\(\int \frac {\cot ^5(e+f x)}{(a+b \tan ^2(e+f x))^2} \, dx\) [229]

Optimal result

Integrand size = 23, antiderivative size = 161 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {(a+2 b) \cot ^2(e+f x)}{2 a^3 f}-\frac {\cot ^4(e+f x)}{4 a^2 f}+\frac {\log (\cos (e+f x))}{(a-b)^2 f}+\frac {\left (a^2+2 a b+3 b^2\right ) \log (\tan (e+f x))}{a^4 f}+\frac {(4 a-3 b) b^3 \log \left (a+b \tan ^2(e+f x)\right )}{2 a^4 (a-b)^2 f}-\frac {b^3}{2 a^3 (a-b) f \left (a+b \tan ^2(e+f x)\right )} \] Output:

1/2*(a+2*b)*cot(f*x+e)^2/a^3/f-1/4*cot(f*x+e)^4/a^2/f+ln(cos(f*x+e))/(a-b) 
^2/f+(a^2+2*a*b+3*b^2)*ln(tan(f*x+e))/a^4/f+1/2*(4*a-3*b)*b^3*ln(a+b*tan(f 
*x+e)^2)/a^4/(a-b)^2/f-1/2*b^3/a^3/(a-b)/f/(a+b*tan(f*x+e)^2)
 

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.75 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {-\frac {(a+2 b) \cot ^2(e+f x)}{a^3}+\frac {\cot ^4(e+f x)}{2 a^2}-\frac {b^4}{a^4 (a-b) \left (b+a \cot ^2(e+f x)\right )}-\frac {(4 a-3 b) b^3 \log \left (b+a \cot ^2(e+f x)\right )}{a^4 (a-b)^2}-\frac {2 \log (\sin (e+f x))}{(a-b)^2}}{2 f} \] Input:

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

Output:

-1/2*(-(((a + 2*b)*Cot[e + f*x]^2)/a^3) + Cot[e + f*x]^4/(2*a^2) - b^4/(a^ 
4*(a - b)*(b + a*Cot[e + f*x]^2)) - ((4*a - 3*b)*b^3*Log[b + a*Cot[e + f*x 
]^2])/(a^4*(a - b)^2) - (2*Log[Sin[e + f*x]])/(a - b)^2)/f
 

Rubi [A] (warning: unable to verify)

Time = 0.62 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4153, 354, 99, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(((a + 2*b)*Cot[e + f*x])/a^3 - Cot[e + f*x]^2/(2*a^2) + ((a^2 + 2*a*b + 3 
*b^2)*Log[Tan[e + f*x]^2])/a^4 - Log[1 + Tan[e + f*x]^2]/(a - b)^2 + ((4*a 
 - 3*b)*b^3*Log[a + b*Tan[e + f*x]^2])/(a^4*(a - b)^2) - b^3/(a^3*(a - b)* 
(a + b*Tan[e + f*x]^2)))/(2*f)
 

Defintions of rubi rules used

rule 99

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

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4153

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], 
 x]}, Simp[c*(ff/f)   Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f 
f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, 
n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio 
nalQ[n]))
 

Maple [A] (verified)

