\(\int \frac {\cot ^4(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [439]

Optimal result

Integrand size = 25, antiderivative size = 236 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{5/2} f}-\frac {b \cot ^3(e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (3 a+b) \cot ^3(e+f x)}{a^2 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(a-b) \left (3 a^2+14 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a^2 (a+b)^4 f}-\frac {\left (a^2-10 a b-3 b^2\right ) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a^2 (a+b)^3 f} \] 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)^3/a/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(3/2)-b*(3*a+b)*cot(f*x+e)^3/a^2/( 
a+b)^2/f/(a+b+b*tan(f*x+e)^2)^(1/2)+1/3*(a-b)*(3*a^2+14*a*b+3*b^2)*cot(f*x 
+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/a^2/(a+b)^4/f-1/3*(a^2-10*a*b-3*b^2)*cot(f* 
x+e)^3*(a+b+b*tan(f*x+e)^2)^(1/2)/a^2/(a+b)^3/f
 

Mathematica [A] (verified)

Time = 6.13 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.99 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right ) (a+2 b+a \cos (2 e+2 f x))^{5/2} \sec ^5(e+f x)}{4 \sqrt {2} a^{5/2} f \left (a+b \sec ^2(e+f x)\right )^{5/2}}-\frac {(a+2 b+a \cos (2 (e+f x)))^3 \sec ^5(e+f x) \left (-4 (a+3 b) \csc (e+f x)+(a+b) \csc ^3(e+f x)+\frac {4 b^3 \left (6 a^2+13 a b+3 b^2+2 a (3 a+b) \cos (2 (e+f x))\right ) \sin (e+f x)}{a^2 (a+2 b+a \cos (2 (e+f x)))^2}\right )}{24 (a+b)^4 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \] Input:

Integrate[Cot[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

Output:

(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)/(4*Sqrt[2]*a^(5/2)*f*(a + b*Sec[ 
e + f*x]^2)^(5/2)) - ((a + 2*b + a*Cos[2*(e + f*x)])^3*Sec[e + f*x]^5*(-4* 
(a + 3*b)*Csc[e + f*x] + (a + b)*Csc[e + f*x]^3 + (4*b^3*(6*a^2 + 13*a*b + 
 3*b^2 + 2*a*(3*a + b)*Cos[2*(e + f*x)])*Sin[e + f*x])/(a^2*(a + 2*b + a*C 
os[2*(e + f*x)])^2)))/(24*(a + b)^4*f*(a + b*Sec[e + f*x]^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4629, 2075, 374, 27, 441, 445, 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.

\(\displaystyle \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 374

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 441

\(\displaystyle \frac {\frac {\frac {\int \frac {\cot ^4(e+f x) \left (a^2-10 b a-3 b^2-4 b (3 a+b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a (a+b)}-\frac {b (3 a+b) \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {\frac {-\frac {\int \frac {\cot ^2(e+f x) \left (2 b \left (a^2-10 b a-3 b^2\right ) \tan ^2(e+f x)+(a-b) \left (3 a^2+14 b a+3 b^2\right )\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{3 (a+b)}-\frac {\left (a^2-10 a b-3 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{a (a+b)}-\frac {b (3 a+b) \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {\frac {-\frac {-\frac {\int \frac {3 (a+b)^4}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a+b}-\frac {(a-b) \left (3 a^2+14 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{3 (a+b)}-\frac {\left (a^2-10 a b-3 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{a (a+b)}-\frac {b (3 a+b) \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {-\frac {-3 (a+b)^3 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-\frac {(a-b) \left (3 a^2+14 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{3 (a+b)}-\frac {\left (a^2-10 a b-3 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{a (a+b)}-\frac {b (3 a+b) \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\frac {-\frac {-3 (a+b)^3 \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-\frac {(a-b) \left (3 a^2+14 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{3 (a+b)}-\frac {\left (a^2-10 a b-3 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{a (a+b)}-\frac {b (3 a+b) \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {-\frac {-\frac {(a-b) \left (3 a^2+14 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}-\frac {3 (a+b)^3 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {a}}}{3 (a+b)}-\frac {\left (a^2-10 a b-3 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{a (a+b)}-\frac {b (3 a+b) \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

Input:

Int[Cot[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

Output:

(-1/3*(b*Cot[e + f*x]^3)/(a*(a + b)*(a + b + b*Tan[e + f*x]^2)^(3/2)) + (- 
((b*(3*a + b)*Cot[e + f*x]^3)/(a*(a + b)*Sqrt[a + b + b*Tan[e + f*x]^2])) 
+ (-1/3*((a^2 - 10*a*b - 3*b^2)*Cot[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x] 
^2])/(a + b) - ((-3*(a + b)^3*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b 
*Tan[e + f*x]^2]])/Sqrt[a] - ((a - b)*(3*a^2 + 14*a*b + 3*b^2)*Cot[e + f*x 
]*Sqrt[a + b + b*Tan[e + f*x]^2])/(a + b))/(3*(a + b)))/(a*(a + b)))/(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. \(1396\) vs. \(2(216)=432\).

