3.4.52 \(\int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx\) [352]

3.4.52.1 Optimal result

Integrand size = 31, antiderivative size = 141 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=-\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{3/2} d}-\frac {(A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{3/2} d}+\frac {2 a (A b-a B)}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \]

output

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

3.4.52.2 Mathematica [A] (verified)

Time = 1.52 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.62 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {\frac {b \left (A \left (b^2-a \sqrt {-b^2}\right )-b \left (a+\sqrt {-b^2}\right ) B\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {-b^2}}}\right )}{\sqrt {-b^2} \sqrt {a-\sqrt {-b^2}}}-\frac {b \left (A \left (b^2+a \sqrt {-b^2}\right )+b \left (-a+\sqrt {-b^2}\right ) B\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+\sqrt {-b^2}}}\right )}{\sqrt {-b^2} \sqrt {a+\sqrt {-b^2}}}+\frac {2 a (A b-a B)}{\sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right ) d} \]

input

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

output

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

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

Time = 0.64 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 4074, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4074

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

input

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

output

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

3.4.52.3.1 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 4074

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

3.4.52.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3687\) vs. \(2(121)=242\).

Time = 0.10 (sec) , antiderivative size = 3688, normalized size of antiderivative = 26.16

method	result	size
parts	\(\text {Expression too large to display}\)	\(3688\)
derivativedivides	\(\text {Expression too large to display}\)	\(7956\)
default	\(\text {Expression too large to display}\)	\(7956\)

input

int(tan(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x,method=_RETURNVER 
BOSE)
 

output

A*(1/4/d/(a^2+b^2)^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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2+1/4/d 
*b^2/(a^2+b^2)^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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-1/2/d/(a^2+b^ 
2)^(5/2)*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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3-1/2/d*b^2/(a^2+b^ 
2)^(5/2)*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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a-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^2+1/d/(a^2+b^2)^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^3-1/d*b^2/(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))+1/d*b^2/(a^2+b^2)^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+2/d*b^2/(a^2+b^2)^(5/ 
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^2+2/d*b^4/(a^2+b^2 
)^(5/2)/(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/d/(a^2+b...
 

3.4.52.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4329 vs. \(2 (115) = 230\).

Time = 0.70 (sec) , antiderivative size = 4329, normalized size of antiderivative = 30.70 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

input

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

output

-1/2*(((a^2*b^2 + b^4)*d*tan(d*x + c) + (a^3*b + a*b^3)*d)*sqrt((6*A*B*a^2 
*b - 2*A*B*b^3 + (A^2 - B^2)*a^3 - 3*(A^2 - B^2)*a*b^2 + (a^6 + 3*a^4*b^2 
+ 3*a^2*b^4 + b^6)*d^2*sqrt(-(4*A^2*B^2*a^6 - 12*(A^3*B - A*B^3)*a^5*b + 3 
*(3*A^4 - 14*A^2*B^2 + 3*B^4)*a^4*b^2 + 40*(A^3*B - A*B^3)*a^3*b^3 - 6*(A^ 
4 - 8*A^2*B^2 + B^4)*a^2*b^4 - 12*(A^3*B - A*B^3)*a*b^5 + (A^4 - 2*A^2*B^2 
 + B^4)*b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 
6*a^2*b^10 + b^12)*d^4)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2))*log(- 
(2*(A^3*B + A*B^3)*a^3 - 3*(A^4 - B^4)*a^2*b - 6*(A^3*B + A*B^3)*a*b^2 + ( 
A^4 - B^4)*b^3)*sqrt(b*tan(d*x + c) + a) + ((B*a^8 - 2*A*a^7*b + 2*B*a^6*b 
^2 - 6*A*a^5*b^3 - 6*A*a^3*b^5 - 2*B*a^2*b^6 - 2*A*a*b^7 - B*b^8)*d^3*sqrt 
(-(4*A^2*B^2*a^6 - 12*(A^3*B - A*B^3)*a^5*b + 3*(3*A^4 - 14*A^2*B^2 + 3*B^ 
4)*a^4*b^2 + 40*(A^3*B - A*B^3)*a^3*b^3 - 6*(A^4 - 8*A^2*B^2 + B^4)*a^2*b^ 
4 - 12*(A^3*B - A*B^3)*a*b^5 + (A^4 - 2*A^2*B^2 + B^4)*b^6)/((a^12 + 6*a^1 
0*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) - 
(2*A^2*B*a^5 - (3*A^3 - 7*A*B^2)*a^4*b - 2*(7*A^2*B - 3*B^3)*a^3*b^2 + 4*( 
A^3 - 4*A*B^2)*a^2*b^3 + 2*(4*A^2*B - B^3)*a*b^4 - (A^3 - A*B^2)*b^5)*d)*s 
qrt((6*A*B*a^2*b - 2*A*B*b^3 + (A^2 - B^2)*a^3 - 3*(A^2 - B^2)*a*b^2 + (a^ 
6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(4*A^2*B^2*a^6 - 12*(A^3*B - A* 
B^3)*a^5*b + 3*(3*A^4 - 14*A^2*B^2 + 3*B^4)*a^4*b^2 + 40*(A^3*B - A*B^3)*a 
^3*b^3 - 6*(A^4 - 8*A^2*B^2 + B^4)*a^2*b^4 - 12*(A^3*B - A*B^3)*a*b^5 +...
 

