Integrand size = 21, antiderivative size = 148 \[ \int \frac {\cot (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {\log (\cos (e+f x))}{(a-b)^3 f}+\frac {\log (\tan (e+f x))}{a^3 f}+\frac {b \left (3 a^2-3 a b+b^2\right ) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^3 (a-b)^3 f}-\frac {b}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {(2 a-b) b}{2 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )} \]
ln(cos(f*x+e))/(a-b)^3/f+ln(tan(f*x+e))/a^3/f+1/2*b*(3*a^2-3*a*b+b^2)*ln(a +b*tan(f*x+e)^2)/a^3/(a-b)^3/f-1/4*b/a/(a-b)/f/(a+b*tan(f*x+e)^2)^2-1/2*(2 *a-b)*b/a^2/(a-b)^2/f/(a+b*tan(f*x+e)^2)
Time = 1.84 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.85 \[ \int \frac {\cot (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {\frac {4 \log (\cos (e+f x))}{(a-b)^3}+\frac {4 \log (\tan (e+f x))+\frac {b \left (2 \left (3 a^2-3 a b+b^2\right ) \log \left (a+b \tan ^2(e+f x)\right )-\frac {a (a-b) \left (a (5 a-3 b)+2 (2 a-b) b \tan ^2(e+f x)\right )}{\left (a+b \tan ^2(e+f x)\right )^2}\right )}{(a-b)^3}}{a^3}}{4 f} \]
((4*Log[Cos[e + f*x]])/(a - b)^3 + (4*Log[Tan[e + f*x]] + (b*(2*(3*a^2 - 3 *a*b + b^2)*Log[a + b*Tan[e + f*x]^2] - (a*(a - b)*(a*(5*a - 3*b) + 2*(2*a - b)*b*Tan[e + f*x]^2))/(a + b*Tan[e + f*x]^2)^2))/(a - b)^3)/a^3)/(4*f)
Time = 0.35 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4153, 354, 93, 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.
| \(\displaystyle \int \frac {\cot (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (e+f x) \left (a+b \tan (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle \frac {\int \frac {\cot (e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^3}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {\int \frac {\cot (e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^3}d\tan ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 93 |
| \(\displaystyle \frac {\int \left (\frac {\left (3 a^2-3 b a+b^2\right ) b^2}{a^3 (a-b)^3 \left (b \tan ^2(e+f x)+a\right )}+\frac {(2 a-b) b^2}{a^2 (a-b)^2 \left (b \tan ^2(e+f x)+a\right )^2}+\frac {b^2}{a (a-b) \left (b \tan ^2(e+f x)+a\right )^3}+\frac {\cot (e+f x)}{a^3}-\frac {1}{(a-b)^3 \left (\tan ^2(e+f x)+1\right )}\right )d\tan ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\log \left (\tan ^2(e+f x)\right )}{a^3}-\frac {b (2 a-b)}{a^2 (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}+\frac {b \left (3 a^2-3 a b+b^2\right ) \log \left (a+b \tan ^2(e+f x)\right )}{a^3 (a-b)^3}-\frac {b}{2 a (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\log \left (\tan ^2(e+f x)+1\right )}{(a-b)^3}}{2 f}\) |
(Log[Tan[e + f*x]^2]/a^3 - Log[1 + Tan[e + f*x]^2]/(a - b)^3 + (b*(3*a^2 - 3*a*b + b^2)*Log[a + b*Tan[e + f*x]^2])/(a^3*(a - b)^3) - b/(2*a*(a - b)* (a + b*Tan[e + f*x]^2)^2) - ((2*a - b)*b)/(a^2*(a - b)^2*(a + b*Tan[e + f* x]^2)))/(2*f)
3.3.40.3.1 Defintions of rubi rules used
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Int[ExpandIntegrand[(e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; Fre eQ[{a, b, c, d, e, f}, x] && IntegerQ[p]
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[((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]))
Time = 0.48 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.01
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{2} \left (-\frac {a \left (2 a^{2}-3 a b +b^{2}\right )}{b \left (a +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (3 a^{2}-3 a b +b^{2}\right ) \ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{b}-\frac {a^{2} \left (a^{2}-2 a b +b^{2}\right )}{2 b \left (a +b \tan \left (f x +e \right )^{2}\right )^{2}}\right )}{2 a^{3} \left (a -b \right )^{3}}+\frac {\ln \left (\tan \left (f x +e \right )\right )}{a^{3}}-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right )^{3}}}{f}\) | \(149\) |
| default | \(\frac {\frac {b^{2} \left (-\frac {a \left (2 a^{2}-3 a b +b^{2}\right )}{b \left (a +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (3 a^{2}-3 a b +b^{2}\right ) \ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{b}-\frac {a^{2} \left (a^{2}-2 a b +b^{2}\right )}{2 b \left (a +b \tan \left (f x +e \right )^{2}\right )^{2}}\right )}{2 a^{3} \left (a -b \right )^{3}}+\frac {\ln \left (\tan \left (f x +e \right )\right )}{a^{3}}-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right )^{3}}}{f}\) | \(149\) |
| norman | \(\frac {\frac {\left (3 a b -2 b^{2}\right ) b \tan \left (f x +e \right )^{2}}{2 a^{2} f \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (5 a b -3 b^{2}\right ) b^{2} \tan \left (f x +e \right )^{4}}{4 a^{3} f \left (a^{2}-2 a b +b^{2}\right )}}{\left (a +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\ln \left (\tan \left (f x +e \right )\right )}{a^{3} f}-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {b \left (3 a^{2}-3 a b +b^{2}\right ) \ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{2 a^{3} f \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}\) | \(211\) |
| parallelrisch | \(\frac {12 \left (a^{2}-a b +\frac {1}{3} b^{2}\right ) \left (\frac {\left (a -b \right )^{2} \cos \left (4 f x +4 e \right )}{4}+\left (a^{2}-b^{2}\right ) \cos \left (2 f x +2 e \right )+\frac {3 a^{2}}{4}+\frac {a b}{2}+\frac {3 b^{2}}{4}\right ) b \ln \left (a +b \tan \left (f x +e \right )^{2}\right )+2 \left (a -b \right )^{2} \left (-\frac {\ln \left (\sec \left (f x +e \right )^{2}\right ) a^{3}}{2}+\left (a -b \right )^{3} \ln \left (\tan \left (f x +e \right )\right )-\frac {3 \left (a -\frac {b}{2}\right ) b^{2}}{2}\right ) \cos \left (4 f x +4 e \right )+8 \left (a -b \right ) \left (-\frac {a^{3} \left (a +b \right ) \ln \left (\sec \left (f x +e \right )^{2}\right )}{2}+\left (a +b \right ) \left (a -b \right )^{3} \ln \left (\tan \left (f x +e \right )\right )-\frac {5 \left (a -\frac {3 b}{5}\right ) b^{3}}{4}\right ) \cos \left (2 f x +2 e \right )+\left (-3 a^{5}-2 a^{4} b -3 a^{3} b^{2}\right ) \ln \left (\sec \left (f x +e \right )^{2}\right )+6 \left (a -b \right ) \left (\left (a -b \right )^{2} \left (a^{2}+\frac {2}{3} a b +b^{2}\right ) \ln \left (\tan \left (f x +e \right )\right )+\frac {a^{2} b^{2}}{2}+\frac {11 a \,b^{3}}{12}-\frac {3 b^{4}}{4}\right )}{8 \left (a -b \right )^{3} a^{3} \left (\frac {\left (a -b \right )^{2} \cos \left (4 f x +4 e \right )}{4}+\left (a^{2}-b^{2}\right ) \cos \left (2 f x +2 e \right )+\frac {3 a^{2}}{4}+\frac {a b}{2}+\frac {3 b^{2}}{4}\right ) f}\) | \(355\) |
| risch | \(\frac {i x}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}-\frac {2 i x}{a^{3}}-\frac {2 i e}{a^{3} f}-\frac {6 i b x}{a \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {6 i b e}{a f \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {6 i b^{2} x}{a^{2} \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {6 i b^{2} e}{a^{2} f \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {2 i b^{3} x}{a^{3} \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {2 i b^{3} e}{a^{3} f \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {2 b^{2} \left (3 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}-4 a b \,{\mathrm e}^{6 i \left (f x +e \right )}+b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+6 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+2 a b \,{\mathrm e}^{4 i \left (f x +e \right )}-2 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+3 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-4 a b \,{\mathrm e}^{2 i \left (f x +e \right )}+b^{2} {\mathrm e}^{2 i \left (f x +e \right )}\right )}{\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 )^{2} f \left (a^{2}-2 a b +b^{2}\right ) \left (-a +b \right ) a^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{a^{3} f}+\frac {3 b \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 f \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {3 b^{2} \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^{2} f \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {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 )}{2 a^{3} f \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}\) | \(674\) |
1/f*(1/2*b^2/a^3/(a-b)^3*(-a*(2*a^2-3*a*b+b^2)/b/(a+b*tan(f*x+e)^2)+(3*a^2 -3*a*b+b^2)/b*ln(a+b*tan(f*x+e)^2)-1/2*a^2*(a^2-2*a*b+b^2)/b/(a+b*tan(f*x+ e)^2)^2)+1/a^3*ln(tan(f*x+e))-1/2/(a-b)^3*ln(1+tan(f*x+e)^2))
Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (142) = 284\).
Time = 0.32 (sec) , antiderivative size = 422, normalized size of antiderivative = 2.85 \[ \int \frac {\cot (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {6 \, a^{3} b^{2} - 3 \, a^{2} b^{3} + {\left (5 \, a^{2} b^{3} - 2 \, a b^{4}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{3} b^{2} + a^{2} b^{3} - a b^{4}\right )} \tan \left (f x + e\right )^{2} + 2 \, {\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3} + {\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} \tan \left (f x + e\right )^{2}\right )} \log \left (\frac {\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left (3 \, a^{4} b - 3 \, a^{3} b^{2} + a^{2} b^{3} + {\left (3 \, a^{2} b^{3} - 3 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{3} b^{2} - 3 \, a^{2} b^{3} + a b^{4}\right )} \tan \left (f x + e\right )^{2}\right )} \log \left (\frac {b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right )}{4 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} - a^{3} b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{7} b - 3 \, a^{6} b^{2} + 3 \, a^{5} b^{3} - a^{4} b^{4}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{8} - 3 \, a^{7} b + 3 \, a^{6} b^{2} - a^{5} b^{3}\right )} f\right )}} \]
1/4*(6*a^3*b^2 - 3*a^2*b^3 + (5*a^2*b^3 - 2*a*b^4)*tan(f*x + e)^4 + 2*(3*a ^3*b^2 + a^2*b^3 - a*b^4)*tan(f*x + e)^2 + 2*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3 + (a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*tan(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*tan(f*x + e)^2)*log(tan(f*x + e)^2/(tan(f *x + e)^2 + 1)) + 2*(3*a^4*b - 3*a^3*b^2 + a^2*b^3 + (3*a^2*b^3 - 3*a*b^4 + b^5)*tan(f*x + e)^4 + 2*(3*a^3*b^2 - 3*a^2*b^3 + a*b^4)*tan(f*x + e)^2)* log((b*tan(f*x + e)^2 + a)/(tan(f*x + e)^2 + 1)))/((a^6*b^2 - 3*a^5*b^3 + 3*a^4*b^4 - a^3*b^5)*f*tan(f*x + e)^4 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*f*tan(f*x + e)^2 + (a^8 - 3*a^7*b + 3*a^6*b^2 - a^5*b^3)*f)
Timed out. \[ \int \frac {\cot (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Timed out} \]
Time = 0.24 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.69 \[ \int \frac {\cot (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {\frac {2 \, {\left (3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a^{6} - 3 \, a^{5} b + 3 \, a^{4} b^{2} - a^{3} b^{3}} + \frac {6 \, a^{2} b^{2} - 3 \, a b^{3} - 2 \, {\left (3 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \sin \left (f x + e\right )^{2}}{a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3} + {\left (a^{7} - 5 \, a^{6} b + 10 \, a^{5} b^{2} - 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} - a^{2} b^{5}\right )} \sin \left (f x + e\right )^{4} - 2 \, {\left (a^{7} - 4 \, a^{6} b + 6 \, a^{5} b^{2} - 4 \, a^{4} b^{3} + a^{3} b^{4}\right )} \sin \left (f x + e\right )^{2}} + \frac {2 \, \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{3}}}{4 \, f} \]
