3.4.67 \(\int \frac {\cot (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx\) [367]

3.4.67.1 Optimal result

Integrand size = 34, antiderivative size = 119 \[ \int \frac {\cot (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=-\frac {2 B \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {B \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b} d}+\frac {B \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b} d} \]

output

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

3.4.67.2 Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.94 \[ \int \frac {\cot (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {B \left (-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b}}\right )}{d} \]

input

Integrate[(Cot[c + d*x]*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(3/ 
2),x]
 

output

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

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

Time = 0.88 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.91, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {2011, 3042, 4057, 25, 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.

\(\displaystyle \int \frac {\cot (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4057

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

input

Int[(Cot[c + d*x]*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(3/2),x]
 

output

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

3.4.67.3.1 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 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

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 4057

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]
 

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]
 

3.4.67.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(373\) vs. \(2(97)=194\).

Time = 0.34 (sec) , antiderivative size = 374, normalized size of antiderivative = 3.14

method	result	size
default	\(\frac {2 B \,b^{2} \left (\frac {\frac {\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{4 \sqrt {a^{2}+b^{2}}}+\frac {-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (\sqrt {a^{2}+b^{2}}-a \right ) \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{4 \sqrt {a^{2}+b^{2}}}}{b^{2}}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{b^{2} \sqrt {a}}\right )}{d}\)	\(374\)

input

int(cot(d*x+c)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x,method=_RETUR 
NVERBOSE)
 

output

2*B/d*b^2*(1/b^2*(1/4/(a^2+b^2)^(1/2)*(1/2*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*l 
n(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))+2*(a-(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)))+1/4/(a^2+b^2)^(1/2)*(-1/2*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*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))+2*((a^2+b^2)^(1/2)-a)/(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))))-1/b^2/a^(1/2)*arctanh((a+b*tan(d*x+c))^(1/2)/a^(1/2)))
 

3.4.67.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 829 vs. \(2 (93) = 186\).

Time = 0.29 (sec) , antiderivative size = 1674, normalized size of antiderivative = 14.07 \[ \int \frac {\cot (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

input

integrate(cot(d*x+c)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x, algori 
thm="fricas")
 

output

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

3.4.67.6 Sympy [F]

\[ \int \frac {\cot (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=B \int \frac {\cot {\left (c + d x \right )}}{\sqrt {a + b \tan {\left (c + d x \right )}}}\, dx \]

input

integrate(cot(d*x+c)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c))**(3/2),x)
 

output

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

3.4.67.7 Maxima [F(-1)]

Timed out. \[ \int \frac {\cot (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]

input

integrate(cot(d*x+c)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x, algori 
thm="maxima")
 

output

Timed out
 

3.4.67.8 Giac [F(-1)]

Timed out. \[ \int \frac {\cot (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]

input

integrate(cot(d*x+c)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x, algori 
thm="giac")
 

output

Timed out
 

3.4.67.9 Mupad [B] (verification not implemented)

Time = 10.11 (sec) , antiderivative size = 2142, normalized size of antiderivative = 18.00 \[ \int \frac {\cot (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

input

int((cot(c + d*x)*(B*a + B*b*tan(c + d*x)))/(a + b*tan(c + d*x))^(3/2),x)
 

output

- atan(((((((32*(16*B*b^10*d^2 + 12*B*a^2*b^8*d^2))/d^3 - (32*(16*b^10*d^4 
 + 24*a^2*b^8*d^4)*(a + b*tan(c + d*x))^(1/2)*(B^2/(4*(a*d^2 - b*d^2*1i))) 
^(1/2))/d^4)*(B^2/(4*(a*d^2 - b*d^2*1i)))^(1/2) + (576*B^2*a*b^8*(a + b*ta 
n(c + d*x))^(1/2))/d^2)*(B^2/(4*(a*d^2 - b*d^2*1i)))^(1/2) - (96*B^3*a*b^8 
)/d^3)*(B^2/(4*(a*d^2 - b*d^2*1i)))^(1/2) - (96*B^4*b^8*(a + b*tan(c + d*x 
))^(1/2))/d^4)*(B^2/(4*(a*d^2 - b*d^2*1i)))^(1/2)*1i - (((((32*(16*B*b^10* 
d^2 + 12*B*a^2*b^8*d^2))/d^3 + (32*(16*b^10*d^4 + 24*a^2*b^8*d^4)*(a + b*t 
an(c + d*x))^(1/2)*(B^2/(4*(a*d^2 - b*d^2*1i)))^(1/2))/d^4)*(B^2/(4*(a*d^2 
 - b*d^2*1i)))^(1/2) - (576*B^2*a*b^8*(a + b*tan(c + d*x))^(1/2))/d^2)*(B^ 
2/(4*(a*d^2 - b*d^2*1i)))^(1/2) - (96*B^3*a*b^8)/d^3)*(B^2/(4*(a*d^2 - b*d 
^2*1i)))^(1/2) + (96*B^4*b^8*(a + b*tan(c + d*x))^(1/2))/d^4)*(B^2/(4*(a*d 
^2 - b*d^2*1i)))^(1/2)*1i)/((((((32*(16*B*b^10*d^2 + 12*B*a^2*b^8*d^2))/d^ 
3 - (32*(16*b^10*d^4 + 24*a^2*b^8*d^4)*(a + b*tan(c + d*x))^(1/2)*(B^2/(4* 
(a*d^2 - b*d^2*1i)))^(1/2))/d^4)*(B^2/(4*(a*d^2 - b*d^2*1i)))^(1/2) + (576 
*B^2*a*b^8*(a + b*tan(c + d*x))^(1/2))/d^2)*(B^2/(4*(a*d^2 - b*d^2*1i)))^( 
1/2) - (96*B^3*a*b^8)/d^3)*(B^2/(4*(a*d^2 - b*d^2*1i)))^(1/2) - (96*B^4*b^ 
8*(a + b*tan(c + d*x))^(1/2))/d^4)*(B^2/(4*(a*d^2 - b*d^2*1i)))^(1/2) + (( 
(((32*(16*B*b^10*d^2 + 12*B*a^2*b^8*d^2))/d^3 + (32*(16*b^10*d^4 + 24*a^2* 
b^8*d^4)*(a + b*tan(c + d*x))^(1/2)*(B^2/(4*(a*d^2 - b*d^2*1i)))^(1/2))/d^ 
4)*(B^2/(4*(a*d^2 - b*d^2*1i)))^(1/2) - (576*B^2*a*b^8*(a + b*tan(c + d...