Integrand size = 25, antiderivative size = 243 \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {(a+b)^2 \left (3 a^2-10 a b+35 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{128 b^{5/2} f}+\frac {(a+b) \left (3 a^2-10 a b+35 b^2\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{128 b^2 f}+\frac {\left (3 a^2-10 a b+35 b^2\right ) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{192 b^2 f}-\frac {(3 a-7 b) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{48 b^2 f}+\frac {\sec ^2(e+f x) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{8 b f} \]
1/128*(a+b)^2*(3*a^2-10*a*b+35*b^2)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan( f*x+e)^2)^(1/2))/b^(5/2)/f+1/128*(a+b)*(3*a^2-10*a*b+35*b^2)*(a+b+b*tan(f* x+e)^2)^(1/2)*tan(f*x+e)/b^2/f+1/192*(3*a^2-10*a*b+35*b^2)*tan(f*x+e)*(a+b +b*tan(f*x+e)^2)^(3/2)/b^2/f-1/48*(3*a-7*b)*tan(f*x+e)*(a+b+b*tan(f*x+e)^2 )^(5/2)/b^2/f+1/8*sec(f*x+e)^2*tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^(5/2)/b/f
Result contains complex when optimal does not.
Time = 11.22 (sec) , antiderivative size = 512, normalized size of antiderivative = 2.11 \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {e^{i (e+f x)} \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \cos ^3(e+f x) \left (-\frac {i \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) \left (-9 a^3 \left (1+e^{2 i (e+f x)}\right )^6+3 a^2 b \left (1+e^{2 i (e+f x)}\right )^4 \left (5+18 e^{2 i (e+f x)}+5 e^{4 i (e+f x)}\right )+a b^2 \left (1+e^{2 i (e+f x)}\right )^2 \left (145+948 e^{2 i (e+f x)}+2758 e^{4 i (e+f x)}+948 e^{6 i (e+f x)}+145 e^{8 i (e+f x)}\right )+b^3 \left (105+910 e^{2 i (e+f x)}+3591 e^{4 i (e+f x)}+8644 e^{6 i (e+f x)}+3591 e^{8 i (e+f x)}+910 e^{10 i (e+f x)}+105 e^{12 i (e+f x)}\right )\right )}{\left (1+e^{2 i (e+f x)}\right )^8}-\frac {3 (a+b)^2 \left (3 a^2-10 a b+35 b^2\right ) \log \left (\frac {-4 \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) f+4 i \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2} f}{1+e^{2 i (e+f x)}}\right )}{\sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}}\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{96 \sqrt {2} b^{5/2} f (a+2 b+a \cos (2 e+2 f x))^{3/2}} \]
(E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2*I)*(e + f*x))]*Cos[e + f*x]^3*(((-I)*Sqrt[b]*(-1 + E^((2*I)*(e + f*x)))*(-9*a^3*(1 + E^((2*I)*(e + f*x)))^6 + 3*a^2*b*(1 + E^((2*I)*(e + f*x)))^4*(5 + 18*E^ ((2*I)*(e + f*x)) + 5*E^((4*I)*(e + f*x))) + a*b^2*(1 + E^((2*I)*(e + f*x) ))^2*(145 + 948*E^((2*I)*(e + f*x)) + 2758*E^((4*I)*(e + f*x)) + 948*E^((6 *I)*(e + f*x)) + 145*E^((8*I)*(e + f*x))) + b^3*(105 + 910*E^((2*I)*(e + f *x)) + 3591*E^((4*I)*(e + f*x)) + 8644*E^((6*I)*(e + f*x)) + 3591*E^((8*I) *(e + f*x)) + 910*E^((10*I)*(e + f*x)) + 105*E^((12*I)*(e + f*x)))))/(1 + E^((2*I)*(e + f*x)))^8 - (3*(a + b)^2*(3*a^2 - 10*a*b + 35*b^2)*Log[(-4*Sq rt[b]*(-1 + E^((2*I)*(e + f*x)))*f + (4*I)*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]*f)/(1 + E^((2*I)*(e + f*x)))])/Sqrt[4*b*E^( (2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2])*(a + b*Sec[e + f*x]^2)^ (3/2))/(96*Sqrt[2]*b^(5/2)*f*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2))
Time = 0.35 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.89, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4634, 318, 25, 299, 211, 211, 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.
