Integrand size = 25, antiderivative size = 196 \[ \int \cos ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {(a+b) \left (5 a^2-2 a b+b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 a^{5/2} f}+\frac {(3 a-b) (5 a+3 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 a^2 f}+\frac {(5 a+b) \cos ^3(e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 a f}+\frac {\cos ^5(e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{6 f} \]
1/16*(a+b)*(5*a^2-2*a*b+b^2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2 )^(1/2))/a^(5/2)/f+1/48*(3*a-b)*(5*a+3*b)*cos(f*x+e)*sin(f*x+e)*(a+b+b*tan (f*x+e)^2)^(1/2)/a^2/f+1/24*(5*a+b)*cos(f*x+e)^3*sin(f*x+e)*(a+b+b*tan(f*x +e)^2)^(1/2)/a/f+1/6*cos(f*x+e)^5*sin(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/f
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 16.88 (sec) , antiderivative size = 1902, normalized size of antiderivative = 9.70 \[ \int \cos ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx =\text {Too large to display} \]
(3*(a + b)*AppellF1[1/2, -2, -1/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2) /(a + b)]*Cos[e + f*x]^10*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*Sqrt[a + b*Se c[e + f*x]^2]*Sin[e + f*x])/(f*(3*(a + b)*AppellF1[1/2, -2, -1/2, 3/2, Sin [e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - (a*AppellF1[3/2, -2, 1/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + 4*(a + b)*AppellF1[3/2, -1, -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, -2, -1/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/ (a + b)]*Cos[e + f*x]^5*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]])/(3*(a + b)*App ellF1[1/2, -2, -1/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - (a *AppellF1[3/2, -2, 1/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + 4*(a + b)*AppellF1[3/2, -1, -1/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2) /(a + b)])*Sin[e + f*x]^2) - (12*(a + b)*AppellF1[1/2, -2, -1/2, 3/2, Sin[ e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^3*Sqrt[a + 2*b + a*Co s[2*(e + f*x)]]*Sin[e + f*x]^2)/(3*(a + b)*AppellF1[1/2, -2, -1/2, 3/2, Si n[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - (a*AppellF1[3/2, -2, 1/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + 4*(a + b)*AppellF1[3/2, -1, -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]^4*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*Sin[e + f*x] *(-1/3*(a*f*AppellF1[3/2, -2, 1/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2) /(a + b)]*Cos[e + f*x]*Sin[e + f*x])/(a + b) - (4*f*AppellF1[3/2, -1, -...
Time = 0.37 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4634, 314, 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.
\(\displaystyle \int \cos ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a+b \sec (e+f x)^2}}{\sec (e+f x)^6}dx\) |
\(\Big \downarrow \) 4634 |
\(\displaystyle \frac {\int \frac {\sqrt {b \tan ^2(e+f x)+a+b}}{\left (\tan ^2(e+f x)+1\right )^4}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 314 |
\(\displaystyle \frac {\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}-\frac {1}{6} \int -\frac {4 b \tan ^2(e+f x)+5 (a+b)}{\left (\tan ^2(e+f x)+1\right )^3 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {1}{6} \int \frac {4 b \tan ^2(e+f x)+5 (a+b)}{\left (\tan ^2(e+f x)+1\right )^3 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {\frac {1}{6} \left (\frac {(5 a+b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}-\frac {\int -\frac {2 b (5 a+b) \tan ^2(e+f x)+(15 a-b) (a+b)}{\left (\tan ^2(e+f x)+1\right )^2 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{4 a}\right )+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {1}{6} \left (\frac {\int \frac {2 b (5 a+b) \tan ^2(e+f x)+(15 a-b) (a+b)}{\left (\tan ^2(e+f x)+1\right )^2 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{4 a}+\frac {(5 a+b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}\right )+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {\frac {1}{6} \left (\frac {\frac {(3 a-b) (5 a+3 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {\int -\frac {3 (a+b) \left (5 a^2-2 b a+b^2\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 a}}{4 a}+\frac {(5 a+b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}\right )+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{6} \left (\frac {\frac {3 (a+b) \left (5 a^2-2 a b+b^2\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 a}+\frac {(3 a-b) (5 a+3 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}}{4 a}+\frac {(5 a+b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}\right )+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\frac {1}{6} \left (\frac {\frac {3 (a+b) \left (5 a^2-2 a b+b^2\right ) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{2 a}+\frac {(3 a-b) (5 a+3 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}}{4 a}+\frac {(5 a+b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}\right )+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {1}{6} \left (\frac {\frac {3 (a+b) \left (5 a^2-2 a b+b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 a^{3/2}}+\frac {(3 a-b) (5 a+3 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}}{4 a}+\frac {(5 a+b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 a \left (\tan ^2(e+f x)+1\right )^2}\right )+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
((Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(6*(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 + b^2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*a^(3/2)) + ((3*a - b)*(5*a + 3*b )*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(2*a*(1 + Tan[e + f*x]^2))) /(4*a))/6)/f
3.3.40.3.1 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[(-x)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*(p + 1))), x] + Simp[1/(2*a* (p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(2*p + 3) + d *(2*(p + q + 1) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && LtQ[0, 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. \(969\) vs. \(2(176)=352\).
