Integrand size = 25, antiderivative size = 157 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{5/2} f}+\frac {\text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{b^{5/2} f}-\frac {(a+b) \tan ^3(e+f x)}{3 a b f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\left (\frac {1}{a^2}-\frac {1}{b^2}\right ) \tan (e+f x)}{f \sqrt {a+b+b \tan ^2(e+f x)}} \]
-arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(5/2)/f+arctanh(b ^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/b^(5/2)/f+(1/a^2-1/b^2)*tan( f*x+e)/f/(a+b+b*tan(f*x+e)^2)^(1/2)-1/3*(a+b)*tan(f*x+e)^3/a/b/f/(a+b+b*ta n(f*x+e)^2)^(3/2)
Leaf count is larger than twice the leaf count of optimal. \(316\) vs. \(2(157)=314\).
Time = 11.15 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.01 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\left (\frac {b^2 \arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )}{\sqrt {a}}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )}{\sqrt {b}}\right ) (a+2 b+a \cos (2 e+2 f x))^{5/2} \sec ^5(e+f x)}{4 \sqrt {2} a^2 b^2 f \left (a+b \sec ^2(e+f x)\right )^{5/2}}+\frac {(a+2 b+a \cos (2 e+2 f x))^3 \sec ^5(e+f x) \left (\frac {-a^2 \sin (e+f x)-2 a b \sin (e+f x)-b^2 \sin (e+f x)}{6 a^2 b f (a+2 b+a \cos (2 e+2 f x))^2}+\frac {-3 a^2 \sin (e+f x)+a b \sin (e+f x)+4 b^2 \sin (e+f x)}{12 a^2 b^2 f (a+2 b+a \cos (2 e+2 f x))}\right )}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \]
-1/4*(((b^2*ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]]) /Sqrt[a] - (a^2*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x] ^2]])/Sqrt[b])*(a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f*x]^5)/(Sqrt[ 2]*a^2*b^2*f*(a + b*Sec[e + f*x]^2)^(5/2)) + ((a + 2*b + a*Cos[2*e + 2*f*x ])^3*Sec[e + f*x]^5*((-(a^2*Sin[e + f*x]) - 2*a*b*Sin[e + f*x] - b^2*Sin[e + f*x])/(6*a^2*b*f*(a + 2*b + a*Cos[2*e + 2*f*x])^2) + (-3*a^2*Sin[e + f* x] + a*b*Sin[e + f*x] + 4*b^2*Sin[e + f*x])/(12*a^2*b^2*f*(a + 2*b + a*Cos [2*e + 2*f*x]))))/(a + b*Sec[e + f*x]^2)^(5/2)
Time = 0.47 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.13, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 4629, 2075, 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.
\(\displaystyle \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (e+f x)^6}{\left (a+b \sec (e+f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4629 |
\(\displaystyle \frac {\int \frac {\tan ^6(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^{5/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \frac {\int \frac {\tan ^6(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 372 |
\(\displaystyle \frac {\frac {\int \frac {3 \tan ^2(e+f x) \left (a \tan ^2(e+f x)+a+b\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 a b}-\frac {(a+b) \tan ^3(e+f x)}{3 a b \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {\tan ^2(e+f x) \left (a \tan ^2(e+f x)+a+b\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{a b}-\frac {(a+b) \tan ^3(e+f x)}{3 a b \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 440 |
\(\displaystyle \frac {\frac {-\frac {\int -\frac {\tan ^2(e+f x) a^2+a^2-b^2}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a b}-\frac {\left (a^2-b^2\right ) \tan (e+f x)}{a b \sqrt {a+b \tan ^2(e+f x)+b}}}{a b}-\frac {(a+b) \tan ^3(e+f x)}{3 a b \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\frac {\int \frac {\tan ^2(e+f x) a^2+a^2-b^2}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a b}-\frac {\left (a^2-b^2\right ) \tan (e+f x)}{a b \sqrt {a+b \tan ^2(e+f x)+b}}}{a b}-\frac {(a+b) \tan ^3(e+f x)}{3 a b \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 398 |
\(\displaystyle \frac {\frac {\frac {a^2 \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-b^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a b}-\frac {\left (a^2-b^2\right ) \tan (e+f x)}{a b \sqrt {a+b \tan ^2(e+f x)+b}}}{a b}-\frac {(a+b) \tan ^3(e+f x)}{3 a b \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\frac {\frac {a^2 \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-b^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a b}-\frac {\left (a^2-b^2\right ) \tan (e+f x)}{a b \sqrt {a+b \tan ^2(e+f x)+b}}}{a b}-\frac {(a+b) \tan ^3(e+f x)}{3 a b \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\frac {\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b}}-b^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a b}-\frac {\left (a^2-b^2\right ) \tan (e+f x)}{a b \sqrt {a+b \tan ^2(e+f x)+b}}}{a b}-\frac {(a+b) \tan ^3(e+f x)}{3 a b \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\frac {\frac {\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b}}-b^2 \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{a b}-\frac {\left (a^2-b^2\right ) \tan (e+f x)}{a b \sqrt {a+b \tan ^2(e+f x)+b}}}{a b}-\frac {(a+b) \tan ^3(e+f x)}{3 a b \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {\frac {\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b}}-\frac {b^2 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {a}}}{a b}-\frac {\left (a^2-b^2\right ) \tan (e+f x)}{a b \sqrt {a+b \tan ^2(e+f x)+b}}}{a b}-\frac {(a+b) \tan ^3(e+f x)}{3 a b \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
(-1/3*((a + b)*Tan[e + f*x]^3)/(a*b*(a + b + b*Tan[e + f*x]^2)^(3/2)) + (( -((b^2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/Sqrt [a]) + (a^2*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]] )/Sqrt[b])/(a*b) - ((a^2 - b^2)*Tan[e + f*x])/(a*b*Sqrt[a + b + b*Tan[e + f*x]^2]))/(a*b))/f
3.5.34.3.1 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[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && ! BinomialMatchQ[{u, v}, x]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f _.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/f Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 )), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
Leaf count of result is larger than twice the leaf count of optimal. \(1499\) vs. \(2(139)=278\).
