\(\int \frac {\csc ^3(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [123]

Optimal result

Integrand size = 25, antiderivative size = 171 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a-4 b) \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 (a+b)^{7/2} f}-\frac {\cot (e+f x) \csc (e+f x)}{2 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {5 b \sec (e+f x)}{6 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(13 a-2 b) b \sec (e+f x)}{6 a (a+b)^3 f \sqrt {a+b \sec ^2(e+f x)}} \] Output:

-1/2*(a-4*b)*arctanh((a+b)^(1/2)*sec(f*x+e)/(a+b*sec(f*x+e)^2)^(1/2))/(a+b 
)^(7/2)/f-1/2*cot(f*x+e)*csc(f*x+e)/(a+b)/f/(a+b*sec(f*x+e)^2)^(3/2)-5/6*b 
*sec(f*x+e)/(a+b)^2/f/(a+b*sec(f*x+e)^2)^(3/2)-1/6*(13*a-2*b)*b*sec(f*x+e) 
/a/(a+b)^3/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 = 1.03 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.88 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left ((a+b) \left (3 a^2+6 a b-2 b^2+\left (3 a^2+2 b^2\right ) \cos (2 (e+f x))\right ) \csc ^2(e+f x)-3 a (a-4 b) (a+2 b+a \cos (2 (e+f x))) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1-\frac {a \sin ^2(e+f x)}{a+b}\right )\right ) \sec ^5(e+f x)}{24 a (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \] Input:

Integrate[Csc[e + f*x]^3/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

Output:

-1/24*((a + 2*b + a*Cos[2*(e + f*x)])*((a + b)*(3*a^2 + 6*a*b - 2*b^2 + (3 
*a^2 + 2*b^2)*Cos[2*(e + f*x)])*Csc[e + f*x]^2 - 3*a*(a - 4*b)*(a + 2*b + 
a*Cos[2*(e + f*x)])*Hypergeometric2F1[-1/2, 1, 1/2, 1 - (a*Sin[e + f*x]^2) 
/(a + b)])*Sec[e + f*x]^5)/(a*(a + b)^3*f*(a + b*Sec[e + f*x]^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4622, 373, 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.

\(\displaystyle \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4622

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

\(\Big \downarrow \) 373

\(\displaystyle \frac {\frac {\sec (e+f x)}{2 (a+b) \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\int \frac {a-4 b \sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{5/2}}d\sec (e+f x)}{2 (a+b)}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\sec (e+f x)}{2 (a+b) \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\frac {5 b \sec (e+f x)}{3 (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\int -\frac {a \left (-10 b \sec ^2(e+f x)+3 a-2 b\right )}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec (e+f x)}{3 a (a+b)}}{2 (a+b)}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\sec (e+f x)}{2 (a+b) \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\frac {\int \frac {a \left (-10 b \sec ^2(e+f x)+3 a-2 b\right )}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec (e+f x)}{3 a (a+b)}+\frac {5 b \sec (e+f x)}{3 (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{2 (a+b)}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\sec (e+f x)}{2 (a+b) \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\frac {\int \frac {-10 b \sec ^2(e+f x)+3 a-2 b}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec (e+f x)}{3 (a+b)}+\frac {5 b \sec (e+f x)}{3 (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{2 (a+b)}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\sec (e+f x)}{2 (a+b) \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\frac {\frac {b (13 a-2 b) \sec (e+f x)}{a (a+b) \sqrt {a+b \sec ^2(e+f x)}}-\frac {\int -\frac {3 a (a-4 b)}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{a (a+b)}}{3 (a+b)}+\frac {5 b \sec (e+f x)}{3 (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{2 (a+b)}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\sec (e+f x)}{2 (a+b) \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\frac {\frac {3 (a-4 b) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{a+b}+\frac {b (13 a-2 b) \sec (e+f x)}{a (a+b) \sqrt {a+b \sec ^2(e+f x)}}}{3 (a+b)}+\frac {5 b \sec (e+f x)}{3 (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{2 (a+b)}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\sec (e+f x)}{2 (a+b) \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\frac {\frac {3 (a-4 b) \int \frac {1}{1-\frac {(a+b) \sec ^2(e+f x)}{b \sec ^2(e+f x)+a}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}}{a+b}+\frac {b (13 a-2 b) \sec (e+f x)}{a (a+b) \sqrt {a+b \sec ^2(e+f x)}}}{3 (a+b)}+\frac {5 b \sec (e+f x)}{3 (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{2 (a+b)}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\sec (e+f x)}{2 (a+b) \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\frac {\frac {3 (a-4 b) \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{(a+b)^{3/2}}+\frac {b (13 a-2 b) \sec (e+f x)}{a (a+b) \sqrt {a+b \sec ^2(e+f x)}}}{3 (a+b)}+\frac {5 b \sec (e+f x)}{3 (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{2 (a+b)}}{f}\)

