3.4.47 \(\int \frac {\cot (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx\) [347]

3.4.47.1 Optimal result

Integrand size = 31, antiderivative size = 131 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=-\frac {2 A \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b} d}+\frac {(A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b} d} \]

output

-2*A*arctanh((a+b*tan(d*x+c))^(1/2)/a^(1/2))/d/a^(1/2)+(A-I*B)*arctanh((a+ 
b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d/(a-I*b)^(1/2)+(A+I*B)*arctanh((a+b*ta 
n(d*x+c))^(1/2)/(a+I*b)^(1/2))/d/(a+I*b)^(1/2)
 

3.4.47.2 Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.30 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=-\frac {\frac {2 A \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {\frac {\left (-A b+\sqrt {-b^2} B\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {-b^2}}}\right )}{\sqrt {a-\sqrt {-b^2}}}-\frac {\left (A b+\sqrt {-b^2} B\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+\sqrt {-b^2}}}\right )}{\sqrt {a+\sqrt {-b^2}}}}{b}}{d} \]

input

Integrate[(Cot[c + d*x]*(A + B*Tan[c + d*x]))/Sqrt[a + b*Tan[c + d*x]],x]
 

output

-(((2*A*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/Sqrt[a] + (((-(A*b) + S 
qrt[-b^2]*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - Sqrt[-b^2]]])/Sqrt[ 
a - Sqrt[-b^2]] - ((A*b + Sqrt[-b^2]*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/S 
qrt[a + Sqrt[-b^2]]])/Sqrt[a + Sqrt[-b^2]])/b)/d)
 

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

Time = 0.94 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.90, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4096, 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.

\(\displaystyle \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx\)

\(\Big \downarrow \) 4096

\(\displaystyle \int \frac {B-A \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+A \int \frac {\cot (c+d x) \left (\tan ^2(c+d x)+1\right )}{\sqrt {a+b \tan (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {B-A \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+A \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx\)

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle A \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx+\frac {(B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}-\frac {(-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {A \int \frac {\cot (c+d x)}{\sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{d}+\frac {(B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}-\frac {(-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 A \int \frac {1}{\frac {a+b \tan (c+d x)}{b}-\frac {a}{b}}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}-\frac {(-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}-\frac {(-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}-\frac {2 A \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}\)

input

Int[(Cot[c + d*x]*(A + B*Tan[c + d*x]))/Sqrt[a + b*Tan[c + d*x]],x]
 

output

((I*A + B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) - ((I*A - 
 B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d) - (2*A*ArcTanh[S 
qrt[a + b*Tan[c + d*x]]/Sqrt[a]])/(Sqrt[a]*d)
 

3.4.47.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 4096

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

rule 4117

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]
 

3.4.47.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(4001\) vs. \(2(107)=214\).

Time = 0.28 (sec) , antiderivative size = 4002, normalized size of antiderivative = 30.55

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(4002\)
default	\(\text {Expression too large to display}\)	\(4002\)

input

int(cot(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2),x,method=_RETURNVER 
BOSE)
 

output

1/d/(a^2+b^2)^(3/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c 
))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*A*a 
^3+1/4/d/(a^2+b^2)^(3/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+ 
b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*A*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2 
-1/4/d/(a^2+b^2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b 
*tan(d*x+c)-a-(a^2+b^2)^(1/2))*A*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a+2/d/(a^2+ 
b^2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2 
*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*A*a^2-1/d/(a^2+b^2 
)^(1/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2* 
(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*A*a-1/d/(a^2+b^ 
2)^(3/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/ 
2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*A*a^3-1/4/d/(a 
^2+b^2)^(3/2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*ta 
n(d*x+c)-a-(a^2+b^2)^(1/2))*A*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2+1/d/(a^2+b 
^2)^(1/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1 
/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*A*a+1/4/d/(a^ 
2+b^2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1 
/2)+(a^2+b^2)^(1/2))*A*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a-2/d/(a^2+b^2)/(2*(a 
^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/ 
2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*A*a^2+2/d*b^3/(a^2+b^2)^(...
 

3.4.47.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1721 vs. \(2 (101) = 202\).

Time = 0.82 (sec) , antiderivative size = 3458, normalized size of antiderivative = 26.40 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]