Time = 13.35 (sec) , antiderivative size = 1397, normalized size of antiderivative = 5.92

method	result	size
default	\(\text {Expression too large to display}\)	\(1397\)

Input:

int(cot(f*x+e)^4/(a+b*sec(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/3/f/(a+b)^4/a^2/(-a)^(1/2)/(a+b*sec(f*x+e)^2)^(5/2)*(((b+a*cos(f*x+e)^2 
)/(1+cos(f*x+e))^2)^(1/2)*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*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^ 
(1/2)-4*sin(f*x+e)*a)*a^6*(-3-3*sec(f*x+e))+((b+a*cos(f*x+e)^2)/(1+cos(f*x 
+e))^2)^(1/2)*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*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin( 
f*x+e)*a)*a^5*b*(-12-12*sec(f*x+e)-6*sec(f*x+e)^2-6*sec(f*x+e)^3)+((b+a*co 
s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*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*cos(f*x+e)^2)/(1+cos(f 
*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a^4*b^2*(-18-18*sec(f*x+e)-24*sec(f*x+e)^2 
-24*sec(f*x+e)^3-3*sec(f*x+e)^4-3*sec(f*x+e)^5)+((b+a*cos(f*x+e)^2)/(1+cos 
(f*x+e))^2)^(1/2)*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*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4* 
sin(f*x+e)*a)*a^3*b^3*(-12-12*sec(f*x+e)-36*sec(f*x+e)^2-36*sec(f*x+e)^3-1 
2*sec(f*x+e)^4-12*sec(f*x+e)^5)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2 
)*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*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a^ 
2*b^4*(-3-3*sec(f*x+e)-24*sec(f*x+e)^2-24*sec(f*x+e)^3-18*sec(f*x+e)^4-18* 
sec(f*x+e)^5)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*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...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 729 vs. \(2 (216) = 432\).

Time = 6.71 (sec) , antiderivative size = 1579, normalized size of antiderivative = 6.69 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(cot(f*x+e)^4/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")
 

Output:

[-1/24*(3*((a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*cos(f*x + e)^ 
6 - a^4*b^2 - 4*a^3*b^3 - 6*a^2*b^4 - 4*a*b^5 - b^6 - (a^6 + 2*a^5*b - 2*a 
^4*b^2 - 8*a^3*b^3 - 7*a^2*b^4 - 2*a*b^5)*cos(f*x + e)^4 - (2*a^5*b + 7*a^ 
4*b^2 + 8*a^3*b^3 + 2*a^2*b^4 - 2*a*b^5 - b^6)*cos(f*x + e)^2)*sqrt(-a)*lo 
g(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*(4*(a^6 + 3*a^5*b + 3*a^3*b^3 + a^2*b^4)*cos(f*x + e)^7 - 3*(a^ 
6 + a^5*b - 8*a^4*b^2 + 8*a^3*b^3 - a^2*b^4 - a*b^5)*cos(f*x + e)^5 - 6*(a 
^5*b + 3*a^4*b^2 - 4*a^3*b^3 + 3*a^2*b^4 + a*b^5)*cos(f*x + e)^3 - (3*a^4* 
b^2 + 11*a^3*b^3 - 11*a^2*b^4 - 3*a*b^5)*cos(f*x + e))*sqrt((a*cos(f*x + e 
)^2 + b)/cos(f*x + e)^2))/(((a^9 + 4*a^8*b + 6*a^7*b^2 + 4*a^6*b^3 + a^5*b 
^4)*f*cos(f*x + e)^6 - (a^9 + 2*a^8*b - 2*a^7*b^2 - 8*a^6*b^3 - 7*a^5*b^4 
- 2*a^4*b^5)*f*cos(f*x + e)^4 - (2*a^8*b + 7*a^7*b^2 + 8*a^6*b^3 + 2*a^5*b 
^4 - 2*a^4*b^5 - a^3*b^6)*f*cos(f*x + e)^2 - (a^7*b^2 + 4*a^6*b^3 + 6*a^5* 
b^4 + 4*a^4*b^5 + a^3*b^6)*f)*sin(f*x + e)), -1/12*(3*((a^6 + 4*a^5*b + 6* 
a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*cos(f*x + e)^6 - a^4*b^2 - 4*a^3*b^3 - 6...
 

Sympy [F]

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

integrate(cot(f*x+e)**4/(a+b*sec(f*x+e)**2)**(5/2),x)
 

Output:

Integral(cot(e + f*x)**4/(a + b*sec(e + f*x)**2)**(5/2), x)
                                                                                    
                                                                                    
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(cot(f*x+e)^4/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate(cot(f*x+e)^4/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")
 

Output:

integrate(cot(f*x + e)^4/(b*sec(f*x + e)^2 + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Hanged} \] Input:

int(cot(e + f*x)^4/(a + b/cos(e + f*x)^2)^(5/2),x)
 

Output:

\text{Hanged}
 

Reduce [F]

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

int(cot(f*x+e)^4/(a+b*sec(f*x+e)^2)^(5/2),x)
 

Output:

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