Time = 1.55 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right )^{2}}-\frac {1}{4 a^{2} \tan \left (f x +e \right )^{4}}-\frac {-2 b -a}{2 a^{3} \tan \left (f x +e \right )^{2}}+\frac {\left (a^{2}+2 a b +3 b^{2}\right ) \ln \left (\tan \left (f x +e \right )\right )}{a^{4}}+\frac {b^{4} \left (\frac {\left (4 a -3 b \right ) \ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{b}-\frac {a \left (a -b \right )}{b \left (a +b \tan \left (f x +e \right )^{2}\right )}\right )}{2 a^{4} \left (a -b \right )^{2}}}{f}\)	\(148\)
default	\(\frac {-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right )^{2}}-\frac {1}{4 a^{2} \tan \left (f x +e \right )^{4}}-\frac {-2 b -a}{2 a^{3} \tan \left (f x +e \right )^{2}}+\frac {\left (a^{2}+2 a b +3 b^{2}\right ) \ln \left (\tan \left (f x +e \right )\right )}{a^{4}}+\frac {b^{4} \left (\frac {\left (4 a -3 b \right ) \ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{b}-\frac {a \left (a -b \right )}{b \left (a +b \tan \left (f x +e \right )^{2}\right )}\right )}{2 a^{4} \left (a -b \right )^{2}}}{f}\)	\(148\)
norman	\(\frac {-\frac {1}{4 a f}+\frac {\left (2 a +3 b \right ) \tan \left (f x +e \right )^{2}}{4 a^{2} f}+\frac {\left (-a^{2} b -a \,b^{2}+3 b^{3}\right ) b \tan \left (f x +e \right )^{6}}{2 a^{4} f \left (a -b \right )}}{\tan \left (f x +e \right )^{4} \left (a +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (a^{2}+2 a b +3 b^{2}\right ) \ln \left (\tan \left (f x +e \right )\right )}{a^{4} f}-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \left (a^{2}-2 a b +b^{2}\right )}+\frac {b^{3} \left (4 a -3 b \right ) \ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{2 a^{4} f \left (a^{2}-2 a b +b^{2}\right )}\)	\(199\)
parallelrisch	\(\frac {8 b^{3} \left (a -\frac {3 b}{4}\right ) \left (a +b \tan \left (f x +e \right )^{2}\right ) \ln \left (a +b \tan \left (f x +e \right )^{2}\right )+\left (-2 \tan \left (f x +e \right )^{2} a^{4} b -2 a^{5}\right ) \ln \left (\sec \left (f x +e \right )^{2}\right )-\left (-4 \left (a -b \right ) \left (a^{2}+2 a b +3 b^{2}\right ) \left (a +b \tan \left (f x +e \right )^{2}\right ) \ln \left (\tan \left (f x +e \right )\right )+\left (2 a^{2} b^{2}+2 a \,b^{3}-6 b^{4}\right ) \tan \left (f x +e \right )^{2}+\cot \left (f x +e \right )^{2} a^{2} \left (a -b \right ) \left (a \cot \left (f x +e \right )^{2}-2 a -3 b \right )\right ) \left (a -b \right )}{4 \left (a -b \right )^{2} a^{4} f \left (a +b \tan \left (f x +e \right )^{2}\right )}\)	\(206\)
risch	\(\frac {i x}{a^{2}-2 a b +b^{2}}+\frac {6 i b^{4} e}{a^{4} f \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 i x}{a^{2}}-\frac {4 i b x}{a^{3}}-\frac {6 i b^{2} e}{a^{4} f}-\frac {4 i b e}{a^{3} f}-\frac {2 i e}{a^{2} f}-\frac {8 i b^{3} x}{a^{3} \left (a^{2}-2 a b +b^{2}\right )}-\frac {6 i b^{2} x}{a^{4}}-\frac {8 i b^{3} e}{a^{3} f \left (a^{2}-2 a b +b^{2}\right )}+\frac {6 i b^{4} x}{a^{4} \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 \left (2 a^{4} {\mathrm e}^{10 i \left (f x +e \right )}-4 a^{3} b \,{\mathrm e}^{10 i \left (f x +e \right )}+4 a \,b^{3} {\mathrm e}^{10 i \left (f x +e \right )}-3 b^{4} {\mathrm e}^{10 i \left (f x +e \right )}+2 a^{4} {\mathrm e}^{8 i \left (f x +e \right )}+2 a^{3} b \,{\mathrm e}^{8 i \left (f x +e \right )}-2 a^{2} b^{2} {\mathrm e}^{8 i \left (f x +e \right )}-10 a \,b^{3} {\mathrm e}^{8 i \left (f x +e \right )}+12 b^{4} {\mathrm e}^{8 i \left (f x +e \right )}-12 a^{3} b \,{\mathrm e}^{6 i \left (f x +e \right )}+12 a^{2} b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+12 a \,b^{3} {\mathrm e}^{6 i \left (f x +e \right )}-18 b^{4} {\mathrm e}^{6 i \left (f x +e \right )}+2 a^{4} {\mathrm e}^{4 i \left (f x +e \right )}+2 a^{3} b \,{\mathrm e}^{4 i \left (f x +e \right )}-2 a^{2} b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-10 a \,b^{3} {\mathrm e}^{4 i \left (f x +e \right )}+12 b^{4} {\mathrm e}^{4 i \left (f x +e \right )}+2 a^{4} {\mathrm e}^{2 i \left (f x +e \right )}-4 a^{3} b \,{\mathrm e}^{2 i \left (f x +e \right )}+4 a \,b^{3} {\mathrm e}^{2 i \left (f x +e \right )}-3 b^{4} {\mathrm e}^{2 i \left (f x +e \right )}\right )}{f \,a^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4} \left (a -b \right )^{2} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}-b \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a -b \right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{a^{2} f}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b}{a^{3} f}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b^{2}}{a^{4} f}+\frac {2 b^{3} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a -b}+1\right )}{a^{3} f \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 b^{4} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a -b}+1\right )}{2 a^{4} f \left (a^{2}-2 a b +b^{2}\right )}\)	\(763\)