3.4.52.6 Sympy [F]

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

input

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

output

Integral((A + B*tan(c + d*x))*tan(c + d*x)/(a + b*tan(c + d*x))**(3/2), x)
 

3.4.52.7 Maxima [F]

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

input

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

output

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

3.4.52.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.4.52.9 Mupad [B] (verification not implemented)

Time = 15.34 (sec) , antiderivative size = 5742, normalized size of antiderivative = 40.72 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

input

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

output

(log(- ((((((96*A^4*a^2*b^4*d^4 - 16*A^4*b^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1 
/2) + 4*A^2*a^3*d^2 - 12*A^2*a*b^2*d^2)/(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 
 + 3*a^4*b^2*d^4))^(1/2)*(32*A*b^12*d^4 + ((((96*A^4*a^2*b^4*d^4 - 16*A^4* 
b^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) + 4*A^2*a^3*d^2 - 12*A^2*a*b^2*d^2)/( 
a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2)*(a + b*tan(c + d 
*x))^(1/2)*(64*a*b^12*d^5 + 320*a^3*b^10*d^5 + 640*a^5*b^8*d^5 + 640*a^7*b 
^6*d^5 + 320*a^9*b^4*d^5 + 64*a^11*b^2*d^5))/4 + 96*A*a^2*b^10*d^4 + 64*A* 
a^4*b^8*d^4 - 64*A*a^6*b^6*d^4 - 96*A*a^8*b^4*d^4 - 32*A*a^10*b^2*d^4))/4 
+ (a + b*tan(c + d*x))^(1/2)*(16*A^2*b^10*d^3 + 32*A^2*a^2*b^8*d^3 - 32*A^ 
2*a^6*b^4*d^3 - 16*A^2*a^8*b^2*d^3))*(((96*A^4*a^2*b^4*d^4 - 16*A^4*b^6*d^ 
4 - 144*A^4*a^4*b^2*d^4)^(1/2) + 4*A^2*a^3*d^2 - 12*A^2*a*b^2*d^2)/(a^6*d^ 
4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2))/4 - 24*A^3*a^3*b^6*d^ 
2 - 24*A^3*a^5*b^4*d^2 - 8*A^3*a^7*b^2*d^2 - 8*A^3*a*b^8*d^2)*(((96*A^4*a^ 
2*b^4*d^4 - 16*A^4*b^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) + 4*A^2*a^3*d^2 - 
12*A^2*a*b^2*d^2)/(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/ 
2))/4 + (log(- ((((-((96*A^4*a^2*b^4*d^4 - 16*A^4*b^6*d^4 - 144*A^4*a^4*b^ 
2*d^4)^(1/2) - 4*A^2*a^3*d^2 + 12*A^2*a*b^2*d^2)/(a^6*d^4 + b^6*d^4 + 3*a^ 
2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2)*(32*A*b^12*d^4 + ((-((96*A^4*a^2*b^4*d^4 
 - 16*A^4*b^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) - 4*A^2*a^3*d^2 + 12*A^2*a* 
b^2*d^2)/(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2)*(a ...