\(\int \frac {\cot ^4(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\) [361]

Optimal result

Integrand size = 23, antiderivative size = 160 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {x}{a^2}-\frac {b^{5/2} (7 a+2 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^2 (a+b)^{7/2} f}+\frac {\left (2 a^2+6 a b-b^2\right ) \cot (e+f x)}{2 a (a+b)^3 f}-\frac {(2 a-3 b) \cot ^3(e+f x)}{6 a (a+b)^2 f}-\frac {b \cot ^3(e+f x)}{2 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )} \] Output:

x/a^2-1/2*b^(5/2)*(7*a+2*b)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/a^2/(a+ 
b)^(7/2)/f+1/2*(2*a^2+6*a*b-b^2)*cot(f*x+e)/a/(a+b)^3/f-1/6*(2*a-3*b)*cot( 
f*x+e)^3/a/(a+b)^2/f-1/2*b*cot(f*x+e)^3/a/(a+b)/f/(a+b+b*tan(f*x+e)^2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 5.44 (sec) , antiderivative size = 1588, normalized size of antiderivative = 9.92 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^4*((48*b^3*(7*a + 2*b)*ArcTan 
[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f* 
x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(a + 2*b + a*Cos[2*(e 
+ f*x)])*(Cos[2*e] - I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^ 
4]) + Csc[e]*Csc[e + f*x]^3*Sec[2*e]*(-6*(a + b)^3*(a + 6*b)*f*x*Cos[f*x] 
+ 3*(a - 4*b)*(a + b)^3*f*x*Cos[3*f*x] + 6*a^4*f*x*Cos[2*e - f*x] + 54*a^3 
*b*f*x*Cos[2*e - f*x] + 126*a^2*b^2*f*x*Cos[2*e - f*x] + 114*a*b^3*f*x*Cos 
[2*e - f*x] + 36*b^4*f*x*Cos[2*e - f*x] + 6*a^4*f*x*Cos[2*e + f*x] + 54*a^ 
3*b*f*x*Cos[2*e + f*x] + 126*a^2*b^2*f*x*Cos[2*e + f*x] + 114*a*b^3*f*x*Co 
s[2*e + f*x] + 36*b^4*f*x*Cos[2*e + f*x] - 6*a^4*f*x*Cos[4*e + f*x] - 54*a 
^3*b*f*x*Cos[4*e + f*x] - 126*a^2*b^2*f*x*Cos[4*e + f*x] - 114*a*b^3*f*x*C 
os[4*e + f*x] - 36*b^4*f*x*Cos[4*e + f*x] - 3*a^4*f*x*Cos[2*e + 3*f*x] + 3 
*a^3*b*f*x*Cos[2*e + 3*f*x] + 27*a^2*b^2*f*x*Cos[2*e + 3*f*x] + 33*a*b^3*f 
*x*Cos[2*e + 3*f*x] + 12*b^4*f*x*Cos[2*e + 3*f*x] + 3*a^4*f*x*Cos[4*e + 3* 
f*x] - 3*a^3*b*f*x*Cos[4*e + 3*f*x] - 27*a^2*b^2*f*x*Cos[4*e + 3*f*x] - 33 
*a*b^3*f*x*Cos[4*e + 3*f*x] - 12*b^4*f*x*Cos[4*e + 3*f*x] - 3*a^4*f*x*Cos[ 
6*e + 3*f*x] + 3*a^3*b*f*x*Cos[6*e + 3*f*x] + 27*a^2*b^2*f*x*Cos[6*e + 3*f 
*x] + 33*a*b^3*f*x*Cos[6*e + 3*f*x] + 12*b^4*f*x*Cos[6*e + 3*f*x] - 3*a^4* 
f*x*Cos[2*e + 5*f*x] - 9*a^3*b*f*x*Cos[2*e + 5*f*x] - 9*a^2*b^2*f*x*Cos[2* 
e + 5*f*x] - 3*a*b^3*f*x*Cos[2*e + 5*f*x] + 3*a^4*f*x*Cos[4*e + 5*f*x] ...
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.14, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 4629, 2075, 374, 445, 27, 445, 397, 216, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 374

