\(\int \frac {\cot ^3(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [419]

Optimal result

Integrand size = 25, antiderivative size = 153 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2} f}+\frac {(2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{5/2} f}-\frac {(a-2 b) b}{2 a (a+b)^2 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\cot ^2(e+f x)}{2 (a+b) f \sqrt {a+b \sec ^2(e+f x)}} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.13, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4627, 354, 114, 27, 169, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4627

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

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 169

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] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n, 2*p]
 

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. \(3945\) vs. \(2(131)=262\).

Time = 1.13 (sec) , antiderivative size = 3946, normalized size of antiderivative = 25.79

method	result	size
default	\(\text {Expression too large to display}\)	\(3946\)

Input:

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

Output:

1/4/f/a^(3/2)/(a+b)^(11/2)/((-a*b)^(1/2)+a)/((-a*b)^(1/2)-a)/(b+a*cos(f*x+ 
e)^2)/(a+b*sec(f*x+e)^2)^(3/2)*(-4*(a+b)^(5/2)*a^(3/2)*b^6*sec(f*x+e)^2+2* 
(a+b)^(5/2)*a^(15/2)*cos(f*x+e)^2*cot(f*x+e)^2+a^(19/2)*((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)+((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)))*(2*cos(f*x+e)^2+2*cos(f 
*x+e))+a^(19/2)*((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)+((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))* 
(-2*cos(f*x+e)^2-2*cos(f*x+e))+(8*cos(f*x+e)^6+6*cos(f*x+e)^4+4*cos(f*x+e) 
^2+2)*(a+b)^(5/2)*a^(9/2)*b^3*sec(f*x+e)^2*csc(f*x+e)^2+(a+b)^(5/2)*((b+a* 
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ 
e))^2)^(1/2)*a^(1/2)*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e 
))^2)^(1/2)+4*cos(f*x+e)*a)*a^4*b^3*(16*cos(f*x+e)^2+16*cos(f*x+e)+48+48*s 
ec(f*x+e)+16*sec(f*x+e)^2+16*sec(f*x+e)^3)+(a+b)^(5/2)*((b+a*cos(f*x+e)^2) 
/(1+cos(f*x+e))^2)^(1/2)*ln(4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)* 
a^(1/2)*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+4 
*cos(f*x+e)*a)*a^3*b^4*(4*cos(f*x+e)^2+4*cos(f*x+e)+32+32*sec(f*x+e)+24*se 
c(f*x+e)^2+24*sec(f*x+e)^3)+(a+b)^(5/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e)) 
^2)^(1/2)*ln(4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)*cos(...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 540 vs. \(2 (131) = 262\).

Time = 1.34 (sec) , antiderivative size = 2347, normalized size of antiderivative = 15.34 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/8*(((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(f*x + e)^4 - a^3*b - 3*a^2* 
b^2 - 3*a*b^3 - b^4 - (a^4 + 2*a^3*b - 2*a*b^3 - b^4)*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*co 
s(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)*sq 
rt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) + ((2*a^4 + 5*a^3*b)*co 
s(f*x + e)^4 - 2*a^3*b - 5*a^2*b^2 - (2*a^4 + 3*a^3*b - 5*a^2*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*((a^4 + a^3*b + 2*a^2*b^2 + 2*a*b^3)*c 
os(f*x + e)^4 + (a^3*b - a^2*b^2 - 2*a*b^3)*cos(f*x + e)^2)*sqrt((a*cos(f* 
x + e)^2 + b)/cos(f*x + e)^2))/((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*f*co 
s(f*x + e)^4 - (a^6 + 2*a^5*b - 2*a^3*b^3 - a^2*b^4)*f*cos(f*x + e)^2 - (a 
^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*f), -1/8*(2*((2*a^4 + 5*a^3*b)*cos 
(f*x + e)^4 - 2*a^3*b - 5*a^2*b^2 - (2*a^4 + 3*a^3*b - 5*a^2*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)) - ((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(f*x + e)^4 - a^3* 
b - 3*a^2*b^2 - 3*a*b^3 - b^4 - (a^4 + 2*a^3*b - 2*a*b^3 - b^4)*cos(f*x...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1062 vs. \(2 (131) = 262\).

Time = 1.40 (sec) , antiderivative size = 1062, normalized size of antiderivative = 6.94 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/8*((((a^7*b*sgn(cos(f*x + e)) + 4*a^6*b^2*sgn(cos(f*x + e)) + 6*a^5*b^3* 
sgn(cos(f*x + e)) + 4*a^4*b^4*sgn(cos(f*x + e)) + a^3*b^5*sgn(cos(f*x + e) 
))*tan(1/2*f*x + 1/2*e)^2/(a^8*b + 5*a^7*b^2 + 10*a^6*b^3 + 10*a^5*b^4 + 5 
*a^4*b^5 + a^3*b^6) - 2*(a^7*b*sgn(cos(f*x + e)) + 2*a^6*b^2*sgn(cos(f*x + 
 e)) + 4*a^5*b^3*sgn(cos(f*x + e)) + 10*a^4*b^4*sgn(cos(f*x + e)) + 11*a^3 
*b^5*sgn(cos(f*x + e)) + 4*a^2*b^6*sgn(cos(f*x + e)))/(a^8*b + 5*a^7*b^2 + 
 10*a^6*b^3 + 10*a^5*b^4 + 5*a^4*b^5 + a^3*b^6))*tan(1/2*f*x + 1/2*e)^2 + 
(a^7*b*sgn(cos(f*x + e)) + 4*a^6*b^2*sgn(cos(f*x + e)) + 14*a^5*b^3*sgn(co 
s(f*x + e)) + 28*a^4*b^4*sgn(cos(f*x + e)) + 25*a^3*b^5*sgn(cos(f*x + e)) 
+ 8*a^2*b^6*sgn(cos(f*x + e)))/(a^8*b + 5*a^7*b^2 + 10*a^6*b^3 + 10*a^5*b^ 
4 + 5*a^4*b^5 + a^3*b^6))/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 
) - 4*(2*a + 5*b)*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)*sgn(cos(f*x + e))) + 2*(2*a + 5*b)*log(abs(-(sqrt(a + b)*ta 
n(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)*sgn(cos(f*x + e))) 
+ 16*arctan(-1/2*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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