Integrand size = 23, antiderivative size = 140 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {b^4}{2 a^2 (a+b)^3 f \left (b+a \cos ^2(e+f x)\right )}+\frac {(a+2 b) \csc ^2(e+f x)}{(a+b)^3 f}-\frac {\csc ^4(e+f x)}{4 (a+b)^2 f}+\frac {b^3 (4 a+b) \log \left (b+a \cos ^2(e+f x)\right )}{2 a^2 (a+b)^4 f}+\frac {\left (a^2+4 a b+6 b^2\right ) \log (\sin (e+f x))}{(a+b)^4 f} \] Output:
1/2*b^4/a^2/(a+b)^3/f/(b+a*cos(f*x+e)^2)+(a+2*b)*csc(f*x+e)^2/(a+b)^3/f-1/ 4*csc(f*x+e)^4/(a+b)^2/f+1/2*b^3*(4*a+b)*ln(b+a*cos(f*x+e)^2)/a^2/(a+b)^4/ f+(a^2+4*a*b+6*b^2)*ln(sin(f*x+e))/(a+b)^4/f
Time = 1.38 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.16 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a+2 b+a \cos (2 (e+f x)))^2 \sec ^4(e+f x) \left (4 (a+b) (a+2 b) \csc ^2(e+f x)-(a+b)^2 \csc ^4(e+f x)+4 \left (a^2+4 a b+6 b^2\right ) \log (\sin (e+f x))+\frac {2 b^3 (4 a+b) \log \left (a+b-a \sin ^2(e+f x)\right )}{a^2}+\frac {2 b^4 (a+b)}{a^2 \left (a+b-a \sin ^2(e+f x)\right )}\right )}{16 (a+b)^4 f \left (a+b \sec ^2(e+f x)\right )^2} \] Input:
Integrate[Cot[e + f*x]^5/(a + b*Sec[e + f*x]^2)^2,x]
Output:
((a + 2*b + a*Cos[2*(e + f*x)])^2*Sec[e + f*x]^4*(4*(a + b)*(a + 2*b)*Csc[ e + f*x]^2 - (a + b)^2*Csc[e + f*x]^4 + 4*(a^2 + 4*a*b + 6*b^2)*Log[Sin[e + f*x]] + (2*b^3*(4*a + b)*Log[a + b - a*Sin[e + f*x]^2])/a^2 + (2*b^4*(a + b))/(a^2*(a + b - a*Sin[e + f*x]^2))))/(16*(a + b)^4*f*(a + b*Sec[e + f* x]^2)^2)
Time = 0.39 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4626, 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.
| \(\displaystyle \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (e+f x)^5 \left (a+b \sec (e+f x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4626 |
| \(\displaystyle -\frac {\int \frac {\cos ^9(e+f x)}{\left (1-\cos ^2(e+f x)\right )^3 \left (a \cos ^2(e+f x)+b\right )^2}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle -\frac {\int \frac {\cos ^8(e+f x)}{\left (1-\cos ^2(e+f x)\right )^3 \left (a \cos ^2(e+f x)+b\right )^2}d\cos ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {\int \left (\frac {b^4}{a (a+b)^3 \left (a \cos ^2(e+f x)+b\right )^2}-\frac {(4 a+b) b^3}{a (a+b)^4 \left (a \cos ^2(e+f x)+b\right )}+\frac {-a^2-4 b a-6 b^2}{(a+b)^4 \left (\cos ^2(e+f x)-1\right )}-\frac {2 (a+2 b)}{(a+b)^3 \left (\cos ^2(e+f x)-1\right )^2}-\frac {1}{(a+b)^2 \left (\cos ^2(e+f x)-1\right )^3}\right )d\cos ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-\frac {b^4}{a^2 (a+b)^3 \left (a \cos ^2(e+f x)+b\right )}-\frac {b^3 (4 a+b) \log \left (a \cos ^2(e+f x)+b\right )}{a^2 (a+b)^4}-\frac {\left (a^2+4 a b+6 b^2\right ) \log \left (1-\cos ^2(e+f x)\right )}{(a+b)^4}-\frac {2 (a+2 b)}{(a+b)^3 \left (1-\cos ^2(e+f x)\right )}+\frac {1}{2 (a+b)^2 \left (1-\cos ^2(e+f x)\right )^2}}{2 f}\) |
Input:
Int[Cot[e + f*x]^5/(a + b*Sec[e + f*x]^2)^2,x]
Output:
-1/2*(1/(2*(a + b)^2*(1 - Cos[e + f*x]^2)^2) - (2*(a + 2*b))/((a + b)^3*(1 - Cos[e + f*x]^2)) - b^4/(a^2*(a + b)^3*(b + a*Cos[e + f*x]^2)) - ((a^2 + 4*a*b + 6*b^2)*Log[1 - Cos[e + f*x]^2])/(a + b)^4 - (b^3*(4*a + b)*Log[b + a*Cos[e + f*x]^2])/(a^2*(a + b)^4))/f
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]))
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]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> Module[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-(f *ff^(m + n*p - 1))^(-1) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff* x)^n)^p/x^(m + n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p]
Time = 7.77 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.44
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{16 \left (a +b \right )^{2} \left (-1+\cos \left (f x +e \right )\right )^{2}}-\frac {7 a +15 b}{16 \left (a +b \right )^{3} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+4 a b +6 b^{2}\right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{4}}+\frac {b^{3} \left (\frac {\left (a +b \right ) b}{a^{2} \left (b +a \cos \left (f x +e \right )^{2}\right )}+\frac {\left (4 a +b \right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{a^{2}}\right )}{2 \left (a +b \right )^{4}}-\frac {1}{16 \left (a +b \right )^{2} \left (1+\cos \left (f x +e \right )\right )^{2}}-\frac {-7 a -15 b}{16 \left (a +b \right )^{3} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+4 a b +6 b^{2}\right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{4}}}{f}\) | \(201\) |
| default | \(\frac {-\frac {1}{16 \left (a +b \right )^{2} \left (-1+\cos \left (f x +e \right )\right )^{2}}-\frac {7 a +15 b}{16 \left (a +b \right )^{3} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+4 a b +6 b^{2}\right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{4}}+\frac {b^{3} \left (\frac {\left (a +b \right ) b}{a^{2} \left (b +a \cos \left (f x +e \right )^{2}\right )}+\frac {\left (4 a +b \right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{a^{2}}\right )}{2 \left (a +b \right )^{4}}-\frac {1}{16 \left (a +b \right )^{2} \left (1+\cos \left (f x +e \right )\right )^{2}}-\frac {-7 a -15 b}{16 \left (a +b \right )^{3} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+4 a b +6 b^{2}\right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{4}}}{f}\) | \(201\) |
| risch | \(-\frac {12 i b^{2} e}{f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {2 i b^{4} x}{a^{2} \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {8 i b^{3} x}{a \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {2 i b^{4} e}{a^{2} f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {8 i a b e}{f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {12 i b^{2} x}{a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}}-\frac {8 i a b x}{a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}}+\frac {i x}{a^{2}}-\frac {8 i b^{3} e}{a f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {2 i a^{2} e}{f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {2 i a^{2} x}{a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}}-\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 )}-b^{4} {\mathrm e}^{10 i \left (f x +e \right )}+2 a^{4} {\mathrm e}^{8 i \left (f x +e \right )}+10 a^{3} b \,{\mathrm e}^{8 i \left (f x +e \right )}+16 a^{2} b^{2} {\mathrm e}^{8 i \left (f x +e \right )}+4 b^{4} {\mathrm e}^{8 i \left (f x +e \right )}-12 a^{3} b \,{\mathrm e}^{6 i \left (f x +e \right )}-24 a^{2} b^{2} {\mathrm e}^{6 i \left (f x +e \right )}-6 b^{4} {\mathrm e}^{6 i \left (f x +e \right )}+2 a^{4} {\mathrm e}^{4 i \left (f x +e \right )}+10 a^{3} b \,{\mathrm e}^{4 i \left (f x +e \right )}+16 a^{2} b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+4 b^{4} {\mathrm e}^{4 i \left (f x +e \right )}+2 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{4}+4 a^{3} b \,{\mathrm e}^{2 i \left (f x +e \right )}-b^{4} {\mathrm e}^{2 i \left (f x +e \right )}\right )}{f \left (a +b \right )^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4} 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 {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a^{2}}{f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}+\frac {4 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a b}{f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}+\frac {6 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b^{2}}{f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}+\frac {2 b^{3} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{a f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}+\frac {b^{4} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a^{2} f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}\) | \(999\) |
Input:
int(cot(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/f*(-1/16/(a+b)^2/(-1+cos(f*x+e))^2-1/16*(7*a+15*b)/(a+b)^3/(-1+cos(f*x+e ))+1/2*(a^2+4*a*b+6*b^2)/(a+b)^4*ln(-1+cos(f*x+e))+1/2*b^3/(a+b)^4*((a+b)* b/a^2/(b+a*cos(f*x+e)^2)+(4*a+b)/a^2*ln(b+a*cos(f*x+e)^2))-1/16/(a+b)^2/(1 +cos(f*x+e))^2-1/16*(-7*a-15*b)/(a+b)^3/(1+cos(f*x+e))+1/2*(a^2+4*a*b+6*b^ 2)/(a+b)^4*ln(1+cos(f*x+e)))
Leaf count of result is larger than twice the leaf count of optimal. 557 vs. \(2 (134) = 268\).
Time = 0.63 (sec) , antiderivative size = 557, normalized size of antiderivative = 3.98 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {3 \, a^{4} b + 10 \, a^{3} b^{2} + 7 \, a^{2} b^{3} + 2 \, a b^{4} + 2 \, b^{5} - 2 \, {\left (2 \, a^{5} + 6 \, a^{4} b + 4 \, a^{3} b^{2} - a b^{4} - b^{5}\right )} \cos \left (f x + e\right )^{4} + {\left (3 \, a^{5} + 6 \, a^{4} b - 5 \, a^{3} b^{2} - 8 \, a^{2} b^{3} - 4 \, a b^{4} - 4 \, b^{5}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left ({\left (4 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (f x + e\right )^{6} + 4 \, a b^{4} + b^{5} - {\left (8 \, a^{2} b^{3} - 2 \, a b^{4} - b^{5}\right )} \cos \left (f x + e\right )^{4} + {\left (4 \, a^{2} b^{3} - 7 \, a b^{4} - 2 \, b^{5}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) + 4 \, {\left ({\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2}\right )} \cos \left (f x + e\right )^{6} + a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - {\left (2 \, a^{5} + 7 \, a^{4} b + 8 \, a^{3} b^{2} - 6 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{4} + {\left (a^{5} + 2 \, a^{4} b - 2 \, a^{3} b^{2} - 12 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (f x + e\right )\right )}{4 \, {\left ({\left (a^{7} + 4 \, a^{6} b + 6 \, a^{5} b^{2} + 4 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{6} - {\left (2 \, a^{7} + 7 \, a^{6} b + 8 \, a^{5} b^{2} + 2 \, a^{4} b^{3} - 2 \, a^{3} b^{4} - a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{7} + 2 \, a^{6} b - 2 \, a^{5} b^{2} - 8 \, a^{4} b^{3} - 7 \, a^{3} b^{4} - 2 \, a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{6} b + 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} + 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}} \] Input:
integrate(cot(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")
Output:
1/4*(3*a^4*b + 10*a^3*b^2 + 7*a^2*b^3 + 2*a*b^4 + 2*b^5 - 2*(2*a^5 + 6*a^4 *b + 4*a^3*b^2 - a*b^4 - b^5)*cos(f*x + e)^4 + (3*a^5 + 6*a^4*b - 5*a^3*b^ 2 - 8*a^2*b^3 - 4*a*b^4 - 4*b^5)*cos(f*x + e)^2 + 2*((4*a^2*b^3 + a*b^4)*c os(f*x + e)^6 + 4*a*b^4 + b^5 - (8*a^2*b^3 - 2*a*b^4 - b^5)*cos(f*x + e)^4 + (4*a^2*b^3 - 7*a*b^4 - 2*b^5)*cos(f*x + e)^2)*log(a*cos(f*x + e)^2 + b) + 4*((a^5 + 4*a^4*b + 6*a^3*b^2)*cos(f*x + e)^6 + a^4*b + 4*a^3*b^2 + 6*a ^2*b^3 - (2*a^5 + 7*a^4*b + 8*a^3*b^2 - 6*a^2*b^3)*cos(f*x + e)^4 + (a^5 + 2*a^4*b - 2*a^3*b^2 - 12*a^2*b^3)*cos(f*x + e)^2)*log(1/2*sin(f*x + e)))/ ((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^6 - (2*a ^7 + 7*a^6*b + 8*a^5*b^2 + 2*a^4*b^3 - 2*a^3*b^4 - a^2*b^5)*f*cos(f*x + e) ^4 + (a^7 + 2*a^6*b - 2*a^5*b^2 - 8*a^4*b^3 - 7*a^3*b^4 - 2*a^2*b^5)*f*cos (f*x + e)^2 + (a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5)*f)
\[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\int \frac {\cot ^{5}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \] Input:
integrate(cot(f*x+e)**5/(a+b*sec(f*x+e)**2)**2,x)
Output:
Integral(cot(e + f*x)**5/(a + b*sec(e + f*x)**2)**2, x)
Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (134) = 268\).
Time = 0.04 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.99 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {2 \, {\left (4 \, a b^{3} + b^{4}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}} + \frac {2 \, {\left (a^{2} + 4 \, a b + 6 \, b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac {2 \, {\left (2 \, a^{4} + 4 \, a^{3} b - b^{4}\right )} \sin \left (f x + e\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2} - {\left (5 \, a^{4} + 13 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \sin \left (f x + e\right )^{2}}{{\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} \sin \left (f x + e\right )^{6} - {\left (a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} \sin \left (f x + e\right )^{4}}}{4 \, f} \] Input:
integrate(cot(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")
Output:
1/4*(2*(4*a*b^3 + b^4)*log(a*sin(f*x + e)^2 - a - b)/(a^6 + 4*a^5*b + 6*a^ 4*b^2 + 4*a^3*b^3 + a^2*b^4) + 2*(a^2 + 4*a*b + 6*b^2)*log(sin(f*x + e)^2) /(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) + (2*(2*a^4 + 4*a^3*b - b^4)* sin(f*x + e)^4 + a^4 + 2*a^3*b + a^2*b^2 - (5*a^4 + 13*a^3*b + 8*a^2*b^2)* 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 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*sin(f*x + e)^4))/f
Leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (134) = 268\).
Time = 0.18 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.20 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {{\left (4 \, a b^{3} + b^{4}\right )} \log \left ({\left | a \cos \left (f x + e\right )^{2} + b \right |}\right )}{2 \, {\left (a^{6} f + 4 \, a^{5} b f + 6 \, a^{4} b^{2} f + 4 \, a^{3} b^{3} f + a^{2} b^{4} f\right )}} + \frac {{\left (a^{2} + 4 \, a b + 6 \, b^{2}\right )} \log \left ({\left | -\cos \left (f x + e\right )^{2} + 1 \right |}\right )}{2 \, {\left (a^{4} f + 4 \, a^{3} b f + 6 \, a^{2} b^{2} f + 4 \, a b^{3} f + b^{4} f\right )}} - \frac {\frac {2 \, {\left (2 \, a^{5} + 6 \, a^{4} b + 4 \, a^{3} b^{2} - a b^{4} - b^{5}\right )} \cos \left (f x + e\right )^{4}}{a} - \frac {{\left (3 \, a^{5} + 6 \, a^{4} b - 5 \, a^{3} b^{2} - 8 \, a^{2} b^{3} - 4 \, a b^{4} - 4 \, b^{5}\right )} \cos \left (f x + e\right )^{2}}{a} - \frac {3 \, a^{4} b + 10 \, a^{3} b^{2} + 7 \, a^{2} b^{3} + 2 \, a b^{4} + 2 \, b^{5}}{a}}{4 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )} {\left (\cos \left (f x + e\right )^{2} - 1\right )}^{2} {\left (a + b\right )}^{4} a f} \] Input:
integrate(cot(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")
Output:
1/2*(4*a*b^3 + b^4)*log(abs(a*cos(f*x + e)^2 + b))/(a^6*f + 4*a^5*b*f + 6* a^4*b^2*f + 4*a^3*b^3*f + a^2*b^4*f) + 1/2*(a^2 + 4*a*b + 6*b^2)*log(abs(- cos(f*x + e)^2 + 1))/(a^4*f + 4*a^3*b*f + 6*a^2*b^2*f + 4*a*b^3*f + b^4*f) - 1/4*(2*(2*a^5 + 6*a^4*b + 4*a^3*b^2 - a*b^4 - b^5)*cos(f*x + e)^4/a - ( 3*a^5 + 6*a^4*b - 5*a^3*b^2 - 8*a^2*b^3 - 4*a*b^4 - 4*b^5)*cos(f*x + e)^2/ a - (3*a^4*b + 10*a^3*b^2 + 7*a^2*b^3 + 2*a*b^4 + 2*b^5)/a)/((a*cos(f*x + e)^2 + b)*(cos(f*x + e)^2 - 1)^2*(a + b)^4*a*f)
Time = 16.26 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.47 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,a+5\,b\right )}{4\,{\left (a+b\right )}^2}-\frac {1}{4\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (a^2\,b+3\,a\,b^2-b^3\right )}{2\,a\,{\left (a+b\right )}^3}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^6+\left (a+b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a^2\,f}+\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a^2+4\,a\,b+6\,b^2\right )}{f\,\left (a^4+4\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3+b^4\right )}+\frac {b^3\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )\,\left (4\,a+b\right )}{2\,a^2\,f\,{\left (a+b\right )}^4} \] Input:
int(cot(e + f*x)^5/(a + b/cos(e + f*x)^2)^2,x)
Output:
((tan(e + f*x)^2*(2*a + 5*b))/(4*(a + b)^2) - 1/(4*(a + b)) + (tan(e + f*x )^4*(3*a*b^2 + a^2*b - b^3))/(2*a*(a + b)^3))/(f*(tan(e + f*x)^4*(a + b) + b*tan(e + f*x)^6)) - log(tan(e + f*x)^2 + 1)/(2*a^2*f) + (log(tan(e + f*x ))*(4*a*b + a^2 + 6*b^2))/(f*(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a^2*b^2)) + (b^3*log(a + b + b*tan(e + f*x)^2)*(4*a + b))/(2*a^2*f*(a + b)^4)
Time = 0.42 (sec) , antiderivative size = 1312, normalized size of antiderivative = 9.37 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:
int(cot(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x)
Output:
( - 64*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**6*a**5 - 256*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**6*a**4*b - 384*log(tan((e + f*x)/2)**2 + 1 )*sin(e + f*x)**6*a**3*b**2 - 256*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x )**6*a**2*b**3 - 64*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**6*a*b**4 + 64*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4*a**5 + 320*log(tan((e + f* x)/2)**2 + 1)*sin(e + f*x)**4*a**4*b + 640*log(tan((e + f*x)/2)**2 + 1)*si n(e + f*x)**4*a**3*b**2 + 640*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4 *a**2*b**3 + 320*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4*a*b**4 + 64* log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4*b**5 + 128*log(sqrt(a + b)*ta n((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x) **6*a**2*b**3 + 32*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*s qrt(a)*tan((e + f*x)/2))*sin(e + f*x)**6*a*b**4 - 128*log(sqrt(a + b)*tan( (e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)** 4*a**2*b**3 - 160*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sq rt(a)*tan((e + f*x)/2))*sin(e + f*x)**4*a*b**4 - 32*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**4* b**5 + 128*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) + 2*sqrt(a)*t an((e + f*x)/2))*sin(e + f*x)**6*a**2*b**3 + 32*log(sqrt(a + b)*tan((e + f *x)/2)**2 + sqrt(a + b) + 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**6*a*b* *4 - 128*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) + 2*sqrt(a)*...