1/4*(2*(3*a^2*b - 3*a*b^2 + b^3)*log(-(a - b)*sin(f*x + e)^2 + a)/(a^6 - 3 *a^5*b + 3*a^4*b^2 - a^3*b^3) + (6*a^2*b^2 - 3*a*b^3 - 2*(3*a^2*b^2 - 4*a* b^3 + b^4)*sin(f*x + e)^2)/(a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3 + (a^7 - 5 *a^6*b + 10*a^5*b^2 - 10*a^4*b^3 + 5*a^3*b^4 - a^2*b^5)*sin(f*x + e)^4 - 2 *(a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4)*sin(f*x + e)^2) + 2*log (sin(f*x + e)^2)/a^3)/f
Time = 1.16 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.72 \[ \int \frac {\cot (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {\frac {2 \, {\left (3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \log \left ({\left | -a \sin \left (f x + e\right )^{2} + b \sin \left (f x + e\right )^{2} + a \right |}\right )}{a^{6} - 3 \, a^{5} b + 3 \, a^{4} b^{2} - a^{3} b^{3}} - \frac {9 \, a^{3} b \sin \left (f x + e\right )^{4} - 18 \, a^{2} b^{2} \sin \left (f x + e\right )^{4} + 12 \, a b^{3} \sin \left (f x + e\right )^{4} - 3 \, b^{4} \sin \left (f x + e\right )^{4} - 18 \, a^{3} b \sin \left (f x + e\right )^{2} + 24 \, a^{2} b^{2} \sin \left (f x + e\right )^{2} - 8 \, a b^{3} \sin \left (f x + e\right )^{2} + 9 \, a^{3} b - 6 \, a^{2} b^{2}}{{\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} {\left (a \sin \left (f x + e\right )^{2} - b \sin \left (f x + e\right )^{2} - a\right )}^{2}} + \frac {2 \, \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{3}}}{4 \, f} \]
1/4*(2*(3*a^2*b - 3*a*b^2 + b^3)*log(abs(-a*sin(f*x + e)^2 + b*sin(f*x + e )^2 + a))/(a^6 - 3*a^5*b + 3*a^4*b^2 - a^3*b^3) - (9*a^3*b*sin(f*x + e)^4 - 18*a^2*b^2*sin(f*x + e)^4 + 12*a*b^3*sin(f*x + e)^4 - 3*b^4*sin(f*x + e) ^4 - 18*a^3*b*sin(f*x + e)^2 + 24*a^2*b^2*sin(f*x + e)^2 - 8*a*b^3*sin(f*x + e)^2 + 9*a^3*b - 6*a^2*b^2)/((a^5 - 2*a^4*b + a^3*b^2)*(a*sin(f*x + e)^ 2 - b*sin(f*x + e)^2 - a)^2) + 2*log(sin(f*x + e)^2)/a^3)/f
Time = 11.55 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.22 \[ \int \frac {\cot (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )}{a^3\,f}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f\,{\left (a-b\right )}^3}-\frac {\frac {5\,a\,b-3\,b^2}{4\,a\,\left (a^2-2\,a\,b+b^2\right )}+\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,a\,b-b^2\right )}{2\,a^2\,\left (a^2-2\,a\,b+b^2\right )}}{f\,\left (a^2+2\,a\,b\,{\mathrm {tan}\left (e+f\,x\right )}^2+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}+\frac {b\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )\,\left (3\,a^2-3\,a\,b+b^2\right )}{2\,a^3\,f\,{\left (a-b\right )}^3} \]
log(tan(e + f*x))/(a^3*f) - log(tan(e + f*x)^2 + 1)/(2*f*(a - b)^3) - ((5* a*b - 3*b^2)/(4*a*(a^2 - 2*a*b + b^2)) + (b*tan(e + f*x)^2*(2*a*b - b^2))/ (2*a^2*(a^2 - 2*a*b + b^2)))/(f*(a^2 + b^2*tan(e + f*x)^4 + 2*a*b*tan(e + f*x)^2)) + (b*log(a + b*tan(e + f*x)^2)*(3*a^2 - 3*a*b + b^2))/(2*a^3*f*(a - b)^3)