Integrand size = 25, antiderivative size = 187 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {3 (a-5 b) (a-b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{8 a^{7/2} f}-\frac {5 (a-b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {(13 a-15 b) b \sec (e+f x)}{8 a^3 f \sqrt {a-b+b \sec ^2(e+f x)}} \]
-3/8*(a-5*b)*(a-b)*arctanh(sec(f*x+e)*a^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))/ a^(7/2)/f-5/8*(a-b)*cot(f*x+e)*csc(f*x+e)/a^2/f/(a-b+b*sec(f*x+e)^2)^(1/2) -1/4*cot(f*x+e)^3*csc(f*x+e)/a/f/(a-b+b*sec(f*x+e)^2)^(1/2)-1/8*(13*a-15*b )*b*sec(f*x+e)/a^3/f/(a-b+b*sec(f*x+e)^2)^(1/2)
Time = 5.50 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.84 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {\frac {\left (\left (-8 a^2+52 a b-60 b^2\right ) \cos (2 (e+f x))+(a-b) (-11 a-45 b+3 (a-5 b) \cos (4 (e+f x)))\right ) \csc ^4(e+f x) \sec (e+f x)}{4 \sqrt {2} a^3 \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}+\frac {3 (a-5 b) (a-b) \cos (e+f x) \left (2 \text {arctanh}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right )+\log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{2 a^{7/2} \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}}}{8 f} \]
((((-8*a^2 + 52*a*b - 60*b^2)*Cos[2*(e + f*x)] + (a - b)*(-11*a - 45*b + 3 *(a - 5*b)*Cos[4*(e + f*x)]))*Csc[e + f*x]^4*Sec[e + f*x])/(4*Sqrt[2]*a^3* Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]) + (3*(a - 5*b)*(a - b)*Cos[e + f*x]*(2*ArcTanh[Tan[(e + f*x)/2]^2 - Sqrt[4*b*Tan[(e + f*x)/ 2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]/Sqrt[a]] + Log[a - 2*b - a*Tan[(e + f*x)/2]^2 + Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2] ^2)^2]])*Sec[(e + f*x)/2]^2*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(2*a^(7/2)*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f*x )/2]^4]))/(8*f)
Time = 0.41 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4147, 25, 372, 402, 25, 27, 402, 27, 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.
| \(\displaystyle \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^5 \left (a+b \tan (e+f x)^2\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int -\frac {\sec ^4(e+f x)}{\left (1-\sec ^2(e+f x)\right )^3 \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\sec ^4(e+f x)}{\left (1-\sec ^2(e+f x)\right )^3 \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\int \frac {4 (a-b) \sec ^2(e+f x)+a-b}{\left (1-\sec ^2(e+f x)\right )^2 \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \sqrt {a+b \sec ^2(e+f x)-b}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int -\frac {(a-b) \left (-10 b \sec ^2(e+f x)+3 a-5 b\right )}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{2 a}+\frac {5 (a-b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \sqrt {a+b \sec ^2(e+f x)-b}}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {5 (a-b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\int \frac {(a-b) \left (-10 b \sec ^2(e+f x)+3 a-5 b\right )}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \sqrt {a+b \sec ^2(e+f x)-b}}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {5 (a-b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {(a-b) \int \frac {-10 b \sec ^2(e+f x)+3 a-5 b}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \sqrt {a+b \sec ^2(e+f x)-b}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {5 (a-b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {(a-b) \left (\frac {b (13 a-15 b) \sec (e+f x)}{a (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\int -\frac {3 (a-5 b) (a-b)}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{a (a-b)}\right )}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \sqrt {a+b \sec ^2(e+f x)-b}}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {5 (a-b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {(a-b) \left (\frac {3 (a-5 b) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{a}+\frac {b (13 a-15 b) \sec (e+f x)}{a (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \sqrt {a+b \sec ^2(e+f x)-b}}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {5 (a-b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {(a-b) \left (\frac {3 (a-5 b) \int \frac {1}{1-\frac {a \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}}{a}+\frac {b (13 a-15 b) \sec (e+f x)}{a (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \sqrt {a+b \sec ^2(e+f x)-b}}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {5 (a-b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {(a-b) \left (\frac {3 (a-5 b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{a^{3/2}}+\frac {b (13 a-15 b) \sec (e+f x)}{a (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \sqrt {a+b \sec ^2(e+f x)-b}}}{f}\) |
(-1/4*Sec[e + f*x]/(a*(1 - Sec[e + f*x]^2)^2*Sqrt[a - b + b*Sec[e + f*x]^2 ]) + ((5*(a - b)*Sec[e + f*x])/(2*a*(1 - Sec[e + f*x]^2)*Sqrt[a - b + b*Se c[e + f*x]^2]) - ((a - b)*((3*(a - 5*b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqr t[a - b + b*Sec[e + f*x]^2]])/a^(3/2) + ((13*a - 15*b)*b*Sec[e + f*x])/(a* (a - b)*Sqrt[a - b + b*Sec[e + f*x]^2])))/(2*a))/(4*a))/f
3.2.33.3.1 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]))* 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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[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[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]
Leaf count of result is larger than twice the leaf count of optimal. \(1845\) vs. \(2(167)=334\).
Time = 0.88 (sec) , antiderivative size = 1846, normalized size of antiderivative = 9.87
1/64/f/a^(9/2)*(a^(7/2)*(-cos(f*x+e)+1)^10*csc(f*x+e)^10+7*a^(7/2)*(-cos(f *x+e)+1)^8*csc(f*x+e)^8-10*a^(5/2)*b*(-cos(f*x+e)+1)^8*csc(f*x+e)^8-8*a^(7 /2)*(-cos(f*x+e)+1)^6*csc(f*x+e)^6+114*a^(5/2)*b*(-cos(f*x+e)+1)^6*csc(f*x +e)^6-120*a^(3/2)*b^2*(-cos(f*x+e)+1)^6*csc(f*x+e)^6+12*ln((a*(-cos(f*x+e) +1)^2*csc(f*x+e)^2+(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2 *csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)^(1/2)*a^(1/2)-a+2*b)/a ^(1/2))*(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e) ^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)^(1/2)*a^3*(-cos(f*x+e)+1)^4*csc(f *x+e)^4-72*ln((a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+(a*(-cos(f*x+e)+1)^4*csc(f *x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e )^2+a)^(1/2)*a^(1/2)-a+2*b)/a^(1/2))*(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a *(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)^(1/2 )*a^2*b*(-cos(f*x+e)+1)^4*csc(f*x+e)^4+60*ln((a*(-cos(f*x+e)+1)^2*csc(f*x+ e)^2+(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+ 4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)^(1/2)*a^(1/2)-a+2*b)/a^(1/2))*(a*(-c os(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f *x+e)+1)^2*csc(f*x+e)^2+a)^(1/2)*a*b^2*(-cos(f*x+e)+1)^4*csc(f*x+e)^4+12*l n(2/(-cos(f*x+e)+1)^2*(-a*(-cos(f*x+e)+1)^2+2*b*(-cos(f*x+e)+1)^2+(a*(-cos (f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x +e)+1)^2*csc(f*x+e)^2+a)^(1/2)*a^(1/2)*sin(f*x+e)^2+a*sin(f*x+e)^2))*(a...
Leaf count of result is larger than twice the leaf count of optimal. 339 vs. \(2 (167) = 334\).
Time = 0.42 (sec) , antiderivative size = 705, normalized size of antiderivative = 3.77 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\left [\frac {3 \, {\left ({\left (a^{3} - 7 \, a^{2} b + 11 \, a b^{2} - 5 \, b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (2 \, a^{3} - 15 \, a^{2} b + 28 \, a b^{2} - 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + a^{2} b - 6 \, a b^{2} + 5 \, b^{3} + {\left (a^{3} - 9 \, a^{2} b + 23 \, a b^{2} - 15 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \log \left (-\frac {2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + 2 \, {\left (3 \, {\left (a^{3} - 6 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (5 \, a^{3} - 31 \, a^{2} b + 30 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (13 \, a^{2} b - 15 \, a b^{2}\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}}}}{16 \, {\left ({\left (a^{5} - a^{4} b\right )} f \cos \left (f x + e\right )^{6} + a^{4} b f - {\left (2 \, a^{5} - 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{5} - 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{2}\right )}}, \frac {3 \, {\left ({\left (a^{3} - 7 \, a^{2} b + 11 \, a b^{2} - 5 \, b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (2 \, a^{3} - 15 \, a^{2} b + 28 \, a b^{2} - 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + a^{2} b - 6 \, a b^{2} + 5 \, b^{3} + {\left (a^{3} - 9 \, a^{2} b + 23 \, a b^{2} - 15 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a}\right ) + {\left (3 \, {\left (a^{3} - 6 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (5 \, a^{3} - 31 \, a^{2} b + 30 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (13 \, a^{2} b - 15 \, a b^{2}\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}}}}{8 \, {\left ({\left (a^{5} - a^{4} b\right )} f \cos \left (f x + e\right )^{6} + a^{4} b f - {\left (2 \, a^{5} - 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{5} - 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{2}\right )}}\right ] \]
[1/16*(3*((a^3 - 7*a^2*b + 11*a*b^2 - 5*b^3)*cos(f*x + e)^6 - (2*a^3 - 15* a^2*b + 28*a*b^2 - 15*b^3)*cos(f*x + e)^4 + a^2*b - 6*a*b^2 + 5*b^3 + (a^3 - 9*a^2*b + 23*a*b^2 - 15*b^3)*cos(f*x + e)^2)*sqrt(a)*log(-2*((a - b)*co s(f*x + e)^2 - 2*sqrt(a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2) *cos(f*x + e) + a + b)/(cos(f*x + e)^2 - 1)) + 2*(3*(a^3 - 6*a^2*b + 5*a*b ^2)*cos(f*x + e)^5 - (5*a^3 - 31*a^2*b + 30*a*b^2)*cos(f*x + e)^3 - (13*a^ 2*b - 15*a*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^5 - a^4*b)*f*cos(f*x + e)^6 + a^4*b*f - (2*a^5 - 3*a^4*b)*f*cos (f*x + e)^4 + (a^5 - 3*a^4*b)*f*cos(f*x + e)^2), 1/8*(3*((a^3 - 7*a^2*b + 11*a*b^2 - 5*b^3)*cos(f*x + e)^6 - (2*a^3 - 15*a^2*b + 28*a*b^2 - 15*b^3)* cos(f*x + e)^4 + a^2*b - 6*a*b^2 + 5*b^3 + (a^3 - 9*a^2*b + 23*a*b^2 - 15* b^3)*cos(f*x + e)^2)*sqrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/a) + (3*(a^3 - 6*a^2*b + 5*a*b^2)*cos(f *x + e)^5 - (5*a^3 - 31*a^2*b + 30*a*b^2)*cos(f*x + e)^3 - (13*a^2*b - 15* a*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(( a^5 - a^4*b)*f*cos(f*x + e)^6 + a^4*b*f - (2*a^5 - 3*a^4*b)*f*cos(f*x + e) ^4 + (a^5 - 3*a^4*b)*f*cos(f*x + e)^2)]
\[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\csc ^{5}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Timed out. \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\csc \left (f x + e\right )^{5}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Hanged} \]