\(\displaystyle \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (e+f x)^6 \left (a+b \sec (e+f x)^2\right )^{3/2}dx\) |
\(\Big \downarrow \) 4634 |
\(\displaystyle \frac {\int \left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^{3/2}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 318 |
\(\displaystyle \frac {\frac {\int -\left (\left ((3 a-7 b) \tan ^2(e+f x)+a-7 b\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}\right )d\tan (e+f x)}{8 b}+\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{8 b}}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{8 b}-\frac {\int \left ((3 a-7 b) \tan ^2(e+f x)+a-7 b\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}d\tan (e+f x)}{8 b}}{f}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{8 b}-\frac {\frac {(3 a-7 b) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 b}-\frac {\left (3 a^2-10 a b+35 b^2\right ) \int \left (b \tan ^2(e+f x)+a+b\right )^{3/2}d\tan (e+f x)}{6 b}}{8 b}}{f}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{8 b}-\frac {\frac {(3 a-7 b) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 b}-\frac {\left (3 a^2-10 a b+35 b^2\right ) \left (\frac {3}{4} (a+b) \int \sqrt {b \tan ^2(e+f x)+a+b}d\tan (e+f x)+\frac {1}{4} \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}\right )}{6 b}}{8 b}}{f}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{8 b}-\frac {\frac {(3 a-7 b) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 b}-\frac {\left (3 a^2-10 a b+35 b^2\right ) \left (\frac {3}{4} (a+b) \left (\frac {1}{2} (a+b) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )+\frac {1}{4} \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}\right )}{6 b}}{8 b}}{f}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{8 b}-\frac {\frac {(3 a-7 b) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 b}-\frac {\left (3 a^2-10 a b+35 b^2\right ) \left (\frac {3}{4} (a+b) \left (\frac {1}{2} (a+b) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )+\frac {1}{4} \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}\right )}{6 b}}{8 b}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{8 b}-\frac {\frac {(3 a-7 b) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 b}-\frac {\left (3 a^2-10 a b+35 b^2\right ) \left (\frac {3}{4} (a+b) \left (\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 \sqrt {b}}+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )+\frac {1}{4} \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}\right )}{6 b}}{8 b}}{f}\) |
((Tan[e + f*x]*(1 + Tan[e + f*x]^2)*(a + b + b*Tan[e + f*x]^2)^(5/2))/(8*b ) - (((3*a - 7*b)*Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(5/2))/(6*b) - ( (3*a^2 - 10*a*b + 35*b^2)*((Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2)) /4 + (3*(a + b)*(((a + b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Ta n[e + f*x]^2]])/(2*Sqrt[b]) + (Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2] )/2))/4))/(6*b))/(8*b))/f
3.3.47.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 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. \(1976\) vs. \(2(219)=438\).
Time = 21.49 (sec) , antiderivative size = 1977, normalized size of antiderivative = 8.14
1/768/f/b^(13/2)*(a+b*sec(f*x+e)^2)^(3/2)/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e ))^2)^(1/2)/(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))*(140*((b+a*cos(f*x+e)^2)/(1+ cos(f*x+e))^2)^(1/2)*b^(15/2)*sin(f*x+e)+105*cos(f*x+e)^3*ln(4*(((b+a*cos( f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x +e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*b^8+105*c os(f*x+e)^3*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos (f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a +b)/(sin(f*x+e)-1))*b^8+140*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^ (15/2)*tan(f*x+e)+290*b^(13/2)*cos(f*x+e)^2*sin(f*x+e)*((b+a*cos(f*x+e)^2) /(1+cos(f*x+e))^2)^(1/2)*a+290*b^(13/2)*cos(f*x+e)*sin(f*x+e)*((b+a*cos(f* x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a+30*b^(11/2)*cos(f*x+e)^2*sin(f*x+e)*((b+ a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2+30*b^(11/2)*cos(f*x+e)*sin(f*x +e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2-18*b^(9/2)*cos(f*x+e)^ 2*sin(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3-18*b^(9/2)*co s(f*x+e)*sin(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3+144*(( b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(13/2)*a*tan(f*x+e)*sec(f*x+e) +144*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(13/2)*a*tan(f*x+e)*sec (f*x+e)^2+12*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(11/2)*a^2*tan( f*x+e)+184*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(13/2)*a*tan(f*x+ e)+96*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(15/2)*tan(f*x+e)*s...
