\(\int \frac {\csc (e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [122]

Optimal result

Integrand size = 23, antiderivative size = 127 \[ \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{(a+b)^{5/2} f}-\frac {b \sec (e+f x)}{3 a (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {b (5 a+2 b) \sec (e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b \sec ^2(e+f x)}} \] Output:

-arctanh((a+b)^(1/2)*sec(f*x+e)/(a+b*sec(f*x+e)^2)^(1/2))/(a+b)^(5/2)/f-1/ 
3*b*sec(f*x+e)/a/(a+b)/f/(a+b*sec(f*x+e)^2)^(3/2)-1/3*b*(5*a+2*b)*sec(f*x+ 
e)/a^2/(a+b)^2/f/(a+b*sec(f*x+e)^2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 3.77 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.85 \[ \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^5(e+f x) \left (a^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},1-\frac {a \sin ^2(e+f x)}{a+b}\right )+(a+b) \left (-2 (2 a+b)+3 a \sin ^2(e+f x)\right )\right )}{6 a^2 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \] Input:

Integrate[Csc[e + f*x]/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

Output:

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^5*(a^2*Hypergeometric2F1[-3/2 
, 1, -1/2, 1 - (a*Sin[e + f*x]^2)/(a + b)] + (a + b)*(-2*(2*a + b) + 3*a*S 
in[e + f*x]^2)))/(6*a^2*(a + b)*f*(a + b*Sec[e + f*x]^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4622, 25, 316, 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.

\(\displaystyle \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (e+f x) \left (a+b \sec (e+f x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 4622

\(\displaystyle \frac {\int -\frac {1}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{5/2}}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{5/2}}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 316

\(\displaystyle \frac {\frac {\int \frac {2 b \sec ^2(e+f x)+b-3 (a+b)}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec (e+f x)}{3 a (a+b)}-\frac {b \sec (e+f x)}{3 a (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {-\frac {\int \frac {3 a^2}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{a (a+b)}-\frac {b (5 a+2 b) \sec (e+f x)}{a (a+b) \sqrt {a+b \sec ^2(e+f x)}}}{3 a (a+b)}-\frac {b \sec (e+f x)}{3 a (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {3 a \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{a+b}-\frac {b (5 a+2 b) \sec (e+f x)}{a (a+b) \sqrt {a+b \sec ^2(e+f x)}}}{3 a (a+b)}-\frac {b \sec (e+f x)}{3 a (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {-\frac {3 a \int \frac {1}{1-\frac {(a+b) \sec ^2(e+f x)}{b \sec ^2(e+f x)+a}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}}{a+b}-\frac {b (5 a+2 b) \sec (e+f x)}{a (a+b) \sqrt {a+b \sec ^2(e+f x)}}}{3 a (a+b)}-\frac {b \sec (e+f x)}{3 a (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {-\frac {3 a \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{(a+b)^{3/2}}-\frac {b (5 a+2 b) \sec (e+f x)}{a (a+b) \sqrt {a+b \sec ^2(e+f x)}}}{3 a (a+b)}-\frac {b \sec (e+f x)}{3 a (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\)

Input:

Int[Csc[e + f*x]/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

Output:

(-1/3*(b*Sec[e + f*x])/(a*(a + b)*(a + b*Sec[e + f*x]^2)^(3/2)) + ((-3*a*A 
rcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(a + b)^(3/ 
2) - (b*(5*a + 2*b)*Sec[e + f*x])/(a*(a + b)*Sqrt[a + b*Sec[e + f*x]^2]))/ 
(3*a*(a + b)))/f
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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])
 

rule 291

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]
 

rule 316

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[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

rule 402

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4622

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])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1319\) vs. \(2(113)=226\).

