\(\int \frac {\cot ^2(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [438]

Optimal result

Integrand size = 25, antiderivative size = 174 \[ \int \frac {\cot ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{5/2} f}-\frac {b \cot (e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (7 a+3 b) \cot (e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {(a-3 b) (3 a+b) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a^2 (a+b)^3 f} \] Output:

-arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(5/2)/f-1/3*b*cot 
(f*x+e)/a/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(3/2)-1/3*b*(7*a+3*b)*cot(f*x+e)/a^ 
2/(a+b)^2/f/(a+b+b*tan(f*x+e)^2)^(1/2)-1/3*(a-3*b)*(3*a+b)*cot(f*x+e)*(a+b 
+b*tan(f*x+e)^2)^(1/2)/a^2/(a+b)^3/f
 

Mathematica [A] (verified)

Time = 4.20 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.42 \[ \int \frac {\cot ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right ) (a+2 b+a \cos (2 e+2 f x))^{5/2} \sec ^5(e+f x)}{4 \sqrt {2} a^{5/2} f \left (a+b \sec ^2(e+f x)\right )^{5/2}}-\frac {(a+2 b+a \cos (2 (e+f x))) \left (3 \left (3 a^4+8 a^3 b+5 a^2 b^2-12 a b^3-4 b^4\right )+4 \left (3 a^4+6 a^3 b+8 a b^3+3 b^4\right ) \cos (2 (e+f x))+a \left (3 a^3+9 a b^2+4 b^3\right ) \cos (4 (e+f x))\right ) \csc (e+f x) \sec ^5(e+f x)}{48 a^2 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \] Input:

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

Output:

-1/4*(ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]]*(a + 2 
*b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f*x]^5)/(Sqrt[2]*a^(5/2)*f*(a + b*S 
ec[e + f*x]^2)^(5/2)) - ((a + 2*b + a*Cos[2*(e + f*x)])*(3*(3*a^4 + 8*a^3* 
b + 5*a^2*b^2 - 12*a*b^3 - 4*b^4) + 4*(3*a^4 + 6*a^3*b + 8*a*b^3 + 3*b^4)* 
Cos[2*(e + f*x)] + a*(3*a^3 + 9*a*b^2 + 4*b^3)*Cos[4*(e + f*x)])*Csc[e + f 
*x]*Sec[e + f*x]^5)/(48*a^2*(a + b)^3*f*(a + b*Sec[e + f*x]^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4629, 2075, 374, 441, 445, 27, 291, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 374

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

\(\Big \downarrow \) 441

\(\displaystyle \frac {\frac {\frac {\int \frac {\cot ^2(e+f x) \left ((a-3 b) (3 a+b)-2 b (7 a+3 b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a (a+b)}-\frac {b (7 a+3 b) \cot (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}-\frac {b \cot (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {\frac {-\frac {\int \frac {3 (a+b)^3}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a+b}-\frac {(a-3 b) (3 a+b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{a (a+b)}-\frac {b (7 a+3 b) \cot (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}-\frac {b \cot (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {-3 (a+b)^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-\frac {(a-3 b) (3 a+b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{a (a+b)}-\frac {b (7 a+3 b) \cot (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}-\frac {b \cot (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\frac {-3 (a+b)^2 \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-\frac {(a-3 b) (3 a+b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{a (a+b)}-\frac {b (7 a+3 b) \cot (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}-\frac {b \cot (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {-\frac {3 (a+b)^2 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {a}}-\frac {(a-3 b) (3 a+b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{a (a+b)}-\frac {b (7 a+3 b) \cot (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}-\frac {b \cot (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

Input:

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

Output:

(-1/3*(b*Cot[e + f*x])/(a*(a + b)*(a + b + b*Tan[e + f*x]^2)^(3/2)) + (-(( 
b*(7*a + 3*b)*Cot[e + f*x])/(a*(a + b)*Sqrt[a + b + b*Tan[e + f*x]^2])) + 
((-3*(a + b)^2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2 
]])/Sqrt[a] - ((a - 3*b)*(3*a + b)*Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x 
]^2])/(a + b))/(a*(a + b)))/(3*a*(a + b)))/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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[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 374

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

rule 441

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a 
+ b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*g*2*(b*c - a*d)*(p + 1))), x] + Si 
mp[1/(a*2*(b*c - a*d)*(p + 1))   Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2 
)^q*Simp[c*(b*e - a*f)*(m + 1) + e*2*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m 
 + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m, q}, 
 x] && LtQ[p, -1]
 

rule 445

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ 
.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p 
+ 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) 
 Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c 
+ a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 
2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
 

rule 2075

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa 
ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi 
alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  ! 
BinomialMatchQ[{u, v}, x]
 

rule 3042

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

rule 4629

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f 
_.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim 
p[ff/f   Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 
)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte 
gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1108\) vs. \(2(156)=312\).

Time = 9.80 (sec) , antiderivative size = 1109, normalized size of antiderivative = 6.37

method	result	size
default	\(\text {Expression too large to display}\)	\(1109\)

Input:

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

Output:

-1/3/f/(a+b)^3/a^2/(-a)^(1/2)/(a+b*sec(f*x+e)^2)^(5/2)*(ln(4*(-a)^(1/2)*(( 
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*co 
s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1 
+cos(f*x+e))^2)^(1/2)*a^5*(3+3*sec(f*x+e))+ln(4*(-a)^(1/2)*((b+a*cos(f*x+e 
)^2)/(1+cos(f*x+e))^2)^(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*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^ 
2)^(1/2)*a^4*b*(9+9*sec(f*x+e)+6*sec(f*x+e)^2+6*sec(f*x+e)^3)+ln(4*(-a)^(1 
/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(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*sin(f*x+e)*a)*((b+a*cos(f*x+e) 
^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b^2*(9+9*sec(f*x+e)+18*sec(f*x+e)^2+18*sec 
(f*x+e)^3+3*sec(f*x+e)^4+3*sec(f*x+e)^5)+ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^ 
2)/(1+cos(f*x+e))^2)^(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*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2) 
^(1/2)*a^2*b^3*(3+3*sec(f*x+e)+18*sec(f*x+e)^2+18*sec(f*x+e)^3+9*sec(f*x+e 
)^4+9*sec(f*x+e)^5)+ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^ 
(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*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b^4*(6*sec(f 
*x+e)^2+6*sec(f*x+e)^3+9*sec(f*x+e)^4+9*sec(f*x+e)^5)+ln(4*(-a)^(1/2)*((b+ 
a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(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*sin(f*x+e)*a)*((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. 488 vs. \(2 (156) = 312\).

Time = 1.84 (sec) , antiderivative size = 1097, normalized size of antiderivative = 6.30 \[ \int \frac {\cot ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/24*(3*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5 + (a^5 + 3*a^4*b + 3*a^3*b^ 
2 + a^2*b^3)*cos(f*x + e)^4 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*co 
s(f*x + e)^2)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos( 
f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a 
^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3 
)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e 
)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a 
*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e 
)^2)*sin(f*x + e))*sin(f*x + e) + 8*((3*a^5 + 9*a^3*b^2 + 4*a^2*b^3)*cos(f 
*x + e)^5 + (6*a^4*b - 9*a^3*b^2 + 4*a^2*b^3 + 3*a*b^4)*cos(f*x + e)^3 + ( 
3*a^3*b^2 - 8*a^2*b^3 - 3*a*b^4)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b) 
/cos(f*x + e)^2))/(((a^8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3)*f*cos(f*x + e)^4 
 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a^4*b^4)*f*cos(f*x + e)^2 + (a^6*b^2 
 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*f)*sin(f*x + e)), 1/12*(3*(a^3*b^2 + 3 
*a^2*b^3 + 3*a*b^4 + b^5 + (a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cos(f*x + 
 e)^4 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(f*x + e)^2)*sqrt(a)* 
arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6 
*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e) 
^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^ 
2)*sin(f*x + e)))*sin(f*x + e) - 4*((3*a^5 + 9*a^3*b^2 + 4*a^2*b^3)*cos...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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