Integrand size = 25, antiderivative size = 242 \[ \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {5 (a+b)^2 (a+7 b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 a^{9/2} f}-\frac {(a+b) (33 a+35 b) \cos (e+f x) \sin (e+f x)}{48 a^3 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(9 a+7 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {b \left (81 a^2+190 a b+105 b^2\right ) \tan (e+f x)}{48 a^4 f \sqrt {a+b+b \tan ^2(e+f x)}} \]
5/16*(a+b)^2*(a+7*b)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2)) /a^(9/2)/f-1/48*(a+b)*(33*a+35*b)*cos(f*x+e)*sin(f*x+e)/a^3/f/(a+b+b*tan(f *x+e)^2)^(1/2)+1/24*(9*a+7*b)*cos(f*x+e)^3*sin(f*x+e)/a^2/f/(a+b+b*tan(f*x +e)^2)^(1/2)+1/6*cos(f*x+e)^3*sin(f*x+e)^3/a/f/(a+b+b*tan(f*x+e)^2)^(1/2)- 1/48*b*(81*a^2+190*a*b+105*b^2)*tan(f*x+e)/a^4/f/(a+b+b*tan(f*x+e)^2)^(1/2 )
Time = 6.30 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.06 \[ \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^3(e+f x) \left (120 (a+b)^2 (a+7 b) \arcsin \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right ) (a+2 b+a \cos (2 (e+f x)))-2 \sqrt {2} \sqrt {a} \sqrt {a+b} \sqrt {\frac {a+2 b+a \cos (2 (e+f x))}{a+b}} \left (37 a^3+439 a^2 b+830 a b^2+420 b^3+a \left (29 a^2+108 a b+70 b^2\right ) \cos (2 (e+f x))-7 a^2 (a+b) \cos (4 (e+f x))+a^3 \cos (6 (e+f x))\right ) \sin (e+f x)\right )}{1536 a^{9/2} \sqrt {a+b} f \left (a+b \sec ^2(e+f x)\right )^{3/2} \sqrt {\frac {a+b-a \sin ^2(e+f x)}{a+b}}} \]
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^3*(120*(a + b)^2*(a + 7*b)*Ar cSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*(a + 2*b + a*Cos[2*(e + f*x)]) - 2*Sqrt[2]*Sqrt[a]*Sqrt[a + b]*Sqrt[(a + 2*b + a*Cos[2*(e + f*x)])/(a + b)] *(37*a^3 + 439*a^2*b + 830*a*b^2 + 420*b^3 + a*(29*a^2 + 108*a*b + 70*b^2) *Cos[2*(e + f*x)] - 7*a^2*(a + b)*Cos[4*(e + f*x)] + a^3*Cos[6*(e + f*x)]) *Sin[e + f*x]))/(1536*a^(9/2)*Sqrt[a + b]*f*(a + b*Sec[e + f*x]^2)^(3/2)*S qrt[(a + b - a*Sin[e + f*x]^2)/(a + b)])
Time = 0.49 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4620, 372, 440, 27, 402, 402, 27, 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.
| \(\displaystyle \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^6}{\left (a+b \sec (e+f x)^2\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4620 |
| \(\displaystyle \frac {\int \frac {\tan ^6(e+f x)}{\left (\tan ^2(e+f x)+1\right )^4 \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\int \frac {\tan ^2(e+f x) \left (2 (b-3 (a+b)) \tan ^2(e+f x)+3 (a+b)\right )}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{6 a}}{f}\) |
| \(\Big \downarrow \) 440 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {\int \frac {(a+b) \left (-4 (6 a+7 b) \tan ^2(e+f x)+9 a+7 b\right )}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{4 a}-\frac {(9 a+7 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}}{6 a}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {(a+b) \int \frac {-4 (6 a+7 b) \tan ^2(e+f x)+9 a+7 b}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{4 a}-\frac {(9 a+7 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}}{6 a}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {(a+b) \left (\frac {(33 a+35 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\int \frac {15 a^2+54 b a+35 b^2-2 b (33 a+35 b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{2 a}\right )}{4 a}-\frac {(9 a+7 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}}{6 a}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {(a+b) \left (\frac {(33 a+35 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {\int \frac {15 (a+b)^2 (a+7 b)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a (a+b)}-\frac {b \left (81 a^2+190 a b+105 b^2\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{2 a}\right )}{4 a}-\frac {(9 a+7 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}}{6 a}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {(a+b) \left (\frac {(33 a+35 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {15 (a+b) (a+7 b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a}-\frac {b \left (81 a^2+190 a b+105 b^2\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{2 a}\right )}{4 a}-\frac {(9 a+7 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}}{6 a}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {(a+b) \left (\frac {(33 a+35 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {15 (a+b) (a+7 b) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{a}-\frac {b \left (81 a^2+190 a b+105 b^2\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{2 a}\right )}{4 a}-\frac {(9 a+7 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}}{6 a}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {(a+b) \left (\frac {(33 a+35 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {15 (a+b) (a+7 b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2}}-\frac {b \left (81 a^2+190 a b+105 b^2\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{2 a}\right )}{4 a}-\frac {(9 a+7 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}}{6 a}}{f}\) |
(Tan[e + f*x]^3/(6*a*(1 + Tan[e + f*x]^2)^3*Sqrt[a + b + b*Tan[e + f*x]^2] ) - (-1/4*((9*a + 7*b)*Tan[e + f*x])/(a*(1 + Tan[e + f*x]^2)^2*Sqrt[a + b + b*Tan[e + f*x]^2]) + ((a + b)*(((33*a + 35*b)*Tan[e + f*x])/(2*a*(1 + Ta n[e + f*x]^2)*Sqrt[a + b + b*Tan[e + f*x]^2]) - ((15*(a + b)*(a + 7*b)*Arc Tan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/a^(3/2) - (b*( 81*a^2 + 190*a*b + 105*b^2)*Tan[e + f*x])/(a*(a + b)*Sqrt[a + b + b*Tan[e + f*x]^2]))/(2*a)))/(4*a))/(6*a))/f
3.2.12.3.1 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[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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
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[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1)/f Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f f^2*x^2)^(m/2 + 1)), 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. \(1161\) vs. \(2(218)=436\).
Time = 9.33 (sec) , antiderivative size = 1162, normalized size of antiderivative = 4.80
-1/48/f/a^4/(-a)^(1/2)*(b+a*cos(f*x+e)^2)*(8*(-a)^(1/2)*a^3*cos(f*x+e)^6*s in(f*x+e)-26*(-a)^(1/2)*a^3*cos(f*x+e)^4*sin(f*x+e)-14*(-a)^(1/2)*a^2*b*co s(f*x+e)^4*sin(f*x+e)+33*(-a)^(1/2)*a^3*cos(f*x+e)^2*sin(f*x+e)+68*(-a)^(1 /2)*a^2*b*cos(f*x+e)^2*sin(f*x+e)+35*(-a)^(1/2)*a*b^2*cos(f*x+e)^2*sin(f*x +e)-15*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x +e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)* a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*cos(f*x+e)-135*ln(4*(-a )^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2 )*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*((b+a*cos(f* x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b*cos(f*x+e)-225*ln(4*(-a)^(1/2)*((b+a *cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f *x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+co s(f*x+e))^2)^(1/2)*a*b^2*cos(f*x+e)-105*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2 )/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+c os(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^ (1/2)*b^3*cos(f*x+e)+81*(-a)^(1/2)*a^2*b*sin(f*x+e)+190*(-a)^(1/2)*a*b^2*s in(f*x+e)+105*(-a)^(1/2)*b^3*sin(f*x+e)-15*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ e))^2)^(1/2)*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*c os(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f *x+e)*a)*a^3-135*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(-a)^...
Time = 5.91 (sec) , antiderivative size = 813, normalized size of antiderivative = 3.36 \[ \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\left [-\frac {15 \, {\left (a^{3} b + 9 \, a^{2} b^{2} + 15 \, a b^{3} + 7 \, b^{4} + {\left (a^{4} + 9 \, a^{3} b + 15 \, a^{2} b^{2} + 7 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a^{4} - a^{3} b\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{4} - 14 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 28 \, a^{3} b + 70 \, a^{2} b^{2} - 28 \, a b^{3} + b^{4} - 32 \, {\left (a^{4} - 7 \, a^{3} b + 7 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a^{3} - a^{2} b\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 14 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 7 \, a^{2} b + 7 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right ) + 8 \, {\left (8 \, a^{4} \cos \left (f x + e\right )^{7} - 2 \, {\left (13 \, a^{4} + 7 \, a^{3} b\right )} \cos \left (f x + e\right )^{5} + {\left (33 \, a^{4} + 68 \, a^{3} b + 35 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (81 \, a^{3} b + 190 \, a^{2} b^{2} + 105 \, 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 )}{384 \, {\left (a^{6} f \cos \left (f x + e\right )^{2} + a^{5} b f\right )}}, -\frac {15 \, {\left (a^{3} b + 9 \, a^{2} b^{2} + 15 \, a b^{3} + 7 \, b^{4} + {\left (a^{4} + 9 \, a^{3} b + 15 \, a^{2} b^{2} + 7 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 8 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} - a^{2} b + a b^{2} - {\left (a^{3} - 3 \, a^{2} b\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) + 4 \, {\left (8 \, a^{4} \cos \left (f x + e\right )^{7} - 2 \, {\left (13 \, a^{4} + 7 \, a^{3} b\right )} \cos \left (f x + e\right )^{5} + {\left (33 \, a^{4} + 68 \, a^{3} b + 35 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (81 \, a^{3} b + 190 \, a^{2} b^{2} + 105 \, 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 )}{192 \, {\left (a^{6} f \cos \left (f x + e\right )^{2} + a^{5} b f\right )}}\right ] \]
[-1/384*(15*(a^3*b + 9*a^2*b^2 + 15*a*b^3 + 7*b^4 + (a^4 + 9*a^3*b + 15*a^ 2*b^2 + 7*a*b^3)*cos(f*x + e)^2)*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)) + 8*(8*a^4*cos(f*x + e)^7 - 2*(13*a^ 4 + 7*a^3*b)*cos(f*x + e)^5 + (33*a^4 + 68*a^3*b + 35*a^2*b^2)*cos(f*x + e )^3 + (81*a^3*b + 190*a^2*b^2 + 105*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^6*f*cos(f*x + e)^2 + a^5*b*f), -1/192*(15*(a^3*b + 9*a^2*b^2 + 15*a*b^3 + 7*b^4 + (a^4 + 9*a^3*b + 15*a^ 2*b^2 + 7*a*b^3)*cos(f*x + e)^2)*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a) *sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2* b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e))) + 4*(8*a^4*cos( f*x + e)^7 - 2*(13*a^4 + 7*a^3*b)*cos(f*x + e)^5 + (33*a^4 + 68*a^3*b + 35 *a^2*b^2)*cos(f*x + e)^3 + (81*a^3*b + 190*a^2*b^2 + 105*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^6*f*cos(f *x + e)^2 + a^5*b*f)]
\[ \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sin ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^6}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x \]