Input:

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

Output:

1/f*(-1/2/(a-b)^2*ln(1+tan(f*x+e)^2)-1/4/a^2/tan(f*x+e)^4-1/2*(-2*b-a)/a^3 
/tan(f*x+e)^2+(a^2+2*a*b+3*b^2)/a^4*ln(tan(f*x+e))+1/2*b^4/a^4/(a-b)^2*((4 
*a-3*b)/b*ln(a+b*tan(f*x+e)^2)-a*(a-b)/b/(a+b*tan(f*x+e)^2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 347 vs. \(2 (153) = 306\).

Time = 0.13 (sec) , antiderivative size = 347, normalized size of antiderivative = 2.16 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {{\left (3 \, a^{4} b - 2 \, a^{3} b^{2} - 5 \, a^{2} b^{3} + 6 \, a b^{4}\right )} \tan \left (f x + e\right )^{6} - a^{5} + 2 \, a^{4} b - a^{3} b^{2} + {\left (3 \, a^{5} - 5 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + 6 \, a b^{4}\right )} \tan \left (f x + e\right )^{4} + {\left (2 \, a^{5} - a^{4} b - 4 \, a^{3} b^{2} + 3 \, a^{2} b^{3}\right )} \tan \left (f x + e\right )^{2} + 2 \, {\left ({\left (a^{4} b - 4 \, a b^{4} + 3 \, b^{5}\right )} \tan \left (f x + e\right )^{6} + {\left (a^{5} - 4 \, a^{2} b^{3} + 3 \, a b^{4}\right )} \tan \left (f x + e\right )^{4}\right )} \log \left (\frac {\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left ({\left (4 \, a b^{4} - 3 \, b^{5}\right )} \tan \left (f x + e\right )^{6} + {\left (4 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \tan \left (f x + e\right )^{4}\right )} \log \left (\frac {b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right )}{4 \, {\left ({\left (a^{6} b - 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \tan \left (f x + e\right )^{6} + {\left (a^{7} - 2 \, a^{6} b + a^{5} b^{2}\right )} f \tan \left (f x + e\right )^{4}\right )}} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.47 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {\frac {2 \, {\left (4 \, a b^{3} - 3 \, b^{4}\right )} \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a^{6} - 2 \, a^{5} b + a^{4} b^{2}} + \frac {2 \, {\left (2 \, a^{4} - 4 \, a^{3} b + 4 \, a b^{3} - 3 \, b^{4}\right )} \sin \left (f x + e\right )^{4} + a^{4} - 2 \, a^{3} b + a^{2} b^{2} - {\left (5 \, a^{4} - 7 \, a^{3} b - a^{2} b^{2} + 3 \, a b^{3}\right )} \sin \left (f x + e\right )^{2}}{{\left (a^{6} - 3 \, a^{5} b + 3 \, a^{4} b^{2} - a^{3} b^{3}\right )} \sin \left (f x + e\right )^{6} - {\left (a^{6} - 2 \, a^{5} b + a^{4} b^{2}\right )} \sin \left (f x + e\right )^{4}} + \frac {2 \, {\left (a^{2} + 2 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{4}}}{4 \, f} \] Input:

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

Output:

1/4*(2*(4*a*b^3 - 3*b^4)*log(-(a - b)*sin(f*x + e)^2 + a)/(a^6 - 2*a^5*b + 
 a^4*b^2) + (2*(2*a^4 - 4*a^3*b + 4*a*b^3 - 3*b^4)*sin(f*x + e)^4 + a^4 - 
2*a^3*b + a^2*b^2 - (5*a^4 - 7*a^3*b - a^2*b^2 + 3*a*b^3)*sin(f*x + e)^2)/ 
((a^6 - 3*a^5*b + 3*a^4*b^2 - a^3*b^3)*sin(f*x + e)^6 - (a^6 - 2*a^5*b + a 
^4*b^2)*sin(f*x + e)^4) + 2*(a^2 + 2*a*b + 3*b^2)*log(sin(f*x + e)^2)/a^4) 
/f
 

