Integrand size = 19, antiderivative size = 161 \[ \int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\frac {\sqrt {-1+\sqrt {2}} \arctan \left (\frac {3-2 \sqrt {2}+\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (-7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{2 f}-\frac {2 \text {arctanh}\left (\sqrt {1+\tan (e+f x)}\right )}{f}+\frac {\sqrt {1+\sqrt {2}} \text {arctanh}\left (\frac {3+2 \sqrt {2}+\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{2 f} \]
-2*arctanh((1+tan(f*x+e))^(1/2))/f+1/2*arctan((3-2*2^(1/2)+(1-2^(1/2))*tan (f*x+e))/(-14+10*2^(1/2))^(1/2)/(1+tan(f*x+e))^(1/2))*(2^(1/2)-1)^(1/2)/f+ 1/2*arctanh((3+2*2^(1/2)+(1+2^(1/2))*tan(f*x+e))/(14+10*2^(1/2))^(1/2)/(1+ tan(f*x+e))^(1/2))*(1+2^(1/2))^(1/2)/f
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.52 \[ \int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=-\frac {2 \text {arctanh}\left (\sqrt {1+\tan (e+f x)}\right )}{f}+\frac {\text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )}{\sqrt {1-i} f}+\frac {\text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )}{\sqrt {1+i} f} \]
(-2*ArcTanh[Sqrt[1 + Tan[e + f*x]]])/f + ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sq rt[1 - I]]/(Sqrt[1 - I]*f) + ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]]/( Sqrt[1 + I]*f)
Time = 0.79 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.14, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {3042, 4057, 25, 3042, 4019, 3042, 4018, 216, 220, 4117, 73, 220}
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.
\(\displaystyle \int \frac {\cot (e+f x)}{\sqrt {\tan (e+f x)+1}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (e+f x) \sqrt {\tan (e+f x)+1}}dx\) |
\(\Big \downarrow \) 4057 |
\(\displaystyle \int -\frac {\tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx+\int \frac {\cot (e+f x) \left (\tan ^2(e+f x)+1\right )}{\sqrt {\tan (e+f x)+1}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\cot (e+f x) \left (\tan ^2(e+f x)+1\right )}{\sqrt {\tan (e+f x)+1}}dx-\int \frac {\tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (e+f x)^2+1}{\tan (e+f x) \sqrt {\tan (e+f x)+1}}dx-\int \frac {\tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx\) |
\(\Big \downarrow \) 4019 |
\(\displaystyle \int \frac {\tan (e+f x)^2+1}{\tan (e+f x) \sqrt {\tan (e+f x)+1}}dx-\frac {\int \frac {1-\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}+\frac {\int \frac {1-\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {1-\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}+\frac {\int \frac {1-\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}+\int \frac {\tan (e+f x)^2+1}{\tan (e+f x) \sqrt {\tan (e+f x)+1}}dx\) |
\(\Big \downarrow \) 4018 |
\(\displaystyle \int \frac {\tan (e+f x)^2+1}{\tan (e+f x) \sqrt {\tan (e+f x)+1}}dx+\frac {\left (3-2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7-5 \sqrt {2}\right )}d\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}-\frac {\left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7+5 \sqrt {2}\right )}d\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \int \frac {\tan (e+f x)^2+1}{\tan (e+f x) \sqrt {\tan (e+f x)+1}}dx-\frac {\left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7+5 \sqrt {2}\right )}d\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}+\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \int \frac {\tan (e+f x)^2+1}{\tan (e+f x) \sqrt {\tan (e+f x)+1}}dx+\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}+\frac {\left (3+2 \sqrt {2}\right ) \text {arctanh}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {7+5 \sqrt {2}} f}\) |
\(\Big \downarrow \) 4117 |
\(\displaystyle \frac {\int \frac {\cot (e+f x)}{\sqrt {\tan (e+f x)+1}}d\tan (e+f x)}{f}+\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}+\frac {\left (3+2 \sqrt {2}\right ) \text {arctanh}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {7+5 \sqrt {2}} f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 \int \cot (e+f x)d\sqrt {\tan (e+f x)+1}}{f}+\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}+\frac {\left (3+2 \sqrt {2}\right ) \text {arctanh}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {7+5 \sqrt {2}} f}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}-\frac {2 \text {arctanh}\left (\sqrt {\tan (e+f x)+1}\right )}{f}+\frac {\left (3+2 \sqrt {2}\right ) \text {arctanh}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {7+5 \sqrt {2}} f}\) |
((3 - 2*Sqrt[2])*ArcTan[(3 - 2*Sqrt[2] + (1 - Sqrt[2])*Tan[e + f*x])/(Sqrt [2*(-7 + 5*Sqrt[2])]*Sqrt[1 + Tan[e + f*x]])])/(2*Sqrt[-7 + 5*Sqrt[2]]*f) - (2*ArcTanh[Sqrt[1 + Tan[e + f*x]]])/f + ((3 + 2*Sqrt[2])*ArcTanh[(3 + 2* Sqrt[2] + (1 + Sqrt[2])*Tan[e + f*x])/(Sqrt[2*(7 + 5*Sqrt[2])]*Sqrt[1 + Ta n[e + f*x]])])/(2*Sqrt[7 + 5*Sqrt[2]]*f)
3.5.3.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*tan[(e_.) + ( f_.)*(x_)]], x_Symbol] :> Simp[-2*(d^2/f) Subst[Int[1/(2*b*c*d - 4*a*d^2 + x^2), x], x, (b*c - 2*a*d - b*d*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0 ] && NeQ[c^2 + d^2, 0] && EqQ[2*a*c*d - b*(c^2 - d^2), 0]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*tan[(e_.) + ( f_.)*(x_)]], x_Symbol] :> With[{q = Rt[a^2 + b^2, 2]}, Simp[1/(2*q) Int[( a*c + b*d + c*q + (b*c - a*d + d*q)*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]], x], x] - Simp[1/(2*q) Int[(a*c + b*d - c*q + (b*c - a*d - d*q)*Tan[e + f *x])/Sqrt[a + b*Tan[e + f*x]], x], x]] /; FreeQ[{a, b, c, d, e, f}, x] && N eQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && NeQ[2*a*c*d - b*(c^2 - d^2), 0] && NiceSqrtQ[a^2 + b^2]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/(c^2 + d^2) Int[(a + b*Tan[e + f*x])^m *(c - d*Tan[e + f*x]), x], x] + Simp[d^2/(c^2 + d^2) Int[(a + b*Tan[e + f *x])^m*((1 + Tan[e + f*x]^2)/(c + d*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d ^2, 0] && !IntegerQ[m]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A/f Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
Leaf count of result is larger than twice the leaf count of optimal. \(2129\) vs. \(2(123)=246\).
Time = 117.36 (sec) , antiderivative size = 2130, normalized size of antiderivative = 13.23
-1/4/f*cot(f*x+e)*(3*(-2+2*2^(1/2))^(1/2)*(1+2^(1/2))^(1/2)*cos(f*x+e)*((c os(f*x+e)+sin(f*x+e))*cos(f*x+e)/(2*2^(1/2)*cos(f*x+e)*sin(f*x+e)-2*sin(f* x+e)^2*2^(1/2)-2*sin(f*x+e)*cos(f*x+e)+2*sin(f*x+e)^2+1))^(1/2)*arctan(1/4 *((4+3*2^(1/2))*(cos(f*x+e)+sin(f*x+e))*cos(f*x+e)/(2*2^(1/2)*cos(f*x+e)*s in(f*x+e)-2*sin(f*x+e)^2*2^(1/2)-2*sin(f*x+e)*cos(f*x+e)+2*sin(f*x+e)^2+1) *(3*2^(1/2)-4))^(1/2)/(2*cos(f*x+e)^2-1)*(4*sin(f*x+e)*cos(f*x+e)-1-tan(f* x+e))*(-2+2*2^(1/2))^(1/2)*(2*2^(1/2)+3)*(3*2^(1/2)-4))*2^(1/2)-7*(-2+2*2^ (1/2))^(1/2)*(1+2^(1/2))^(1/2)*arctan(1/4*((4+3*2^(1/2))*(cos(f*x+e)+sin(f *x+e))*cos(f*x+e)/(2*2^(1/2)*cos(f*x+e)*sin(f*x+e)-2*sin(f*x+e)^2*2^(1/2)- 2*sin(f*x+e)*cos(f*x+e)+2*sin(f*x+e)^2+1)*(3*2^(1/2)-4))^(1/2)/(2*cos(f*x+ e)^2-1)*(4*sin(f*x+e)*cos(f*x+e)-1-tan(f*x+e))*(-2+2*2^(1/2))^(1/2)*(2*2^( 1/2)+3)*(3*2^(1/2)-4))*((cos(f*x+e)+sin(f*x+e))*cos(f*x+e)/(2*2^(1/2)*cos( f*x+e)*sin(f*x+e)-2*sin(f*x+e)^2*2^(1/2)-2*sin(f*x+e)*cos(f*x+e)+2*sin(f*x +e)^2+1))^(1/2)*2^(1/2)*sin(f*x+e)-4*(-2+2*2^(1/2))^(1/2)*(1+2^(1/2))^(1/2 )*cos(f*x+e)*((cos(f*x+e)+sin(f*x+e))*cos(f*x+e)/(2*2^(1/2)*cos(f*x+e)*sin (f*x+e)-2*sin(f*x+e)^2*2^(1/2)-2*sin(f*x+e)*cos(f*x+e)+2*sin(f*x+e)^2+1))^ (1/2)*arctan(1/4*((4+3*2^(1/2))*(cos(f*x+e)+sin(f*x+e))*cos(f*x+e)/(2*2^(1 /2)*cos(f*x+e)*sin(f*x+e)-2*sin(f*x+e)^2*2^(1/2)-2*sin(f*x+e)*cos(f*x+e)+2 *sin(f*x+e)^2+1)*(3*2^(1/2)-4))^(1/2)/(2*cos(f*x+e)^2-1)*(4*sin(f*x+e)*cos (f*x+e)-1-tan(f*x+e))*(-2+2*2^(1/2))^(1/2)*(2*2^(1/2)+3)*(3*2^(1/2)-4))...
Leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (122) = 244\).
Time = 0.26 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.09 \[ \int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=-\frac {\sqrt {\frac {1}{2}} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \sqrt {\frac {1}{2}} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \sqrt {\frac {1}{2}} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + \sqrt {\frac {1}{2}} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 2 \, \log \left (\sqrt {\tan \left (f x + e\right ) + 1} + 1\right ) - 2 \, \log \left (\sqrt {\tan \left (f x + e\right ) + 1} - 1\right )}{2 \, f} \]
-1/2*(sqrt(1/2)*f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2)*log(sqrt(1/2)*(f^3*sqrt (-1/f^4) - f)*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2) + sqrt(tan(f*x + e) + 1)) - sqrt(1/2)*f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2)*log(-sqrt(1/2)*(f^3*sqrt(-1/ f^4) - f)*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2) + sqrt(tan(f*x + e) + 1)) - sqr t(1/2)*f*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2)*log(sqrt(1/2)*(f^3*sqrt(-1/f^4) + f)*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2) + sqrt(tan(f*x + e) + 1)) + sqrt(1 /2)*f*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2)*log(-sqrt(1/2)*(f^3*sqrt(-1/f^4) + f)*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2) + sqrt(tan(f*x + e) + 1)) + 2*log(sq rt(tan(f*x + e) + 1) + 1) - 2*log(sqrt(tan(f*x + e) + 1) - 1))/f
\[ \int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\int \frac {\cot {\left (e + f x \right )}}{\sqrt {\tan {\left (e + f x \right )} + 1}}\, dx \]
\[ \int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\int { \frac {\cot \left (f x + e\right )}{\sqrt {\tan \left (f x + e\right ) + 1}} \,d x } \]
\[ \int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\int { \frac {\cot \left (f x + e\right )}{\sqrt {\tan \left (f x + e\right ) + 1}} \,d x } \]
Time = 0.18 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.53 \[ \int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=-\frac {2\,\mathrm {atanh}\left (\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )}{f}+2\,\mathrm {atanh}\left (2\,f\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}+2\,\mathrm {atanh}\left (2\,f\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}} \]