Integrand size = 25, antiderivative size = 167 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {(a+5 b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 a^{7/2} f}-\frac {\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {5 b \tan (e+f x)}{6 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (13 a+15 b) \tan (e+f x)}{6 a^3 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}} \]
1/2*(a+5*b)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(7/2)/ f-1/6*b*(13*a+15*b)*tan(f*x+e)/a^3/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(1/2)-1/2* cos(f*x+e)*sin(f*x+e)/a/f/(a+b+b*tan(f*x+e)^2)^(3/2)-5/6*b*tan(f*x+e)/a^2/ f/(a+b+b*tan(f*x+e)^2)^(3/2)
Leaf count is larger than twice the leaf count of optimal. \(983\) vs. \(2(167)=334\).
Time = 8.40 (sec) , antiderivative size = 983, normalized size of antiderivative = 5.89 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a+2 b+a \cos (2 e+2 f x))^{5/2} \csc (e+f x) \sec ^5(e+f x) \left (\frac {\sin ^2(e+f x)}{a+b}+\frac {(a+2 b+a \cos (2 (e+f x))) \sin ^2(e+f x)}{(a+b)^2}-\frac {12 \sin ^4(e+f x)}{a+b}+\frac {16 \left (a+b-a \sin ^2(e+f x)\right ) \left (1-\frac {a \sin ^2(e+f x)}{a+b}\right ) \left (-\frac {6 a (a+b) \sin ^2(e+f x)}{a+2 b+a \cos (2 (e+f x))}+\frac {a^2 (a+b) \sin ^4(e+f x)}{\left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {3 \sqrt {a} \sqrt {a+b} \arcsin \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right ) \sin (e+f x)}{\sqrt {\frac {a+b-a \sin ^2(e+f x)}{a+b}}}\right )}{a^3}\right )}{256 \sqrt {2} f \left (a+b \sec ^2(e+f x)\right )^{5/2} \left (a+b-a \sin ^2(e+f x)\right )^{3/2}}-\frac {(a+2 b+a \cos (2 e+2 f x))^{5/2} \csc (e+f x) \sec ^5(e+f x) \left (\frac {\sin ^2(e+f x)}{a+b}+\frac {(a+2 b+a \cos (2 (e+f x))) \sin ^2(e+f x)}{(a+b)^2}-\frac {24 \sin ^4(e+f x)}{a+b}+\frac {96 \sin ^6(e+f x)}{a}+\frac {80 \left (a+b-a \sin ^2(e+f x)\right ) \left (1-\frac {a \sin ^2(e+f x)}{a+b}\right ) \left (-\frac {6 a (a+b) \sin ^2(e+f x)}{a+2 b+a \cos (2 (e+f x))}+\frac {a^2 (a+b) \sin ^4(e+f x)}{\left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {3 \sqrt {a} \sqrt {a+b} \arcsin \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right ) \sin (e+f x)}{\sqrt {\frac {a+b-a \sin ^2(e+f x)}{a+b}}}\right )}{a^3}-\frac {160 \left (a+b-a \sin ^2(e+f x)\right ) \left (1-\frac {a \sin ^2(e+f x)}{a+b}\right ) \left (-\frac {6 a (a+b)^2 \sin ^2(e+f x)}{a+2 b+a \cos (2 (e+f x))}+\frac {3 \sqrt {a} (a+b)^{3/2} \arcsin \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right ) \sin (e+f x)}{\sqrt {\frac {a+b-a \sin ^2(e+f x)}{a+b}}}+\frac {a^2 \sin ^4(e+f x)}{\left (-1+\frac {a \sin ^2(e+f x)}{a+b}\right )^2}\right )}{a^4}\right )}{768 \sqrt {2} f \left (a+b \sec ^2(e+f x)\right )^{5/2} \left (a+b-a \sin ^2(e+f x)\right )^{3/2}}+\frac {5 (2 a+3 b+a \cos (2 (e+f x))) (a+2 b+a \cos (2 e+2 f x))^{5/2} \sec ^4(e+f x) \tan (e+f x)}{384 (a+b)^2 f (a+2 b+a \cos (2 (e+f x)))^{3/2} \left (a+b \sec ^2(e+f x)\right )^{5/2}}-\frac {(b+(3 a+2 b) \cos (2 (e+f x))) (a+2 b+a \cos (2 e+2 f x))^{5/2} \sec ^4(e+f x) \tan (e+f x)}{384 (a+b)^2 f (a+2 b+a \cos (2 (e+f x)))^{3/2} \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]
