Integrand size = 23, antiderivative size = 192 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {b^5}{4 a^3 (a+b)^3 f \left (b+a \cos ^2(e+f x)\right )^2}+\frac {b^4 (5 a+2 b)}{2 a^3 (a+b)^4 f \left (b+a \cos ^2(e+f x)\right )}+\frac {(2 a+5 b) \csc ^2(e+f x)}{2 (a+b)^4 f}-\frac {\csc ^4(e+f x)}{4 (a+b)^3 f}+\frac {b^3 \left (10 a^2+5 a b+b^2\right ) \log \left (b+a \cos ^2(e+f x)\right )}{2 a^3 (a+b)^5 f}+\frac {\left (a^2+5 a b+10 b^2\right ) \log (\sin (e+f x))}{(a+b)^5 f} \]
-1/4*b^5/a^3/(a+b)^3/f/(b+a*cos(f*x+e)^2)^2+1/2*b^4*(5*a+2*b)/a^3/(a+b)^4/ f/(b+a*cos(f*x+e)^2)+1/2*(2*a+5*b)*csc(f*x+e)^2/(a+b)^4/f-1/4*csc(f*x+e)^4 /(a+b)^3/f+1/2*b^3*(10*a^2+5*a*b+b^2)*ln(b+a*cos(f*x+e)^2)/a^3/(a+b)^5/f+( a^2+5*a*b+10*b^2)*ln(sin(f*x+e))/(a+b)^5/f
Time = 5.53 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.08 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(a+2 b+a \cos (2 (e+f x)))^3 \sec ^6(e+f x) \left (2 (a+b) (2 a+5 b) \csc ^2(e+f x)-(a+b)^2 \csc ^4(e+f x)+4 \left (a^2+5 a b+10 b^2\right ) \log (\sin (e+f x))+\frac {2 b^3 \left (10 a^2+5 a b+b^2\right ) \log \left (a+b-a \sin ^2(e+f x)\right )}{a^3}-\frac {b^5 (a+b)^2}{a^3 \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {2 b^4 (a+b) (5 a+2 b)}{a^3 \left (a+b-a \sin ^2(e+f x)\right )}\right )}{32 (a+b)^5 f \left (a+b \sec ^2(e+f x)\right )^3} \]
((a + 2*b + a*Cos[2*(e + f*x)])^3*Sec[e + f*x]^6*(2*(a + b)*(2*a + 5*b)*Cs c[e + f*x]^2 - (a + b)^2*Csc[e + f*x]^4 + 4*(a^2 + 5*a*b + 10*b^2)*Log[Sin [e + f*x]] + (2*b^3*(10*a^2 + 5*a*b + b^2)*Log[a + b - a*Sin[e + f*x]^2])/ a^3 - (b^5*(a + b)^2)/(a^3*(a + b - a*Sin[e + f*x]^2)^2) + (2*b^4*(a + b)* (5*a + 2*b))/(a^3*(a + b - a*Sin[e + f*x]^2))))/(32*(a + b)^5*f*(a + b*Sec [e + f*x]^2)^3)
Time = 0.43 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.01, 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.
| \(\displaystyle \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (e+f x)^5 \left (a+b \sec (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4626 |
| \(\displaystyle -\frac {\int \frac {\cos ^{11}(e+f x)}{\left (1-\cos ^2(e+f x)\right )^3 \left (a \cos ^2(e+f x)+b\right )^3}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle -\frac {\int \frac {\cos ^{10}(e+f x)}{\left (1-\cos ^2(e+f x)\right )^3 \left (a \cos ^2(e+f x)+b\right )^3}d\cos ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {\int \left (-\frac {b^5}{a^2 (a+b)^3 \left (a \cos ^2(e+f x)+b\right )^3}+\frac {(5 a+2 b) b^4}{a^2 (a+b)^4 \left (a \cos ^2(e+f x)+b\right )^2}-\frac {\left (10 a^2+5 b a+b^2\right ) b^3}{a^2 (a+b)^5 \left (a \cos ^2(e+f x)+b\right )}+\frac {-a^2-5 b a-10 b^2}{(a+b)^5 \left (\cos ^2(e+f x)-1\right )}+\frac {-2 a-5 b}{(a+b)^4 \left (\cos ^2(e+f x)-1\right )^2}-\frac {1}{(a+b)^3 \left (\cos ^2(e+f x)-1\right )^3}\right )d\cos ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {b^5}{2 a^3 (a+b)^3 \left (a \cos ^2(e+f x)+b\right )^2}-\frac {b^4 (5 a+2 b)}{a^3 (a+b)^4 \left (a \cos ^2(e+f x)+b\right )}-\frac {\left (a^2+5 a b+10 b^2\right ) \log \left (1-\cos ^2(e+f x)\right )}{(a+b)^5}-\frac {b^3 \left (10 a^2+5 a b+b^2\right ) \log \left (a \cos ^2(e+f x)+b\right )}{a^3 (a+b)^5}-\frac {2 a+5 b}{(a+b)^4 \left (1-\cos ^2(e+f x)\right )}+\frac {1}{2 (a+b)^3 \left (1-\cos ^2(e+f x)\right )^2}}{2 f}\) |
-1/2*(1/(2*(a + b)^3*(1 - Cos[e + f*x]^2)^2) - (2*a + 5*b)/((a + b)^4*(1 - Cos[e + f*x]^2)) + b^5/(2*a^3*(a + b)^3*(b + a*Cos[e + f*x]^2)^2) - (b^4* (5*a + 2*b))/(a^3*(a + b)^4*(b + a*Cos[e + f*x]^2)) - ((a^2 + 5*a*b + 10*b ^2)*Log[1 - Cos[e + f*x]^2])/(a + b)^5 - (b^3*(10*a^2 + 5*a*b + b^2)*Log[b + a*Cos[e + f*x]^2])/(a^3*(a + b)^5))/f
3.4.68.3.1 Defintions of rubi rules used
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 = 53.50 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.32
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{16 \left (a +b \right )^{3} \left (-1+\cos \left (f x +e \right )\right )^{2}}-\frac {7 a +19 b}{16 \left (a +b \right )^{4} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+5 a b +10 b^{2}\right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{5}}-\frac {1}{16 \left (a +b \right )^{3} \left (1+\cos \left (f x +e \right )\right )^{2}}-\frac {-7 a -19 b}{16 \left (a +b \right )^{4} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+5 a b +10 b^{2}\right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{5}}+\frac {b^{3} \left (\frac {\left (10 a^{2}+5 a b +b^{2}\right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{a^{3}}-\frac {b^{2} \left (a^{2}+2 a b +b^{2}\right )}{2 a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}+\frac {b \left (5 a^{2}+7 a b +2 b^{2}\right )}{a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )}\right )}{2 \left (a +b \right )^{5}}}{f}\) | \(254\) |
| default | \(\frac {-\frac {1}{16 \left (a +b \right )^{3} \left (-1+\cos \left (f x +e \right )\right )^{2}}-\frac {7 a +19 b}{16 \left (a +b \right )^{4} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+5 a b +10 b^{2}\right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{5}}-\frac {1}{16 \left (a +b \right )^{3} \left (1+\cos \left (f x +e \right )\right )^{2}}-\frac {-7 a -19 b}{16 \left (a +b \right )^{4} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+5 a b +10 b^{2}\right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{5}}+\frac {b^{3} \left (\frac {\left (10 a^{2}+5 a b +b^{2}\right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{a^{3}}-\frac {b^{2} \left (a^{2}+2 a b +b^{2}\right )}{2 a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}+\frac {b \left (5 a^{2}+7 a b +2 b^{2}\right )}{a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )}\right )}{2 \left (a +b \right )^{5}}}{f}\) | \(254\) |
| risch | \(\text {Expression too large to display}\) | \(1676\) |
1/f*(-1/16/(a+b)^3/(-1+cos(f*x+e))^2-1/16*(7*a+19*b)/(a+b)^4/(-1+cos(f*x+e ))+1/2*(a^2+5*a*b+10*b^2)/(a+b)^5*ln(-1+cos(f*x+e))-1/16/(a+b)^3/(1+cos(f* x+e))^2-1/16*(-7*a-19*b)/(a+b)^4/(1+cos(f*x+e))+1/2*(a^2+5*a*b+10*b^2)/(a+ b)^5*ln(1+cos(f*x+e))+1/2*b^3/(a+b)^5*((10*a^2+5*a*b+b^2)/a^3*ln(b+a*cos(f *x+e)^2)-1/2*b^2*(a^2+2*a*b+b^2)/a^3/(b+a*cos(f*x+e)^2)^2+1/a^3*b*(5*a^2+7 *a*b+2*b^2)/(b+a*cos(f*x+e)^2)))
Leaf count of result is larger than twice the leaf count of optimal. 859 vs. \(2 (182) = 364\).
Time = 1.68 (sec) , antiderivative size = 859, normalized size of antiderivative = 4.47 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {3 \, a^{5} b^{2} + 12 \, a^{4} b^{3} + 9 \, a^{3} b^{4} + 9 \, a^{2} b^{5} + 12 \, a b^{6} + 3 \, b^{7} - 2 \, {\left (2 \, a^{7} + 7 \, a^{6} b + 5 \, a^{5} b^{2} - 5 \, a^{3} b^{4} - 7 \, a^{2} b^{5} - 2 \, a b^{6}\right )} \cos \left (f x + e\right )^{6} + {\left (3 \, a^{7} + 4 \, a^{6} b - 19 \, a^{5} b^{2} - 20 \, a^{4} b^{3} - 20 \, a^{3} b^{4} - 19 \, a^{2} b^{5} + 4 \, a b^{6} + 3 \, b^{7}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{6} b + 10 \, a^{5} b^{2} + 2 \, a^{4} b^{3} - 2 \, a^{2} b^{5} - 10 \, a b^{6} - 3 \, b^{7}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left ({\left (10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} + a^{2} b^{5}\right )} \cos \left (f x + e\right )^{8} + 10 \, a^{2} b^{5} + 5 \, a b^{6} + b^{7} - 2 \, {\left (10 \, a^{4} b^{3} - 5 \, a^{3} b^{4} - 4 \, a^{2} b^{5} - a b^{6}\right )} \cos \left (f x + e\right )^{6} + {\left (10 \, a^{4} b^{3} - 35 \, a^{3} b^{4} - 9 \, a^{2} b^{5} + a b^{6} + b^{7}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (10 \, a^{3} b^{4} - 5 \, a^{2} b^{5} - 4 \, a b^{6} - b^{7}\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^{7} + 5 \, a^{6} b + 10 \, a^{5} b^{2}\right )} \cos \left (f x + e\right )^{8} + a^{5} b^{2} + 5 \, a^{4} b^{3} + 10 \, a^{3} b^{4} - 2 \, {\left (a^{7} + 4 \, a^{6} b + 5 \, a^{5} b^{2} - 10 \, a^{4} b^{3}\right )} \cos \left (f x + e\right )^{6} + {\left (a^{7} + a^{6} b - 9 \, a^{5} b^{2} - 35 \, a^{4} b^{3} + 10 \, a^{3} b^{4}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b + 4 \, a^{5} b^{2} + 5 \, a^{4} b^{3} - 10 \, a^{3} b^{4}\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^{10} + 5 \, a^{9} b + 10 \, a^{8} b^{2} + 10 \, a^{7} b^{3} + 5 \, a^{6} b^{4} + a^{5} b^{5}\right )} f \cos \left (f x + e\right )^{8} - 2 \, {\left (a^{10} + 4 \, a^{9} b + 5 \, a^{8} b^{2} - 5 \, a^{6} b^{4} - 4 \, a^{5} b^{5} - a^{4} b^{6}\right )} f \cos \left (f x + e\right )^{6} + {\left (a^{10} + a^{9} b - 9 \, a^{8} b^{2} - 25 \, a^{7} b^{3} - 25 \, a^{6} b^{4} - 9 \, a^{5} b^{5} + a^{4} b^{6} + a^{3} b^{7}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{9} b + 4 \, a^{8} b^{2} + 5 \, a^{7} b^{3} - 5 \, a^{5} b^{5} - 4 \, a^{4} b^{6} - a^{3} b^{7}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{8} b^{2} + 5 \, a^{7} b^{3} + 10 \, a^{6} b^{4} + 10 \, a^{5} b^{5} + 5 \, a^{4} b^{6} + a^{3} b^{7}\right )} f\right )}} \]
1/4*(3*a^5*b^2 + 12*a^4*b^3 + 9*a^3*b^4 + 9*a^2*b^5 + 12*a*b^6 + 3*b^7 - 2 *(2*a^7 + 7*a^6*b + 5*a^5*b^2 - 5*a^3*b^4 - 7*a^2*b^5 - 2*a*b^6)*cos(f*x + e)^6 + (3*a^7 + 4*a^6*b - 19*a^5*b^2 - 20*a^4*b^3 - 20*a^3*b^4 - 19*a^2*b ^5 + 4*a*b^6 + 3*b^7)*cos(f*x + e)^4 + 2*(3*a^6*b + 10*a^5*b^2 + 2*a^4*b^3 - 2*a^2*b^5 - 10*a*b^6 - 3*b^7)*cos(f*x + e)^2 + 2*((10*a^4*b^3 + 5*a^3*b ^4 + a^2*b^5)*cos(f*x + e)^8 + 10*a^2*b^5 + 5*a*b^6 + b^7 - 2*(10*a^4*b^3 - 5*a^3*b^4 - 4*a^2*b^5 - a*b^6)*cos(f*x + e)^6 + (10*a^4*b^3 - 35*a^3*b^4 - 9*a^2*b^5 + a*b^6 + b^7)*cos(f*x + e)^4 + 2*(10*a^3*b^4 - 5*a^2*b^5 - 4 *a*b^6 - b^7)*cos(f*x + e)^2)*log(a*cos(f*x + e)^2 + b) + 4*((a^7 + 5*a^6* b + 10*a^5*b^2)*cos(f*x + e)^8 + a^5*b^2 + 5*a^4*b^3 + 10*a^3*b^4 - 2*(a^7 + 4*a^6*b + 5*a^5*b^2 - 10*a^4*b^3)*cos(f*x + e)^6 + (a^7 + a^6*b - 9*a^5 *b^2 - 35*a^4*b^3 + 10*a^3*b^4)*cos(f*x + e)^4 + 2*(a^6*b + 4*a^5*b^2 + 5* a^4*b^3 - 10*a^3*b^4)*cos(f*x + e)^2)*log(1/2*sin(f*x + e)))/((a^10 + 5*a^ 9*b + 10*a^8*b^2 + 10*a^7*b^3 + 5*a^6*b^4 + a^5*b^5)*f*cos(f*x + e)^8 - 2* (a^10 + 4*a^9*b + 5*a^8*b^2 - 5*a^6*b^4 - 4*a^5*b^5 - a^4*b^6)*f*cos(f*x + e)^6 + (a^10 + a^9*b - 9*a^8*b^2 - 25*a^7*b^3 - 25*a^6*b^4 - 9*a^5*b^5 + a^4*b^6 + a^3*b^7)*f*cos(f*x + e)^4 + 2*(a^9*b + 4*a^8*b^2 + 5*a^7*b^3 - 5 *a^5*b^5 - 4*a^4*b^6 - a^3*b^7)*f*cos(f*x + e)^2 + (a^8*b^2 + 5*a^7*b^3 + 10*a^6*b^4 + 10*a^5*b^5 + 5*a^4*b^6 + a^3*b^7)*f)
Timed out. \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 454 vs. \(2 (182) = 364\).
Time = 0.23 (sec) , antiderivative size = 454, normalized size of antiderivative = 2.36 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {2 \, {\left (10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{8} + 5 \, a^{7} b + 10 \, a^{6} b^{2} + 10 \, a^{5} b^{3} + 5 \, a^{4} b^{4} + a^{3} b^{5}} + \frac {2 \, {\left (a^{2} + 5 \, a b + 10 \, b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}} + \frac {2 \, {\left (2 \, a^{6} + 5 \, a^{5} b - 5 \, a^{2} b^{4} - 2 \, a b^{5}\right )} \sin \left (f x + e\right )^{6} - a^{6} - 3 \, a^{5} b - 3 \, a^{4} b^{2} - a^{3} b^{3} - {\left (9 \, a^{6} + 29 \, a^{5} b + 20 \, a^{4} b^{2} - 10 \, a^{2} b^{4} - 13 \, a b^{5} - 3 \, b^{6}\right )} \sin \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{6} + 11 \, a^{5} b + 13 \, a^{4} b^{2} + 5 \, a^{3} b^{3}\right )} \sin \left (f x + e\right )^{2}}{{\left (a^{9} + 4 \, a^{8} b + 6 \, a^{7} b^{2} + 4 \, a^{6} b^{3} + a^{5} b^{4}\right )} \sin \left (f x + e\right )^{8} - 2 \, {\left (a^{9} + 5 \, a^{8} b + 10 \, a^{7} b^{2} + 10 \, a^{6} b^{3} + 5 \, a^{5} b^{4} + a^{4} b^{5}\right )} \sin \left (f x + e\right )^{6} + {\left (a^{9} + 6 \, a^{8} b + 15 \, a^{7} b^{2} + 20 \, a^{6} b^{3} + 15 \, a^{5} b^{4} + 6 \, a^{4} b^{5} + a^{3} b^{6}\right )} \sin \left (f x + e\right )^{4}}}{4 \, f} \]
1/4*(2*(10*a^2*b^3 + 5*a*b^4 + b^5)*log(a*sin(f*x + e)^2 - a - b)/(a^8 + 5 *a^7*b + 10*a^6*b^2 + 10*a^5*b^3 + 5*a^4*b^4 + a^3*b^5) + 2*(a^2 + 5*a*b + 10*b^2)*log(sin(f*x + e)^2)/(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5* a*b^4 + b^5) + (2*(2*a^6 + 5*a^5*b - 5*a^2*b^4 - 2*a*b^5)*sin(f*x + e)^6 - a^6 - 3*a^5*b - 3*a^4*b^2 - a^3*b^3 - (9*a^6 + 29*a^5*b + 20*a^4*b^2 - 10 *a^2*b^4 - 13*a*b^5 - 3*b^6)*sin(f*x + e)^4 + 2*(3*a^6 + 11*a^5*b + 13*a^4 *b^2 + 5*a^3*b^3)*sin(f*x + e)^2)/((a^9 + 4*a^8*b + 6*a^7*b^2 + 4*a^6*b^3 + a^5*b^4)*sin(f*x + e)^8 - 2*(a^9 + 5*a^8*b + 10*a^7*b^2 + 10*a^6*b^3 + 5 *a^5*b^4 + a^4*b^5)*sin(f*x + e)^6 + (a^9 + 6*a^8*b + 15*a^7*b^2 + 20*a^6* b^3 + 15*a^5*b^4 + 6*a^4*b^5 + a^3*b^6)*sin(f*x + e)^4))/f
Leaf count of result is larger than twice the leaf count of optimal. 1912 vs. \(2 (182) = 364\).
Time = 0.53 (sec) , antiderivative size = 1912, normalized size of antiderivative = 9.96 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]
1/64*(32*(10*a^2*b^3 + 5*a*b^4 + b^5)*log(a + b + 2*a*(cos(f*x + e) - 1)/( cos(f*x + e) + 1) - 2*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/(a^8 + 5*a^7*b + 10*a^6*b^2 + 10*a^5*b^3 + 5*a^4*b^4 + a^3*b^5) + 3 2*(a^2 + 5*a*b + 10*b^2)*log(abs(-cos(f*x + e) + 1)/abs(cos(f*x + e) + 1)) /(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5) - (12*a^3*(cos( f*x + e) - 1)/(cos(f*x + e) + 1) + 60*a^2*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 84*a*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 36*b^3*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a^3*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1 )^2 + 3*a^2*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 3*a*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b^3*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/(a^6 + 6*a^5*b + 15*a^4*b^2 + 20*a^3*b^3 + 15*a^2*b^4 + 6*a*b^5 + b^6) - (a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4 + 16*a^7*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 80*a^6*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 144*a^5*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 112*a^4*b^3*(co s(f*x + e) - 1)/(cos(f*x + e) + 1) + 32*a^3*b^4*(cos(f*x + e) - 1)/(cos(f* x + e) + 1) + 70*a^7*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 312*a^6*b *(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 436*a^5*b^2*(cos(f*x + e) - 1 )^2/(cos(f*x + e) + 1)^2 + 536*a^4*b^3*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 822*a^3*b^4*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 672*a^...
Time = 22.49 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.70 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a^2+5\,a\,b+10\,b^2\right )}{f\,\left (a^5+5\,a^4\,b+10\,a^3\,b^2+10\,a^2\,b^3+5\,a\,b^4+b^5\right )}-\frac {\frac {1}{4\,\left (a+b\right )}-\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (a+3\,b\right )}{2\,{\left (a+b\right )}^2}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (-4\,a^3\,b-15\,a^2\,b^2+9\,a\,b^3+2\,b^4\right )}{4\,a^2\,\left (a+b\right )\,\left (a^2+2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (-a^3\,b^2-4\,a^2\,b^3+4\,a\,b^4+b^5\right )}{2\,a^2\,{\left (a+b\right )}^2\,\left (a^2+2\,a\,b+b^2\right )}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^4\,\left (a^2+2\,a\,b+b^2\right )+{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^8\right )}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a^3\,f}+\frac {b^3\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )\,\left (10\,a^2+5\,a\,b+b^2\right )}{2\,a^3\,f\,{\left (a+b\right )}^5} \]
(log(tan(e + f*x))*(5*a*b + a^2 + 10*b^2))/(f*(5*a*b^4 + 5*a^4*b + a^5 + b ^5 + 10*a^2*b^3 + 10*a^3*b^2)) - (1/(4*(a + b)) - (tan(e + f*x)^2*(a + 3*b ))/(2*(a + b)^2) + (tan(e + f*x)^4*(9*a*b^3 - 4*a^3*b + 2*b^4 - 15*a^2*b^2 ))/(4*a^2*(a + b)*(2*a*b + a^2 + b^2)) + (tan(e + f*x)^6*(4*a*b^4 + b^5 - 4*a^2*b^3 - a^3*b^2))/(2*a^2*(a + b)^2*(2*a*b + a^2 + b^2)))/(f*(tan(e + f *x)^4*(2*a*b + a^2 + b^2) + tan(e + f*x)^6*(2*a*b + 2*b^2) + b^2*tan(e + f *x)^8)) - log(tan(e + f*x)^2 + 1)/(2*a^3*f) + (b^3*log(a + b + b*tan(e + f *x)^2)*(5*a*b + 10*a^2 + b^2))/(2*a^3*f*(a + b)^5)