Integrand size = 25, antiderivative size = 218 \[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {3 \sqrt {b} (a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f}-\frac {3 \left (a^2+8 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{8 \sqrt {a+b} f}+\frac {3 (a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {3 (a+2 b) \csc ^2(e+f x) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {\cot (e+f x) \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 f} \]
-1/4*cot(f*x+e)*csc(f*x+e)^3*(a+b*sec(f*x+e)^2)^(3/2)/f+3/2*(a+2*b)*arctan h(sec(f*x+e)*b^(1/2)/(a+b*sec(f*x+e)^2)^(1/2))*b^(1/2)/f-3/8*(a^2+8*a*b+8* b^2)*arctanh(sec(f*x+e)*(a+b)^(1/2)/(a+b*sec(f*x+e)^2)^(1/2))/f/(a+b)^(1/2 )+3/8*(a+4*b)*sec(f*x+e)*(a+b*sec(f*x+e)^2)^(1/2)/f-3/8*(a+2*b)*csc(f*x+e) ^2*sec(f*x+e)*(a+b*sec(f*x+e)^2)^(1/2)/f
Time = 2.60 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.20 \[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {\left (b+a \cos ^2(e+f x)\right ) \left (-12 b^{3/2} \left (a^2+3 a b+2 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {b}}\right ) \cos ^2(e+f x)+3 b \sqrt {a+b} \left (a^2+8 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {a+b}}\right ) \cos ^2(e+f x)+\frac {b (a+b) \sqrt {a+2 b+a \cos (2 (e+f x))} (11 a+4 b+8 (a+3 b) \cos (2 (e+f x))-3 (a+4 b) \cos (4 (e+f x))) \csc ^4(e+f x)}{8 \sqrt {2}}\right ) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 \sqrt {2} b (a+b) f (a+2 b+a \cos (2 (e+f x)))^{3/2}} \]
-1/2*((b + a*Cos[e + f*x]^2)*(-12*b^(3/2)*(a^2 + 3*a*b + 2*b^2)*ArcTanh[Sq rt[a + b - a*Sin[e + f*x]^2]/Sqrt[b]]*Cos[e + f*x]^2 + 3*b*Sqrt[a + b]*(a^ 2 + 8*a*b + 8*b^2)*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[a + b]]*Cos [e + f*x]^2 + (b*(a + b)*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(11*a + 4*b + 8*(a + 3*b)*Cos[2*(e + f*x)] - 3*(a + 4*b)*Cos[4*(e + f*x)])*Csc[e + f*x]^ 4)/(8*Sqrt[2]))*Sec[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(Sqrt[2]*b*(a + b )*f*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2))
Time = 0.47 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 4622, 25, 369, 27, 439, 25, 444, 27, 398, 224, 219, 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.
| \(\displaystyle \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^{3/2}}{\sin (e+f x)^5}dx\) |
| \(\Big \downarrow \) 4622 |
| \(\displaystyle \frac {\int -\frac {\sec ^4(e+f x) \left (b \sec ^2(e+f x)+a\right )^{3/2}}{\left (1-\sec ^2(e+f x)\right )^3}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\sec ^4(e+f x) \left (b \sec ^2(e+f x)+a\right )^{3/2}}{\left (1-\sec ^2(e+f x)\right )^3}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 369 |
| \(\displaystyle \frac {\frac {1}{4} \int \frac {3 \sec ^2(e+f x) \sqrt {b \sec ^2(e+f x)+a} \left (2 b \sec ^2(e+f x)+a\right )}{\left (1-\sec ^2(e+f x)\right )^2}d\sec (e+f x)-\frac {\sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3}{4} \int \frac {\sec ^2(e+f x) \sqrt {b \sec ^2(e+f x)+a} \left (2 b \sec ^2(e+f x)+a\right )}{\left (1-\sec ^2(e+f x)\right )^2}d\sec (e+f x)-\frac {\sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 439 |
| \(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \int -\frac {\sec ^2(e+f x) \left (2 b (a+4 b) \sec ^2(e+f x)+a (a+6 b)\right )}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)+\frac {(a+2 b) \sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 \left (1-\sec ^2(e+f x)\right )}\right )-\frac {\sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {3}{4} \left (\frac {(a+2 b) \sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 \left (1-\sec ^2(e+f x)\right )}-\frac {1}{2} \int \frac {\sec ^2(e+f x) \left (2 b (a+4 b) \sec ^2(e+f x)+a (a+6 b)\right )}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)\right )-\frac {\sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 444 |
| \(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \left ((a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}-\frac {\int \frac {2 b \left (4 b (a+2 b) \sec ^2(e+f x)+a (a+4 b)\right )}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{2 b}\right )+\frac {(a+2 b) \sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 \left (1-\sec ^2(e+f x)\right )}\right )-\frac {\sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \left ((a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}-\int \frac {4 b (a+2 b) \sec ^2(e+f x)+a (a+4 b)}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)\right )+\frac {(a+2 b) \sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 \left (1-\sec ^2(e+f x)\right )}\right )-\frac {\sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \left (-\left (a^2+8 a b+8 b^2\right ) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)+4 b (a+2 b) \int \frac {1}{\sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)+(a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}\right )+\frac {(a+2 b) \sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 \left (1-\sec ^2(e+f x)\right )}\right )-\frac {\sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \left (-\left (a^2+8 a b+8 b^2\right ) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)+4 b (a+2 b) \int \frac {1}{1-\frac {b \sec ^2(e+f x)}{b \sec ^2(e+f x)+a}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}+(a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}\right )+\frac {(a+2 b) \sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 \left (1-\sec ^2(e+f x)\right )}\right )-\frac {\sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \left (-\left (a^2+8 a b+8 b^2\right ) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)+4 \sqrt {b} (a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )+(a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}\right )+\frac {(a+2 b) \sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 \left (1-\sec ^2(e+f x)\right )}\right )-\frac {\sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \left (-\left (a^2+8 a b+8 b^2\right ) \int \frac {1}{1-\frac {(a+b) \sec ^2(e+f x)}{b \sec ^2(e+f x)+a}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}+4 \sqrt {b} (a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )+(a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}\right )+\frac {(a+2 b) \sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 \left (1-\sec ^2(e+f x)\right )}\right )-\frac {\sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \left (-\frac {\left (a^2+8 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{\sqrt {a+b}}+4 \sqrt {b} (a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )+(a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}\right )+\frac {(a+2 b) \sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 \left (1-\sec ^2(e+f x)\right )}\right )-\frac {\sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\) |
(-1/4*(Sec[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2))/(1 - Sec[e + f*x]^2)^2 + (3*(((a + 2*b)*Sec[e + f*x]^3*Sqrt[a + b*Sec[e + f*x]^2])/(2*(1 - Sec[e + f*x]^2)) + (4*Sqrt[b]*(a + 2*b)*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]] - ((a^2 + 8*a*b + 8*b^2)*ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/Sqrt[a + b] + (a + 4*b)*Sec[e + f*x]*Sq rt[a + b*Sec[e + f*x]^2])/2))/4)/f
3.1.85.3.1 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], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 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/(2* b*(p + 1))), x] - Simp[e^2/(2*b*(p + 1)) Int[(e*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(m - 1) + d*(m + 2*q - 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 0 ] && GtQ[m, 1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) , x_Symbol] :> Simp[f/b Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ b Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} , x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*b*g*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(2*b*e*( p + 1) + (b*e - a*f)*(m + 1)) + d*(2*b*e*(p + 1) + (b*e - a*f)*(m + 2*q + 1 ))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && G tQ[q, 0] && !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[f*g*(g*x)^(m - 1)*(a + b*x^2)^ (p + 1)*((c + d*x^2)^(q + 1)/(b*d*(m + 2*(p + q + 1) + 1))), x] - Simp[g^2/ (b*d*(m + 2*(p + q + 1) + 1)) Int[(g*x)^(m - 2)*(a + b*x^2)^p*(c + d*x^2) ^q*Simp[a*f*c*(m - 1) + (a*f*d*(m + 2*q + 1) + b*(f*c*(m + 2*p + 1) - e*d*( m + 2*(p + q + 1) + 1)))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && GtQ[m, 1]
Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + ( f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Si mp[1/(f*ff^m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p /x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4])
Leaf count of result is larger than twice the leaf count of optimal. \(5130\) vs. \(2(190)=380\).
Time = 6.61 (sec) , antiderivative size = 5131, normalized size of antiderivative = 23.54
Time = 0.47 (sec) , antiderivative size = 1511, normalized size of antiderivative = 6.93 \[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]
[1/16*(3*((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^5 - 2*(a^2 + 8*a*b + 8*b^2)*c os(f*x + e)^3 + (a^2 + 8*a*b + 8*b^2)*cos(f*x + e))*sqrt(a + b)*log(2*(a*c os(f*x + e)^2 - 2*sqrt(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)* cos(f*x + e) + a + 2*b)/(cos(f*x + e)^2 - 1)) + 12*((a^2 + 3*a*b + 2*b^2)* cos(f*x + e)^5 - 2*(a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 + (a^2 + 3*a*b + 2 *b^2)*cos(f*x + e))*sqrt(b)*log((a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*cos( f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) + 2*(3 *(a^2 + 5*a*b + 4*b^2)*cos(f*x + e)^4 - (5*a^2 + 23*a*b + 18*b^2)*cos(f*x + e)^2 + 4*a*b + 4*b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a + b)*f*cos(f*x + e)^5 - 2*(a + b)*f*cos(f*x + e)^3 + (a + b)*f*cos(f*x + e) ), 1/8*(3*((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^5 - 2*(a^2 + 8*a*b + 8*b^2)* cos(f*x + e)^3 + (a^2 + 8*a*b + 8*b^2)*cos(f*x + e))*sqrt(-a - b)*arctan(s qrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a + b)) + 6*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^5 - 2*(a^2 + 3*a*b + 2*b^2)*co s(f*x + e)^3 + (a^2 + 3*a*b + 2*b^2)*cos(f*x + e))*sqrt(b)*log((a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e ) + 2*b)/cos(f*x + e)^2) + (3*(a^2 + 5*a*b + 4*b^2)*cos(f*x + e)^4 - (5*a^ 2 + 23*a*b + 18*b^2)*cos(f*x + e)^2 + 4*a*b + 4*b^2)*sqrt((a*cos(f*x + e)^ 2 + b)/cos(f*x + e)^2))/((a + b)*f*cos(f*x + e)^5 - 2*(a + b)*f*cos(f*x + e)^3 + (a + b)*f*cos(f*x + e)), -1/16*(24*((a^2 + 3*a*b + 2*b^2)*cos(f*...
Timed out. \[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
\[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \csc \left (f x + e\right )^{5} \,d x } \]
Exception generated. \[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{%%%{262144,[6,6]%%%},[12]%%%}+%%%{%%{[%%%{1572864,[6,6]%%% },0]:[1,0
Timed out. \[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}}{{\sin \left (e+f\,x\right )}^5} \,d x \]