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

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {-\frac {\int \frac {3 \cot ^2(e+f x) \left (2 a^2+6 b a-b^2+(2 a-3 b) b \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{3 (a+b)}-\frac {(2 a-3 b) \cot ^3(e+f x)}{3 (a+b)}}{2 a (a+b)}-\frac {b \cot ^3(e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {\int \frac {\cot ^2(e+f x) \left (2 a^2+6 b a-b^2+(2 a-3 b) b \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{a+b}-\frac {(2 a-3 b) \cot ^3(e+f x)}{3 (a+b)}}{2 a (a+b)}-\frac {b \cot ^3(e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {-\frac {-\frac {\int \frac {2 a^3+8 b a^2+12 b^2 a+b^3+b \left (2 a^2+6 b a-b^2\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{a+b}-\frac {\left (2 a^2+6 a b-b^2\right ) \cot (e+f x)}{a+b}}{a+b}-\frac {(2 a-3 b) \cot ^3(e+f x)}{3 (a+b)}}{2 a (a+b)}-\frac {b \cot ^3(e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {-\frac {-\frac {\frac {2 (a+b)^3 \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}-\frac {b^3 (7 a+2 b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{a+b}-\frac {\left (2 a^2+6 a b-b^2\right ) \cot (e+f x)}{a+b}}{a+b}-\frac {(2 a-3 b) \cot ^3(e+f x)}{3 (a+b)}}{2 a (a+b)}-\frac {b \cot ^3(e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {-\frac {-\frac {\frac {2 (a+b)^3 \arctan (\tan (e+f x))}{a}-\frac {b^3 (7 a+2 b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{a+b}-\frac {\left (2 a^2+6 a b-b^2\right ) \cot (e+f x)}{a+b}}{a+b}-\frac {(2 a-3 b) \cot ^3(e+f x)}{3 (a+b)}}{2 a (a+b)}-\frac {b \cot ^3(e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {-\frac {-\frac {\left (2 a^2+6 a b-b^2\right ) \cot (e+f x)}{a+b}-\frac {\frac {2 (a+b)^3 \arctan (\tan (e+f x))}{a}-\frac {b^{5/2} (7 a+2 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}}{a+b}}{a+b}-\frac {(2 a-3 b) \cot ^3(e+f x)}{3 (a+b)}}{2 a (a+b)}-\frac {b \cot ^3(e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\)

Input:

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

Output:

((-1/3*((2*a - 3*b)*Cot[e + f*x]^3)/(a + b) - (-(((2*(a + b)^3*ArcTan[Tan[ 
e + f*x]])/a - (b^(5/2)*(7*a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + 
 b]])/(a*Sqrt[a + b]))/(a + b)) - ((2*a^2 + 6*a*b - b^2)*Cot[e + f*x])/(a 
+ b))/(a + b))/(2*a*(a + b)) - (b*Cot[e + f*x]^3)/(2*a*(a + b)*(a + b + b* 
Tan[e + f*x]^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 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 218

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

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 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

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 [A] (verified)

