Integrand size = 25, antiderivative size = 248 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\left (5 a^2+10 a b+b^2\right ) \cos (e+f x)}{5 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac {2 (5 a-b) \cos ^3(e+f x)}{15 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {4 b \left (5 a^2+10 a b+b^2\right ) \sec (e+f x)}{15 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {8 b \left (5 a^2+10 a b+b^2\right ) \sec (e+f x)}{15 (a-b)^5 f \sqrt {a-b+b \sec ^2(e+f x)}} \]
-1/5*(5*a^2+10*a*b+b^2)*cos(f*x+e)/(a-b)^3/f/(a-b+b*sec(f*x+e)^2)^(3/2)+2/ 15*(5*a-b)*cos(f*x+e)^3/(a-b)^2/f/(a-b+b*sec(f*x+e)^2)^(3/2)-1/5*cos(f*x+e )^5/(a-b)/f/(a-b+b*sec(f*x+e)^2)^(3/2)-4/15*b*(5*a^2+10*a*b+b^2)*sec(f*x+e )/(a-b)^4/f/(a-b+b*sec(f*x+e)^2)^(3/2)-8/15*b*(5*a^2+10*a*b+b^2)*sec(f*x+e )/(a-b)^5/f/(a-b+b*sec(f*x+e)^2)^(1/2)
Time = 2.91 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.19 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\cos (e+f x) \left (425 a^4+4700 a^3 b+6134 a^2 b^2+4700 a b^3+425 b^4+48 \left (11 a^4+106 a^3 b-106 a b^3-11 b^4\right ) \cos (2 (e+f x))+12 (a-b)^2 \left (7 a^2+50 a b+7 b^2\right ) \cos (4 (e+f x))-16 a^4 \cos (6 (e+f x))+32 a^3 b \cos (6 (e+f x))-32 a b^3 \cos (6 (e+f x))+16 b^4 \cos (6 (e+f x))+3 a^4 \cos (8 (e+f x))-12 a^3 b \cos (8 (e+f x))+18 a^2 b^2 \cos (8 (e+f x))-12 a b^3 \cos (8 (e+f x))+3 b^4 \cos (8 (e+f x))\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{480 \sqrt {2} (a-b)^5 f (a+b+(a-b) \cos (2 (e+f x)))^2} \]
-1/480*(Cos[e + f*x]*(425*a^4 + 4700*a^3*b + 6134*a^2*b^2 + 4700*a*b^3 + 4 25*b^4 + 48*(11*a^4 + 106*a^3*b - 106*a*b^3 - 11*b^4)*Cos[2*(e + f*x)] + 1 2*(a - b)^2*(7*a^2 + 50*a*b + 7*b^2)*Cos[4*(e + f*x)] - 16*a^4*Cos[6*(e + f*x)] + 32*a^3*b*Cos[6*(e + f*x)] - 32*a*b^3*Cos[6*(e + f*x)] + 16*b^4*Cos [6*(e + f*x)] + 3*a^4*Cos[8*(e + f*x)] - 12*a^3*b*Cos[8*(e + f*x)] + 18*a^ 2*b^2*Cos[8*(e + f*x)] - 12*a*b^3*Cos[8*(e + f*x)] + 3*b^4*Cos[8*(e + f*x) ])*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(Sqrt[2]*(a - b)^5*f*(a + b + (a - b)*Cos[2*(e + f*x)])^2)
Time = 0.39 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4147, 365, 25, 359, 245, 209, 208}
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.
| \(\displaystyle \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^5}{\left (a+b \tan (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int \frac {\cos ^6(e+f x) \left (1-\sec ^2(e+f x)\right )^2}{\left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 365 |
| \(\displaystyle \frac {\frac {\int -\frac {\cos ^4(e+f x) \left (2 (5 a-b)-5 (a-b) \sec ^2(e+f x)\right )}{\left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {\cos ^4(e+f x) \left (2 (5 a-b)-5 (a-b) \sec ^2(e+f x)\right )}{\left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {-\frac {-\frac {\left (5 a^2+10 a b+b^2\right ) \int \frac {\cos ^2(e+f x)}{\left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{a-b}-\frac {2 (5 a-b) \cos ^3(e+f x)}{3 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 245 |
| \(\displaystyle \frac {-\frac {-\frac {\left (5 a^2+10 a b+b^2\right ) \left (-\frac {4 b \int \frac {1}{\left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{a-b}-\frac {\cos (e+f x)}{(a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}\right )}{a-b}-\frac {2 (5 a-b) \cos ^3(e+f x)}{3 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 209 |
| \(\displaystyle \frac {-\frac {-\frac {\left (5 a^2+10 a b+b^2\right ) \left (-\frac {4 b \left (\frac {2 \int \frac {1}{\left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{3 (a-b)}+\frac {\sec (e+f x)}{3 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}\right )}{a-b}-\frac {\cos (e+f x)}{(a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}\right )}{a-b}-\frac {2 (5 a-b) \cos ^3(e+f x)}{3 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 208 |
| \(\displaystyle \frac {-\frac {-\frac {\left (5 a^2+10 a b+b^2\right ) \left (-\frac {4 b \left (\frac {2 \sec (e+f x)}{3 (a-b)^2 \sqrt {a+b \sec ^2(e+f x)-b}}+\frac {\sec (e+f x)}{3 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}\right )}{a-b}-\frac {\cos (e+f x)}{(a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}\right )}{a-b}-\frac {2 (5 a-b) \cos ^3(e+f x)}{3 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
(-1/5*Cos[e + f*x]^5/((a - b)*(a - b + b*Sec[e + f*x]^2)^(3/2)) - ((-2*(5* a - b)*Cos[e + f*x]^3)/(3*(a - b)*(a - b + b*Sec[e + f*x]^2)^(3/2)) - ((5* a^2 + 10*a*b + b^2)*(-(Cos[e + f*x]/((a - b)*(a - b + b*Sec[e + f*x]^2)^(3 /2))) - (4*b*(Sec[e + f*x]/(3*(a - b)*(a - b + b*Sec[e + f*x]^2)^(3/2)) + (2*Sec[e + f*x])/(3*(a - b)^2*Sqrt[a - b + b*Sec[e + f*x]^2])))/(a - b)))/ (a - b))/(5*(a - b)))/f
3.2.40.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x _Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] - Simp[1/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 )), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( m - 1)/2]
Time = 3.57 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.54
| method | result | size |
| default | \(\frac {a^{7} \left (3 a^{4} \cos \left (f x +e \right )^{8}-12 a^{3} b \cos \left (f x +e \right )^{8}+18 a^{2} b^{2} \cos \left (f x +e \right )^{8}+12 \cos \left (f x +e \right )^{6} \sin \left (f x +e \right )^{2} a \,b^{3}+3 \cos \left (f x +e \right )^{4} \sin \left (f x +e \right )^{4} b^{4}-10 a^{4} \cos \left (f x +e \right )^{6}+32 a^{3} b \cos \left (f x +e \right )^{6}-36 a^{2} b^{2} \cos \left (f x +e \right )^{6}-4 \cos \left (f x +e \right )^{4} \sin \left (f x +e \right )^{2} a \,b^{3}+4 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{4} b^{4}+15 a^{4} \cos \left (f x +e \right )^{4}-42 a^{2} b^{2} \cos \left (f x +e \right )^{4}-28 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2} a \,b^{3}+8 \sin \left (f x +e \right )^{4} b^{4}+60 a^{3} b \cos \left (f x +e \right )^{2}+60 a^{2} b^{2} \cos \left (f x +e \right )^{2}+80 \sin \left (f x +e \right )^{2} a \,b^{3}+40 a^{2} b^{2}\right ) \left (a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}\right ) \left (a -b \right )^{2} \sec \left (f x +e \right )^{5}}{15 f \left (\sqrt {-b \left (a -b \right )}+a -b \right )^{7} \left (\sqrt {-b \left (a -b \right )}-a +b \right )^{7} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}\) | \(382\) |
1/15/f*a^7/((-b*(a-b))^(1/2)+a-b)^7/((-b*(a-b))^(1/2)-a+b)^7*(3*a^4*cos(f* x+e)^8-12*a^3*b*cos(f*x+e)^8+18*a^2*b^2*cos(f*x+e)^8+12*cos(f*x+e)^6*sin(f *x+e)^2*a*b^3+3*cos(f*x+e)^4*sin(f*x+e)^4*b^4-10*a^4*cos(f*x+e)^6+32*a^3*b *cos(f*x+e)^6-36*a^2*b^2*cos(f*x+e)^6-4*cos(f*x+e)^4*sin(f*x+e)^2*a*b^3+4* cos(f*x+e)^2*sin(f*x+e)^4*b^4+15*a^4*cos(f*x+e)^4-42*a^2*b^2*cos(f*x+e)^4- 28*cos(f*x+e)^2*sin(f*x+e)^2*a*b^3+8*sin(f*x+e)^4*b^4+60*a^3*b*cos(f*x+e)^ 2+60*a^2*b^2*cos(f*x+e)^2+80*sin(f*x+e)^2*a*b^3+40*a^2*b^2)*(a*cos(f*x+e)^ 2+b*sin(f*x+e)^2)*(a-b)^2/(a+b*tan(f*x+e)^2)^(5/2)*sec(f*x+e)^5
Time = 0.56 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.49 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {{\left (3 \, {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{9} - 2 \, {\left (5 \, a^{4} - 16 \, a^{3} b + 18 \, a^{2} b^{2} - 8 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{7} + 3 \, {\left (5 \, a^{4} - 14 \, a^{2} b^{2} + 8 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{5} + 12 \, {\left (5 \, a^{3} b + 5 \, a^{2} b^{2} - 9 \, a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{3} + 8 \, {\left (5 \, a^{2} b^{2} + 10 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left ({\left (a^{7} - 7 \, a^{6} b + 21 \, a^{5} b^{2} - 35 \, a^{4} b^{3} + 35 \, a^{3} b^{4} - 21 \, a^{2} b^{5} + 7 \, a b^{6} - b^{7}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b - 6 \, a^{5} b^{2} + 15 \, a^{4} b^{3} - 20 \, a^{3} b^{4} + 15 \, a^{2} b^{5} - 6 \, a b^{6} + b^{7}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} - 5 \, a^{4} b^{3} + 10 \, a^{3} b^{4} - 10 \, a^{2} b^{5} + 5 \, a b^{6} - b^{7}\right )} f\right )}} \]
-1/15*(3*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^9 - 2*(5 *a^4 - 16*a^3*b + 18*a^2*b^2 - 8*a*b^3 + b^4)*cos(f*x + e)^7 + 3*(5*a^4 - 14*a^2*b^2 + 8*a*b^3 + b^4)*cos(f*x + e)^5 + 12*(5*a^3*b + 5*a^2*b^2 - 9*a *b^3 - b^4)*cos(f*x + e)^3 + 8*(5*a^2*b^2 + 10*a*b^3 + b^4)*cos(f*x + e))* sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a^7 - 7*a^6*b + 21*a^5 *b^2 - 35*a^4*b^3 + 35*a^3*b^4 - 21*a^2*b^5 + 7*a*b^6 - b^7)*f*cos(f*x + e )^4 + 2*(a^6*b - 6*a^5*b^2 + 15*a^4*b^3 - 20*a^3*b^4 + 15*a^2*b^5 - 6*a*b^ 6 + b^7)*f*cos(f*x + e)^2 + (a^5*b^2 - 5*a^4*b^3 + 10*a^3*b^4 - 10*a^2*b^5 + 5*a*b^6 - b^7)*f)
Timed out. \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 534 vs. \(2 (228) = 456\).
Time = 0.33 (sec) , antiderivative size = 534, normalized size of antiderivative = 2.15 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {15 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {3 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5} - 20 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} b \cos \left (f x + e\right )^{3} + 90 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )}{a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}} - \frac {10 \, {\left ({\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 9 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} + \frac {5 \, {\left (12 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b^{3} \cos \left (f x + e\right )^{2} - b^{4}\right )}}{{\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}} + \frac {10 \, {\left (9 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b^{2} \cos \left (f x + e\right )^{2} - b^{3}\right )}}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}} + \frac {5 \, {\left (6 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b \cos \left (f x + e\right )^{2} - b^{2}\right )}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}}}{15 \, f} \]
-1/15*(15*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)/(a^3 - 3*a^2*b + 3*a *b^2 - b^3) + (3*(a - b + b/cos(f*x + e)^2)^(5/2)*cos(f*x + e)^5 - 20*(a - b + b/cos(f*x + e)^2)^(3/2)*b*cos(f*x + e)^3 + 90*sqrt(a - b + b/cos(f*x + e)^2)*b^2*cos(f*x + e))/(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b ^4 - b^5) - 10*((a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3 - 9*sqrt(a - b + b/cos(f*x + e)^2)*b*cos(f*x + e))/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a* b^3 + b^4) + 5*(12*(a - b + b/cos(f*x + e)^2)*b^3*cos(f*x + e)^2 - b^4)/(( a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*(a - b + b/cos(f* x + e)^2)^(3/2)*cos(f*x + e)^3) + 10*(9*(a - b + b/cos(f*x + e)^2)*b^2*cos (f*x + e)^2 - b^3)/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*(a - b + b /cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3) + 5*(6*(a - b + b/cos(f*x + e)^2)*b *cos(f*x + e)^2 - b^2)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3))/f
\[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{5}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^5}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]