Integrand size = 25, antiderivative size = 143 \[ \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {a (3 a+4 b) \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{8 (a+b)^{3/2} f}+\frac {(3 a+4 b) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{8 (a+b) f}+\frac {\sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{4 (a+b) f} \]
1/8*a*(3*a+4*b)*arctanh(sin(f*x+e)*(a+b)^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))/( a+b)^(3/2)/f+1/4*sec(f*x+e)^3*(a+b*sin(f*x+e)^2)^(3/2)*tan(f*x+e)/(a+b)/f+ 1/8*(3*a+4*b)*sec(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)/(a+b)/f
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 14.16 (sec) , antiderivative size = 669, normalized size of antiderivative = 4.68 \[ \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=-\frac {\sec ^3(e+f x) \left (1+\frac {b \sin ^2(e+f x)}{a}\right ) \tan (e+f x) \left (-15 a \arcsin \left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right )-10 b \arcsin \left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \sin ^2(e+f x)-30 a \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{3/2}-20 b \sin ^2(e+f x) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{3/2}-32 a \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{5/2}-32 b \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{5/2}+32 a \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{7/2}+32 b \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{7/2}+15 a \sqrt {-\frac {(a+b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{a^2}}+10 b \sin ^2(e+f x) \sqrt {-\frac {(a+b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{a^2}}\right )}{40 f \sqrt {a+b \sin ^2(e+f x)} \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{3/2}} \]
-1/40*(Sec[e + f*x]^3*(1 + (b*Sin[e + f*x]^2)/a)*Tan[e + f*x]*(-15*a*ArcSi n[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]] - 10*b*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Sin[e + f*x]^2 - 30*a*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2) - 20*b*Sin[e + f*x]^2*Sq rt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/ a))^(3/2) - 32*a*Hypergeometric2F1[2, 4, 7/2, -(((a + b)*Tan[e + f*x]^2)/a )]*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x ]^2)/a))^(5/2) - 32*b*Hypergeometric2F1[2, 4, 7/2, -(((a + b)*Tan[e + f*x] ^2)/a)]*Sin[e + f*x]^2*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-( ((a + b)*Tan[e + f*x]^2)/a))^(5/2) + 32*a*Hypergeometric2F1[2, 4, 7/2, -(( (a + b)*Tan[e + f*x]^2)/a)]*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a ]*(-(((a + b)*Tan[e + f*x]^2)/a))^(7/2) + 32*b*Hypergeometric2F1[2, 4, 7/2 , -(((a + b)*Tan[e + f*x]^2)/a)]*Sin[e + f*x]^2*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(7/2) + 15*a*Sqrt[-( ((a + b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)*Tan[e + f*x]^2)/a^2)] + 10* b*Sin[e + f*x]^2*Sqrt[-(((a + b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)*Tan [e + f*x]^2)/a^2)]))/(f*Sqrt[a + b*Sin[e + f*x]^2]*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2))
Time = 0.32 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3669, 296, 292, 291, 219}
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.
| \(\displaystyle \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a+b \sin (e+f x)^2}}{\cos (e+f x)^5}dx\) |
| \(\Big \downarrow \) 3669 |
| \(\displaystyle \frac {\int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\left (1-\sin ^2(e+f x)\right )^3}d\sin (e+f x)}{f}\) |
| \(\Big \downarrow \) 296 |
| \(\displaystyle \frac {\frac {(3 a+4 b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\left (1-\sin ^2(e+f x)\right )^2}d\sin (e+f x)}{4 (a+b)}+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 (a+b) \left (1-\sin ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 292 |
| \(\displaystyle \frac {\frac {(3 a+4 b) \left (\frac {1}{2} a \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}\right )}{4 (a+b)}+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 (a+b) \left (1-\sin ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {(3 a+4 b) \left (\frac {1}{2} a \int \frac {1}{1-\frac {(a+b) \sin ^2(e+f x)}{b \sin ^2(e+f x)+a}}d\frac {\sin (e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}\right )}{4 (a+b)}+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 (a+b) \left (1-\sin ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {(3 a+4 b) \left (\frac {a \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 \sqrt {a+b}}+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}\right )}{4 (a+b)}+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 (a+b) \left (1-\sin ^2(e+f x)\right )^2}}{f}\) |
((Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(3/2))/(4*(a + b)*(1 - Sin[e + f*x]^ 2)^2) + ((3*a + 4*b)*((a*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin [e + f*x]^2]])/(2*Sqrt[a + b]) + (Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2]) /(2*(1 - Sin[e + f*x]^2))))/(4*(a + b)))/f
3.4.28.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[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] :> Si mp[(-x)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*(p + 1))), x] - Simp[c*(q/( a*(p + 1))) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1), x], x] /; FreeQ[ {a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && EqQ[2*(p + q + 1) + 1, 0] && Gt Q[q, 0] && NeQ[p, -1]
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[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d)) Int[ (a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] || !LtQ[q, -1 ]) && NeQ[p, -1]
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f S ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] /ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(569\) vs. \(2(127)=254\).
