\(\int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [407]

Optimal result

Integrand size = 25, antiderivative size = 166 \[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} f}-\frac {\left (8 a^2+20 a b+15 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 (a+b)^{5/2} f}+\frac {(4 a+7 b) \cot ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 (a+b)^2 f}-\frac {\cot ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 (a+b) f} \] Output:

arctanh((a+b*sec(f*x+e)^2)^(1/2)/a^(1/2))/a^(1/2)/f-1/8*(8*a^2+20*a*b+15*b 
^2)*arctanh((a+b*sec(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/f+1/8*(4*a+7 
*b)*cot(f*x+e)^2*(a+b*sec(f*x+e)^2)^(1/2)/(a+b)^2/f-1/4*cot(f*x+e)^4*(a+b* 
sec(f*x+e)^2)^(1/2)/(a+b)/f
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.17, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4627, 25, 354, 114, 27, 168, 27, 174, 73, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4627

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 168

\(\displaystyle -\frac {\frac {\frac {(4 a+7 b) \sqrt {a+b \sec ^2(e+f x)}}{(a+b) \left (1-\sec ^2(e+f x)\right )}-\frac {\int -\frac {\cos (e+f x) \left (8 (a+b)^2+b (4 a+7 b) \sec ^2(e+f x)\right )}{2 \left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)}{a+b}}{4 (a+b)}+\frac {\sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\frac {\int \frac {\cos (e+f x) \left (8 (a+b)^2+b (4 a+7 b) \sec ^2(e+f x)\right )}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)}{2 (a+b)}+\frac {(4 a+7 b) \sqrt {a+b \sec ^2(e+f x)}}{(a+b) \left (1-\sec ^2(e+f x)\right )}}{4 (a+b)}+\frac {\sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\)

\(\Big \downarrow \) 174

\(\displaystyle -\frac {\frac {\frac {\left (8 a^2+20 a b+15 b^2\right ) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)+8 (a+b)^2 \int \frac {\cos (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)}{2 (a+b)}+\frac {(4 a+7 b) \sqrt {a+b \sec ^2(e+f x)}}{(a+b) \left (1-\sec ^2(e+f x)\right )}}{4 (a+b)}+\frac {\sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {\frac {\frac {\frac {2 \left (8 a^2+20 a b+15 b^2\right ) \int \frac {1}{\frac {a+b}{b}-\frac {\sec ^4(e+f x)}{b}}d\sqrt {b \sec ^2(e+f x)+a}}{b}+\frac {16 (a+b)^2 \int \frac {1}{\frac {\sec ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sec ^2(e+f x)+a}}{b}}{2 (a+b)}+\frac {(4 a+7 b) \sqrt {a+b \sec ^2(e+f x)}}{(a+b) \left (1-\sec ^2(e+f x)\right )}}{4 (a+b)}+\frac {\sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {\frac {\frac {\frac {2 \left (8 a^2+20 a b+15 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}-\frac {16 (a+b)^2 \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a}}}{2 (a+b)}+\frac {(4 a+7 b) \sqrt {a+b \sec ^2(e+f x)}}{(a+b) \left (1-\sec ^2(e+f x)\right )}}{4 (a+b)}+\frac {\sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\)

Input:

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

Output:

-1/2*(Sqrt[a + b*Sec[e + f*x]^2]/(2*(a + b)*(1 - Sec[e + f*x]^2)^2) + (((- 
16*(a + b)^2*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]])/Sqrt[a] + (2*(8* 
a^2 + 20*a*b + 15*b^2)*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a + b]])/Sq 
rt[a + b])/(2*(a + b)) + ((4*a + 7*b)*Sqrt[a + b*Sec[e + f*x]^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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 114

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e 
 - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

rule 168

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
 

rule 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 3042

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

rule 4627

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si 
mp[1/f   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), 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] || IGtQ[p, 0] || Integers 
Q[2*n, p])
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1977\) vs. \(2(144)=288\).

Time = 0.49 (sec) , antiderivative size = 1978, normalized size of antiderivative = 11.92

method	result	size
default	\(\text {Expression too large to display}\)	\(1978\)

Input:

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

Output:

-1/16/f/(a+b)^(9/2)/a^(1/2)*((8*cos(f*x+e)+8)*sin(f*x+e)^4*((b+a*cos(f*x+e 
)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(9/2)*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)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2) 
^(1/2)*(a+b)^(1/2)+cos(f*x+e)*a+b)/(cos(f*x+e)-1))+(-8*cos(f*x+e)-8)*sin(f 
*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(9/2)*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)+((b 
+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-cos(f*x+e)*a+b)/(1+co 
s(f*x+e)))+(36*cos(f*x+e)+36)*sin(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ 
e))^2)^(1/2)*a^(7/2)*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)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/ 
2)+cos(f*x+e)*a+b)/(cos(f*x+e)-1))*b+(-36*cos(f*x+e)-36)*sin(f*x+e)^4*((b+ 
a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(7/2)*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)+((b+a*cos(f*x+e 
)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-cos(f*x+e)*a+b)/(1+cos(f*x+e)))*b 
+(63*cos(f*x+e)+63)*sin(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/ 
2)*a^(5/2)*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)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+cos(f*x 
+e)*a+b)/(cos(f*x+e)-1))*b^2+(-63*cos(f*x+e)-63)*sin(f*x+e)^4*((b+a*cos(f* 
x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(5/2)*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)+((b+a*cos(f*x+e)^2)/...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 518 vs. \(2 (144) = 288\).

Time = 1.15 (sec) , antiderivative size = 2257, normalized size of antiderivative = 13.60 \[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/32*(4*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e)^4 + a^3 + 3*a^2*b + 
 3*a*b^2 + b^3 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e)^2)*sqrt(a) 
*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b*cos(f*x + e)^6 + 160*a^2*b^2*cos(f 
*x + e)^4 + 32*a*b^3*cos(f*x + e)^2 + b^4 + 8*(16*a^3*cos(f*x + e)^8 + 24* 
a^2*b*cos(f*x + e)^6 + 10*a*b^2*cos(f*x + e)^4 + b^3*cos(f*x + e)^2)*sqrt( 
a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) + ((8*a^3 + 20*a^2*b + 15* 
a*b^2)*cos(f*x + e)^4 + 8*a^3 + 20*a^2*b + 15*a*b^2 - 2*(8*a^3 + 20*a^2*b 
+ 15*a*b^2)*cos(f*x + e)^2)*sqrt(a + b)*log(2*((8*a^2 + 8*a*b + b^2)*cos(f 
*x + e)^4 + 2*(4*a*b + 3*b^2)*cos(f*x + e)^2 + b^2 - 4*((2*a + b)*cos(f*x 
+ e)^4 + b*cos(f*x + e)^2)*sqrt(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x 
 + e)^2))/(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)) - 4*(3*(2*a^3 + 5*a^2*b 
 + 3*a*b^2)*cos(f*x + e)^4 - (4*a^3 + 11*a^2*b + 7*a*b^2)*cos(f*x + e)^2)* 
sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^4 + 3*a^3*b + 3*a^2*b^2 + 
 a*b^3)*f*cos(f*x + e)^4 - 2*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*cos(f*x 
 + e)^2 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f), 1/16*(((8*a^3 + 20*a^2*b 
 + 15*a*b^2)*cos(f*x + e)^4 + 8*a^3 + 20*a^2*b + 15*a*b^2 - 2*(8*a^3 + 20* 
a^2*b + 15*a*b^2)*cos(f*x + e)^2)*sqrt(-a - b)*arctan(1/2*((2*a + b)*cos(f 
*x + e)^2 + b)*sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(( 
a^2 + a*b)*cos(f*x + e)^2 + a*b + b^2)) + 2*((a^3 + 3*a^2*b + 3*a*b^2 + b^ 
3)*cos(f*x + e)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 - 2*(a^3 + 3*a^2*b + ...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(cot(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. 924 vs. \(2 (144) = 288\).

Time = 1.05 (sec) , antiderivative size = 924, normalized size of antiderivative = 5.57 \[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Too large to display} \] Input:

integrate(cot(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) - (11*a + 17*b)/(a^2 + 2*a*b + b^2)) - 1 
28*arctan(-1/2*(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) + sqrt(a + b))/sqrt(-a))/sqrt(-a) + 8*(8*a^2 
+ 20*a*b + 15*b^2)*arctan(-(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*ta 
n(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)) - 4*(8*a^2 + 20*a*b + 15*b^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))*(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*(3*a^2 + 2*a*b - 4*b^2) - 7*(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*(2*a^3 + 4*a^2*b - 3*a*b^2 - 5*b^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*...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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