Time = 7.74 (sec) , antiderivative size = 970, normalized size of antiderivative = 4.95
1/48/f/a^2/(-a)^(1/2)*(8*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^ (1/2)*a^2*cos(f*x+e)^5*sin(f*x+e)+8*(-a)^(1/2)*sin(f*x+e)*cos(f*x+e)^4*((b +a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2+10*(-a)^(1/2)*((b+a*cos(f*x+e )^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*cos(f*x+e)^3*sin(f*x+e)+2*(-a)^(1/2)*((b+ a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b*cos(f*x+e)^3*sin(f*x+e)+10*(-a )^(1/2)*sin(f*x+e)*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2 )*a^2+2*(-a)^(1/2)*sin(f*x+e)*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ e))^2)^(1/2)*a*b+15*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2) *a^2*cos(f*x+e)*sin(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e)) ^2)^(1/2)*a*b*cos(f*x+e)*sin(f*x+e)-3*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+co s(f*x+e))^2)^(1/2)*b^2*cos(f*x+e)*sin(f*x+e)+15*(-a)^(1/2)*sin(f*x+e)*((b+ a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2+4*(-a)^(1/2)*sin(f*x+e)*((b+a* cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b-3*(-a)^(1/2)*sin(f*x+e)*((b+a*co s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2+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)*a^3+9*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-3*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+3*ln(4*(-a)^(1/2)*((...
Time = 1.17 (sec) , antiderivative size = 641, normalized size of antiderivative = 3.27 \[ \int \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 - a b^{2} + 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} + 2 \, {\left (5 \, a^{3} + a^{2} b\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{3} + 4 \, a^{2} b - 3 \, 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^{3} f}, -\frac {3 \, {\left (5 \, a^{3} + 3 \, a^{2} b - a b^{2} + 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} + 2 \, {\left (5 \, a^{3} + a^{2} b\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{3} + 4 \, a^{2} b - 3 \, 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^{3} f}\right ] \]
[-1/384*(3*(5*a^3 + 3*a^2*b - a*b^2 + 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*co s(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) - 8*(8*a^3*cos(f*x + e)^5 + 2*(5*a^3 + a^2*b)*cos(f*x + e)^3 + (15*a^3 + 4*a^2*b - 3*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^3*f), - 1/192*(3*(5*a^3 + 3*a^2*b - a*b^2 + 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 + 2*(5*a^3 + a^2*b)*cos(f*x + e)^3 + (15*a^3 + 4*a^2*b - 3*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^3*f)]
Timed out. \[ \int \cos ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\text {Timed out} \]
\[ \int \cos ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \cos \left (f x + e\right )^{6} \,d x } \]
\[ \int \cos ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \cos \left (f x + e\right )^{6} \,d x } \]
Timed out. \[ \int \cos ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int {\cos \left (e+f\,x\right )}^6\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}} \,d x \]