Integrand size = 27, antiderivative size = 170 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{a+b \tan (e+f x)} \, dx=\frac {\sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(i a+b) f}-\frac {\sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(i a-b) f}-\frac {2 \sqrt {b} \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\left (a^2+b^2\right ) f} \]
arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))*(c-I*d)^(1/2)/(I*a+b)/f-arct anh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))*(c+I*d)^(1/2)/(I*a-b)/f-2*arctan h(b^(1/2)*(c+d*tan(f*x+e))^(1/2)/(-a*d+b*c)^(1/2))*b^(1/2)*(-a*d+b*c)^(1/2 )/(a^2+b^2)/f
Time = 0.31 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{a+b \tan (e+f x)} \, dx=\frac {(-i a+b) \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+(i a+b) \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )-2 \sqrt {b} \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\left (a^2+b^2\right ) f} \]
(((-I)*a + b)*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d] ] + (I*a + b)*Sqrt[c + I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d] ] - 2*Sqrt[b]*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/S qrt[b*c - a*d]])/((a^2 + b^2)*f)
Time = 1.08 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.93, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 4055, 3042, 4022, 3042, 4020, 25, 73, 221, 4117, 73, 221}
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.
| \(\displaystyle \int \frac {\sqrt {c+d \tan (e+f x)}}{a+b \tan (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {c+d \tan (e+f x)}}{a+b \tan (e+f x)}dx\) |
| \(\Big \downarrow \) 4055 |
| \(\displaystyle \frac {b (b c-a d) \int \frac {\tan ^2(e+f x)+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {\int \frac {a c+b d-(b c-a d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a c+b d-(b c-a d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {b (b c-a d) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {b (b c-a d) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {\frac {1}{2} (a-i b) (c+i d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a+i b) (c-i d) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b (b c-a d) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {\frac {1}{2} (a-i b) (c+i d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a+i b) (c-i d) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {b (b c-a d) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {\frac {i (a+i b) (c-i d) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (a-i b) (c+i d) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}}{a^2+b^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b (b c-a d) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {\frac {i (a-i b) (c+i d) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {i (a+i b) (c-i d) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}}{a^2+b^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {b (b c-a d) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {\frac {(a-i b) (c+i d) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(a+i b) (c-i d) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}}{a^2+b^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {b (b c-a d) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {\frac {(a+i b) \sqrt {c-i d} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {(a-i b) \sqrt {c+i d} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{a^2+b^2}\) |
| \(\Big \downarrow \) 4117 |
| \(\displaystyle \frac {b (b c-a d) \int \frac {1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{f \left (a^2+b^2\right )}+\frac {\frac {(a+i b) \sqrt {c-i d} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {(a-i b) \sqrt {c+i d} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{a^2+b^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {2 b (b c-a d) \int \frac {1}{a+\frac {b (c+d \tan (e+f x))}{d}-\frac {b c}{d}}d\sqrt {c+d \tan (e+f x)}}{d f \left (a^2+b^2\right )}+\frac {\frac {(a+i b) \sqrt {c-i d} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {(a-i b) \sqrt {c+i d} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{a^2+b^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {2 \sqrt {b} \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{f \left (a^2+b^2\right )}+\frac {\frac {(a+i b) \sqrt {c-i d} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {(a-i b) \sqrt {c+i d} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{a^2+b^2}\) |
(((a + I*b)*Sqrt[c - I*d]*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f + ((a - I* b)*Sqrt[c + I*d]*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/f)/(a^2 + b^2) - (2*S qrt[b]*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/((a^2 + b^2)*f)
3.13.32.3.1 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[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*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[Sqrt[(a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/(c^2 + d^2) Int[Simp[a*c + b*d + (b*c - a*d)*Tan[e + f*x], x]/Sqrt[a + b*Tan[e + f*x]], x], x] - Simp[d*((b*c - a* d)/(c^2 + d^2)) Int[(1 + Tan[e + f*x]^2)/(Sqrt[a + b*Tan[e + f*x]]*(c + d *Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && N eQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
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. \(1163\) vs. \(2(142)=284\).
Time = 0.85 (sec) , antiderivative size = 1164, normalized size of antiderivative = 6.85
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1164\) |
| default | \(\text {Expression too large to display}\) | \(1164\) |
-1/4/f/d/(a^2+b^2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^( 1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^( 1/2)*a+1/4/f/d/(a^2+b^2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+ d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c+1 /4/f/(a^2+b^2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2) +2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b+1/f/(a^2+b^2) /(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^ 2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*b+1/f* d/(a^2+b^2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2) +(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*a-1/f/(a^2+ b^2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^ 2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*b*c+1/4/f/d/(a^2+b ^2)*ln(d*tan(f*x+e)+c-(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2) +(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a-1/4/f/d/ (a^2+b^2)*ln(d*tan(f*x+e)+c-(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c) ^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c-1/4/f/(a^2+b^2)* ln(d*tan(f*x+e)+c-(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^ 2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b+1/f/(a^2+b^2)/(2*(c^2+d^2)^( 1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(2*(c^2+d^2)^(1/2)+2*c)^( 1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*b+1/f*d/(a^2+b^2)/...
Leaf count of result is larger than twice the leaf count of optimal. 1958 vs. \(2 (136) = 272\).
Time = 0.51 (sec) , antiderivative size = 3934, normalized size of antiderivative = 23.14 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{a+b \tan (e+f x)} \, dx=\text {Too large to display} \]
[1/2*((a^2 + b^2)*f*sqrt(-((a^4 + 2*a^2*b^2 + b^4)*f^2*sqrt(-(4*a^2*b^2*c^ 2 - 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*f^4)) + 2*a*b*d + (a^2 - b^2)*c)/((a^4 + 2 *a^2*b^2 + b^4)*f^2))*log((2*a*b*c - (a^2 - b^2)*d)*sqrt(d*tan(f*x + e) + c) + ((a^5 + 2*a^3*b^2 + a*b^4)*f^3*sqrt(-(4*a^2*b^2*c^2 - 4*(a^3*b - a*b^ 3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^ 2*b^6 + b^8)*f^4)) + (2*a*b^2*c - (a^2*b - b^3)*d)*f)*sqrt(-((a^4 + 2*a^2* b^2 + b^4)*f^2*sqrt(-(4*a^2*b^2*c^2 - 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2 *b^2 + b^4)*d^2)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*f^4)) + 2*a*b*d + (a^2 - b^2)*c)/((a^4 + 2*a^2*b^2 + b^4)*f^2))) - (a^2 + b^2)*f*s qrt(-((a^4 + 2*a^2*b^2 + b^4)*f^2*sqrt(-(4*a^2*b^2*c^2 - 4*(a^3*b - a*b^3) *c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2* b^6 + b^8)*f^4)) + 2*a*b*d + (a^2 - b^2)*c)/((a^4 + 2*a^2*b^2 + b^4)*f^2)) *log((2*a*b*c - (a^2 - b^2)*d)*sqrt(d*tan(f*x + e) + c) - ((a^5 + 2*a^3*b^ 2 + a*b^4)*f^3*sqrt(-(4*a^2*b^2*c^2 - 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2 *b^2 + b^4)*d^2)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*f^4)) + (2*a*b^2*c - (a^2*b - b^3)*d)*f)*sqrt(-((a^4 + 2*a^2*b^2 + b^4)*f^2*sqrt(- (4*a^2*b^2*c^2 - 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2)/((a^ 8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*f^4)) + 2*a*b*d + (a^2 - b^2) *c)/((a^4 + 2*a^2*b^2 + b^4)*f^2))) - (a^2 + b^2)*f*sqrt(((a^4 + 2*a^2*...
\[ \int \frac {\sqrt {c+d \tan (e+f x)}}{a+b \tan (e+f x)} \, dx=\int \frac {\sqrt {c + d \tan {\left (e + f x \right )}}}{a + b \tan {\left (e + f x \right )}}\, dx \]
Exception generated. \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{a+b \tan (e+f x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
\[ \int \frac {\sqrt {c+d \tan (e+f x)}}{a+b \tan (e+f x)} \, dx=\int { \frac {\sqrt {d \tan \left (f x + e\right ) + c}}{b \tan \left (f x + e\right ) + a} \,d x } \]
Time = 11.30 (sec) , antiderivative size = 11975, normalized size of antiderivative = 70.44 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{a+b \tan (e+f x)} \, dx=\text {Too large to display} \]
atan(((((((32*(12*a*b^7*d^11*f^4 - 12*b^8*c*d^10*f^4 + 24*a^3*b^5*d^11*f^4 + 12*a^5*b^3*d^11*f^4 - 12*b^8*c^3*d^8*f^4 - 24*a^2*b^6*c^3*d^8*f^4 + 24* a^3*b^5*c^2*d^9*f^4 - 12*a^4*b^4*c^3*d^8*f^4 + 12*a^5*b^3*c^2*d^9*f^4 + 12 *a*b^7*c^2*d^9*f^4 - 24*a^2*b^6*c*d^10*f^4 - 12*a^4*b^4*c*d^10*f^4))/f^5 - (32*(c + d*tan(e + f*x))^(1/2)*(-(c + d*1i)/(4*(a^2*f^2 - b^2*f^2 + a*b*f ^2*2i)))^(1/2)*(16*b^9*d^10*f^4 + 16*a^2*b^7*d^10*f^4 - 16*a^4*b^5*d^10*f^ 4 - 16*a^6*b^3*d^10*f^4 + 24*b^9*c^2*d^8*f^4 + 40*a^2*b^7*c^2*d^8*f^4 + 8* a^4*b^5*c^2*d^8*f^4 - 8*a^6*b^3*c^2*d^8*f^4 + 8*a*b^8*c*d^9*f^4 + 24*a^3*b ^6*c*d^9*f^4 + 24*a^5*b^4*c*d^9*f^4 + 8*a^7*b^2*c*d^9*f^4))/f^4)*(-(c + d* 1i)/(4*(a^2*f^2 - b^2*f^2 + a*b*f^2*2i)))^(1/2) - (32*(c + d*tan(e + f*x)) ^(1/2)*(14*a*b^6*d^11*f^2 - 6*b^7*c*d^10*f^2 - 20*a^3*b^4*d^11*f^2 - 2*a^5 *b^2*d^11*f^2 - 18*b^7*c^3*d^8*f^2 + 12*a^2*b^5*c^3*d^8*f^2 - 12*a^3*b^4*c ^2*d^9*f^2 - 2*a^4*b^3*c^3*d^8*f^2 + 2*a^5*b^2*c^2*d^9*f^2 + 18*a*b^6*c^2* d^9*f^2 + 36*a^2*b^5*c*d^10*f^2 + 10*a^4*b^3*c*d^10*f^2))/f^4)*(-(c + d*1i )/(4*(a^2*f^2 - b^2*f^2 + a*b*f^2*2i)))^(1/2) + (32*(13*a^2*b^4*d^12*f^2 + a^4*b^2*d^12*f^2 + 3*b^6*c^2*d^10*f^2 + 3*b^6*c^4*d^8*f^2 + 12*a^2*b^4*c^ 2*d^10*f^2 - a^2*b^4*c^4*d^8*f^2 + a^4*b^2*c^2*d^10*f^2 - 16*a*b^5*c*d^11* f^2 - 16*a*b^5*c^3*d^9*f^2))/f^5)*(-(c + d*1i)/(4*(a^2*f^2 - b^2*f^2 + a*b *f^2*2i)))^(1/2) - (32*(c + d*tan(e + f*x))^(1/2)*(b^5*d^12 - 2*a^2*b^3*d^ 12 + 3*b^5*c^4*d^8 - 4*a*b^4*c^3*d^9 + 2*a^2*b^3*c^2*d^10 + 4*a*b^4*c*d...