Integrand size = 25, antiderivative size = 133 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{b^{5/2} f}-\frac {a \sec ^2(e+f x) \tan (e+f x)}{3 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (3 a+5 b) \tan (e+f x)}{3 b^2 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}} \]
arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/b^(5/2)/f-1/3*a*(3* a+5*b)*tan(f*x+e)/b^2/(a+b)^2/f/(a+b+b*tan(f*x+e)^2)^(1/2)-1/3*a*sec(f*x+e )^2*tan(f*x+e)/b/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 10.35 (sec) , antiderivative size = 592, normalized size of antiderivative = 4.45 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {(a+2 b+a \cos (2 e+2 f x))^{5/2} \sec ^6(e+f x) \sqrt {1-\frac {2 a \sin ^2(e+f x)}{2 a+2 b}} \tan (e+f x) \left (-\frac {24 b \cos ^2(e+f x) \, _3F_2\left (2,2,2;1,\frac {9}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x) \left (-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}\right )^{5/2}}{a+b}-\frac {\sec ^6(e+f x) \left (24 b^3 \operatorname {Hypergeometric2F1}\left (2,2,\frac {9}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^6(e+f x) \left (4 b^2+a b \left (8-7 \sin ^2(e+f x)\right )+a^2 \left (4-7 \sin ^2(e+f x)+3 \sin ^4(e+f x)\right )\right ) \sqrt {-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}}+35 (a+b) \cos ^2(e+f x) \left (15 b^2+10 a b \left (3-2 \sin ^2(e+f x)\right )+a^2 \left (15-20 \sin ^2(e+f x)+8 \sin ^4(e+f x)\right )\right ) \left (-3 \arcsin \left (\sqrt {-\frac {b \tan ^2(e+f x)}{a+b}}\right ) \left (a+b-a \sin ^2(e+f x)\right )^2-(a+b) \cos ^2(e+f x) \left (-3 a \cos ^2(e+f x)-b \left (3+\sin ^2(e+f x)\right )\right ) \sqrt {-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}}\right )\right )}{(a+b)^5}\right )}{315 (2 a+2 b)^2 f \left (a+b \sec ^2(e+f x)\right )^{5/2} \sqrt {\frac {a+b \sec ^2(e+f x)}{a+b}} \sqrt {2 a+2 b-2 a \sin ^2(e+f x)} \left (1-\frac {a \sin ^2(e+f x)}{a+b}\right )^{3/2} \left (-\frac {b \tan ^2(e+f x)}{a+b}\right )^{5/2}} \]
((a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f*x]^6*Sqrt[1 - (2*a*Sin[e + f*x]^2)/(2*a + 2*b)]*Tan[e + f*x]*((-24*b*Cos[e + f*x]^2*HypergeometricPF Q[{2, 2, 2}, {1, 9/2}, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^2*(-((b *Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)*Tan[e + f*x]^2)/(a + b)^2))^(5/ 2))/(a + b) - (Sec[e + f*x]^6*(24*b^3*Hypergeometric2F1[2, 2, 9/2, -((b*Ta n[e + f*x]^2)/(a + b))]*Sin[e + f*x]^6*(4*b^2 + a*b*(8 - 7*Sin[e + f*x]^2) + a^2*(4 - 7*Sin[e + f*x]^2 + 3*Sin[e + f*x]^4))*Sqrt[-((b*Sec[e + f*x]^2 *(a + b - a*Sin[e + f*x]^2)*Tan[e + f*x]^2)/(a + b)^2)] + 35*(a + b)*Cos[e + f*x]^2*(15*b^2 + 10*a*b*(3 - 2*Sin[e + f*x]^2) + a^2*(15 - 20*Sin[e + f *x]^2 + 8*Sin[e + f*x]^4))*(-3*ArcSin[Sqrt[-((b*Tan[e + f*x]^2)/(a + b))]] *(a + b - a*Sin[e + f*x]^2)^2 - (a + b)*Cos[e + f*x]^2*(-3*a*Cos[e + f*x]^ 2 - b*(3 + Sin[e + f*x]^2))*Sqrt[-((b*Sec[e + f*x]^2*(a + b - a*Sin[e + f* x]^2)*Tan[e + f*x]^2)/(a + b)^2)])))/(a + b)^5))/(315*(2*a + 2*b)^2*f*(a + b*Sec[e + f*x]^2)^(5/2)*Sqrt[(a + b*Sec[e + f*x]^2)/(a + b)]*Sqrt[2*a + 2 *b - 2*a*Sin[e + f*x]^2]*(1 - (a*Sin[e + f*x]^2)/(a + b))^(3/2)*(-((b*Tan[ e + f*x]^2)/(a + b)))^(5/2))
Time = 0.32 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4634, 315, 298, 224, 219}
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.
| \(\displaystyle \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (e+f x)^6}{\left (a+b \sec (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4634 |
| \(\displaystyle \frac {\int \frac {\left (\tan ^2(e+f x)+1\right )^2}{\left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 315 |
| \(\displaystyle \frac {\frac {\int \frac {3 (a+b) \tan ^2(e+f x)+a+3 b}{\left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 b (a+b)}-\frac {a \tan (e+f x) \left (\tan ^2(e+f x)+1\right )}{3 b (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle \frac {\frac {\frac {3 (a+b) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{b}-\frac {a (3 a+5 b) \tan (e+f x)}{b (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 b (a+b)}-\frac {a \tan (e+f x) \left (\tan ^2(e+f x)+1\right )}{3 b (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\frac {3 (a+b) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{b}-\frac {a (3 a+5 b) \tan (e+f x)}{b (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 b (a+b)}-\frac {a \tan (e+f x) \left (\tan ^2(e+f x)+1\right )}{3 b (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {3 (a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{b^{3/2}}-\frac {a (3 a+5 b) \tan (e+f x)}{b (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 b (a+b)}-\frac {a \tan (e+f x) \left (\tan ^2(e+f x)+1\right )}{3 b (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
(-1/3*(a*Tan[e + f*x]*(1 + Tan[e + f*x]^2))/(b*(a + b)*(a + b + b*Tan[e + f*x]^2)^(3/2)) + ((3*(a + b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b *Tan[e + f*x]^2]])/b^(3/2) - (a*(3*a + 5*b)*Tan[e + f*x])/(b*(a + b)*Sqrt[ a + b + b*Tan[e + f*x]^2]))/(3*b*(a + b)))/f
3.3.89.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) )^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ [m/2] && IntegerQ[n/2]
Leaf count of result is larger than twice the leaf count of optimal. \(2354\) vs. \(2(119)=238\).
