Integrand size = 25, antiderivative size = 171 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{5/2} f}+\frac {\text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{b^{5/2} f}-\frac {a \tan ^3(e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (a-2 b) \tan (e+f x)}{(a-b)^2 b^2 f \sqrt {a+b \tan ^2(e+f x)}} \]
-arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(5/2)/f+arc tanh(b^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/b^(5/2)/f-a*(a-2*b)*tan( f*x+e)/(a-b)^2/b^2/f/(a+b*tan(f*x+e)^2)^(1/2)-1/3*a*tan(f*x+e)^3/(a-b)/b/f /(a+b*tan(f*x+e)^2)^(3/2)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 4.90 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.73 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)} \left (a^2 (a-b) (2 a b+(3 a-7 b) (a+b+(a-b) \cos (2 (e+f x)))) \sin (2 (e+f x))-\frac {3 a^2 b \left (\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}\right )^{3/2} \left (\left (a^2-3 a b+2 b^2\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )+b^2 \operatorname {EllipticPi}\left (-\frac {b}{a-b},\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )\right ) \sin ^2(e+f x) \sin (2 (e+f x))}{\sqrt {2}}\right )}{3 \sqrt {2} a (a-b)^3 b^2 f (a+b+(a-b) \cos (2 (e+f x)))^2} \]
-1/3*(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(a^2*(a - b) *(2*a*b + (3*a - 7*b)*(a + b + (a - b)*Cos[2*(e + f*x)]))*Sin[2*(e + f*x)] - (3*a^2*b*(((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)^(3/2)* ((a^2 - 3*a*b + 2*b^2)*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1] + b^2*EllipticPi[-(b/(a - b)), ArcS in[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1 ])*Sin[e + f*x]^2*Sin[2*(e + f*x)])/Sqrt[2]))/(Sqrt[2]*a*(a - b)^3*b^2*f*( a + b + (a - b)*Cos[2*(e + f*x)])^2)
Time = 0.43 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.15, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4153, 372, 27, 440, 25, 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.
| \(\displaystyle \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (e+f x)^6}{\left (a+b \tan (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle \frac {\int \frac {\tan ^6(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\int \frac {3 \tan ^2(e+f x) \left ((a-b) \tan ^2(e+f x)+a\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{3 b (a-b)}-\frac {a \tan ^3(e+f x)}{3 b (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\tan ^2(e+f x) \left ((a-b) \tan ^2(e+f x)+a\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{b (a-b)}-\frac {a \tan ^3(e+f x)}{3 b (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 440 |
| \(\displaystyle \frac {\frac {-\frac {\int -\frac {(a-b)^2 \tan ^2(e+f x)+a (a-2 b)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{b (a-b)}-\frac {a (a-2 b) \tan (e+f x)}{b (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{b (a-b)}-\frac {a \tan ^3(e+f x)}{3 b (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {(a-b)^2 \tan ^2(e+f x)+a (a-2 b)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{b (a-b)}-\frac {a (a-2 b) \tan (e+f x)}{b (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{b (a-b)}-\frac {a \tan ^3(e+f x)}{3 b (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle \frac {\frac {\frac {(a-b)^2 \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)-b^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{b (a-b)}-\frac {a (a-2 b) \tan (e+f x)}{b (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{b (a-b)}-\frac {a \tan ^3(e+f x)}{3 b (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\frac {(a-b)^2 \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}-b^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{b (a-b)}-\frac {a (a-2 b) \tan (e+f x)}{b (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{b (a-b)}-\frac {a \tan ^3(e+f x)}{3 b (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {\frac {(a-b)^2 \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {b}}-b^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{b (a-b)}-\frac {a (a-2 b) \tan (e+f x)}{b (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{b (a-b)}-\frac {a \tan ^3(e+f x)}{3 b (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {\frac {(a-b)^2 \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {b}}-b^2 \int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}}{b (a-b)}-\frac {a (a-2 b) \tan (e+f x)}{b (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{b (a-b)}-\frac {a \tan ^3(e+f x)}{3 b (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {(a-b)^2 \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {b}}-\frac {b^2 \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b}}}{b (a-b)}-\frac {a (a-2 b) \tan (e+f x)}{b (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{b (a-b)}-\frac {a \tan ^3(e+f x)}{3 b (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\) |
(-1/3*(a*Tan[e + f*x]^3)/((a - b)*b*(a + b*Tan[e + f*x]^2)^(3/2)) + ((-((b ^2*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/Sqrt[a - b]) + ((a - b)^2*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2 ]])/Sqrt[b])/((a - b)*b) - (a*(a - 2*b)*Tan[e + f*x])/((a - b)*b*Sqrt[a + b*Tan[e + f*x]^2]))/((a - b)*b))/f
3.4.52.3.1 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[((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[g*(b*e - a*f)*(g*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] - Simp[ g^2/(2*b*(b*c - a*d)*(p + 1)) Int[(g*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f)*(m - 1) + (d*(b*e - a*f)*(m + 2*q + 1) - b*2*(c *f - d*e)*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, q}, x] && LtQ[p, -1] && 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]))
Leaf count of result is larger than twice the leaf count of optimal. \(381\) vs. \(2(153)=306\).
Time = 0.08 (sec) , antiderivative size = 382, normalized size of antiderivative = 2.23
| method | result | size |
| derivativedivides | \(\frac {\tan \left (f x +e \right )}{3 f a \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 \tan \left (f x +e \right )}{3 f \,a^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}-\frac {\tan \left (f x +e \right )^{3}}{3 f b \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {\tan \left (f x +e \right )}{f \,b^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{f \,b^{\frac {5}{2}}}+\frac {\tan \left (f x +e \right )}{3 f b \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {\tan \left (f x +e \right )}{3 f a b \sqrt {a +b \tan \left (f x +e \right )^{2}}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{f \left (a -b \right )^{3} b^{2}}+\frac {b \tan \left (f x +e \right )}{3 a \left (a -b \right ) f \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 b \tan \left (f x +e \right )}{3 f \left (a -b \right ) a^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {b \tan \left (f x +e \right )}{f \left (a -b \right )^{2} a \sqrt {a +b \tan \left (f x +e \right )^{2}}}\) | \(382\) |
| default | \(\frac {\tan \left (f x +e \right )}{3 f a \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 \tan \left (f x +e \right )}{3 f \,a^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}-\frac {\tan \left (f x +e \right )^{3}}{3 f b \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {\tan \left (f x +e \right )}{f \,b^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{f \,b^{\frac {5}{2}}}+\frac {\tan \left (f x +e \right )}{3 f b \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {\tan \left (f x +e \right )}{3 f a b \sqrt {a +b \tan \left (f x +e \right )^{2}}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{f \left (a -b \right )^{3} b^{2}}+\frac {b \tan \left (f x +e \right )}{3 a \left (a -b \right ) f \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 b \tan \left (f x +e \right )}{3 f \left (a -b \right ) a^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {b \tan \left (f x +e \right )}{f \left (a -b \right )^{2} a \sqrt {a +b \tan \left (f x +e \right )^{2}}}\) | \(382\) |
1/3/f*tan(f*x+e)/a/(a+b*tan(f*x+e)^2)^(3/2)+2/3/f/a^2*tan(f*x+e)/(a+b*tan( f*x+e)^2)^(1/2)-1/3/f*tan(f*x+e)^3/b/(a+b*tan(f*x+e)^2)^(3/2)-1/f/b^2*tan( f*x+e)/(a+b*tan(f*x+e)^2)^(1/2)+1/f/b^(5/2)*ln(b^(1/2)*tan(f*x+e)+(a+b*tan (f*x+e)^2)^(1/2))+1/3/f/b*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(3/2)-1/3/f/a/b*ta n(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2)-1/f/(a-b)^3*(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))+1/3*b*ta n(f*x+e)/a/(a-b)/f/(a+b*tan(f*x+e)^2)^(3/2)+2/3/f*b/(a-b)/a^2*tan(f*x+e)/( a+b*tan(f*x+e)^2)^(1/2)+1/f*b/(a-b)^2*tan(f*x+e)/a/(a+b*tan(f*x+e)^2)^(1/2 )
Leaf count of result is larger than twice the leaf count of optimal. 410 vs. \(2 (153) = 306\).
Time = 1.48 (sec) , antiderivative size = 1714, normalized size of antiderivative = 10.02 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]
[1/6*(3*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3 + (a^3*b^2 - 3*a^2*b^3 + 3*a* b^4 - b^5)*tan(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*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) - 3*(b^5*tan(f*x + e)^4 + 2*a*b^4*tan(f*x + e)^2 + a^2*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)) - 2*((4* a^3*b^2 - 11*a^2*b^3 + 7*a*b^4)*tan(f*x + e)^3 + 3*(a^4*b - 3*a^3*b^2 + 2* a^2*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/((a^3*b^5 - 3*a^2*b^6 + 3*a*b^7 - b^8)*f*tan(f*x + e)^4 + 2*(a^4*b^4 - 3*a^3*b^5 + 3*a^2*b^6 - a* b^7)*f*tan(f*x + e)^2 + (a^5*b^3 - 3*a^4*b^4 + 3*a^3*b^5 - a^2*b^6)*f), -1 /6*(6*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3 + (a^3*b^2 - 3*a^2*b^3 + 3*a*b^ 4 - b^5)*tan(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*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))) + 3*(b^5*tan(f*x + e)^4 + 2*a*b^4*tan(f*x + e)^2 + a^2*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)) + 2*((4*a^3*b^2 - 11*a^2*b^3 + 7*a*b^4)*tan(f*x + e)^3 + 3*(a^4*b - 3*a^3*b^2 + 2*a^2*b^3)*tan(f*x + e ))*sqrt(b*tan(f*x + e)^2 + a))/((a^3*b^5 - 3*a^2*b^6 + 3*a*b^7 - b^8)*f*ta n(f*x + e)^4 + 2*(a^4*b^4 - 3*a^3*b^5 + 3*a^2*b^6 - a*b^7)*f*tan(f*x + e)^ 2 + (a^5*b^3 - 3*a^4*b^4 + 3*a^3*b^5 - a^2*b^6)*f), -1/6*(6*(b^5*tan(f*...
\[ \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\tan ^{6}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{6}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^6}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]