Time = 1.59 (sec) , antiderivative size = 570, normalized size of antiderivative = 3.99
| method | result | size |
| default | \(\frac {2 \left (a +b \right )^{\frac {3}{2}} \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, b \left (3 a +4 b \right ) \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )+a \left (3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{3}+10 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2} b +11 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a \,b^{2}+4 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b^{3}-3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{3}-10 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2} b -11 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a \,b^{2}-4 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b^{3}\right ) \left (\cos ^{4}\left (f x +e \right )\right )+2 \left (a +b \right )^{\frac {3}{2}} {\left (a +b -b \left (\cos ^{2}\left (f x +e \right )\right )\right )}^{\frac {3}{2}} \left (3 a +4 b \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+4 \left (a +b \right )^{\frac {5}{2}} {\left (a +b -b \left (\cos ^{2}\left (f x +e \right )\right )\right )}^{\frac {3}{2}} \sin \left (f x +e \right )}{16 \left (a +b \right )^{\frac {3}{2}} \cos \left (f x +e \right )^{4} \left (a^{2}+2 a b +b^{2}\right ) f}\) | \(570\) |
1/16*(2*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*b*(3*a+4*b)*cos(f*x+e)^4*si n(f*x+e)+a*(3*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+ b*sin(f*x+e)+a))*a^3+10*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e) ^2)^(1/2)+b*sin(f*x+e)+a))*a^2*b+11*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b- b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a*b^2+4*ln(2/(sin(f*x+e)-1)*((a+b)^ (1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b^3-3*ln(2/(1+sin(f*x+e) )*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^3-10*ln(2/(1+ sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2*b -11*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+ e)+a))*a*b^2-4*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2) -b*sin(f*x+e)+a))*b^3)*cos(f*x+e)^4+2*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/ 2)*(3*a+4*b)*cos(f*x+e)^2*sin(f*x+e)+4*(a+b)^(5/2)*(a+b-b*cos(f*x+e)^2)^(3 /2)*sin(f*x+e))/(a+b)^(3/2)/cos(f*x+e)^4/(a^2+2*a*b+b^2)/f
Time = 0.55 (sec) , antiderivative size = 443, normalized size of antiderivative = 3.10 \[ \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\left [\frac {{\left (3 \, a^{2} + 4 \, a b\right )} \sqrt {a + b} \cos \left (f x + e\right )^{4} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) + 4 \, {\left ({\left (3 \, a^{2} + 5 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{32 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4}}, -\frac {{\left (3 \, a^{2} + 4 \, a b\right )} \sqrt {-a - b} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{4} - 2 \, {\left ({\left (3 \, a^{2} + 5 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{16 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4}}\right ] \]
[1/32*((3*a^2 + 4*a*b)*sqrt(a + b)*cos(f*x + e)^4*log(((a^2 + 8*a*b + 8*b^ 2)*cos(f*x + e)^4 - 8*(a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^2 - 4*((a + 2*b)* cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b)*si n(f*x + e) + 8*a^2 + 16*a*b + 8*b^2)/cos(f*x + e)^4) + 4*((3*a^2 + 5*a*b + 2*b^2)*cos(f*x + e)^2 + 2*a^2 + 4*a*b + 2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^4), -1/16*((3*a^2 + 4*a*b)*sqrt(-a - b)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sq rt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(((a*b + b^2)*cos(f*x + e)^2 - a^2 - 2*a*b - b^2)*sin(f*x + e)))*cos(f*x + e)^4 - 2*((3*a^2 + 5*a*b + 2*b ^2)*cos(f*x + e)^2 + 2*a^2 + 4*a*b + 2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b )*sin(f*x + e))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^4)]
\[ \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \sec ^{5}{\left (e + f x \right )}\, dx \]
\[ \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{5} \,d x } \]
\[ \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{5} \,d x } \]
Timed out. \[ \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int \frac {\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}}{{\cos \left (e+f\,x\right )}^5} \,d x \]