Time = 9.86 (sec) , antiderivative size = 1500, normalized size of antiderivative = 9.55
-1/6/f/b^(7/2)/(-a)^(1/2)/a^2*(12*(-a)^(1/2)*b^(7/2)*a*(1-cos(f*x+e))^5*cs c(f*x+e)^5+12*(-a)^(1/2)*b^(9/2)*(1-cos(f*x+e))^5*csc(f*x+e)^5-40*(1-cos(f *x+e))^3*a*(-a)^(1/2)*b^(7/2)*csc(f*x+e)^3+24*(-a)^(1/2)*b^(9/2)*(1-cos(f* x+e))^3*csc(f*x+e)^3-12*(-a)^(1/2)*b^(3/2)*a^3*(1-cos(f*x+e))^5*csc(f*x+e) ^5-12*(-a)^(1/2)*b^(5/2)*a^2*(1-cos(f*x+e))^5*csc(f*x+e)^5-40*(1-cos(f*x+e ))^3*a^2*(-a)^(1/2)*b^(5/2)*csc(f*x+e)^3-6*ln(4*((-a)^(1/2)*(a*(1-cos(f*x+ e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*cs c(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)-2*a*(csc(f*x+e)-co t(f*x+e)))/((1-cos(f*x+e))^2*csc(f*x+e)^2+1))*b^(7/2)*(a*(1-cos(f*x+e))^4* csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+ e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(3/2)+12*(-a)^(1/2)*b^(7/2)*a* (csc(f*x+e)-cot(f*x+e))+12*(-a)^(1/2)*b^(9/2)*(csc(f*x+e)-cot(f*x+e))+24*( -a)^(1/2)*b^(3/2)*a^3*(1-cos(f*x+e))^3*csc(f*x+e)^3+3*(-a)^(1/2)*ln(4*(-a* (1-cos(f*x+e))^2*csc(f*x+e)^2-b*(1-cos(f*x+e))^2*csc(f*x+e)^2+b^(1/2)*(a*( 1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f* x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)-2*a*(csc (f*x+e)-cot(f*x+e))-a-b)/((1-cos(f*x+e))^2*csc(f*x+e)^2+2*csc(f*x+e)-2*cot (f*x+e)+1))*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e) ^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b )^(3/2)*a^2*b+3*(-a)^(1/2)*ln(4*(a*(1-cos(f*x+e))^2*csc(f*x+e)^2+b*(1-c...
Leaf count of result is larger than twice the leaf count of optimal. 433 vs. \(2 (139) = 278\).
Time = 2.15 (sec) , antiderivative size = 2035, normalized size of antiderivative = 12.96 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]
[-1/24*(3*(a^2*b^3*cos(f*x + e)^4 + 2*a*b^4*cos(f*x + e)^2 + b^5)*sqrt(-a) *log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28 *a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 - 8*( 16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^ 2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) - 6 *(a^5*cos(f*x + e)^4 + 2*a^4*b*cos(f*x + e)^2 + a^3*b^2)*sqrt(b)*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 + 4*((a - b) *cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/co s(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4) + 8*((3*a^4*b - a^3*b^ 2 - 4*a^2*b^3)*cos(f*x + e)^3 + (4*a^3*b^2 + a^2*b^3 - 3*a*b^4)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^5*b^3*f*c os(f*x + e)^4 + 2*a^4*b^4*f*cos(f*x + e)^2 + a^3*b^5*f), 1/24*(12*(a^5*cos (f*x + e)^4 + 2*a^4*b*cos(f*x + e)^2 + a^3*b^2)*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b )/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e))) - 3*(a^2*b^3* cos(f*x + e)^4 + 2*a*b^4*cos(f*x + e)^2 + b^5)*sqrt(-a)*log(128*a^4*cos(f* x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2 *b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 3...
\[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\tan ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^6}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \]