\(\int \frac {\csc ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [98]

Optimal result

Integrand size = 25, antiderivative size = 138 \[ \int \frac {\csc ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{8 (a+b)^{5/2} f}-\frac {(5 a+2 b) \cot (e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 (a+b)^2 f}-\frac {\cot ^3(e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 (a+b) f} \] Output:

-3/8*a^2*arctanh((a+b)^(1/2)*sec(f*x+e)/(a+b*sec(f*x+e)^2)^(1/2))/(a+b)^(5 
/2)/f-1/8*(5*a+2*b)*cot(f*x+e)*csc(f*x+e)*(a+b*sec(f*x+e)^2)^(1/2)/(a+b)^2 
/f-1/4*cot(f*x+e)^3*csc(f*x+e)*(a+b*sec(f*x+e)^2)^(1/2)/(a+b)/f
 

Mathematica [C] (verified)

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

Time = 0.12 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.57 \[ \int \frac {\csc ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {a^2 (a+2 b+a \cos (2 (e+f x))) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1-\frac {a \sin ^2(e+f x)}{a+b}\right ) \sec (e+f x)}{2 (a+b)^3 f \sqrt {a+b \sec ^2(e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4622, 25, 372, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4622

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(-1/4*(Sec[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/((a + b)*(1 - Sec[e + f*x] 
^2)^2) + ((-3*a^2*ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f* 
x]^2]])/(2*(a + b)^(3/2)) + ((5*a + 2*b)*Sec[e + f*x]*Sqrt[a + b*Sec[e + f 
*x]^2])/(2*(a + b)*(1 - Sec[e + f*x]^2)))/(4*(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 372

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]
 

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. \(974\) vs. \(2(122)=244\).

Time = 1.44 (sec) , antiderivative size = 975, normalized size of antiderivative = 7.07

method	result	size
default	\(\text {Expression too large to display}\)	\(975\)

Input:

int(csc(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/16/f/(a+b)^(9/2)*(cos(f*x+e)^2*(-6*cos(f*x+e)^2+10)*(a+b)^(5/2)*a^2+(-2 
*cos(f*x+e)^2+10)*(a+b)^(5/2)*a*b+4*(a+b)^(5/2)*b^2+(3*cos(f*x+e)+3)*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^4+(6*cos(f*x+e)+6)*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^3*b+(3*cos(f*x+e)+3)*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^2*b^2+(3* 
cos(f*x+e)+3)*sin(f*x+e)^4*((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+c 
os(f*x+e)))*a^4+(6*cos(f*x+e)+6)*sin(f*x+e)^4*((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)+co 
s(f*x+e)*a+b)/(-1+cos(f*x+e)))*a^3*b+(3*cos(f*x+e)+3)*sin(f*x+e)^4*((b+a*c 
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*((a+b)^(1/2)*((b+a*cos(f*x+e...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (122) = 244\).

Time = 0.18 (sec) , antiderivative size = 500, normalized size of antiderivative = 3.62 \[ \int \frac {\csc ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\left [\frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {a + b} \log \left (\frac {2 \, {\left (a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a + b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + 2 \, b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + 2 \, {\left (3 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} + 7 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{16 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f\right )}}, \frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-a - b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a \cos \left (f x + e\right )^{2} + b}\right ) + {\left (3 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} + 7 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{8 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f\right )}}\right ] \] Input:

integrate(csc(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[1/16*(3*(a^2*cos(f*x + e)^4 - 2*a^2*cos(f*x + e)^2 + a^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^2 + a*b)*co 
s(f*x + e)^3 - (5*a^2 + 7*a*b + 2*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^ 
2 + b)/cos(f*x + e)^2))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^4 
- 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^2 + (a^3 + 3*a^2*b + 3* 
a*b^2 + b^3)*f), 1/8*(3*(a^2*cos(f*x + e)^4 - 2*a^2*cos(f*x + e)^2 + a^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^2 + a*b)*cos(f*x + e)^3 - 
(5*a^2 + 7*a*b + 2*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x 
+ e)^2))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^4 - 2*(a^3 + 3*a^ 
2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^2 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f) 
]
 

Sympy [F]

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

integrate(csc(f*x+e)**5/(a+b*sec(f*x+e)**2)**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate(csc(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 785 vs. \(2 (122) = 244\).

Time = 0.77 (sec) , antiderivative size = 785, normalized size of antiderivative = 5.69 \[ \int \frac {\csc ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Too large to display} \] Input:

integrate(csc(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

-1/64*(sqrt(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)*((a + b)*tan(1/2* 
f*x + 1/2*e)^2/(a^2 + 2*a*b + b^2) + 3*(3*a + b)/(a^2 + 2*a*b + b^2)) - 24 
*a^2*arctan(-(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*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1 
/2*f*x + 1/2*e)^2 + a + b))/sqrt(-a - b))/((a^2 + 2*a*b + b^2)*sqrt(-a - b 
)) - 12*a^2*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*a*tan(1/2*f*x + 1/2*e)^2 + 2* 
b*tan(1/2*f*x + 1/2*e)^2 + a + b))*sqrt(a + b) - a + b))/((a^2 + 2*a*b + b 
^2)*sqrt(a + b)) + 4*(2*(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*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*a^2 - 2*a*b - b^2) - 3*(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*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + 
a + b))^2*(a^2 + 2*a*b + b^2)*sqrt(a + b) - 2*(3*a^3 + a^2*b - 2*a*b^2)*(s 
qrt(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*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) 
^2 + a + b)) + (5*a^3 + 11*a^2*b + 7*a*b^2 + b^3)*sqrt(a + b))/(((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*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 +...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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