-1/256*((a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Csc[e + f*x]*Sec[e + f*x]^5*( Sin[e + f*x]^2/(a + b) + ((a + 2*b + a*Cos[2*(e + f*x)])*Sin[e + f*x]^2)/( a + b)^2 - (12*Sin[e + f*x]^4)/(a + b) + (16*(a + b - a*Sin[e + f*x]^2)*(1 - (a*Sin[e + f*x]^2)/(a + b))*((-6*a*(a + b)*Sin[e + f*x]^2)/(a + 2*b + a *Cos[2*(e + f*x)]) + (a^2*(a + b)*Sin[e + f*x]^4)/(a + b - a*Sin[e + f*x]^ 2)^2 + (3*Sqrt[a]*Sqrt[a + b]*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*S in[e + f*x])/Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b)]))/a^3))/(Sqrt[2]*f*( a + b*Sec[e + f*x]^2)^(5/2)*(a + b - a*Sin[e + f*x]^2)^(3/2)) - ((a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Csc[e + f*x]*Sec[e + f*x]^5*(Sin[e + f*x]^2/(a + b) + ((a + 2*b + a*Cos[2*(e + f*x)])*Sin[e + f*x]^2)/(a + b)^2 - (24*Si n[e + f*x]^4)/(a + b) + (96*Sin[e + f*x]^6)/a + (80*(a + b - a*Sin[e + f*x ]^2)*(1 - (a*Sin[e + f*x]^2)/(a + b))*((-6*a*(a + b)*Sin[e + f*x]^2)/(a + 2*b + a*Cos[2*(e + f*x)]) + (a^2*(a + b)*Sin[e + f*x]^4)/(a + b - a*Sin[e + f*x]^2)^2 + (3*Sqrt[a]*Sqrt[a + b]*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*Sin[e + f*x])/Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b)]))/a^3 - (160* (a + b - a*Sin[e + f*x]^2)*(1 - (a*Sin[e + f*x]^2)/(a + b))*((-6*a*(a + b) ^2*Sin[e + f*x]^2)/(a + 2*b + a*Cos[2*(e + f*x)]) + (3*Sqrt[a]*(a + b)^(3/ 2)*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*Sin[e + f*x])/Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b)] + (a^2*Sin[e + f*x]^4)/(-1 + (a*Sin[e + f*x]^2) /(a + b))^2))/a^4))/(768*Sqrt[2]*f*(a + b*Sec[e + f*x]^2)^(5/2)*(a + b ...
Time = 0.37 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4620, 373, 402, 27, 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.
| \(\displaystyle \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^2}{\left (a+b \sec (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4620 |
| \(\displaystyle \frac {\int \frac {\tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 373 |
| \(\displaystyle \frac {\frac {\int \frac {-4 b \tan ^2(e+f x)+a+b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {(a+b) \left (-10 b \tan ^2(e+f x)+3 a+5 b\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 a (a+b)}-\frac {5 b \tan (e+f x)}{3 a \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {-10 b \tan ^2(e+f x)+3 a+5 b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 a}-\frac {5 b \tan (e+f x)}{3 a \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int \frac {3 (a+b) (a+5 b)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a (a+b)}-\frac {b (13 a+15 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a}-\frac {5 b \tan (e+f x)}{3 a \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\frac {3 (a+5 b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a}-\frac {b (13 a+15 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a}-\frac {5 b \tan (e+f x)}{3 a \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {\frac {3 (a+5 b) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{a}-\frac {b (13 a+15 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a}-\frac {5 b \tan (e+f x)}{3 a \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {3 (a+5 b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2}}-\frac {b (13 a+15 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a}-\frac {5 b \tan (e+f x)}{3 a \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
(-1/2*Tan[e + f*x]/(a*(1 + Tan[e + f*x]^2)*(a + b + b*Tan[e + f*x]^2)^(3/2 )) + ((-5*b*Tan[e + f*x])/(3*a*(a + b + b*Tan[e + f*x]^2)^(3/2)) + ((3*(a + 5*b)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/a^(3 /2) - (b*(13*a + 15*b)*Tan[e + f*x])/(a*(a + b)*Sqrt[a + b + b*Tan[e + f*x ]^2]))/(3*a))/(2*a))/f
3.2.27.3.1 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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1)) Int[(e *x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 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[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1)/f Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f f^2*x^2)^(m/2 + 1)), 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. \(1591\) vs. \(2(147)=294\).