Giac [A] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.74 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {{\left (4 \, a b^{4} - 3 \, b^{5}\right )} \log \left ({\left | b \tan \left (f x + e\right )^{2} + a \right |}\right )}{2 \, {\left (a^{6} b f - 2 \, a^{5} b^{2} f + a^{4} b^{3} f\right )}} - \frac {\log \left (\tan \left (f x + e\right )^{2} + 1\right )}{2 \, {\left (a^{2} f - 2 \, a b f + b^{2} f\right )}} - \frac {4 \, a b^{4} \tan \left (f x + e\right )^{2} - 3 \, b^{5} \tan \left (f x + e\right )^{2} + 5 \, a^{2} b^{3} - 4 \, a b^{4}}{2 \, {\left (a^{6} f - 2 \, a^{5} b f + a^{4} b^{2} f\right )} {\left (b \tan \left (f x + e\right )^{2} + a\right )}} + \frac {{\left (a^{2} + 2 \, a b + 3 \, b^{2}\right )} \log \left (\tan \left (f x + e\right )^{2}\right )}{2 \, a^{4} f} - \frac {3 \, a^{2} \tan \left (f x + e\right )^{4} + 6 \, a b \tan \left (f x + e\right )^{4} + 9 \, b^{2} \tan \left (f x + e\right )^{4} - 2 \, a^{2} \tan \left (f x + e\right )^{2} - 4 \, a b \tan \left (f x + e\right )^{2} + a^{2}}{4 \, a^{4} f \tan \left (f x + e\right )^{4}} \] Input:

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

Output:

1/2*(4*a*b^4 - 3*b^5)*log(abs(b*tan(f*x + e)^2 + a))/(a^6*b*f - 2*a^5*b^2* 
f + a^4*b^3*f) - 1/2*log(tan(f*x + e)^2 + 1)/(a^2*f - 2*a*b*f + b^2*f) - 1 
/2*(4*a*b^4*tan(f*x + e)^2 - 3*b^5*tan(f*x + e)^2 + 5*a^2*b^3 - 4*a*b^4)/( 
(a^6*f - 2*a^5*b*f + a^4*b^2*f)*(b*tan(f*x + e)^2 + a)) + 1/2*(a^2 + 2*a*b 
 + 3*b^2)*log(tan(f*x + e)^2)/(a^4*f) - 1/4*(3*a^2*tan(f*x + e)^4 + 6*a*b* 
tan(f*x + e)^4 + 9*b^2*tan(f*x + e)^4 - 2*a^2*tan(f*x + e)^2 - 4*a*b*tan(f 
*x + e)^2 + a^2)/(a^4*f*tan(f*x + e)^4)
 

Mupad [B] (verification not implemented)

Time = 8.62 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.19 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,a+3\,b\right )}{4\,a^2}-\frac {1}{4\,a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (a^2\,b+a\,b^2-3\,b^3\right )}{2\,a^3\,\left (a-b\right )}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^6+a\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f\,{\left (a-b\right )}^2}+\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a^2+2\,a\,b+3\,b^2\right )}{a^4\,f}+\frac {\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )\,\left (4\,a\,b^3-3\,b^4\right )}{f\,\left (2\,a^6-4\,a^5\,b+2\,a^4\,b^2\right )} \] Input:

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

Output:

((tan(e + f*x)^2*(2*a + 3*b))/(4*a^2) - 1/(4*a) + (tan(e + f*x)^4*(a*b^2 + 
 a^2*b - 3*b^3))/(2*a^3*(a - b)))/(f*(a*tan(e + f*x)^4 + b*tan(e + f*x)^6) 
) - log(tan(e + f*x)^2 + 1)/(2*f*(a - b)^2) + (log(tan(e + f*x))*(2*a*b + 
a^2 + 3*b^2))/(a^4*f) + (log(a + b*tan(e + f*x)^2)*(4*a*b^3 - 3*b^4))/(f*( 
2*a^6 - 4*a^5*b + 2*a^4*b^2))
 

Reduce [F]

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

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

Output:

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