input

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

output

[-1/2*(a*d*sqrt(((a^2 + b^2)*d^2*sqrt(-(4*A^2*B^2*a^2 - 4*(A^3*B - A*B^3)* 
a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + 2*A*B* 
b + (A^2 - B^2)*a)/((a^2 + b^2)*d^2))*log((2*(A^3*B + A*B^3)*a - (A^4 - B^ 
4)*b)*sqrt(b*tan(d*x + c) + a) + ((B*a^3 - A*a^2*b + B*a*b^2 - A*b^3)*d^3* 
sqrt(-(4*A^2*B^2*a^2 - 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2 
)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - (2*A^2*B*a^2 - (A^3 - 3*A*B^2)*a*b - (A 
^2*B - B^3)*b^2)*d)*sqrt(((a^2 + b^2)*d^2*sqrt(-(4*A^2*B^2*a^2 - 4*(A^3*B 
- A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) 
 + 2*A*B*b + (A^2 - B^2)*a)/((a^2 + b^2)*d^2))) - a*d*sqrt(((a^2 + b^2)*d^ 
2*sqrt(-(4*A^2*B^2*a^2 - 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b 
^2)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + 2*A*B*b + (A^2 - B^2)*a)/((a^2 + b^2) 
*d^2))*log((2*(A^3*B + A*B^3)*a - (A^4 - B^4)*b)*sqrt(b*tan(d*x + c) + a) 
- ((B*a^3 - A*a^2*b + B*a*b^2 - A*b^3)*d^3*sqrt(-(4*A^2*B^2*a^2 - 4*(A^3*B 
 - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^4) 
) - (2*A^2*B*a^2 - (A^3 - 3*A*B^2)*a*b - (A^2*B - B^3)*b^2)*d)*sqrt(((a^2 
+ b^2)*d^2*sqrt(-(4*A^2*B^2*a^2 - 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 
 + B^4)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + 2*A*B*b + (A^2 - B^2)*a)/((a 
^2 + b^2)*d^2))) - a*d*sqrt(-((a^2 + b^2)*d^2*sqrt(-(4*A^2*B^2*a^2 - 4*(A^ 
3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d 
^4)) - 2*A*B*b - (A^2 - B^2)*a)/((a^2 + b^2)*d^2))*log((2*(A^3*B + A*B^...
 

3.4.47.6 Sympy [F]

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

input

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

output

Integral((A + B*tan(c + d*x))*cot(c + d*x)/sqrt(a + b*tan(c + d*x)), x)
 

3.4.47.7 Maxima [F]

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

input

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

output

integrate((B*tan(d*x + c) + A)*cot(d*x + c)/sqrt(b*tan(d*x + c) + a), x)
 

3.4.47.8 Giac [F(-1)]

Timed out. \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.4.47.9 Mupad [B] (verification not implemented)

Time = 12.12 (sec) , antiderivative size = 7099, normalized size of antiderivative = 54.19 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]

input

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

output

- atan(((((32*(12*A^2*B*b^9*d^2 + 3*A^3*a*b^8*d^2 - 9*A*B^2*a*b^8*d^2))/d^ 
5 - (((32*(16*A*b^10*d^4 - 4*B*a*b^9*d^4 + 12*A*a^2*b^8*d^4))/d^5 - (32*(1 
6*b^10*d^4 + 24*a^2*b^8*d^4)*(a + b*tan(c + d*x))^(1/2)*(-(((8*A^2*a*d^2 - 
 8*B^2*a*d^2 + 16*A*B*b*d^2)^2/4 - (16*a^2*d^4 + 16*b^2*d^4)*(A^4 + 2*A^2* 
B^2 + B^4))^(1/2) - 4*A^2*a*d^2 + 4*B^2*a*d^2 - 8*A*B*b*d^2)/(16*(a^2*d^4 
+ b^2*d^4)))^(1/2))/d^4)*(-(((8*A^2*a*d^2 - 8*B^2*a*d^2 + 16*A*B*b*d^2)^2/ 
4 - (16*a^2*d^4 + 16*b^2*d^4)*(A^4 + 2*A^2*B^2 + B^4))^(1/2) - 4*A^2*a*d^2 
 + 4*B^2*a*d^2 - 8*A*B*b*d^2)/(16*(a^2*d^4 + b^2*d^4)))^(1/2) + (32*(a + b 
*tan(c + d*x))^(1/2)*(16*A*B*b^9*d^2 + 18*A^2*a*b^8*d^2 - 10*B^2*a*b^8*d^2 
))/d^4)*(-(((8*A^2*a*d^2 - 8*B^2*a*d^2 + 16*A*B*b*d^2)^2/4 - (16*a^2*d^4 + 
 16*b^2*d^4)*(A^4 + 2*A^2*B^2 + B^4))^(1/2) - 4*A^2*a*d^2 + 4*B^2*a*d^2 - 
8*A*B*b*d^2)/(16*(a^2*d^4 + b^2*d^4)))^(1/2))*(-(((8*A^2*a*d^2 - 8*B^2*a*d 
^2 + 16*A*B*b*d^2)^2/4 - (16*a^2*d^4 + 16*b^2*d^4)*(A^4 + 2*A^2*B^2 + B^4) 
)^(1/2) - 4*A^2*a*d^2 + 4*B^2*a*d^2 - 8*A*B*b*d^2)/(16*(a^2*d^4 + b^2*d^4) 
))^(1/2) + (32*(3*A^4*b^8 + B^4*b^8)*(a + b*tan(c + d*x))^(1/2))/d^4)*(-(( 
(8*A^2*a*d^2 - 8*B^2*a*d^2 + 16*A*B*b*d^2)^2/4 - (16*a^2*d^4 + 16*b^2*d^4) 
*(A^4 + 2*A^2*B^2 + B^4))^(1/2) - 4*A^2*a*d^2 + 4*B^2*a*d^2 - 8*A*B*b*d^2) 
/(16*(a^2*d^4 + b^2*d^4)))^(1/2)*1i - (((32*(12*A^2*B*b^9*d^2 + 3*A^3*a*b^ 
8*d^2 - 9*A*B^2*a*b^8*d^2))/d^5 - (((32*(16*A*b^10*d^4 - 4*B*a*b^9*d^4 + 1 
2*A*a^2*b^8*d^4))/d^5 + (32*(16*b^10*d^4 + 24*a^2*b^8*d^4)*(a + b*tan(c...