Time = 4.13 (sec) , antiderivative size = 566, normalized size of antiderivative = 2.33 \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\left [\frac {3 \, {\left (3 \, a^{4} - 4 \, a^{3} b + 18 \, a^{2} b^{2} + 60 \, a b^{3} + 35 \, b^{4}\right )} \sqrt {b} \cos \left (f x + e\right )^{7} \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 (9 \, a^{3} b - 15 \, a^{2} b^{2} - 145 \, a b^{3} - 105 \, b^{4}\right )} \cos \left (f x + e\right )^{6} - 2 \, {\left (3 \, a^{2} b^{2} + 46 \, a b^{3} + 35 \, b^{4}\right )} \cos \left (f x + e\right )^{4} - 48 \, b^{4} - 8 \, {\left (9 \, a b^{3} + 7 \, b^{4}\right )} \cos \left (f x + e\right )^{2}\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 )}{1536 \, b^{3} f \cos \left (f x + e\right )^{7}}, \frac {3 \, {\left (3 \, a^{4} - 4 \, a^{3} b + 18 \, a^{2} b^{2} + 60 \, a b^{3} + 35 \, b^{4}\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 ) \cos \left (f x + e\right )^{7} - 2 \, {\left ({\left (9 \, a^{3} b - 15 \, a^{2} b^{2} - 145 \, a b^{3} - 105 \, b^{4}\right )} \cos \left (f x + e\right )^{6} - 2 \, {\left (3 \, a^{2} b^{2} + 46 \, a b^{3} + 35 \, b^{4}\right )} \cos \left (f x + e\right )^{4} - 48 \, b^{4} - 8 \, {\left (9 \, a b^{3} + 7 \, b^{4}\right )} \cos \left (f x + e\right )^{2}\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 )}{768 \, b^{3} f \cos \left (f x + e\right )^{7}}\right ] \]
[1/1536*(3*(3*a^4 - 4*a^3*b + 18*a^2*b^2 + 60*a*b^3 + 35*b^4)*sqrt(b)*cos( f*x + e)^7*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*co s(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4) - 4*((9*a^3*b - 15*a^2*b^2 - 145*a*b^3 - 105*b^4)*cos(f*x + e)^6 - 2*(3*a^2* b^2 + 46*a*b^3 + 35*b^4)*cos(f*x + e)^4 - 48*b^4 - 8*(9*a*b^3 + 7*b^4)*cos (f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(b^ 3*f*cos(f*x + e)^7), 1/768*(3*(3*a^4 - 4*a^3*b + 18*a^2*b^2 + 60*a*b^3 + 3 5*b^4)*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sq rt(-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)))*cos(f*x + e)^7 - 2*((9*a^3*b - 15*a^2*b^2 - 145*a*b^3 - 105*b^4)*cos(f*x + e)^6 - 2*(3*a^2*b^2 + 46*a*b^3 + 35*b^4)*cos(f*x + e) ^4 - 48*b^4 - 8*(9*a*b^3 + 7*b^4)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(b^3*f*cos(f*x + e)^7)]
\[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \sec ^{6}{\left (e + f x \right )}\, dx \]
Time = 0.21 (sec) , antiderivative size = 415, normalized size of antiderivative = 1.71 \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\frac {48 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {5}{2}} \tan \left (f x + e\right )^{3}}{b} + \frac {9 \, {\left (a + b\right )}^{3} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {5}{2}}} + \frac {9 \, {\left (a + b\right )}^{3} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} - \frac {48 \, {\left (a + b\right )}^{2} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} - \frac {48 \, {\left (a + b\right )}^{2} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} + \frac {144 \, {\left (a + b\right )} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} + 144 \, {\left (a + b\right )} \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + 96 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \tan \left (f x + e\right ) + 144 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right ) - \frac {24 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {5}{2}} {\left (a + b\right )} \tan \left (f x + e\right )}{b^{2}} + \frac {6 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{2} \tan \left (f x + e\right )}{b^{2}} + \frac {9 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{3} \tan \left (f x + e\right )}{b^{2}} + \frac {128 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {5}{2}} \tan \left (f x + e\right )}{b} - \frac {32 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \tan \left (f x + e\right )}{b} - \frac {48 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2} \tan \left (f x + e\right )}{b}}{384 \, f} \]
1/384*(48*(b*tan(f*x + e)^2 + a + b)^(5/2)*tan(f*x + e)^3/b + 9*(a + b)^3* a*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/b^(5/2) + 9*(a + b)^3*arcsinh(b* tan(f*x + e)/sqrt((a + b)*b))/b^(3/2) - 48*(a + b)^2*a*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/b^(3/2) - 48*(a + b)^2*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/sqrt(b) + 144*(a + b)*a*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/ sqrt(b) + 144*(a + b)*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) + 96 *(b*tan(f*x + e)^2 + a + b)^(3/2)*tan(f*x + e) + 144*sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)*tan(f*x + e) - 24*(b*tan(f*x + e)^2 + a + b)^(5/2)*(a + b)*tan(f*x + e)/b^2 + 6*(b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)^2*tan(f*x + e)/b^2 + 9*sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^3*tan(f*x + e)/b^2 + 128*(b*tan(f*x + e)^2 + a + b)^(5/2)*tan(f*x + e)/b - 32*(b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)*tan(f*x + e)/b - 48*sqrt(b*tan(f*x + e)^2 + a + b) *(a + b)^2*tan(f*x + e)/b)/f
\[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )^{6} \,d x } \]
Timed out. \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}}{{\cos \left (e+f\,x\right )}^6} \,d x \]