Integrand size = 25, antiderivative size = 204 \[ \int \frac {\cos ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {(a-b) \left (5 a^2+2 a b+5 b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 a^{7/2} f}+\frac {\left (15 a^2-14 a b+15 b^2\right ) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 a^3 f}+\frac {5 (a-b) \cos ^3(e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 a^2 f}+\frac {\cos ^5(e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{6 a f} \]
1/16*(a-b)*(5*a^2+2*a*b+5*b^2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e) ^2)^(1/2))/a^(7/2)/f+1/48*(15*a^2-14*a*b+15*b^2)*cos(f*x+e)*sin(f*x+e)*(a+ b+b*tan(f*x+e)^2)^(1/2)/a^3/f+5/24*(a-b)*cos(f*x+e)^3*sin(f*x+e)*(a+b+b*ta n(f*x+e)^2)^(1/2)/a^2/f+1/6*cos(f*x+e)^5*sin(f*x+e)*(a+b+b*tan(f*x+e)^2)^( 1/2)/a/f
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 15.44 (sec) , antiderivative size = 1739, normalized size of antiderivative = 8.52 \[ \int \frac {\cos ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx =\text {Too large to display} \]
(3*(a + b)*AppellF1[1/2, -3, 1/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/ (a + b)]*Cos[e + f*x]^12*Sin[e + f*x])/(f*Sqrt[a + 2*b + a*Cos[2*(e + f*x) ]]*Sqrt[a + b*Sec[e + f*x]^2]*(3*(a + b)*AppellF1[1/2, -3, 1/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*AppellF1[3/2, -3, 3/2, 5/2, Si n[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 6*(a + b)*AppellF1[3/2, -2, 1/ 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, -3, 1/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^7)/(Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(3*(a + b)*Appell F1[1/2, -3, 1/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*App ellF1[3/2, -3, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 6*( a + b)*AppellF1[3/2, -2, 1/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e + f*x]^2)) - (18*(a + b)*AppellF1[1/2, -3, 1/2, 3/2, Sin[e + f *x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^5*Sin[e + f*x]^2)/(Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(3*(a + b)*AppellF1[1/2, -3, 1/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a*AppellF1[3/2, -3, 3/2, 5/2, Sin[ e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 6*(a + b)*AppellF1[3/2, -2, 1/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e + f*x]^2)) + (3*( a + b)*Cos[e + f*x]^6*Sin[e + f*x]*((a*f*AppellF1[3/2, -3, 3/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]*Sin[e + f*x])/(3*(a + b)) - 2*f*AppellF1[3/2, -2, 1/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2...
Time = 0.37 (sec) , antiderivative size = 226, 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, 4634, 316, 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.
| \(\displaystyle \int \frac {\cos ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (e+f x)^6 \sqrt {a+b \sec (e+f x)^2}}dx\) |
| \(\Big \downarrow \) 4634 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right )^4 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}-\frac {\int -\frac {4 b \tan ^2(e+f x)+5 a-b}{\left (\tan ^2(e+f x)+1\right )^3 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{6 a}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {4 b \tan ^2(e+f x)+5 a-b}{\left (\tan ^2(e+f x)+1\right )^3 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{6 a}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {5 (a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}-\frac {\int -\frac {15 a^2-4 b a+5 b^2+10 (a-b) b \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{4 a}}{6 a}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {15 a^2-4 b a+5 b^2+10 (a-b) b \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{4 a}+\frac {5 (a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{6 a}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\frac {\left (15 a^2-14 a b+15 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {\int -\frac {3 (a-b) \left (5 a^2+2 b a+5 b^2\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 a}}{4 a}+\frac {5 (a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{6 a}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\frac {3 (a-b) \left (5 a^2+2 a b+5 b^2\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 a}+\frac {\left (15 a^2-14 a b+15 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}}{4 a}+\frac {5 (a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{6 a}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {\frac {3 (a-b) \left (5 a^2+2 a b+5 b^2\right ) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{2 a}+\frac {\left (15 a^2-14 a b+15 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}}{4 a}+\frac {5 (a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{6 a}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {\left (15 a^2-14 a b+15 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}+\frac {3 (a-b) \left (5 a^2+2 a b+5 b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 a^{3/2}}}{4 a}+\frac {5 (a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}}{6 a}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 a \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
((Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(6*a*(1 + Tan[e + f*x]^2)^3 ) + ((5*(a - b)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(4*a*(1 + Tan [e + f*x]^2)^2) + ((3*(a - b)*(5*a^2 + 2*a*b + 5*b^2)*ArcTan[(Sqrt[a]*Tan[ e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*a^(3/2)) + ((15*a^2 - 14*a*b + 15*b^2)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(2*a*(1 + Tan[e + f*x]^2)))/(4*a))/(6*a))/f
3.3.69.3.1 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. \(1135\) vs. \(2(184)=368\).