Time = 6.33 (sec) , antiderivative size = 2355, normalized size of antiderivative = 17.71
-1/6/f/(a+b)^2/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)*(-36*b^(5/2)*a^2*(1-cos(f*x+e))^5*csc(f*x+e)^5-24*b^(7/2)*a*( 1-cos(f*x+e))^5*csc(f*x+e)^5-12*b^(3/2)*a^3*(1-cos(f*x+e))^5*csc(f*x+e)^5+ 8*b^(5/2)*a^2*(1-cos(f*x+e))^3*csc(f*x+e)^3-48*(1-cos(f*x+e))^3*b^(7/2)*a* csc(f*x+e)^3+24*b^(3/2)*a^3*(1-cos(f*x+e))^3*csc(f*x+e)^3-36*b^(5/2)*a^2*( csc(f*x+e)-cot(f*x+e))-24*b^(7/2)*a*(csc(f*x+e)-cot(f*x+e))+3*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-c os(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+6*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-co s(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*b^2+3*ln(4*(-a*(1-cos(f*x+e))^2*csc(f*x...
Leaf count of result is larger than twice the leaf count of optimal. 328 vs. \(2 (119) = 238\).
Time = 0.50 (sec) , antiderivative size = 688, normalized size of antiderivative = 5.17 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left ({\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{2} b^{2} + 2 \, a b^{3} + b^{4} + 2 \, {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {b} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \, {\left ({\left (3 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (2 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{12 \, {\left ({\left (a^{4} b^{3} + 2 \, a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{3} b^{4} + 2 \, a^{2} b^{5} + a b^{6}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} b^{5} + 2 \, a b^{6} + b^{7}\right )} f\right )}}, \frac {3 \, {\left ({\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{2} b^{2} + 2 \, a b^{3} + b^{4} + 2 \, {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) - 2 \, {\left ({\left (3 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (2 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{6 \, {\left ({\left (a^{4} b^{3} + 2 \, a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{3} b^{4} + 2 \, a^{2} b^{5} + a b^{6}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} b^{5} + 2 \, a b^{6} + b^{7}\right )} f\right )}}\right ] \]
[1/12*(3*((a^4 + 2*a^3*b + a^2*b^2)*cos(f*x + e)^4 + a^2*b^2 + 2*a*b^3 + b ^4 + 2*(a^3*b + 2*a^2*b^2 + a*b^3)*cos(f*x + e)^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)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4) - 4*((3*a^3*b + 5*a^2*b^2)*c os(f*x + e)^3 + 2*(2*a^2*b^2 + 3*a*b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e) ^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^4*b^3 + 2*a^3*b^4 + a^2*b^5)*f*c os(f*x + e)^4 + 2*(a^3*b^4 + 2*a^2*b^5 + a*b^6)*f*cos(f*x + e)^2 + (a^2*b^ 5 + 2*a*b^6 + b^7)*f), 1/6*(3*((a^4 + 2*a^3*b + a^2*b^2)*cos(f*x + e)^4 + a^2*b^2 + 2*a*b^3 + b^4 + 2*(a^3*b + 2*a^2*b^2 + a*b^3)*cos(f*x + e)^2)*sq rt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sq rt((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))) - 2*((3*a^3*b + 5*a^2*b^2)*cos(f*x + e)^3 + 2*(2*a^2*b^2 + 3*a* b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e ))/((a^4*b^3 + 2*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^4 + 2*(a^3*b^4 + 2*a^2* b^5 + a*b^6)*f*cos(f*x + e)^2 + (a^2*b^5 + 2*a*b^6 + b^7)*f)]
\[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sec ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (119) = 238\).
Time = 0.20 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.07 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {{\left (\frac {3 \, \tan \left (f x + e\right )^{2}}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b^{2}} + \frac {2}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b}\right )} \tan \left (f x + e\right ) - \frac {3 \, \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {5}{2}}} - \frac {2 \, \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2}} - \frac {\tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}} + \frac {3 \, \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} b^{2}} - \frac {2 \, a \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} b^{2}} + \frac {2 \, \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b} - \frac {4 \, \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} b}}{3 \, f} \]
-1/3*((3*tan(f*x + e)^2/((b*tan(f*x + e)^2 + a + b)^(3/2)*b) + 2*a/((b*tan (f*x + e)^2 + a + b)^(3/2)*b^2) + 2/((b*tan(f*x + e)^2 + a + b)^(3/2)*b))* tan(f*x + e) - 3*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/b^(5/2) - 2*tan(f *x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^2) - tan(f*x + e)/((b*tan( f*x + e)^2 + a + b)^(3/2)*(a + b)) + 3*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*b^2) - 2*a*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)* b^2) + 2*tan(f*x + e)/((b*tan(f*x + e)^2 + a + b)^(3/2)*b) - 4*tan(f*x + e )/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)*b))/f
\[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\sec \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{{\cos \left (e+f\,x\right )}^6\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \]