Integrand size = 25, antiderivative size = 271 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {\left (5 a^3-9 a^2 b+15 a b^2-35 b^3\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 a^{9/2} f}+\frac {\left (15 a^2-22 a b+35 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(5 a-7 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {b \left (15 a^3-17 a^2 b+25 a b^2+105 b^3\right ) \tan (e+f x)}{48 a^4 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}} \]
1/16*(5*a^3-9*a^2*b+15*a*b^2-35*b^3)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan( f*x+e)^2)^(1/2))/a^(9/2)/f+1/48*(15*a^2-22*a*b+35*b^2)*cos(f*x+e)*sin(f*x+ e)/a^3/f/(a+b+b*tan(f*x+e)^2)^(1/2)+1/24*(5*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)^5*sin(f*x+e)/a/f/(a+b+b*t an(f*x+e)^2)^(1/2)+1/48*b*(15*a^3-17*a^2*b+25*a*b^2+105*b^3)*tan(f*x+e)/a^ 4/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 17.72 (sec) , antiderivative size = 2068, normalized size of antiderivative = 7.63 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Result too large to show} \]
(3*(a + b)*AppellF1[1/2, -4, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/ (a + b)]*Cos[e + f*x]^14*Sin[e + f*x])/(2*f*Sqrt[a + 2*b + a*Cos[2*(e + f* x)]]*(a + b*Sec[e + f*x]^2)^(3/2)*(a + b - a*Sin[e + f*x]^2)*(3*(a + b)*Ap pellF1[1/2, -4, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (3 *a*AppellF1[3/2, -4, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 8*(a + b)*AppellF1[3/2, -3, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2 )/(a + b)])*Sin[e + f*x]^2)*((3*a*(a + b)*AppellF1[1/2, -4, 3/2, 3/2, Sin[ e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^9*Sin[e + f*x]^2)/(Sq rt[a + 2*b + a*Cos[2*(e + f*x)]]*(a + b - a*Sin[e + f*x]^2)^2*(3*(a + b)*A ppellF1[1/2, -4, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + ( 3*a*AppellF1[3/2, -4, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b) ] - 8*(a + b)*AppellF1[3/2, -3, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^ 2)/(a + b)])*Sin[e + f*x]^2)) + (3*(a + b)*AppellF1[1/2, -4, 3/2, 3/2, Sin [e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^9)/(2*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(a + b - a*Sin[e + f*x]^2)*(3*(a + b)*AppellF1[1/2, - 4, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (3*a*AppellF1[3 /2, -4, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 8*(a + b)* AppellF1[3/2, -3, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*S in[e + f*x]^2)) - (12*(a + b)*AppellF1[1/2, -4, 3/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^7*Sin[e + f*x]^2)/(Sqrt[a + 2*...
Time = 0.46 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 4634, 316, 25, 402, 25, 402, 25, 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.
| \(\displaystyle \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (e+f x)^6 \left (a+b \sec (e+f x)^2\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4634 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right )^4 \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\int -\frac {6 b \tan ^2(e+f x)+5 a-b}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{6 a}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {6 b \tan ^2(e+f x)+5 a-b}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {(5 a-7 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\int -\frac {15 a^2-2 b a+7 b^2+4 (5 a-7 b) b \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{4 a}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {15 a^2-2 b a+7 b^2+4 (5 a-7 b) b \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{4 a}+\frac {(5 a-7 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\frac {\left (15 a^2-22 a b+35 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\int -\frac {15 a^3+3 b a^2+b^2 a-35 b^3+2 b \left (15 a^2-22 b a+35 b^2\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{2 a}}{4 a}+\frac {(5 a-7 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int \frac {15 a^3+3 b a^2+b^2 a-35 b^3+2 b \left (15 a^2-22 b a+35 b^2\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{2 a}+\frac {\left (15 a^2-22 a b+35 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}}{4 a}+\frac {(5 a-7 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {\int \frac {3 (a+b) \left (5 a^3-9 b a^2+15 b^2 a-35 b^3\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a (a+b)}+\frac {b \left (15 a^3-17 a^2 b+25 a b^2+105 b^3\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{2 a}+\frac {\left (15 a^2-22 a b+35 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}}{4 