3.2.12 \(\int \frac {(a+b \tan (e+f x)) (A+B \tan (e+f x)+C \tan ^2(e+f x))}{\sqrt {c+d \tan (e+f x)}} \, dx\) [112]

3.2.12.1 Optimal result

Integrand size = 45, antiderivative size = 194 \[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {(i a+b) (A-i B-C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d} f}+\frac {(i a-b) (A+i B-C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d} f}-\frac {2 (2 b c C-3 b B d-3 a C d) \sqrt {c+d \tan (e+f x)}}{3 d^2 f}+\frac {2 b C \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f} \]

output

-(I*a+b)*(A-I*B-C)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f/(c-I*d) 
^(1/2)+(I*a-b)*(A+I*B-C)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/f/( 
c+I*d)^(1/2)-2/3*(-3*B*b*d-3*C*a*d+2*C*b*c)*(c+d*tan(f*x+e))^(1/2)/d^2/f+2 
/3*b*C*(c+d*tan(f*x+e))^(1/2)*tan(f*x+e)/d/f
 

3.2.12.2 Mathematica [A] (verified)

Time = 1.63 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.99 \[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {2 \left (-\frac {3 i (a-i b) (A-i B-C) d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 \sqrt {c-i d}}+\frac {3 i (a+i b) (A+i B-C) d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 \sqrt {c+i d}}+\frac {(-2 b c C+3 b B d+3 a C d) \sqrt {c+d \tan (e+f x)}}{d}+b C \tan (e+f x) \sqrt {c+d \tan (e+f x)}\right )}{3 d f} \]

input

Integrate[((a + b*Tan[e + f*x])*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2))/S 
qrt[c + d*Tan[e + f*x]],x]
 

output

(2*((((-3*I)/2)*(a - I*b)*(A - I*B - C)*d*ArcTanh[Sqrt[c + d*Tan[e + f*x]] 
/Sqrt[c - I*d]])/Sqrt[c - I*d] + (((3*I)/2)*(a + I*b)*(A + I*B - C)*d*ArcT 
anh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/Sqrt[c + I*d] + ((-2*b*c*C + 
3*b*B*d + 3*a*C*d)*Sqrt[c + d*Tan[e + f*x]])/d + b*C*Tan[e + f*x]*Sqrt[c + 
 d*Tan[e + f*x]]))/(3*d*f)
 

3.2.12.3 Rubi [A] (warning: unable to verify)

Time = 0.96 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.95, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 4120, 27, 3042, 4113, 3042, 4022, 3042, 4020, 25, 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.

\(\displaystyle \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan (e+f x)^2\right )}{\sqrt {c+d \tan (e+f x)}}dx\)

\(\Big \downarrow \) 4120

\(\displaystyle \frac {2 b C \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {2 \int \frac {(2 b c C-3 a d C-3 b B d) \tan ^2(e+f x)-3 (A b-C b+a B) d \tan (e+f x)+2 b c C-3 a A d}{2 \sqrt {c+d \tan (e+f x)}}dx}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 b C \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\int \frac {(2 b c C-3 a d C-3 b B d) \tan ^2(e+f x)-3 (A b-C b+a B) d \tan (e+f x)+2 b c C-3 a A d}{\sqrt {c+d \tan (e+f x)}}dx}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b C \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\int \frac {(2 b c C-3 a d C-3 b B d) \tan (e+f x)^2-3 (A b-C b+a B) d \tan (e+f x)+2 b c C-3 a A d}{\sqrt {c+d \tan (e+f x)}}dx}{3 d}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {2 b C \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\int \frac {3 (b B-a (A-C)) d-3 (A b-C b+a B) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 (-3 a C d-3 b B d+2 b c C) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b C \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\int \frac {3 (b B-a (A-C)) d-3 (A b-C b+a B) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 (-3 a C d-3 b B d+2 b c C) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 b C \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {-\frac {3}{2} d (a+i b) (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {3}{2} d (a-i b) (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 (-3 a C d-3 b B d+2 b c C) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b C \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {-\frac {3}{2} d (a+i b) (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {3}{2} d (a-i b) (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 (-3 a C d-3 b B d+2 b c C) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 b C \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {-\frac {3 i d (a-i b) (A-i B-C) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {3 i d (a+i b) (A+i B-C) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 (-3 a C d-3 b B d+2 b c C) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 b C \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\frac {3 i d (a-i b) (A-i B-C) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {3 i d (a+i b) (A+i B-C) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 (-3 a C d-3 b B d+2 b c C) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 b C \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {-\frac {3 (a+i b) (A+i B-C) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {3 (a-i b) (A-i B-C) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (-3 a C d-3 b B d+2 b c C) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 b C \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {-\frac {3 d (a-i b) (A-i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}-\frac {3 d (a+i b) (A+i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}+\frac {2 (-3 a C d-3 b B d+2 b c C) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}\)

input

Int[((a + b*Tan[e + f*x])*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2))/Sqrt[c 
+ d*Tan[e + f*x]],x]
 

output

(2*b*C*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(3*d*f) - ((-3*(a - I*b)*(A 
- I*B - C)*d*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) - (3*(a 
 + I*b)*(A + I*B - C)*d*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d] 
*f) + (2*(2*b*c*C - 3*b*B*d - 3*a*C*d)*Sqrt[c + d*Tan[e + f*x]])/(d*f))/(3 
*d)
 

3.2.12.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

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]
 

rule 221

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4020

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]
 

rule 4022

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]
 

rule 4113

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + 
 b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si 
mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && 
NeQ[A*b^2 - a*b*B + a^2*C, 0] &&  !LeQ[m, -1]
 

rule 4120

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) 
*(x_)])^(n_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[b*C*Tan[e + f*x]*((c + d*Tan[e + f*x])^(n + 
 1)/(d*f*(n + 2))), x] - Simp[1/(d*(n + 2))   Int[(c + d*Tan[e + f*x])^n*Si 
mp[b*c*C - a*A*d*(n + 2) - (A*b + a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C* 
d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, 
b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && 
  !LtQ[n, -1]
 

3.2.12.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3852\) vs. \(2(166)=332\).

Time = 0.14 (sec) , antiderivative size = 3853, normalized size of antiderivative = 19.86

method	result	size
parts	\(\text {Expression too large to display}\)	\(3853\)
derivativedivides	\(\text {Expression too large to display}\)	\(4138\)
default	\(\text {Expression too large to display}\)	\(4138\)

input

int((a+b*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(1/2 
),x,method=_RETURNVERBOSE)
 

output

A*a*(-1/4/f/d/(c^2+d^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^2-1/ 
4/f*d/(c^2+d^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)+1/4/f/d/(c^2+d 
^2)^(3/2)*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^3+1/4/f*d/(c^2+d^2 
)^(3/2)*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-1/f/d/(c^2+d^2)^(1/2 
)/(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^2-1/f*d/(c^2+d^2)^( 
1/2)/(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))+1/f/d/(c^2+d^2)^(3 
/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^4+3/f*d/(c^2+d^2) 
^(3/2)/(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+2/f*d^3/(c^2 
+d^2)^(3/2)/(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))+1/4/f/d/(c^ 
2+d^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+1/4/f*d/(c^2+d^2...
 

3.2.12.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 13473 vs. \(2 (159) = 318\).

Time = 1.75 (sec) , antiderivative size = 13473, normalized size of antiderivative = 69.45 \[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]

input

integrate((a+b*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e) 
)^(1/2),x, algorithm="fricas")
 

output

Too large to include
 

3.2.12.6 Sympy [F]

\[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right ) \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \]

input

integrate((a+b*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f*x+e)**2)/(c+d*tan(f*x+e 
))**(1/2),x)
 

output

Integral((a + b*tan(e + f*x))*(A + B*tan(e + f*x) + C*tan(e + f*x)**2)/sqr 
t(c + d*tan(e + f*x)), x)
 

3.2.12.7 Maxima [F]

\[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=\int { \frac {{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} {\left (b \tan \left (f x + e\right ) + a\right )}}{\sqrt {d \tan \left (f x + e\right ) + c}} \,d x } \]

input

integrate((a+b*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e) 
)^(1/2),x, algorithm="maxima")
 

output

integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*(b*tan(f*x + e) + a)/sqr 
t(d*tan(f*x + e) + c), x)
 

3.2.12.8 Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Timed out} \]

input

integrate((a+b*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e) 
)^(1/2),x, algorithm="giac")
 

output

Timed out
 

3.2.12.9 Mupad [B] (verification not implemented)

Time = 21.65 (sec) , antiderivative size = 16400, normalized size of antiderivative = 84.54 \[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]

input

int(((a + b*tan(e + f*x))*(A + B*tan(e + f*x) + C*tan(e + f*x)^2))/(c + d* 
tan(e + f*x))^(1/2),x)
 

output

((2*B*b*d - 6*C*b*c)/(d^2*f) + (4*C*b*c)/(d^2*f))*(c + d*tan(e + f*x))^(1/ 
2) - atan(((((8*(4*C*a*d^3*f^2 - 4*A*a*d^3*f^2 + 4*B*a*c*d^2*f^2))/f^3 - 6 
4*c*d^2*(c + d*tan(e + f*x))^(1/2)*((((8*A^2*a^2*c*f^2 - 8*B^2*a^2*c*f^2 + 
 8*C^2*a^2*c*f^2 + 16*A*B*a^2*d*f^2 - 16*A*C*a^2*c*f^2 - 16*B*C*a^2*d*f^2) 
^2/4 - (16*c^2*f^4 + 16*d^2*f^4)*(A^4*a^4 + B^4*a^4 + C^4*a^4 - 4*A*C^3*a^ 
4 - 4*A^3*C*a^4 + 2*A^2*B^2*a^4 + 6*A^2*C^2*a^4 + 2*B^2*C^2*a^4 - 4*A*B^2* 
C*a^4))^(1/2) - 4*A^2*a^2*c*f^2 + 4*B^2*a^2*c*f^2 - 4*C^2*a^2*c*f^2 - 8*A* 
B*a^2*d*f^2 + 8*A*C*a^2*c*f^2 + 8*B*C*a^2*d*f^2)/(16*(c^2*f^4 + d^2*f^4))) 
^(1/2))*((((8*A^2*a^2*c*f^2 - 8*B^2*a^2*c*f^2 + 8*C^2*a^2*c*f^2 + 16*A*B*a 
^2*d*f^2 - 16*A*C*a^2*c*f^2 - 16*B*C*a^2*d*f^2)^2/4 - (16*c^2*f^4 + 16*d^2 
*f^4)*(A^4*a^4 + B^4*a^4 + C^4*a^4 - 4*A*C^3*a^4 - 4*A^3*C*a^4 + 2*A^2*B^2 
*a^4 + 6*A^2*C^2*a^4 + 2*B^2*C^2*a^4 - 4*A*B^2*C*a^4))^(1/2) - 4*A^2*a^2*c 
*f^2 + 4*B^2*a^2*c*f^2 - 4*C^2*a^2*c*f^2 - 8*A*B*a^2*d*f^2 + 8*A*C*a^2*c*f 
^2 + 8*B*C*a^2*d*f^2)/(16*(c^2*f^4 + d^2*f^4)))^(1/2) - (16*(c + d*tan(e + 
 f*x))^(1/2)*(A^2*a^2*d^2 - B^2*a^2*d^2 + C^2*a^2*d^2 - 2*A*C*a^2*d^2))/f^ 
2)*((((8*A^2*a^2*c*f^2 - 8*B^2*a^2*c*f^2 + 8*C^2*a^2*c*f^2 + 16*A*B*a^2*d* 
f^2 - 16*A*C*a^2*c*f^2 - 16*B*C*a^2*d*f^2)^2/4 - (16*c^2*f^4 + 16*d^2*f^4) 
*(A^4*a^4 + B^4*a^4 + C^4*a^4 - 4*A*C^3*a^4 - 4*A^3*C*a^4 + 2*A^2*B^2*a^4 
+ 6*A^2*C^2*a^4 + 2*B^2*C^2*a^4 - 4*A*B^2*C*a^4))^(1/2) - 4*A^2*a^2*c*f^2 
+ 4*B^2*a^2*c*f^2 - 4*C^2*a^2*c*f^2 - 8*A*B*a^2*d*f^2 + 8*A*C*a^2*c*f^2...