Time = 9.68 (sec) , antiderivative size = 1136, normalized size of antiderivative = 5.57
1/48/f/(-a)^(1/2)/a^3/(a+b*sec(f*x+e)^2)^(1/2)*(8*cos(f*x+e)^5*sin(f*x+e)* (-a)^(1/2)*a^3+10*cos(f*x+e)^3*sin(f*x+e)*(-a)^(1/2)*a^3-2*cos(f*x+e)^3*si n(f*x+e)*(-a)^(1/2)*a^2*b+15*cos(f*x+e)*sin(f*x+e)*(-a)^(1/2)*a^3-4*cos(f* x+e)*sin(f*x+e)*(-a)^(1/2)*a^2*b+5*cos(f*x+e)*sin(f*x+e)*(-a)^(1/2)*a*b^2+ 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^3-9*((b+a*cos(f*x+e)^2)/(1+co s(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+9*((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*b^2-15*(( 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)*b^3+15*(-a)^(1/2)*a^2*b*tan(f*x+e)-1 4*(-a)^(1/2)*a*b^2*tan(f*x+e)+15*(-a)^(1/2)*b^3*tan(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^3*sec(f*x+e)-9*((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...
Time = 1.14 (sec) , antiderivative size = 643, normalized size of antiderivative = 3.15 \[ \int \frac {\cos ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\left [\frac {3 \, {\left (5 \, a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - 5 \, b^{3}\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^{3} \cos \left (f x + e\right )^{5} + 10 \, {\left (a^{3} - a^{2} b\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{3} - 14 \, a^{2} b + 15 \, a b^{2}\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 \, a^{4} f}, -\frac {3 \, {\left (5 \, a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - 5 \, b^{3}\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^{3} \cos \left (f x + e\right )^{5} + 10 \, {\left (a^{3} - a^{2} b\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{3} - 14 \, a^{2} b + 15 \, a b^{2}\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 \, a^{4} f}\right ] \]
[1/384*(3*(5*a^3 - 3*a^2*b + 3*a*b^2 - 5*b^3)*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^3*cos(f*x + e) ^5 + 10*(a^3 - a^2*b)*cos(f*x + e)^3 + (15*a^3 - 14*a^2*b + 15*a*b^2)*cos( f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^4*f ), -1/192*(3*(5*a^3 - 3*a^2*b + 3*a*b^2 - 5*b^3)*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^3*cos(f*x + e)^5 + 10*(a^3 - a^2*b)*cos(f*x + e)^3 + (15*a^3 - 1 4*a^2*b + 15*a*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e) ^2)*sin(f*x + e))/(a^4*f)]
\[ \int \frac {\cos ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\cos ^{6}{\left (e + f x \right )}}{\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \]
\[ \int \frac {\cos ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\cos \left (f x + e\right )^{6}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]
\[ \int \frac {\cos ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\cos \left (f x + e\right )^{6}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\cos ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^6}{\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \]