Integrand size = 25, antiderivative size = 182 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{3/2} f}-\frac {(3 a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 b^{5/2} f}-\frac {a \tan ^3(e+f x)}{(a-b) b f \sqrt {a+b \tan ^2(e+f x)}}+\frac {(3 a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 (a-b) b^2 f} \]
-arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(3/2)/f-1/2 *(3*a+2*b)*arctanh(b^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/b^(5/2)/f+ 1/2*(3*a-b)*(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/(a-b)/b^2/f-a*tan(f*x+e)^3 /(a-b)/b/f/(a+b*tan(f*x+e)^2)^(1/2)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 6.46 (sec) , antiderivative size = 787, normalized size of antiderivative = 4.32 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {-\frac {b \left (3 a^2-a b-b^2\right ) \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{a (a+b+(a-b) \cos (2 (e+f x)))}-\frac {4 b^3 \sqrt {1+\cos (2 (e+f x))} \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{4 a \sqrt {1+\cos (2 (e+f x))} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}-\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticPi}\left (-\frac {b}{a-b},\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{2 (a-b) \sqrt {1+\cos (2 (e+f x))} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}\right )}{\sqrt {a+b+(a-b) \cos (2 (e+f x))}}}{(a-b) b^2 f}+\frac {\sqrt {\frac {a+b+a \cos (2 (e+f x))-b \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (-\frac {a^2 \sin (2 (e+f x))}{(a-b) b^2 (-a-b-a \cos (2 (e+f x))+b \cos (2 (e+f x)))}+\frac {\tan (e+f x)}{2 b^2}\right )}{f} \]
-((-((b*(3*a^2 - a*b - b^2)*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + C os[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*(a + b + (a - b)*Cos[2*(e + f*x)]))) - (4*b^3*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)]*Ellipt icF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqr t[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 + Co s[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[Sqr t[((a + b + (a - b)*Cos[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)*b^2*f)) + (Sqrt[(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(-((a^2*Sin[2*(e + f*x)])/((a - b)*b^2*(-a - b - a*Cos[2*(e + f*x )] + b*Cos[2*(e + f*x)]))) + Tan[e + f*x]/(2*b^2)))/f
Time = 0.42 (sec) , antiderivative size = 192, 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, 372, 444, 398, 224, 219, 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.
| \(\displaystyle \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (e+f x)^6}{\left (a+b \tan (e+f x)^2\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle \frac {\int \frac {\tan ^6(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\int \frac {\tan ^2(e+f x) \left ((3 a-b) \tan ^2(e+f x)+3 a\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{b (a-b)}-\frac {a \tan ^3(e+f x)}{b (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
| \(\Big \downarrow \) 444 |
| \(\displaystyle \frac {\frac {\frac {(3 a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {\int \frac {(a-b) (3 a+2 b) \tan ^2(e+f x)+a (3 a-b)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{2 b}}{b (a-b)}-\frac {a \tan ^3(e+f x)}{b (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle \frac {\frac {\frac {(3 a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {2 b^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)+(a-b) (3 a+2 b) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{2 b}}{b (a-b)}-\frac {a \tan ^3(e+f x)}{b (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\frac {(3 a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {2 b^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)+(a-b) (3 a+2 b) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}}{2 b}}{b (a-b)}-\frac {a \tan ^3(e+f x)}{b (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {(3 a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {2 b^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)+\frac {(a-b) (3 a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {b}}}{2 b}}{b (a-b)}-\frac {a \tan ^3(e+f x)}{b (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {(3 a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {2 b^2 \int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}+\frac {(a-b) (3 a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {b}}}{2 b}}{b (a-b)}-\frac {a \tan ^3(e+f x)}{b (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {(3 a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {\frac {2 b^2 \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b}}+\frac {(a-b) (3 a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {b}}}{2 b}}{b (a-b)}-\frac {a \tan ^3(e+f x)}{b (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
(-((a*Tan[e + f*x]^3)/((a - b)*b*Sqrt[a + b*Tan[e + f*x]^2])) + (-1/2*((2* b^2*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/Sqrt[a - b] + ((a - b)*(3*a + 2*b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[ e + f*x]^2]])/Sqrt[b])/b + ((3*a - b)*Tan[e + f*x]*Sqrt[a + b*Tan[e + f*x] ^2])/(2*b))/((a - b)*b))/f
3.4.39.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 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[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) , x_Symbol] :> Simp[f/b Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ b Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} , x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[f*g*(g*x)^(m - 1)*(a + b*x^2)^ (p + 1)*((c + d*x^2)^(q + 1)/(b*d*(m + 2*(p + q + 1) + 1))), x] - Simp[g^2/ (b*d*(m + 2*(p + q + 1) + 1)) Int[(g*x)^(m - 2)*(a + b*x^2)^p*(c + d*x^2) ^q*Simp[a*f*c*(m - 1) + (a*f*d*(m + 2*q + 1) + b*(f*c*(m + 2*p + 1) - e*d*( m + 2*(p + q + 1) + 1)))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && GtQ[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]))
Time = 0.08 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.48
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (f x +e \right )}{a \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {\tan \left (f x +e \right )^{3}}{2 b \sqrt {a +b \tan \left (f x +e \right )^{2}}}-\frac {3 a \left (-\frac {\tan \left (f x +e \right )}{b \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{b^{\frac {3}{2}}}\right )}{2 b}+\frac {\tan \left (f x +e \right )}{b \sqrt {a +b \tan \left (f x +e \right )^{2}}}-\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{b^{\frac {3}{2}}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{\left (a -b \right )^{2} b^{2}}+\frac {b \tan \left (f x +e \right )}{\left (a -b \right ) a \sqrt {a +b \tan \left (f x +e \right )^{2}}}}{f}\) | \(270\) |
| default | \(\frac {\frac {\tan \left (f x +e \right )}{a \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {\tan \left (f x +e \right )^{3}}{2 b \sqrt {a +b \tan \left (f x +e \right )^{2}}}-\frac {3 a \left (-\frac {\tan \left (f x +e \right )}{b \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{b^{\frac {3}{2}}}\right )}{2 b}+\frac {\tan \left (f x +e \right )}{b \sqrt {a +b \tan \left (f x +e \right )^{2}}}-\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{b^{\frac {3}{2}}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{\left (a -b \right )^{2} b^{2}}+\frac {b \tan \left (f x +e \right )}{\left (a -b \right ) a \sqrt {a +b \tan \left (f x +e \right )^{2}}}}{f}\) | \(270\) |
1/f*(tan(f*x+e)/a/(a+b*tan(f*x+e)^2)^(1/2)+1/2*tan(f*x+e)^3/b/(a+b*tan(f*x +e)^2)^(1/2)-3/2*a/b*(-tan(f*x+e)/b/(a+b*tan(f*x+e)^2)^(1/2)+1/b^(3/2)*ln( b^(1/2)*tan(f*x+e)+(a+b*tan(f*x+e)^2)^(1/2)))+tan(f*x+e)/b/(a+b*tan(f*x+e) ^2)^(1/2)-1/b^(3/2)*ln(b^(1/2)*tan(f*x+e)+(a+b*tan(f*x+e)^2)^(1/2))-1/(a-b )^2*(b^4*(a-b))^(1/2)/b^2*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/(a+b*tan(f*x+ e)^2)^(1/2)*tan(f*x+e))+b/(a-b)*tan(f*x+e)/a/(a+b*tan(f*x+e)^2)^(1/2))
Time = 1.32 (sec) , antiderivative size = 1207, normalized size of antiderivative = 6.63 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \]
[1/4*((3*a^4 - 4*a^3*b - a^2*b^2 + 2*a*b^3 + (3*a^3*b - 4*a^2*b^2 - a*b^3 + 2*b^4)*tan(f*x + e)^2)*sqrt(b)*log(2*b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(b)*tan(f*x + e) + a) + 2*(b^4*tan(f*x + e)^2 + a*b^3)*sq rt(-a + b)*log(-((a - 2*b)*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*s qrt(-a + b)*tan(f*x + e) - a)/(tan(f*x + e)^2 + 1)) + 2*((a^2*b^2 - 2*a*b^ 3 + b^4)*tan(f*x + e)^3 + (3*a^3*b - 4*a^2*b^2 + a*b^3)*tan(f*x + e))*sqrt (b*tan(f*x + e)^2 + a))/((a^2*b^4 - 2*a*b^5 + b^6)*f*tan(f*x + e)^2 + (a^3 *b^3 - 2*a^2*b^4 + a*b^5)*f), 1/2*((3*a^4 - 4*a^3*b - a^2*b^2 + 2*a*b^3 + (3*a^3*b - 4*a^2*b^2 - a*b^3 + 2*b^4)*tan(f*x + e)^2)*sqrt(-b)*arctan(sqrt (b*tan(f*x + e)^2 + a)*sqrt(-b)/(b*tan(f*x + e))) + (b^4*tan(f*x + e)^2 + a*b^3)*sqrt(-a + b)*log(-((a - 2*b)*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e) ^2 + a)*sqrt(-a + b)*tan(f*x + e) - a)/(tan(f*x + e)^2 + 1)) + ((a^2*b^2 - 2*a*b^3 + b^4)*tan(f*x + e)^3 + (3*a^3*b - 4*a^2*b^2 + a*b^3)*tan(f*x + e ))*sqrt(b*tan(f*x + e)^2 + a))/((a^2*b^4 - 2*a*b^5 + b^6)*f*tan(f*x + e)^2 + (a^3*b^3 - 2*a^2*b^4 + a*b^5)*f), -1/4*(4*(b^4*tan(f*x + e)^2 + a*b^3)* sqrt(a - b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) - (3*a^4 - 4*a^3*b - a^2*b^2 + 2*a*b^3 + (3*a^3*b - 4*a^2*b^2 - a*b^3 + 2 *b^4)*tan(f*x + e)^2)*sqrt(b)*log(2*b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(b)*tan(f*x + e) + a) - 2*((a^2*b^2 - 2*a*b^3 + b^4)*tan(f*x + e)^3 + (3*a^3*b - 4*a^2*b^2 + a*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + ...
\[ \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\tan ^{6}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Timed out. \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^6}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]