3.2.3 \(\int \frac {A+B \cot (c+d x)}{(a+b \cot (c+d x))^{5/2}} \, dx\) [103]

3.2.3.1 Optimal result

Integrand size = 25, antiderivative size = 185 \[ \int \frac {A+B \cot (c+d x)}{(a+b \cot (c+d x))^{5/2}} \, dx=\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \cot (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{5/2} d}-\frac {(i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \cot (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{5/2} d}+\frac {2 (A b-a B)}{3 \left (a^2+b^2\right ) d (a+b \cot (c+d x))^{3/2}}+\frac {2 \left (2 a A b-a^2 B+b^2 B\right )}{\left (a^2+b^2\right )^2 d \sqrt {a+b \cot (c+d x)}} \]

output

(I*A+B)*arctanh((a+b*cot(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(5/2)/d-(I*A 
-B)*arctanh((a+b*cot(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(5/2)/d+2/3*(A*b 
-B*a)/(a^2+b^2)/d/(a+b*cot(d*x+c))^(3/2)+2*(2*A*a*b-B*a^2+B*b^2)/(a^2+b^2) 
^2/d/(a+b*cot(d*x+c))^(1/2)
 

3.2.3.2 Mathematica [A] (verified)

Time = 3.76 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.72 \[ \int \frac {A+B \cot (c+d x)}{(a+b \cot (c+d x))^{5/2}} \, dx=-\frac {\frac {3 \left (2 a b \left (A \sqrt {-b^2}+b B\right )+a^2 \left (A b-\sqrt {-b^2} B\right )+b^2 \left (-A b+\sqrt {-b^2} B\right )\right ) \text {arctanh}\left (\frac {\sqrt {a+b \cot (c+d x)}}{\sqrt {a-\sqrt {-b^2}}}\right )}{\sqrt {-b^2} \sqrt {a-\sqrt {-b^2}}}+\frac {3 \left (2 a b \left (A \sqrt {-b^2}-b B\right )-a^2 \left (A b+\sqrt {-b^2} B\right )+b^2 \left (A b+\sqrt {-b^2} B\right )\right ) \text {arctanh}\left (\frac {\sqrt {a+b \cot (c+d x)}}{\sqrt {a+\sqrt {-b^2}}}\right )}{\sqrt {-b^2} \sqrt {a+\sqrt {-b^2}}}+\frac {2 \left (a^2+b^2\right ) (-A b+a B)}{(a+b \cot (c+d x))^{3/2}}+\frac {6 \left (-2 a A b+a^2 B-b^2 B\right )}{\sqrt {a+b \cot (c+d x)}}}{3 \left (a^2+b^2\right )^2 d} \]

input

Integrate[(A + B*Cot[c + d*x])/(a + b*Cot[c + d*x])^(5/2),x]
 

output

-1/3*((3*(2*a*b*(A*Sqrt[-b^2] + b*B) + a^2*(A*b - Sqrt[-b^2]*B) + b^2*(-(A 
*b) + Sqrt[-b^2]*B))*ArcTanh[Sqrt[a + b*Cot[c + d*x]]/Sqrt[a - Sqrt[-b^2]] 
])/(Sqrt[-b^2]*Sqrt[a - Sqrt[-b^2]]) + (3*(2*a*b*(A*Sqrt[-b^2] - b*B) - a^ 
2*(A*b + Sqrt[-b^2]*B) + b^2*(A*b + Sqrt[-b^2]*B))*ArcTanh[Sqrt[a + b*Cot[ 
c + d*x]]/Sqrt[a + Sqrt[-b^2]]])/(Sqrt[-b^2]*Sqrt[a + Sqrt[-b^2]]) + (2*(a 
^2 + b^2)*(-(A*b) + a*B))/(a + b*Cot[c + d*x])^(3/2) + (6*(-2*a*A*b + a^2* 
B - b^2*B))/Sqrt[a + b*Cot[c + d*x]])/((a^2 + b^2)^2*d)
 

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

Time = 0.99 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4012, 3042, 4012, 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.

\(\displaystyle \int \frac {A+B \cot (c+d x)}{(a+b \cot (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A-B \tan \left (c+d x+\frac {\pi }{2}\right )}{\left (a-b \tan \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\)

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

\(\displaystyle \frac {\frac {\int \frac {A a^2+2 b B a-A b^2-\left (-B a^2+2 A b a+b^2 B\right ) \cot (c+d x)}{\sqrt {a+b \cot (c+d x)}}dx}{a^2+b^2}+\frac {2 \left (a^2 (-B)+2 a A b+b^2 B\right )}{d \left (a^2+b^2\right ) \sqrt {a+b \cot (c+d x)}}}{a^2+b^2}+\frac {2 (A b-a B)}{3 d \left (a^2+b^2\right ) (a+b \cot (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {A a^2+2 b B a-A b^2-\left (B a^2-2 A b a-b^2 B\right ) \tan \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a-b \tan \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2+b^2}+\frac {2 \left (a^2 (-B)+2 a A b+b^2 B\right )}{d \left (a^2+b^2\right ) \sqrt {a+b \cot (c+d x)}}}{a^2+b^2}+\frac {2 (A b-a B)}{3 d \left (a^2+b^2\right ) (a+b \cot (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 (A b-a B)}{3 d \left (a^2+b^2\right ) (a+b \cot (c+d x))^{3/2}}+\frac {\frac {2 \left (a^2 (-B)+2 a A b+b^2 B\right )}{d \left (a^2+b^2\right ) \sqrt {a+b \cot (c+d x)}}+\frac {\frac {1}{2} (a-i b)^2 (A+i B) \int \frac {1-i \cot (c+d x)}{\sqrt {a+b \cot (c+d x)}}dx+\frac {1}{2} (a+i b)^2 (A-i B) \int \frac {i \cot (c+d x)+1}{\sqrt {a+b \cot (c+d x)}}dx}{a^2+b^2}}{a^2+b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 (A b-a B)}{3 d \left (a^2+b^2\right ) (a+b \cot (c+d x))^{3/2}}+\frac {\frac {2 \left (a^2 (-B)+2 a A b+b^2 B\right )}{d \left (a^2+b^2\right ) \sqrt {a+b \cot (c+d x)}}+\frac {\frac {1}{2} (a-i b)^2 (A+i B) \int \frac {i \tan \left (c+d x+\frac {\pi }{2}\right )+1}{\sqrt {a-b \tan \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {1}{2} (a+i b)^2 (A-i B) \int \frac {1-i \tan \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a-b \tan \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2+b^2}}{a^2+b^2}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 (A b-a B)}{3 d \left (a^2+b^2\right ) (a+b \cot (c+d x))^{3/2}}+\frac {\frac {2 \left (a^2 (-B)+2 a A b+b^2 B\right )}{d \left (a^2+b^2\right ) \sqrt {a+b \cot (c+d x)}}+\frac {\frac {i (a-i b)^2 (A+i B) \int -\frac {1}{(i \cot (c+d x)+1) \sqrt {a+b \cot (c+d x)}}d(-i \cot (c+d x))}{2 d}-\frac {i (a+i b)^2 (A-i B) \int -\frac {1}{(1-i \cot (c+d x)) \sqrt {a+b \cot (c+d x)}}d(i \cot (c+d x))}{2 d}}{a^2+b^2}}{a^2+b^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 (A b-a B)}{3 d \left (a^2+b^2\right ) (a+b \cot (c+d x))^{3/2}}+\frac {\frac {2 \left (a^2 (-B)+2 a A b+b^2 B\right )}{d \left (a^2+b^2\right ) \sqrt {a+b \cot (c+d x)}}+\frac {\frac {i (a+i b)^2 (A-i B) \int \frac {1}{(1-i \cot (c+d x)) \sqrt {a+b \cot (c+d x)}}d(i \cot (c+d x))}{2 d}-\frac {i (a-i b)^2 (A+i B) \int \frac {1}{(i \cot (c+d x)+1) \sqrt {a+b \cot (c+d x)}}d(-i \cot (c+d x))}{2 d}}{a^2+b^2}}{a^2+b^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 (A b-a B)}{3 d \left (a^2+b^2\right ) (a+b \cot (c+d x))^{3/2}}+\frac {\frac {2 \left (a^2 (-B)+2 a A b+b^2 B\right )}{d \left (a^2+b^2\right ) \sqrt {a+b \cot (c+d x)}}+\frac {-\frac {(a-i b)^2 (A+i B) \int \frac {1}{-\frac {i \cot ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \cot (c+d x)}}{b d}-\frac {(a+i b)^2 (A-i B) \int \frac {1}{\frac {i \cot ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \cot (c+d x)}}{b d}}{a^2+b^2}}{a^2+b^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 (A b-a B)}{3 d \left (a^2+b^2\right ) (a+b \cot (c+d x))^{3/2}}+\frac {\frac {2 \left (a^2 (-B)+2 a A b+b^2 B\right )}{d \left (a^2+b^2\right ) \sqrt {a+b \cot (c+d x)}}+\frac {-\frac {(a-i b)^2 (A+i B) \arctan \left (\frac {\cot (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}-\frac {(a+i b)^2 (A-i B) \arctan \left (\frac {\cot (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}}{a^2+b^2}}{a^2+b^2}\)