a}+\frac {(5 a-7 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {3 \left (5 a^3-9 a^2 b+15 a b^2-35 b^3\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a}+\frac {b \left (15 a^3-17 a^2 b+25 a b^2+105 b^3\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{2 a}+\frac {\left (15 a^2-22 a b+35 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}}{4 a}+\frac {(5 a-7 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {3 \left (5 a^3-9 a^2 b+15 a b^2-35 b^3\right ) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{a}+\frac {b \left (15 a^3-17 a^2 b+25 a b^2+105 b^3\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{2 a}+\frac {\left (15 a^2-22 a b+35 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}}{4 a}+\frac {(5 a-7 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {\left (15 a^2-22 a b+35 b^2\right ) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}+\frac {\frac {b \left (15 a^3-17 a^2 b+25 a b^2+105 b^3\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}+\frac {3 \left (5 a^3-9 a^2 b+15 a b^2-35 b^3\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2}}}{2 a}}{4 a}+\frac {(5 a-7 b) \tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}}{6 a}+\frac {\tan (e+f x)}{6 a \left (\tan ^2(e+f x)+1\right )^3 \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\) |
(Tan[e + f*x]/(6*a*(1 + Tan[e + f*x]^2)^3*Sqrt[a + b + b*Tan[e + f*x]^2]) + (((5*a - 7*b)*Tan[e + f*x])/(4*a*(1 + Tan[e + f*x]^2)^2*Sqrt[a + b + b*T an[e + f*x]^2]) + (((15*a^2 - 22*a*b + 35*b^2)*Tan[e + f*x])/(2*a*(1 + Tan [e + f*x]^2)*Sqrt[a + b + b*Tan[e + f*x]^2]) + ((3*(5*a^3 - 9*a^2*b + 15*a *b^2 - 35*b^3)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2 ]])/a^(3/2) + (b*(15*a^3 - 17*a^2*b + 25*a*b^2 + 105*b^3)*Tan[e + f*x])/(a *(a + b)*Sqrt[a + b + b*Tan[e + f*x]^2]))/(2*a))/(4*a))/(6*a))/f
3.3.82.3.1 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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 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[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. \(1498\) vs. \(2(247)=494\).
Time = 9.78 (sec) , antiderivative size = 1499, normalized size of antiderivative = 5.53
1/48/f/a^4/(-a)^(1/2)/(a+b)*(b+a*cos(f*x+e)^2)*(8*(-a)^(1/2)*a^4*cos(f*x+e )^6*sin(f*x+e)+8*(-a)^(1/2)*a^3*b*cos(f*x+e)^6*sin(f*x+e)+10*(-a)^(1/2)*a^ 4*cos(f*x+e)^4*sin(f*x+e)-4*(-a)^(1/2)*a^3*b*cos(f*x+e)^4*sin(f*x+e)-14*(- a)^(1/2)*a^2*b^2*cos(f*x+e)^4*sin(f*x+e)+15*(-a)^(1/2)*a^4*cos(f*x+e)^2*si n(f*x+e)-7*(-a)^(1/2)*a^3*b*cos(f*x+e)^2*sin(f*x+e)+13*(-a)^(1/2)*a^2*b^2* cos(f*x+e)^2*sin(f*x+e)+35*(-a)^(1/2)*a*b^3*cos(f*x+e)^2*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)*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)*a^4*cos(f*x+e)-12*((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)*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)*a^3*b*cos(f*x+e)+18*((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)*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)*a^2*b^2*cos(f*x+e)-60*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*l n(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)*a*b^3 *cos(f*x+e)-105*((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)*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^4*cos(f*x+e)+...
Time = 4.39 (sec) , antiderivative size = 941, normalized size of antiderivative = 3.47 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \]
[1/384*(3*(5*a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 20*a*b^4 - 35*b^5 + (5*a^5 - 4*a^4*b + 6*a^3*b^2 - 20*a^2*b^3 - 35*a*b^4)*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))*s qrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 8*(8*( a^5 + a^4*b)*cos(f*x + e)^7 + 2*(5*a^5 - 2*a^4*b - 7*a^3*b^2)*cos(f*x + e) ^5 + (15*a^5 - 7*a^4*b + 13*a^3*b^2 + 35*a^2*b^3)*cos(f*x + e)^3 + (15*a^4 *b - 17*a^3*b^2 + 25*a^2*b^3 + 105*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^7 + a^6*b)*f*cos(f*x + e)^2 + (a^6*b + a^5*b^2)*f), -1/192*(3*(5*a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 20*a*b^ 4 - 35*b^5 + (5*a^5 - 4*a^4*b + 6*a^3*b^2 - 20*a^2*b^3 - 35*a*b^4)*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^5 + a^4*b)*cos(f*x + e)^7 + 2 *(5*a^5 - 2*a^4*b - 7*a^3*b^2)*cos(f*x + e)^5 + (15*a^5 - 7*a^4*b + 13*a^3 *b^2 + 35*a^2*b^3)*cos(f*x + e)^3 + (15*a^4*b - 17*a^3*b^2 + 25*a^2*b^3...
\[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\cos ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^6}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x \]