Integrand size = 25, antiderivative size = 170 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b} f}-\frac {\left (15 a^2+10 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^3 f}+\frac {(5 a+4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a f} \]
-arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/f/(a-b)^(1/2)-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
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 17.38 (sec) , antiderivative size = 794, normalized size of antiderivative = 4.67 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\sqrt {\frac {a+b+a \cos (2 (e+f x))-b \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (\frac {\left (-23 a^2 \cos (e+f x)-14 a b \cos (e+f x)-8 b^2 \cos (e+f x)\right ) \csc (e+f x)}{15 a^3}+\frac {(11 a \cos (e+f x)+4 b \cos (e+f x)) \csc ^3(e+f x)}{15 a^2}-\frac {\cot (e+f x) \csc ^4(e+f x)}{5 a}\right )}{f}+\frac {b \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{a f (a+b+(a-b) \cos (2 (e+f x)))}+\frac {4 b \sqrt {1+\cos (2 (e+f x))} \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{4 a \sqrt {1+\cos (2 (e+f x))} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}-\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticPi}\left (-\frac {b}{a-b},\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{2 (a-b) \sqrt {1+\cos (2 (e+f x))} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}\right )}{f \sqrt {a+b+(a-b) \cos (2 (e+f x))}} \]
(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] - 14*a*b*Cos[e + f*x] - 8*b^2*Cos[e + f*x])* Csc[e + f*x])/(15*a^3) + ((11*a*Cos[e + f*x] + 4*b*Cos[e + f*x])*Csc[e + f *x]^3)/(15*a^2) - (Cot[e + f*x]*Csc[e + f*x]^4)/(5*a)))/f + (b*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)/(a*f*(a + b + (a - b)*Cos[2*(e + f*x)])) + (4*b*Sqr t[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)]*E llipticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Cs c[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)]])))/(f*Sqrt[a + b + (a ...
Time = 0.39 (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 |
| \(\displaystyle \frac {-\frac {-\frac {-\frac {15 a^2 \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b}}-\frac {\left (15 a^2+10 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{3 a}-\frac {(5 a+4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}}{5 a}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a}}{f}\) |
(-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
3.4.32.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]
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]
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]
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]))
Leaf count of result is larger than twice the leaf count of optimal. \(483\) vs. \(2(152)=304\).
Time = 6.59 (sec) , antiderivative size = 484, normalized size of antiderivative = 2.85
| method | result | size |
| default | \(\frac {\left (15 \sin \left (f x +e \right )^{5} \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a^{3} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {a -b}}\right ) \cos \left (f x +e \right )+15 \sin \left (f x +e \right )^{5} \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a^{3} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {a -b}}\right )-8 \sin \left (f x +e \right )^{6} \sqrt {a -b}\, b^{3}+6 \sin \left (f x +e \right )^{4} \cos \left (f x +e \right )^{2} \sqrt {a -b}\, a \,b^{2}-9 \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )^{4} \sqrt {a -b}\, a^{2} b -23 \cos \left (f x +e \right )^{6} \sqrt {a -b}\, a^{3}-10 \sin \left (f x +e \right )^{4} \sqrt {a -b}\, a \,b^{2}+25 \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )^{2} \sqrt {a -b}\, a^{2} b +35 \cos \left (f x +e \right )^{4} \sqrt {a -b}\, a^{3}-15 \sin \left (f x +e \right )^{2} \sqrt {a -b}\, a^{2} b -15 \cos \left (f x +e \right )^{2} \sqrt {a -b}\, a^{3}\right ) \sec \left (f x +e \right ) \csc \left (f x +e \right )}{15 f \,a^{3} \sqrt {a -b}\, \left (\cos \left (f x +e \right )+1\right )^{2} \left (\cos \left (f x +e \right )-1\right )^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}\) | \(484\) |
1/15/f/a^3/(a-b)^(1/2)*(15*sin(f*x+e)^5*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/( cos(f*x+e)+1)^2)^(1/2)*a^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)*(cot(f*x+e)+csc(f*x+e)))*cos(f*x+e)+15*sin( f*x+e)^5*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^3*arct an(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)* (cot(f*x+e)+csc(f*x+e)))-8*sin(f*x+e)^6*(a-b)^(1/2)*b^3+6*sin(f*x+e)^4*cos (f*x+e)^2*(a-b)^(1/2)*a*b^2-9*sin(f*x+e)^2*cos(f*x+e)^4*(a-b)^(1/2)*a^2*b- 23*cos(f*x+e)^6*(a-b)^(1/2)*a^3-10*sin(f*x+e)^4*(a-b)^(1/2)*a*b^2+25*sin(f *x+e)^2*cos(f*x+e)^2*(a-b)^(1/2)*a^2*b+35*cos(f*x+e)^4*(a-b)^(1/2)*a^3-15* sin(f*x+e)^2*(a-b)^(1/2)*a^2*b-15*cos(f*x+e)^2*(a-b)^(1/2)*a^3)/(cos(f*x+e )+1)^2/(cos(f*x+e)-1)^2/(a+b*tan(f*x+e)^2)^(1/2)*sec(f*x+e)*csc(f*x+e)
Time = 0.33 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.57 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\left [-\frac {15 \, a^{3} \sqrt {-a + b} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{5} + 4 \, {\left ({\left (15 \, a^{3} - 5 \, a^{2} b - 2 \, a b^{2} - 8 \, b^{3}\right )} \tan \left (f x + e\right )^{4} + 3 \, a^{3} - 3 \, a^{2} b - {\left (5 \, a^{3} - a^{2} b - 4 \, a b^{2}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{60 \, {\left (a^{4} - a^{3} b\right )} f \tan \left (f x + e\right )^{5}}, -\frac {15 \, \sqrt {a - b} a^{3} \arctan \left (-\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) \tan \left (f x + e\right )^{5} + 2 \, {\left ({\left (15 \, a^{3} - 5 \, a^{2} b - 2 \, a b^{2} - 8 \, b^{3}\right )} \tan \left (f x + e\right )^{4} + 3 \, a^{3} - 3 \, a^{2} b - {\left (5 \, a^{3} - a^{2} b - 4 \, a b^{2}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{30 \, {\left (a^{4} - a^{3} b\right )} f \tan \left (f x + e\right )^{5}}\right ] \]
[-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)]
\[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {\cot ^{6}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \]
Timed out. \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\cot \left (f x + e\right )^{6}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {{\mathrm {cot}\left (e+f\,x\right )}^6}{\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}} \,d x \]