input

Int[(A + B*Cot[c + d*x])/(a + b*Cot[c + d*x])^(5/2),x]
 

output

(2*(A*b - a*B))/(3*(a^2 + b^2)*d*(a + b*Cot[c + d*x])^(3/2)) + ((-(((a + I 
*b)^2*(A - I*B)*ArcTan[Cot[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d)) - ( 
(a - I*b)^2*(A + I*B)*ArcTan[Cot[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d 
))/(a^2 + b^2) + (2*(2*a*A*b - a^2*B + b^2*B))/((a^2 + b^2)*d*Sqrt[a + b*C 
ot[c + d*x]]))/(a^2 + b^2)
 

3.2.3.3.1 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 4012

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

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]
 

3.2.3.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(4471\) vs. \(2(161)=322\).

Time = 0.16 (sec) , antiderivative size = 4472, normalized size of antiderivative = 24.17

method	result	size
parts	\(\text {Expression too large to display}\)	\(4472\)
derivativedivides	\(\text {Expression too large to display}\)	\(12836\)
default	\(\text {Expression too large to display}\)	\(12836\)

input

int((A+B*cot(d*x+c))/(a+b*cot(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

output

A*(2/3/d*b/(a^2+b^2)/(a+b*cot(d*x+c))^(3/2)+2/d*b/(a^2+b^2)^3/(2*(a^2+b^2) 
^(1/2)-2*a)^(1/2)*arctan((2*(a+b*cot(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^3-4/d*b/(a^2+b^2)^(7/2)/(2*(a^2+b 
^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*cot(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^4-4/d*b/(a^2+b^2)^(7/2)/(2*(a^ 
2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*cot(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^4+1/2/d*b/(a^2+b^2)^(7/2)*l 
n(b*cot(d*x+c)+a-(a+b*cot(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/(a^2+b^2)^(7/2)*ln( 
b*cot(d*x+c)+a+(a+b*cot(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^3-1/4/d/b/(a^2+b^2)^(7/2)*ln(b* 
cot(d*x+c)+a-(a+b*cot(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^5+3/4/d*b^3/(a^2+b^2)^(7/2)*ln(b* 
cot(d*x+c)+a-(a+b*cot(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/d*b^3/(a^2+b^2)^3/(2*(a^2+b^2)^ 
(1/2)-2*a)^(1/2)*arctan((2*(a+b*cot(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+1/4/d/b/(a^2+b^2)^3*ln(b*cot(d*x+c 
)+a-(a+b*cot(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^4+1/d/b/(a^2+b^2)^(5/2)/(2*(a^2+b^2)^(1/2) 
-2*a)^(1/2)*arctan((2*(a+b*cot(d*x+c))^(1/2)-(2*(a^2+b^2)^(1/2)+2*a)^(1...
 

3.2.3.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 7422 vs. \(2 (155) = 310\).

Time = 3.60 (sec) , antiderivative size = 7422, normalized size of antiderivative = 40.12 \[ \int \frac {A+B \cot (c+d x)}{(a+b \cot (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

input

integrate((A+B*cot(d*x+c))/(a+b*cot(d*x+c))^(5/2),x, algorithm="fricas")
 

output

Too large to include
 

3.2.3.6 Sympy [F]

\[ \int \frac {A+B \cot (c+d x)}{(a+b \cot (c+d x))^{5/2}} \, dx=\int \frac {A + B \cot {\left (c + d x \right )}}{\left (a + b \cot {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

input

integrate((A+B*cot(d*x+c))/(a+b*cot(d*x+c))**(5/2),x)
 

output

Integral((A + B*cot(c + d*x))/(a + b*cot(c + d*x))**(5/2), x)
 

3.2.3.7 Maxima [F]

\[ \int \frac {A+B \cot (c+d x)}{(a+b \cot (c+d x))^{5/2}} \, dx=\int { \frac {B \cot \left (d x + c\right ) + A}{{\left (b \cot \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate((A+B*cot(d*x+c))/(a+b*cot(d*x+c))^(5/2),x, algorithm="maxima")
 

output

integrate((B*cot(d*x + c) + A)/(b*cot(d*x + c) + a)^(5/2), x)
 