Time = 7.01 (sec) , antiderivative size = 1592, normalized size of antiderivative = 9.53
-1/6/f/(a+b)/a^3/(-a)^(1/2)*(b+a*cos(f*x+e)^2)*(3*(-a)^(1/2)*a^3*cos(f*x+e )^4*sin(f*x+e)+3*(-a)^(1/2)*a^2*b*cos(f*x+e)^4*sin(f*x+e)-3*cos(f*x+e)^3*( (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^3-18*((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*si n(f*x+e)*a)*a^2*b*cos(f*x+e)^3-15*((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*b^2*cos(f*x+e)^3+18*(-a)^(1/2)*a^2*b*cos(f*x+e)^2*sin(f*x+e)+20*(-a)^(1/ 2)*a*b^2*cos(f*x+e)^2*sin(f*x+e)-3*cos(f*x+e)^2*((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^3-18*((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*cos(f*x +e)^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*b^2*cos(f*x+e)^2-3*l...
Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (147) = 294\).
Time = 2.09 (sec) , antiderivative size = 879, normalized size of antiderivative = 5.26 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left ({\left (a^{4} + 6 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{2} b^{2} + 6 \, a b^{3} + 5 \, b^{4} + 2 \, {\left (a^{3} b + 6 \, a^{2} b^{2} + 5 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\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 (3 \, {\left (a^{4} + a^{3} b\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (9 \, a^{3} b + 10 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (13 \, a^{2} b^{2} + 15 \, a b^{3}\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 )}{48 \, {\left ({\left (a^{7} + a^{6} b\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b + a^{5} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} + a^{4} b^{3}\right )} f\right )}}, -\frac {3 \, {\left ({\left (a^{4} + 6 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{2} b^{2} + 6 \, a b^{3} + 5 \, b^{4} + 2 \, {\left (a^{3} b + 6 \, a^{2} b^{2} + 5 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\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 (3 \, {\left (a^{4} + a^{3} b\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (9 \, a^{3} b + 10 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (13 \, a^{2} b^{2} + 15 \, a b^{3}\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 )}{24 \, {\left ({\left (a^{7} + a^{6} b\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b + a^{5} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} + a^{4} b^{3}\right )} f\right )}}\right ] \]
[-1/48*(3*((a^4 + 6*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^2*b^2 + 6*a*b^3 + 5*b^4 + 2*(a^3*b + 6*a^2*b^2 + 5*a*b^3)*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*c os(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*(3*(a^4 + a^3*b)*cos(f*x + e)^5 + 2*(9*a^3*b + 10*a^2*b^2)*cos(f*x + e)^3 + (13*a ^2*b^2 + 15*a*b^3)*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)^4 + 2*(a^6*b + a^5*b^2)*f*c os(f*x + e)^2 + (a^5*b^2 + a^4*b^3)*f), -1/24*(3*((a^4 + 6*a^3*b + 5*a^2*b ^2)*cos(f*x + e)^4 + a^2*b^2 + 6*a*b^3 + 5*b^4 + 2*(a^3*b + 6*a^2*b^2 + 5* a*b^3)*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*c os(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*(3*(a^4 + a^3*b)*cos( f*x + e)^5 + 2*(9*a^3*b + 10*a^2*b^2)*cos(f*x + e)^3 + (13*a^2*b^2 + 15*a* b^3)*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)^4 + 2*(a^6*b + a^5*b^2)*f*cos(f*x + e)...
\[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sin ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^2}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \]