b -b}{a \left (\cos \left (f x +e \right )+1\right )}}\, \operatorname {EllipticF}\left (\sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), \sqrt {\frac {8 i b^{\frac {3}{2}} \sqrt {a -b}-4 i \sqrt {b}\, \sqrt {a -b}\, a +a^{2}-8 a b +8 b^{2}}{a^{2}}}\right ) a^{3} \left (30+30 \sec \left (f x +e \right )\right )}{15}-\frac {\left (23 \cos \left (f x +e \right )^{4}-35 \cos \left (f x +e \right )^{2}+15\right ) \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, a^{3} \cot \left (f x +e \right ) \csc \left (f x +e \right )^{4}}{15}-\frac {\left (9 \cos \left (f x +e \right )^{4}-25 \cos \left (f x +e \right )^{2}+15\right ) \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, a^{2} b \sec \left (f x +e \right ) \csc \left (f x +e \right )^{3}}{15}-\frac {\sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, a \,b^{2} \left (-6 \cot \left (f x +e \right )+10 \sec \left (f x +e \right ) \csc \left (f x +e \right )\right )}{15}-\frac {8 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, b^{3} \tan \left (f x +e \right )}{15}}{f \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, a^{3} \sqrt {a +b \tan \left (f x +e \right )^{2}}}\)	\(773\)

Input:

int(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/f*(-1/15*(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)*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))*a^3*(-60 
-60*sec(f*x+e))-1/15*(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)*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))*a^3*(30+30*sec(f*x+e))-1/15*(23*cos(f*x+e 
)^4-35*cos(f*x+e)^2+15)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^3*cot( 
f*x+e)*csc(f*x+e)^4-1/15*(9*cos(f*x+e)^4-25*cos(f*x+e)^2+15)*((2*I*b^(1/2) 
*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2*b*sec(f*x+e)*csc(f*x+e)^3-1/15*((2*I*b^(1 
/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b^2*(-6*cot(f*x+e)+10*sec(f*x+e)*csc(f*x 
+e))-8/15*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^3*tan(f*x+e))/((2*I* 
b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/a^3/(a+b*tan(f*x+e)^2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.57 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\left [-\frac {15 \, a^{3} \sqrt {-a + b} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{5} + 4 \, {\left ({\left (15 \, a^{3} - 5 \, a^{2} b - 2 \, a b^{2} - 8 \, b^{3}\right )} \tan \left (f x + e\right )^{4} + 3 \, a^{3} - 3 \, a^{2} b - {\left (5 \, a^{3} - a^{2} b - 4 \, a b^{2}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{60 \, {\left (a^{4} - a^{3} b\right )} f \tan \left (f x + e\right )^{5}}, -\frac {15 \, \sqrt {a - b} a^{3} \arctan \left (-\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) \tan \left (f x + e\right )^{5} + 2 \, {\left ({\left (15 \, a^{3} - 5 \, a^{2} b - 2 \, a b^{2} - 8 \, b^{3}\right )} \tan \left (f x + e\right )^{4} + 3 \, a^{3} - 3 \, a^{2} b - {\left (5 \, a^{3} - a^{2} b - 4 \, a b^{2}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{30 \, {\left (a^{4} - a^{3} b\right )} f \tan \left (f x + e\right )^{5}}\right ] \] Input:

integrate(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[-1/60*(15*a^3*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*ta 
n(f*x + e))*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b))/(tan(f*x + e)^4 + 2*t 
an(f*x + e)^2 + 1))*tan(f*x + e)^5 + 4*((15*a^3 - 5*a^2*b - 2*a*b^2 - 8*b^ 
3)*tan(f*x + e)^4 + 3*a^3 - 3*a^2*b - (5*a^3 - a^2*b - 4*a*b^2)*tan(f*x + 
e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^4 - a^3*b)*f*tan(f*x + e)^5), -1/30* 
(15*sqrt(a - b)*a^3*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^3 - 5*a^ 
2*b - 2*a*b^2 - 8*b^3)*tan(f*x + e)^4 + 3*a^3 - 3*a^2*b - (5*a^3 - a^2*b - 
 4*a*b^2)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^4 - a^3*b)*f*tan 
(f*x + e)^5)]
 

Sympy [F]

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

integrate(cot(f*x+e)**6/(a+b*tan(f*x+e)**2)**(1/2),x)
 

Output:

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

Maxima [F(-1)]

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

integrate(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(cot(e + f*x)^6/(a + b*tan(e + f*x)^2)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int(cot(e + f*x)^6/(a + b*tan(e + f*x)^2)^(1/2), x)
 

Reduce [F]

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

int(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(1/2),x)
 

Output:

int(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(1/2),x)