3.2.3.8 Giac [F]

\[ \int \frac {A+B \cot (c+d x)}{(a+b \cot (c+d x))^{5/2}} \, dx=\int { \frac {B \cot \left (d x + c\right ) + A}{{\left (b \cot \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate((A+B*cot(d*x+c))/(a+b*cot(d*x+c))^(5/2),x, algorithm="giac")
 

output

integrate((B*cot(d*x + c) + A)/(b*cot(d*x + c) + a)^(5/2), x)
 

3.2.3.9 Mupad [B] (verification not implemented)

Time = 31.32 (sec) , antiderivative size = 9453, normalized size of antiderivative = 51.10 \[ \int \frac {A+B \cot (c+d x)}{(a+b \cot (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

input

int((A + B*cot(c + d*x))/(a + b*cot(c + d*x))^(5/2),x)
 

output

(log((((a + b*cot(c + d*x))^(1/2)*(320*A^2*a^4*b^14*d^3 - 16*A^2*b^18*d^3 
+ 1024*A^2*a^6*b^12*d^3 + 1440*A^2*a^8*b^10*d^3 + 1024*A^2*a^10*b^8*d^3 + 
320*A^2*a^12*b^6*d^3 - 16*A^2*a^16*b^2*d^3) + ((((320*A^4*a^2*b^8*d^4 - 16 
*A^4*b^10*d^4 - 1760*A^4*a^4*b^6*d^4 + 1600*A^4*a^6*b^4*d^4 - 400*A^4*a^8* 
b^2*d^4)^(1/2) - 4*A^2*a^5*d^2 + 40*A^2*a^3*b^2*d^2 - 20*A^2*a*b^4*d^2)/(a 
^10*d^4 + b^10*d^4 + 5*a^2*b^8*d^4 + 10*a^4*b^6*d^4 + 10*a^6*b^4*d^4 + 5*a 
^8*b^2*d^4))^(1/2)*(896*A*a^6*b^15*d^4 - ((((320*A^4*a^2*b^8*d^4 - 16*A^4* 
b^10*d^4 - 1760*A^4*a^4*b^6*d^4 + 1600*A^4*a^6*b^4*d^4 - 400*A^4*a^8*b^2*d 
^4)^(1/2) - 4*A^2*a^5*d^2 + 40*A^2*a^3*b^2*d^2 - 20*A^2*a*b^4*d^2)/(a^10*d 
^4 + b^10*d^4 + 5*a^2*b^8*d^4 + 10*a^4*b^6*d^4 + 10*a^6*b^4*d^4 + 5*a^8*b^ 
2*d^4))^(1/2)*(a + b*cot(c + d*x))^(1/2)*(64*a*b^22*d^5 + 640*a^3*b^20*d^5 
 + 2880*a^5*b^18*d^5 + 7680*a^7*b^16*d^5 + 13440*a^9*b^14*d^5 + 16128*a^11 
*b^12*d^5 + 13440*a^13*b^10*d^5 + 7680*a^15*b^8*d^5 + 2880*a^17*b^6*d^5 + 
640*a^19*b^4*d^5 + 64*a^21*b^2*d^5))/4 - 160*A*a^2*b^19*d^4 - 128*A*a^4*b^ 
17*d^4 - 32*A*b^21*d^4 + 3136*A*a^8*b^13*d^4 + 4928*A*a^10*b^11*d^4 + 4480 
*A*a^12*b^9*d^4 + 2432*A*a^14*b^7*d^4 + 736*A*a^16*b^5*d^4 + 96*A*a^18*b^3 
*d^4))/4)*(((320*A^4*a^2*b^8*d^4 - 16*A^4*b^10*d^4 - 1760*A^4*a^4*b^6*d^4 
+ 1600*A^4*a^6*b^4*d^4 - 400*A^4*a^8*b^2*d^4)^(1/2) - 4*A^2*a^5*d^2 + 40*A 
^2*a^3*b^2*d^2 - 20*A^2*a*b^4*d^2)/(a^10*d^4 + b^10*d^4 + 5*a^2*b^8*d^4 + 
10*a^4*b^6*d^4 + 10*a^6*b^4*d^4 + 5*a^8*b^2*d^4))^(1/2))/4 - 96*A^3*a^3...