Time = 5.81 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {-\frac {b^{3} \left (\frac {a \tan \left (f x +e \right )}{2 a +2 b +2 b \tan \left (f x +e \right )^{2}}+\frac {\left (7 a +2 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \sqrt {\left (a +b \right ) b}}\right )}{a^{2} \left (a +b \right )^{3}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{2}}-\frac {1}{3 \left (a +b \right )^{2} \tan \left (f x +e \right )^{3}}-\frac {-a -3 b}{\left (a +b \right )^{3} \tan \left (f x +e \right )}}{f}\)	\(124\)
default	\(\frac {-\frac {b^{3} \left (\frac {a \tan \left (f x +e \right )}{2 a +2 b +2 b \tan \left (f x +e \right )^{2}}+\frac {\left (7 a +2 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \sqrt {\left (a +b \right ) b}}\right )}{a^{2} \left (a +b \right )^{3}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{2}}-\frac {1}{3 \left (a +b \right )^{2} \tan \left (f x +e \right )^{3}}-\frac {-a -3 b}{\left (a +b \right )^{3} \tan \left (f x +e \right )}}{f}\)	\(124\)
risch	\(\frac {x}{a^{2}}-\frac {i \left (-12 a^{4} {\mathrm e}^{8 i \left (f x +e \right )}-24 a^{3} b \,{\mathrm e}^{8 i \left (f x +e \right )}+3 a \,b^{3} {\mathrm e}^{8 i \left (f x +e \right )}+6 b^{4} {\mathrm e}^{8 i \left (f x +e \right )}-12 a^{4} {\mathrm e}^{6 i \left (f x +e \right )}-60 a^{3} b \,{\mathrm e}^{6 i \left (f x +e \right )}-96 a^{2} b^{2} {\mathrm e}^{6 i \left (f x +e \right )}-6 a \,b^{3} {\mathrm e}^{6 i \left (f x +e \right )}-18 b^{4} {\mathrm e}^{6 i \left (f x +e \right )}+4 a^{4} {\mathrm e}^{4 i \left (f x +e \right )}+76 a^{3} b \,{\mathrm e}^{4 i \left (f x +e \right )}+144 a^{2} b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+18 b^{4} {\mathrm e}^{4 i \left (f x +e \right )}-4 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{4}-36 a^{3} b \,{\mathrm e}^{2 i \left (f x +e \right )}-80 a^{2} b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+6 a \,b^{3} {\mathrm e}^{2 i \left (f x +e \right )}-6 b^{4} {\mathrm e}^{2 i \left (f x +e \right )}-8 a^{4}-20 b \,a^{3}-3 a \,b^{3}\right )}{3 f \left (a +b \right )^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{3} a^{2} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}+\frac {7 \sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{4 \left (a +b \right )^{4} f a}+\frac {\sqrt {-\left (a +b \right ) b}\, b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 \left (a +b \right )^{4} f \,a^{2}}-\frac {7 \sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{4 \left (a +b \right )^{4} f a}-\frac {\sqrt {-\left (a +b \right ) b}\, b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 \left (a +b \right )^{4} f \,a^{2}}\)	\(585\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 446 vs. \(2 (144) = 288\).

Time = 0.14 (sec) , antiderivative size = 979, normalized size of antiderivative = 6.12 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/24*(4*(8*a^4 + 20*a^3*b + 3*a*b^3)*cos(f*x + e)^5 - 8*(3*a^4 + 5*a^3*b 
- 10*a^2*b^2 + 3*a*b^3)*cos(f*x + e)^3 + 3*((7*a^2*b^2 + 2*a*b^3)*cos(f*x 
+ e)^4 - 7*a*b^3 - 2*b^4 - (7*a^2*b^2 - 5*a*b^3 - 2*b^4)*cos(f*x + e)^2)*s 
qrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b 
^2)*cos(f*x + e)^2 + 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2) 
*cos(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 
2*a*b*cos(f*x + e)^2 + b^2))*sin(f*x + e) - 12*(2*a^3*b + 6*a^2*b^2 - a*b^ 
3)*cos(f*x + e) + 24*((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*x*cos(f*x + e) 
^4 - (a^4 + 2*a^3*b - 2*a*b^3 - b^4)*f*x*cos(f*x + e)^2 - (a^3*b + 3*a^2*b 
^2 + 3*a*b^3 + b^4)*f*x)*sin(f*x + e))/(((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3* 
b^3)*f*cos(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)*sin(f*x + e)), 1/12*(2 
*(8*a^4 + 20*a^3*b + 3*a*b^3)*cos(f*x + e)^5 - 4*(3*a^4 + 5*a^3*b - 10*a^2 
*b^2 + 3*a*b^3)*cos(f*x + e)^3 + 3*((7*a^2*b^2 + 2*a*b^3)*cos(f*x + e)^4 - 
 7*a*b^3 - 2*b^4 - (7*a^2*b^2 - 5*a*b^3 - 2*b^4)*cos(f*x + e)^2)*sqrt(b/(a 
 + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a + b))/(b*cos(f* 
x + e)*sin(f*x + e)))*sin(f*x + e) - 6*(2*a^3*b + 6*a^2*b^2 - a*b^3)*cos(f 
*x + e) + 12*((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*x*cos(f*x + e)^4 - (a^ 
4 + 2*a^3*b - 2*a*b^3 - b^4)*f*x*cos(f*x + e)^2 - (a^3*b + 3*a^2*b^2 + 3*a 
*b^3 + b^4)*f*x)*sin(f*x + e))/(((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*...
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.47 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {3 \, {\left (7 \, a b^{3} + 2 \, b^{4}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} \sqrt {{\left (a + b\right )} b}} - \frac {3 \, {\left (2 \, a^{2} b + 6 \, a b^{2} - b^{3}\right )} \tan \left (f x + e\right )^{4} - 2 \, a^{3} - 4 \, a^{2} b - 2 \, a b^{2} + 2 \, {\left (3 \, a^{3} + 11 \, a^{2} b + 8 \, a b^{2}\right )} \tan \left (f x + e\right )^{2}}{{\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \tan \left (f x + e\right )^{5} + {\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} \tan \left (f x + e\right )^{3}} - \frac {6 \, {\left (f x + e\right )}}{a^{2}}}{6 \, f} \] Input:

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

Output:

-1/6*(3*(7*a*b^3 + 2*b^4)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^5 + 3 
*a^4*b + 3*a^3*b^2 + a^2*b^3)*sqrt((a + b)*b)) - (3*(2*a^2*b + 6*a*b^2 - b 
^3)*tan(f*x + e)^4 - 2*a^3 - 4*a^2*b - 2*a*b^2 + 2*(3*a^3 + 11*a^2*b + 8*a 
*b^2)*tan(f*x + e)^2)/((a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*tan(f*x + e 
)^5 + (a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*tan(f*x + e)^3) - 6* 
(f*x + e)/a^2)/f
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.32 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {3 \, b^{3} \tan \left (f x + e\right )}{{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}} + \frac {3 \, {\left (7 \, a b^{3} + 2 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} \sqrt {a b + b^{2}}} - \frac {6 \, {\left (f x + e\right )}}{a^{2}} - \frac {2 \, {\left (3 \, a \tan \left (f x + e\right )^{2} + 9 \, b \tan \left (f x + e\right )^{2} - a - b\right )}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{3}}}{6 \, f} \] Input:

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

Output:

-1/6*(3*b^3*tan(f*x + e)/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*(b*tan(f*x + 
 e)^2 + a + b)) + 3*(7*a*b^3 + 2*b^4)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) 
 + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/((a^5 + 3*a^4*b + 3*a^3*b^2 + a 
^2*b^3)*sqrt(a*b + b^2)) - 6*(f*x + e)/a^2 - 2*(3*a*tan(f*x + e)^2 + 9*b*t 
an(f*x + e)^2 - a - b)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*tan(f*x + e)^3))/f
 

Mupad [B] (verification not implemented)

