Integrand size = 25, antiderivative size = 261 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\left (3 a^2-10 a b+35 b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 a^{9/2} f}+\frac {(3 a-7 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {b \left (9 a^2-18 a b-35 b^2\right ) \tan (e+f x)}{24 a^3 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {b \left (9 a^3-15 a^2 b-145 a b^2-105 b^3\right ) \tan (e+f x)}{24 a^4 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}} \]
1/8*(3*a^2-10*a*b+35*b^2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^( 1/2))/a^(9/2)/f+1/24*b*(9*a^3-15*a^2*b-145*a*b^2-105*b^3)*tan(f*x+e)/a^4/( a+b)^2/f/(a+b+b*tan(f*x+e)^2)^(1/2)+1/8*(3*a-7*b)*cos(f*x+e)*sin(f*x+e)/a^ 2/f/(a+b+b*tan(f*x+e)^2)^(3/2)+1/4*cos(f*x+e)^3*sin(f*x+e)/a/f/(a+b+b*tan( f*x+e)^2)^(3/2)+1/24*b*(9*a^2-18*a*b-35*b^2)*tan(f*x+e)/a^3/(a+b)/f/(a+b+b *tan(f*x+e)^2)^(3/2)
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 18.42 (sec) , antiderivative size = 1777, normalized size of antiderivative = 6.81 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \]
(3*(a + b)*AppellF1[1/2, -4, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/ (a + b)]*Cos[e + f*x]^12*Sin[e + f*x])/(4*Sqrt[2]*f*(a + b*Sec[e + f*x]^2) ^(5/2)*(a + b - a*Sin[e + f*x]^2)^(5/2)*(3*(a + b)*AppellF1[1/2, -4, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (5*a*AppellF1[3/2, -4, 7/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 8*(a + b)*AppellF1 [3/2, -3, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e + f *x]^2)*((15*a*(a + b)*AppellF1[1/2, -4, 5/2, 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)/(4*Sqrt[2]*(a + b - a*S in[e + f*x]^2)^(7/2)*(3*(a + b)*AppellF1[1/2, -4, 5/2, 3/2, Sin[e + f*x]^2 , (a*Sin[e + f*x]^2)/(a + b)] + (5*a*AppellF1[3/2, -4, 7/2, 5/2, Sin[e + f *x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 8*(a + b)*AppellF1[3/2, -3, 5/2, 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, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]* Cos[e + f*x]^9)/(4*Sqrt[2]*(a + b - a*Sin[e + f*x]^2)^(5/2)*(3*(a + b)*App ellF1[1/2, -4, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (5* a*AppellF1[3/2, -4, 7/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 8*(a + b)*AppellF1[3/2, -3, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2) /(a + b)])*Sin[e + f*x]^2)) - (3*Sqrt[2]*(a + b)*AppellF1[1/2, -4, 5/2, 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)/((a + b - a*Sin[e + f*x]^2)^(5/2)*(3*(a + b)*AppellF1[1/2, -4, 5/2,...
Time = 0.47 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.10, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4634, 316, 25, 402, 25, 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.
| \(\displaystyle \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (e+f x)^4 \left (a+b \sec (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4634 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac {\int -\frac {6 b \tan ^2(e+f x)+3 a-b}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{4 a}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {6 b \tan ^2(e+f x)+3 a-b}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {(3 a-7 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac {\int -\frac {3 a^2+2 b a+7 b^2+4 (3 a-7 b) b \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{2 a}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {3 a^2+2 b a+7 b^2+4 (3 a-7 b) b \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{2 a}+\frac {(3 a-7 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int \frac {9 a^3-3 b a^2+39 b^2 a+35 b^3+2 b \left (9 a^2-18 b a-35 b^2\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 a (a+b)}+\frac {b \left (9 a^2-18 a b-35 b^2\right ) \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}+\frac {(3 a-7 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {\int \frac {3 (a+b)^2 \left (3 a^2-10 b a+35 b^2\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a (a+b)}+\frac {b \left (9 a^3-15 a^2 b-145 a b^2-105 b^3\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}+\frac {b \left (9 a^2-18 a b-35 b^2\right ) \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}+\frac {(3 a-7 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {3 (a+b) \left (3 a^2-10 a b+35 b^2\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a}+\frac {b \left (9 a^3-15 a^2 b-145 a b^2-105 b^3\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}+\frac {b \left (9 a^2-18 a b-35 b^2\right ) \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}+\frac {(3 a-7 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {3 (a+b) \left (3 a^2-10 a b+35 b^2\right ) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{a}+\frac {b \left (9 a^3-15 a^2 b-145 a b^2-105 b^3\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}+\frac {b \left (9 a^2-18 a b-35 b^2\right ) \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}+\frac {(3 a-7 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {b \left (9 a^2-18 a b-35 b^2\right ) \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}+\frac {\frac {3 (a+b) \left (3 a^2-10 a b+35 b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2}}+\frac {b \left (9 a^3-15 a^2 b-145 a b^2-105 b^3\right ) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}}{2 a}+\frac {(3 a-7 b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{4 a}+\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
(Tan[e + f*x]/(4*a*(1 + Tan[e + f*x]^2)^2*(a + b + b*Tan[e + f*x]^2)^(3/2) ) + (((3*a - 7*b)*Tan[e + f*x])/(2*a*(1 + Tan[e + f*x]^2)*(a + b + b*Tan[e + f*x]^2)^(3/2)) + ((b*(9*a^2 - 18*a*b - 35*b^2)*Tan[e + f*x])/(3*a*(a + b)*(a + b + b*Tan[e + f*x]^2)^(3/2)) + ((3*(a + b)*(3*a^2 - 10*a*b + 35*b^ 2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/a^(3/2) + (b*(9*a^3 - 15*a^2*b - 145*a*b^2 - 105*b^3)*Tan[e + f*x])/(a*(a + b)*Sqr t[a + b + b*Tan[e + f*x]^2]))/(3*a*(a + b)))/(2*a))/(4*a))/f
3.3.94.3.1 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. \(2769\) vs. \(2(237)=474\).
Time = 8.14 (sec) , antiderivative size = 2770, normalized size of antiderivative = 10.61
1/24/f/a^4/(-a)^(1/2)/(a+b)^2*(b+a*cos(f*x+e)^2)*(-105*(-a)^(1/2)*b^5*sin( f*x+e)+180*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^3*cos(f*x+e)^2+10 5*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*b^4*cos(f*x+e)^2+18*(-a)^(1/2) *a^4*b*cos(f*x+e)^2*sin(f*x+e)-24*(-a)^(1/2)*a^3*b^2*cos(f*x+e)^2*sin(f*x+ e)-198*(-a)^(1/2)*a^2*b^3*cos(f*x+e)^2*sin(f*x+e)-140*(-a)^(1/2)*a*b^4*cos (f*x+e)^2*sin(f*x+e)+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)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^4*b*cos(f *x+e)-12*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*b^2*cos(f*x+e)+54*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*c os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b^3*cos(f*x+e)+180*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...
Leaf count of result is larger than twice the leaf count of optimal. 533 vs. \(2 (237) = 474\).
Time = 5.42 (sec) , antiderivative size = 1187, normalized size of antiderivative = 4.55 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]
[-1/192*(3*(3*a^4*b^2 - 4*a^3*b^3 + 18*a^2*b^4 + 60*a*b^5 + 35*b^6 + (3*a^ 6 - 4*a^5*b + 18*a^4*b^2 + 60*a^3*b^3 + 35*a^2*b^4)*cos(f*x + e)^4 + 2*(3* a^5*b - 4*a^4*b^2 + 18*a^3*b^3 + 60*a^2*b^4 + 35*a*b^5)*cos(f*x + e)^2)*sq rt(-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*(6*(a^6 + 2*a^5*b + a^4*b^2)*cos(f*x + e)^7 + 3*(3*a^6 - a^5*b - 11 *a^4*b^2 - 7*a^3*b^3)*cos(f*x + e)^5 + 2*(9*a^5*b - 12*a^4*b^2 - 99*a^3*b^ 3 - 70*a^2*b^4)*cos(f*x + e)^3 + (9*a^4*b^2 - 15*a^3*b^3 - 145*a^2*b^4 - 1 05*a*b^5)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f* x + e))/((a^9 + 2*a^8*b + a^7*b^2)*f*cos(f*x + e)^4 + 2*(a^8*b + 2*a^7*b^2 + a^6*b^3)*f*cos(f*x + e)^2 + (a^7*b^2 + 2*a^6*b^3 + a^5*b^4)*f), -1/96*( 3*(3*a^4*b^2 - 4*a^3*b^3 + 18*a^2*b^4 + 60*a*b^5 + 35*b^6 + (3*a^6 - 4*a^5 *b + 18*a^4*b^2 + 60*a^3*b^3 + 35*a^2*b^4)*cos(f*x + e)^4 + 2*(3*a^5*b - 4 *a^4*b^2 + 18*a^3*b^3 + 60*a^2*b^4 + 35*a*b^5)*cos(f*x + e)^2)*sqrt(a)*arc tan(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)...
\[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\cos ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{4}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{4}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^4}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \]