3.1.95 \(\int (a+b \cot (c+d x))^{5/2} (A+B \cot (c+d x)) \, dx\) [95]

3.1.95.1 Optimal result

Integrand size = 25, antiderivative size = 188 \[ \int (a+b \cot (c+d x))^{5/2} (A+B \cot (c+d x)) \, dx=\frac {(a-i b)^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \cot (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{5/2} (i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \cot (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 \left (2 a A b+a^2 B-b^2 B\right ) \sqrt {a+b \cot (c+d x)}}{d}-\frac {2 (A b+a B) (a+b \cot (c+d x))^{3/2}}{3 d}-\frac {2 B (a+b \cot (c+d x))^{5/2}}{5 d} \]

output

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

3.1.95.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(379\) vs. \(2(188)=376\).

Time = 1.91 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.02 \[ \int (a+b \cot (c+d x))^{5/2} (A+B \cot (c+d x)) \, dx=-\frac {2 \left (\frac {\sqrt {a-\sqrt {-b^2}} \left (-3 a^2 b \left (A \sqrt {-b^2}+b B\right )+b^3 \left (A \sqrt {-b^2}+b B\right )+a^3 \left (A b-\sqrt {-b^2} B\right )+3 a b^2 \left (-A b+\sqrt {-b^2} B\right )\right ) \text {arctanh}\left (\frac {\sqrt {a+b \cot (c+d x)}}{\sqrt {a-\sqrt {-b^2}}}\right )}{2 \left (b^2+a \sqrt {-b^2}\right )}+\frac {\left (b^3 \left (A \sqrt {-b^2}-b B\right )+3 a^2 b \left (-A \sqrt {-b^2}+b B\right )-a^3 \left (A b+\sqrt {-b^2} B\right )+3 a b^2 \left (A b+\sqrt {-b^2} B\right )\right ) \text {arctanh}\left (\frac {\sqrt {a+b \cot (c+d x)}}{\sqrt {a+\sqrt {-b^2}}}\right )}{2 \sqrt {-b^2} \sqrt {a+\sqrt {-b^2}}}+\left (2 a A b+a^2 B-b^2 B\right ) \sqrt {a+b \cot (c+d x)}+\frac {1}{3} (A b+a B) (a+b \cot (c+d x))^{3/2}+\frac {1}{5} B (a+b \cot (c+d x))^{5/2}\right )}{d} \]

input

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

output

(-2*((Sqrt[a - Sqrt[-b^2]]*(-3*a^2*b*(A*Sqrt[-b^2] + b*B) + b^3*(A*Sqrt[-b 
^2] + b*B) + a^3*(A*b - Sqrt[-b^2]*B) + 3*a*b^2*(-(A*b) + Sqrt[-b^2]*B))*A 
rcTanh[Sqrt[a + b*Cot[c + d*x]]/Sqrt[a - Sqrt[-b^2]]])/(2*(b^2 + a*Sqrt[-b 
^2])) + ((b^3*(A*Sqrt[-b^2] - b*B) + 3*a^2*b*(-(A*Sqrt[-b^2]) + b*B) - a^3 
*(A*b + Sqrt[-b^2]*B) + 3*a*b^2*(A*b + Sqrt[-b^2]*B))*ArcTanh[Sqrt[a + b*C 
ot[c + d*x]]/Sqrt[a + Sqrt[-b^2]]])/(2*Sqrt[-b^2]*Sqrt[a + Sqrt[-b^2]]) + 
(2*a*A*b + a^2*B - b^2*B)*Sqrt[a + b*Cot[c + d*x]] + ((A*b + a*B)*(a + b*C 
ot[c + d*x])^(3/2))/3 + (B*(a + b*Cot[c + d*x])^(5/2))/5))/d
 

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

Time = 1.05 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.91, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4011, 3042, 4011, 3042, 4011, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

\(\displaystyle \int (a+b \cot (c+d x))^{3/2} (a A-b B+(A b+a B) \cot (c+d x))dx-\frac {2 B (a+b \cot (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a-b \tan \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (a A-b B-(A b+a B) \tan \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 B (a+b \cot (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \int \sqrt {a+b \cot (c+d x)} \left (A a^2-2 b B a-A b^2+\left (B a^2+2 A b a-b^2 B\right ) \cot (c+d x)\right )dx-\frac {2 (a B+A b) (a+b \cot (c+d x))^{3/2}}{3 d}-\frac {2 B (a+b \cot (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {a-b \tan \left (c+d x+\frac {\pi }{2}\right )} \left (A a^2-2 b B a-A b^2-\left (B a^2+2 A b a-b^2 B\right ) \tan \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 (a B+A b) (a+b \cot (c+d x))^{3/2}}{3 d}-\frac {2 B (a+b \cot (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \int \frac {A a^3-3 b B a^2-3 A b^2 a+b^3 B+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \cot (c+d x)}{\sqrt {a+b \cot (c+d x)}}dx-\frac {2 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \cot (c+d x)}}{d}-\frac {2 (a B+A b) (a+b \cot (c+d x))^{3/2}}{3 d}-\frac {2 B (a+b \cot (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A a^3-3 b B a^2-3 A b^2 a+b^3 B-\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a-b \tan \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \cot (c+d x)}}{d}-\frac {2 (a B+A b) (a+b \cot (c+d x))^{3/2}}{3 d}-\frac {2 B (a+b \cot (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {1}{2} (a-i b)^3 (A-i B) \int \frac {i \cot (c+d x)+1}{\sqrt {a+b \cot (c+d x)}}dx+\frac {1}{2} (a+i b)^3 (A+i B) \int \frac {1-i \cot (c+d x)}{\sqrt {a+b \cot (c+d x)}}dx-\frac {2 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \cot (c+d x)}}{d}-\frac {2 (a B+A b) (a+b \cot (c+d x))^{3/2}}{3 d}-\frac {2 B (a+b \cot (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} (a-i b)^3 (A-i B) \int \frac {1-i \tan \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a-b \tan \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {1}{2} (a+i b)^3 (A+i B) \int \frac {i \tan \left (c+d x+\frac {\pi }{2}\right )+1}{\sqrt {a-b \tan \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \cot (c+d x)}}{d}-\frac {2 (a B+A b) (a+b \cot (c+d x))^{3/2}}{3 d}-\frac {2 B (a+b \cot (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 4020

\(\displaystyle -\frac {i (a-i b)^3 (A-i B) \int -\frac {1}{(1-i \cot (c+d x)) \sqrt {a+b \cot (c+d x)}}d(i \cot (c+d x))}{2 d}+\frac {i (a+i b)^3 (A+i B) \int -\frac {1}{(i \cot (c+d x)+1) \sqrt {a+b \cot (c+d x)}}d(-i \cot (c+d x))}{2 d}-\frac {2 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \cot (c+d x)}}{d}-\frac {2 (a B+A b) (a+b \cot (c+d x))^{3/2}}{3 d}-\frac {2 B (a+b \cot (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {i (a-i b)^3 (A-i B) \int \frac {1}{(1-i \cot (c+d x)) \sqrt {a+b \cot (c+d x)}}d(i \cot (c+d x))}{2 d}-\frac {i (a+i b)^3 (A+i B) \int \frac {1}{(i \cot (c+d x)+1) \sqrt {a+b \cot (c+d x)}}d(-i \cot (c+d x))}{2 d}-\frac {2 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \cot (c+d x)}}{d}-\frac {2 (a B+A b) (a+b \cot (c+d x))^{3/2}}{3 d}-\frac {2 B (a+b \cot (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {(a-i b)^3 (A-i B) \int \frac {1}{\frac {i \cot ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \cot (c+d x)}}{b d}-\frac {(a+i b)^3 (A+i B) \int \frac {1}{-\frac {i \cot ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \cot (c+d x)}}{b d}-\frac {2 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \cot (c+d x)}}{d}-\frac {2 (a B+A b) (a+b \cot (c+d x))^{3/2}}{3 d}-\frac {2 B (a+b \cot (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {2 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \cot (c+d x)}}{d}-\frac {(a-i b)^{5/2} (A-i B) \arctan \left (\frac {\cot (c+d x)}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{5/2} (A+i B) \arctan \left (\frac {\cot (c+d x)}{\sqrt {a+i b}}\right )}{d}-\frac {2 (a B+A b) (a+b \cot (c+d x))^{3/2}}{3 d}-\frac {2 B (a+b \cot (c+d x))^{5/2}}{5 d}\)

input

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

output

-(((a - I*b)^(5/2)*(A - I*B)*ArcTan[Cot[c + d*x]/Sqrt[a - I*b]])/d) - ((a 
+ I*b)^(5/2)*(A + I*B)*ArcTan[Cot[c + d*x]/Sqrt[a + I*b]])/d - (2*(2*a*A*b 
 + a^2*B - b^2*B)*Sqrt[a + b*Cot[c + d*x]])/d - (2*(A*b + a*B)*(a + b*Cot[ 
c + d*x])^(3/2))/(3*d) - (2*B*(a + b*Cot[c + d*x])^(5/2))/(5*d)
 

3.1.95.3.1 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 4011

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

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]
 

3.1.95.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2391\) vs. \(2(160)=320\).

Time = 0.17 (sec) , antiderivative size = 2392, normalized size of antiderivative = 12.72

method	result	size
parts	\(\text {Expression too large to display}\)	\(2392\)
derivativedivides	\(\text {Expression too large to display}\)	\(2405\)
default	\(\text {Expression too large to display}\)	\(2405\)

input

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

output

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

3.1.95.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5073 vs. \(2 (154) = 308\).

Time = 1.22 (sec) , antiderivative size = 5073, normalized size of antiderivative = 26.98 \[ \int (a+b \cot (c+d x))^{5/2} (A+B \cot (c+d x)) \, dx=\text {Too large to display} \]

input

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

output

Too large to include
 

3.1.95.6 Sympy [F]

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

input

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

output

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

3.1.95.7 Maxima [F]

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

input

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

output

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

3.1.95.8 Giac [F]

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

input

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

output

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

3.1.95.9 Mupad [B] (verification not implemented)

Time = 45.24 (sec) , antiderivative size = 3864, normalized size of antiderivative = 20.55 \[ \int (a+b \cot (c+d x))^{5/2} (A+B \cot (c+d x)) \, dx=\text {Too large to display} \]

input

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

output

log((8*B^3*a*b^2*(a^2 - 3*b^2)*(a^2 + b^2)^3)/d^3 - ((((((-B^4*b^2*d^4*(5* 
a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + B^2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 + 5*B^ 
2*a*b^4*d^2)/d^4)^(1/2)*(32*B*a^4*b^2 - 32*B*b^6 + 32*a*b^2*d*(((-B^4*b^2* 
d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + B^2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 
 + 5*B^2*a*b^4*d^2)/d^4)^(1/2)*(a + b*cot(c + d*x))^(1/2)))/(2*d) - (16*B^ 
2*b^2*(a + b*cot(c + d*x))^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^ 
2)*(((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + B^2*a^5*d^2 - 10* 
B^2*a^3*b^2*d^2 + 5*B^2*a*b^4*d^2)/d^4)^(1/2))/2)*((20*B^4*a^2*b^8*d^4 - B 
^4*b^10*d^4 - 110*B^4*a^4*b^6*d^4 + 100*B^4*a^6*b^4*d^4 - 25*B^4*a^8*b^2*d 
^4)^(1/2)/(4*d^4) + (B^2*a^5)/(4*d^2) - (5*B^2*a^3*b^2)/(2*d^2) + (5*B^2*a 
*b^4)/(4*d^2))^(1/2) - log(((((((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2 
)^(1/2) + B^2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 + 5*B^2*a*b^4*d^2)/d^4)^(1/2)*( 
32*B*b^6 - 32*B*a^4*b^2 + 32*a*b^2*d*(((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2 
*b^2)^2)^(1/2) + B^2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 + 5*B^2*a*b^4*d^2)/d^4)^ 
(1/2)*(a + b*cot(c + d*x))^(1/2)))/(2*d) - (16*B^2*b^2*(a + b*cot(c + d*x) 
)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2)*(((-B^4*b^2*d^4*(5*a^4 
 + b^4 - 10*a^2*b^2)^2)^(1/2) + B^2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 + 5*B^2*a 
*b^4*d^2)/d^4)^(1/2))/2 + (8*B^3*a*b^2*(a^2 - 3*b^2)*(a^2 + b^2)^3)/d^3)*( 
((20*B^4*a^2*b^8*d^4 - B^4*b^10*d^4 - 110*B^4*a^4*b^6*d^4 + 100*B^4*a^6*b^ 
4*d^4 - 25*B^4*a^8*b^2*d^4)^(1/2) + B^2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 + ...