Time = 21.96 (sec) , antiderivative size = 4987, normalized size of antiderivative = 31.17 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((tan(e + f*x)^2*(3*a + 8*b))/(3*(a + b)^2) - 1/(3*(a + b)) + (tan(e + f*x 
)^4*(6*a*b^2 + 2*a^2*b - b^3))/(2*a*(a + b)^3))/(f*(tan(e + f*x)^3*(a + b) 
 + b*tan(e + f*x)^5)) + atan((560*a^3*b^16*tan(e + f*x))/(560*a^3*b^16 + 7 
280*a^4*b^15 + 42560*a^5*b^14 + 149184*a^6*b^13 + 351904*a^7*b^12 + 593440 
*a^8*b^11 + 741120*a^9*b^10 + 699840*a^10*b^9 + 505008*a^11*b^8 + 278768*a 
^12*b^7 + 116480*a^13*b^6 + 35840*a^14*b^5 + 7680*a^15*b^4 + 1024*a^16*b^3 
 + 64*a^17*b^2) + (7280*a^4*b^15*tan(e + f*x))/(560*a^3*b^16 + 7280*a^4*b^ 
15 + 42560*a^5*b^14 + 149184*a^6*b^13 + 351904*a^7*b^12 + 593440*a^8*b^11 
+ 741120*a^9*b^10 + 699840*a^10*b^9 + 505008*a^11*b^8 + 278768*a^12*b^7 + 
116480*a^13*b^6 + 35840*a^14*b^5 + 7680*a^15*b^4 + 1024*a^16*b^3 + 64*a^17 
*b^2) + (42560*a^5*b^14*tan(e + f*x))/(560*a^3*b^16 + 7280*a^4*b^15 + 4256 
0*a^5*b^14 + 149184*a^6*b^13 + 351904*a^7*b^12 + 593440*a^8*b^11 + 741120* 
a^9*b^10 + 699840*a^10*b^9 + 505008*a^11*b^8 + 278768*a^12*b^7 + 116480*a^ 
13*b^6 + 35840*a^14*b^5 + 7680*a^15*b^4 + 1024*a^16*b^3 + 64*a^17*b^2) + ( 
149184*a^6*b^13*tan(e + f*x))/(560*a^3*b^16 + 7280*a^4*b^15 + 42560*a^5*b^ 
14 + 149184*a^6*b^13 + 351904*a^7*b^12 + 593440*a^8*b^11 + 741120*a^9*b^10 
 + 699840*a^10*b^9 + 505008*a^11*b^8 + 278768*a^12*b^7 + 116480*a^13*b^6 + 
 35840*a^14*b^5 + 7680*a^15*b^4 + 1024*a^16*b^3 + 64*a^17*b^2) + (351904*a 
^7*b^12*tan(e + f*x))/(560*a^3*b^16 + 7280*a^4*b^15 + 42560*a^5*b^14 + 149 
184*a^6*b^13 + 351904*a^7*b^12 + 593440*a^8*b^11 + 741120*a^9*b^10 + 69...
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 993, normalized size of antiderivative = 6.21 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 21*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/s 
qrt(b))*sin(e + f*x)**5*a**2*b**2 - 6*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b 
)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**5*a*b**3 + 21*sqrt(b) 
*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e 
+ f*x)**3*a**2*b**2 + 27*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f* 
x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**3*a*b**3 + 6*sqrt(b)*sqrt(a + b)*a 
tan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**3*b**4 
 - 21*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sq 
rt(b))*sin(e + f*x)**5*a**2*b**2 - 6*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b) 
*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**5*a*b**3 + 21*sqrt(b)* 
sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + 
 f*x)**3*a**2*b**2 + 27*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x 
)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**3*a*b**3 + 6*sqrt(b)*sqrt(a + b)*at 
an((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**3*b**4 
+ 8*cos(e + f*x)*sin(e + f*x)**4*a**5 + 28*cos(e + f*x)*sin(e + f*x)**4*a* 
*4*b + 20*cos(e + f*x)*sin(e + f*x)**4*a**3*b**2 + 3*cos(e + f*x)*sin(e + 
f*x)**4*a**2*b**3 + 3*cos(e + f*x)*sin(e + f*x)**4*a*b**4 - 10*cos(e + f*x 
)*sin(e + f*x)**2*a**5 - 40*cos(e + f*x)*sin(e + f*x)**2*a**4*b - 50*cos(e 
 + f*x)*sin(e + f*x)**2*a**3*b**2 - 20*cos(e + f*x)*sin(e + f*x)**2*a**2*b 
**3 + 2*cos(e + f*x)*a**5 + 6*cos(e + f*x)*a**4*b + 6*cos(e + f*x)*a**3...