Input:

Int[Csc[e + f*x]^3/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

Output:

(Sec[e + f*x]/(2*(a + b)*(1 - Sec[e + f*x]^2)*(a + b*Sec[e + f*x]^2)^(3/2) 
) - ((5*b*Sec[e + f*x])/(3*(a + b)*(a + b*Sec[e + f*x]^2)^(3/2)) + ((3*(a 
- 4*b)*ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(a 
+ b)^(3/2) + ((13*a - 2*b)*b*Sec[e + f*x])/(a*(a + b)*Sqrt[a + b*Sec[e + f 
*x]^2]))/(3*(a + b)))/(2*(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 373

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]
 

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] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(2073\) vs. \(2(151)=302\).

Time = 202.14 (sec) , antiderivative size = 2074, normalized size of antiderivative = 12.13

method	result	size
default	\(\text {Expression too large to display}\)	\(2074\)

Input:

int(csc(f*x+e)^3/(a+b*sec(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/12/f/a/(2*(-a*b)^(1/2)+a-b)^3/(2*(-a*b)^(1/2)-a+b)^3/(a+b)^(13/2)*(a^6+ 
6*a^5*b+15*a^4*b^2+20*a^3*b^3+15*a^2*b^4+6*a*b^5+b^6)*(b+a*cos(f*x+e)^2)*( 
-6*a^3*(a+b)^(7/2)*cos(f*x+e)^4+12*cos(f*x+e)^2*(2*cos(f*x+e)^2-3)*a^2*(a+ 
b)^(7/2)*b+2*(-13+10*cos(f*x+e)^2)*b^2*(a+b)^(7/2)*a+4*sin(f*x+e)^2*b^3*(a 
+b)^(7/2)+3*cos(f*x+e)^2*(-1-cos(f*x+e))*sin(f*x+e)^2*((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^6+3*(-1-cos(f*x+e))*sin 
(f*x+e)^4*((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*b+3*(9*cos(f*x+e)^3+9*cos(f*x+e)^2+cos(f*x+e)+1)*sin(f*x+e)^2*((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^4*b^2+ 
3*(11*cos(f*x+e)^3+11*cos(f*x+e)^2+9*cos(f*x+e)+9)*sin(f*x+e)^2*((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^3*b^3+3*(11+4 
*cos(f*x+e)^3+4*cos(f*x+e)^2+11*cos(f*x+e))*sin(f*x+e)^2*((b+a*cos(f*x+...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 470 vs. \(2 (151) = 302\).

Time = 0.29 (sec) , antiderivative size = 950, normalized size of antiderivative = 5.56 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(csc(f*x+e)^3/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")
 

Output:

[-1/12*(3*((a^4 - 4*a^3*b)*cos(f*x + e)^6 - (a^4 - 6*a^3*b + 8*a^2*b^2)*co 
s(f*x + e)^4 - a^2*b^2 + 4*a*b^3 - (2*a^3*b - 9*a^2*b^2 + 4*a*b^3)*cos(f*x 
 + e)^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*(a^4 - 3*a^3*b - 4*a^2*b^2)*cos(f*x + e)^5 + 2*(9*a^3*b + 4*a^2* 
b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^3 + (13*a^2*b^2 + 11*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 + 4*a^6*b + 6 
*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^6 - (a^7 + 2*a^6*b - 2*a^5* 
b^2 - 8*a^4*b^3 - 7*a^3*b^4 - 2*a^2*b^5)*f*cos(f*x + e)^4 - (2*a^6*b + 7*a 
^5*b^2 + 8*a^4*b^3 + 2*a^3*b^4 - 2*a^2*b^5 - a*b^6)*f*cos(f*x + e)^2 - (a^ 
5*b^2 + 4*a^4*b^3 + 6*a^3*b^4 + 4*a^2*b^5 + a*b^6)*f), 1/6*(3*((a^4 - 4*a^ 
3*b)*cos(f*x + e)^6 - (a^4 - 6*a^3*b + 8*a^2*b^2)*cos(f*x + e)^4 - a^2*b^2 
 + 4*a*b^3 - (2*a^3*b - 9*a^2*b^2 + 4*a*b^3)*cos(f*x + e)^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*(a^4 - 3*a^3*b - 4*a^2*b^2)*cos(f*x + e)^5 
 + 2*(9*a^3*b + 4*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^3 + (13*a^2*b^2 + 
11*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 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^6 - ( 
a^7 + 2*a^6*b - 2*a^5*b^2 - 8*a^4*b^3 - 7*a^3*b^4 - 2*a^2*b^5)*f*cos(f*x + 
 e)^4 - (2*a^6*b + 7*a^5*b^2 + 8*a^4*b^3 + 2*a^3*b^4 - 2*a^2*b^5 - a*b^...
 