Time = 222.00 (sec) , antiderivative size = 1320, normalized size of antiderivative = 10.39

method	result	size
default	\(\text {Expression too large to display}\)	\(1320\)

Input:

int(csc(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/6/f/a^2/(2*(-a*b)^(1/2)-a+b)^2/(2*(-a*b)^(1/2)+a-b)^2/(a+b)^(9/2)*(a^4+ 
4*a^3*b+6*a^2*b^2+4*a*b^3+b^4)*(b+a*cos(f*x+e)^2)*(3*cos(f*x+e)^2*(1+cos(f 
*x+e))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*((a+b)^(1/2)*((b+ 
a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f 
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+cos(f*x+e)*a+b)/(-1+cos(f*x+e)))*a^5+3*(2 
*cos(f*x+e)^3+2*cos(f*x+e)^2+cos(f*x+e)+1)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ 
e))^2)^(1/2)*ln(-4*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2 
)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+cos(f 
*x+e)*a+b)/(-1+cos(f*x+e)))*a^4*b+3*(cos(f*x+e)^3+cos(f*x+e)^2+2*cos(f*x+e 
)+2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*((a+b)^(1/2)*((b+a* 
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f*x 
+e)^2)/(1+cos(f*x+e))^2)^(1/2)+cos(f*x+e)*a+b)/(-1+cos(f*x+e)))*a^3*b^2+3* 
(1+cos(f*x+e))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*((a+b)^(1 
/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+(a+b)^(1/2)*((b 
+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+cos(f*x+e)*a+b)/(-1+cos(f*x+e)))* 
a^2*b^3+3*cos(f*x+e)^2*(1+cos(f*x+e))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2 
)^(1/2)*ln(2/(a+b)^(1/2)*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2 
)^(1/2)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2) 
-cos(f*x+e)*a+b)/(1+cos(f*x+e)))*a^5+3*(2*cos(f*x+e)^3+2*cos(f*x+e)^2+cos( 
f*x+e)+1)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(2/(a+b)^(1/2)*...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (113) = 226\).

Time = 0.21 (sec) , antiderivative size = 601, normalized size of antiderivative = 4.73 \[ \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (a^{4} \cos \left (f x + e\right )^{4} + 2 \, a^{3} b \cos \left (f x + e\right )^{2} + a^{2} b^{2}\right )} \sqrt {a + b} \log \left (\frac {2 \, {\left (a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a + b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + 2 \, b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left (3 \, {\left (2 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (5 \, a^{2} b^{2} + 7 \, a b^{3} + 2 \, b^{4}\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}}}}{6 \, {\left ({\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b + 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}, \frac {3 \, {\left (a^{4} \cos \left (f x + e\right )^{4} + 2 \, a^{3} b \cos \left (f x + e\right )^{2} + a^{2} b^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-a - b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a \cos \left (f x + e\right )^{2} + b}\right ) - {\left (3 \, {\left (2 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (5 \, a^{2} b^{2} + 7 \, a b^{3} + 2 \, b^{4}\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}}}}{3 \, {\left ({\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b + 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}\right ] \] Input:

integrate(csc(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")
 

Output:

[1/6*(3*(a^4*cos(f*x + e)^4 + 2*a^3*b*cos(f*x + e)^2 + a^2*b^2)*sqrt(a + b 
)*log(2*(a*cos(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)) - 2*(3*(2*a^3*b 
+ 3*a^2*b^2 + a*b^3)*cos(f*x + e)^3 + (5*a^2*b^2 + 7*a*b^3 + 2*b^4)*cos(f* 
x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^7 + 3*a^6*b + 3*a 
^5*b^2 + a^4*b^3)*f*cos(f*x + e)^4 + 2*(a^6*b + 3*a^5*b^2 + 3*a^4*b^3 + a^ 
3*b^4)*f*cos(f*x + e)^2 + (a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5)*f), 
1/3*(3*(a^4*cos(f*x + e)^4 + 2*a^3*b*cos(f*x + e)^2 + a^2*b^2)*sqrt(-a - b 
)*arctan(sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x 
+ e)/(a*cos(f*x + e)^2 + b)) - (3*(2*a^3*b + 3*a^2*b^2 + a*b^3)*cos(f*x + 
e)^3 + (5*a^2*b^2 + 7*a*b^3 + 2*b^4)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 
+ b)/cos(f*x + e)^2))/((a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*f*cos(f*x + e 
)^4 + 2*(a^6*b + 3*a^5*b^2 + 3*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^2 + (a^5* 
b^2 + 3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5)*f)]
 

Sympy [F]

\[ \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\csc {\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(csc(f*x+e)/(a+b*sec(f*x+e)**2)**(5/2),x)
 

Output:

Integral(csc(e + f*x)/(a + b*sec(e + f*x)**2)**(5/2), x)
 

Maxima [F]

\[ \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\csc \left (f x + e\right )}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(csc(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")
 

Output:

integrate(csc(f*x + e)/(b*sec(f*x + e)^2 + a)^(5/2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1097 vs. \(2 (113) = 226\).

Time = 0.98 (sec) , antiderivative size = 1097, normalized size of antiderivative = 8.64 \[ \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(csc(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")
 

Output:

-1/6*(4*((((3*a^6*b^3 + 16*a^5*b^4 + 35*a^4*b^5 + 40*a^3*b^6 + 25*a^2*b^7 
+ 8*a*b^8 + b^9)*tan(1/2*f*x + 1/2*e)^2/(a^8*b^2*sgn(cos(f*x + e)) + 6*a^7 
*b^3*sgn(cos(f*x + e)) + 15*a^6*b^4*sgn(cos(f*x + e)) + 20*a^5*b^5*sgn(cos 
(f*x + e)) + 15*a^4*b^6*sgn(cos(f*x + e)) + 6*a^3*b^7*sgn(cos(f*x + e)) + 
a^2*b^8*sgn(cos(f*x + e))) - 3*(a^6*b^3 + 2*a^5*b^4 - 3*a^4*b^5 - 12*a^3*b 
^6 - 13*a^2*b^7 - 6*a*b^8 - b^9)/(a^8*b^2*sgn(cos(f*x + e)) + 6*a^7*b^3*sg 
n(cos(f*x + e)) + 15*a^6*b^4*sgn(cos(f*x + e)) + 20*a^5*b^5*sgn(cos(f*x + 
e)) + 15*a^4*b^6*sgn(cos(f*x + e)) + 6*a^3*b^7*sgn(cos(f*x + e)) + a^2*b^8 
*sgn(cos(f*x + e))))*tan(1/2*f*x + 1/2*e)^2 - 3*(a^6*b^3 + 2*a^5*b^4 - 3*a 
^4*b^5 - 12*a^3*b^6 - 13*a^2*b^7 - 6*a*b^8 - b^9)/(a^8*b^2*sgn(cos(f*x + e 
)) + 6*a^7*b^3*sgn(cos(f*x + e)) + 15*a^6*b^4*sgn(cos(f*x + e)) + 20*a^5*b 
^5*sgn(cos(f*x + e)) + 15*a^4*b^6*sgn(cos(f*x + e)) + 6*a^3*b^7*sgn(cos(f* 
x + e)) + a^2*b^8*sgn(cos(f*x + e))))*tan(1/2*f*x + 1/2*e)^2 + (3*a^6*b^3 
+ 16*a^5*b^4 + 35*a^4*b^5 + 40*a^3*b^6 + 25*a^2*b^7 + 8*a*b^8 + b^9)/(a^8* 
b^2*sgn(cos(f*x + e)) + 6*a^7*b^3*sgn(cos(f*x + e)) + 15*a^6*b^4*sgn(cos(f 
*x + e)) + 20*a^5*b^5*sgn(cos(f*x + e)) + 15*a^4*b^6*sgn(cos(f*x + e)) + 6 
*a^3*b^7*sgn(cos(f*x + e)) + a^2*b^8*sgn(cos(f*x + e))))/(a*tan(1/2*f*x + 
1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan 
(1/2*f*x + 1/2*e)^2 + a + b)^(3/2) - 3*log(abs(-sqrt(a + b)*tan(1/2*f*x + 
1/2*e)^2 + sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{\sin \left (e+f\,x\right )\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \] Input:

int(1/(sin(e + f*x)*(a + b/cos(e + f*x)^2)^(5/2)),x)
 

Output:

int(1/(sin(e + f*x)*(a + b/cos(e + f*x)^2)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \csc \left (f x +e \right )}{\sec \left (f x +e \right )^{6} b^{3}+3 \sec \left (f x +e \right )^{4} a \,b^{2}+3 \sec \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:

int(csc(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x)
 

Output:

int((sqrt(sec(e + f*x)**2*b + a)*csc(e + f*x))/(sec(e + f*x)**6*b**3 + 3*s 
ec(e + f*x)**4*a*b**2 + 3*sec(e + f*x)**2*a**2*b + a**3),x)