Sympy [F]

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

integrate(csc(f*x+e)**3/(a+b*sec(f*x+e)**2)**(5/2),x)
 

Output:

Integral(csc(e + f*x)**3/(a + b*sec(e + f*x)**2)**(5/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(csc(f*x+e)^3/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1980 vs. \(2 (151) = 302\).

Time = 1.44 (sec) , antiderivative size = 1980, normalized size of antiderivative = 11.58 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(csc(f*x+e)^3/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")
 

Output:

-1/24*(((((3*(a^12*b^2*sgn(cos(f*x + e)) + 10*a^11*b^3*sgn(cos(f*x + e)) + 
 45*a^10*b^4*sgn(cos(f*x + e)) + 120*a^9*b^5*sgn(cos(f*x + e)) + 210*a^8*b 
^6*sgn(cos(f*x + e)) + 252*a^7*b^7*sgn(cos(f*x + e)) + 210*a^6*b^8*sgn(cos 
(f*x + e)) + 120*a^5*b^9*sgn(cos(f*x + e)) + 45*a^4*b^10*sgn(cos(f*x + e)) 
 + 10*a^3*b^11*sgn(cos(f*x + e)) + a^2*b^12*sgn(cos(f*x + e)))*tan(1/2*f*x 
 + 1/2*e)^2/(a^13*b^2 + 11*a^12*b^3 + 55*a^11*b^4 + 165*a^10*b^5 + 330*a^9 
*b^6 + 462*a^8*b^7 + 462*a^7*b^8 + 330*a^6*b^9 + 165*a^5*b^10 + 55*a^4*b^1 
1 + 11*a^3*b^12 + a^2*b^13) - 4*(3*a^12*b^2*sgn(cos(f*x + e)) + 12*a^11*b^ 
3*sgn(cos(f*x + e)) - 25*a^10*b^4*sgn(cos(f*x + e)) - 270*a^9*b^5*sgn(cos( 
f*x + e)) - 810*a^8*b^6*sgn(cos(f*x + e)) - 1344*a^7*b^7*sgn(cos(f*x + e)) 
 - 1386*a^6*b^8*sgn(cos(f*x + e)) - 900*a^5*b^9*sgn(cos(f*x + e)) - 345*a^ 
4*b^10*sgn(cos(f*x + e)) - 60*a^3*b^11*sgn(cos(f*x + e)) + 3*a^2*b^12*sgn( 
cos(f*x + e)) + 2*a*b^13*sgn(cos(f*x + e)))/(a^13*b^2 + 11*a^12*b^3 + 55*a 
^11*b^4 + 165*a^10*b^5 + 330*a^9*b^6 + 462*a^8*b^7 + 462*a^7*b^8 + 330*a^6 
*b^9 + 165*a^5*b^10 + 55*a^4*b^11 + 11*a^3*b^12 + a^2*b^13))*tan(1/2*f*x + 
 1/2*e)^2 + 6*(3*a^12*b^2*sgn(cos(f*x + e)) + 14*a^11*b^3*sgn(cos(f*x + e) 
) + 27*a^10*b^4*sgn(cos(f*x + e)) + 68*a^9*b^5*sgn(cos(f*x + e)) + 262*a^8 
*b^6*sgn(cos(f*x + e)) + 644*a^7*b^7*sgn(cos(f*x + e)) + 910*a^6*b^8*sgn(c 
os(f*x + e)) + 752*a^5*b^9*sgn(cos(f*x + e)) + 343*a^4*b^10*sgn(cos(f*x + 
e)) + 62*a^3*b^11*sgn(cos(f*x + e)) - 9*a^2*b^12*sgn(cos(f*x + e)) - 4*...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^3\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \] Input:

int(1/(sin(e + f*x)^3*(a + b/cos(e + f*x)^2)^(5/2)),x)
 

Output:

int(1/(sin(e + f*x)^3*(a + b/cos(e + f*x)^2)^(5/2)), x)
 

Reduce [F]

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

int(csc(f*x+e)^3/(a+b*sec(f*x+e)^2)^(5/2),x)
 

Output:

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