\(\int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [132]

Optimal result

Integrand size = 24, antiderivative size = 159 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7}{4 a^2 d \sqrt {a+i a \tan (c+d x)}} \] Output:

-2*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/a^(5/2)/d+1/8*arctanh(1/2*(a+ 
I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/a^(5/2)/d+1/5/d/(a+I*a*tan( 
d*x+c))^(5/2)+1/2/a/d/(a+I*a*tan(d*x+c))^(3/2)+7/4/a^2/d/(a+I*a*tan(d*x+c) 
)^(1/2)
 

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.90 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {-\frac {80 \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {5 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{a^{5/2}}+\frac {8}{(a+i a \tan (c+d x))^{5/2}}+\frac {20}{a (a+i a \tan (c+d x))^{3/2}}+\frac {70}{a^2 \sqrt {a+i a \tan (c+d x)}}}{40 d} \] Input:

Integrate[Cot[c + d*x]/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

((-80*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/a^(5/2) + (5*Sqrt[2]*Ar 
cTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/a^(5/2) + 8/(a + I*a* 
Tan[c + d*x])^(5/2) + 20/(a*(a + I*a*Tan[c + d*x])^(3/2)) + 70/(a^2*Sqrt[a 
 + I*a*Tan[c + d*x]]))/(40*d)
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.09, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.708, Rules used = {3042, 4042, 27, 3042, 4079, 27, 3042, 4079, 27, 3042, 4083, 3042, 3961, 219, 4082, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (c+d x) (a+i a \tan (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 4042

\(\displaystyle \frac {\int \frac {5 \cot (c+d x) (2 a-i a \tan (c+d x))}{2 (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\cot (c+d x) (2 a-i a \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{2 a^2}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {2 a-i a \tan (c+d x)}{\tan (c+d x) (i \tan (c+d x) a+a)^{3/2}}dx}{2 a^2}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\int \frac {3 \cot (c+d x) \left (4 a^2-3 i a^2 \tan (c+d x)\right )}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}+\frac {a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\cot (c+d x) \left (4 a^2-3 i a^2 \tan (c+d x)\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}+\frac {a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {4 a^2-3 i a^2 \tan (c+d x)}{\tan (c+d x) \sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}+\frac {a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\frac {\int \frac {1}{2} \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (8 a^3-7 i a^3 \tan (c+d x)\right )dx}{a^2}+\frac {7 a^2}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (8 a^3-7 i a^3 \tan (c+d x)\right )dx}{2 a^2}+\frac {7 a^2}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (8 a^3-7 i a^3 \tan (c+d x)\right )}{\tan (c+d x)}dx}{2 a^2}+\frac {7 a^2}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4083

\(\displaystyle \frac {\frac {\frac {i a^3 \int \sqrt {i \tan (c+d x) a+a}dx+8 a^2 \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 a^2}+\frac {7 a^2}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {i a^3 \int \sqrt {i \tan (c+d x) a+a}dx+8 a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{2 a^2}+\frac {7 a^2}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {\frac {\frac {\frac {2 a^4 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+8 a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{2 a^2}+\frac {7 a^2}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {8 a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx+\frac {\sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a^2}+\frac {7 a^2}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {\frac {\frac {\frac {8 a^4 \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}+\frac {\sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a^2}+\frac {7 a^2}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\frac {\frac {\sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {16 i a^3 \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}}{2 a^2}+\frac {7 a^2}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {7 a^2}{d \sqrt {a+i a \tan (c+d x)}}+\frac {\frac {\sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {16 a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}}{2 a^2}}{2 a^2}+\frac {a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}\)

Input:

Int[Cot[c + d*x]/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

1/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + (a/(d*(a + I*a*Tan[c + d*x])^(3/2)) 
 + (((-16*a^(7/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + (Sqrt[2 
]*a^(7/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d)/(2*a^2 
) + (7*a^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]))/(2*a^2))/(2*a^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

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 3961

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) 
  Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a 
, b, c, d}, x] && EqQ[a^2 + b^2, 0]
 

rule 4042

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

rule 4079

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

rule 4082

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

rule 4083

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

Maple [A] (verified)

Time = 1.54 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (-\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{a^{\frac {9}{2}}}+\frac {7}{8 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {1}{4 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{10 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {9}{2}}}\right )}{d}\)	\(122\)
default	\(\frac {2 a^{2} \left (-\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{a^{\frac {9}{2}}}+\frac {7}{8 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {1}{4 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{10 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {9}{2}}}\right )}{d}\)	\(122\)

Input:

int(cot(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/d*a^2*(-1/a^(9/2)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))+7/8/a^4/(a+I 
*a*tan(d*x+c))^(1/2)+1/4/a^3/(a+I*a*tan(d*x+c))^(3/2)+1/10/a^2/(a+I*a*tan( 
d*x+c))^(5/2)+1/16/a^(9/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^ 
(1/2)/a^(1/2)))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 511 vs. \(2 (120) = 240\).

Time = 0.10 (sec) , antiderivative size = 511, normalized size of antiderivative = 3.21 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (5 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 5 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 20 \, a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, \sqrt {2} {\left (a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + 20 \, a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, \sqrt {2} {\left (a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (41 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 48 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{40 \, a^{3} d} \] Input:

integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

1/40*(5*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(4*(sqrt( 
2)*sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I* 
c) + 1))*sqrt(1/(a^5*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 5*sqrt 
(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/2 
)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sq 
rt(1/(a^5*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 20*a^3*d*sqrt(1/( 
a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(16*(3*a^2*e^(2*I*d*x + 2*I*c) + 2*sqrt(2 
)*(a^4*d*e^(3*I*d*x + 3*I*c) + a^4*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 
 2*I*c) + 1))*sqrt(1/(a^5*d^2)) + a^2)*e^(-2*I*d*x - 2*I*c)) + 20*a^3*d*sq 
rt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(16*(3*a^2*e^(2*I*d*x + 2*I*c) - 2* 
sqrt(2)*(a^4*d*e^(3*I*d*x + 3*I*c) + a^4*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I 
*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) + a^2)*e^(-2*I*d*x - 2*I*c)) + sqrt( 
2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(41*e^(6*I*d*x + 6*I*c) + 48*e^(4*I*d 
*x + 4*I*c) + 8*e^(2*I*d*x + 2*I*c) + 1))*e^(-5*I*d*x - 5*I*c)/(a^3*d)
 

Sympy [F]

\[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\cot {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))**(5/2),x)
 

Output:

Integral(cot(c + d*x)/(I*a*(tan(c + d*x) - I))**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.01 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\frac {5 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {5}{2}}} - \frac {80 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}}} - \frac {4 \, {\left (35 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} + 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 4 \, a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2}}}{80 \, d} \] Input:

integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

-1/80*(5*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt 
(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a)))/a^(5/2) - 80*log((sqrt(I*a*tan( 
d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))/a^(5/2) - 
 4*(35*(I*a*tan(d*x + c) + a)^2 + 10*(I*a*tan(d*x + c) + a)*a + 4*a^2)/((I 
*a*tan(d*x + c) + a)^(5/2)*a^2))/d
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.83 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2\,\mathrm {atanh}\left (\frac {a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {a^5}}\right )}{d\,\sqrt {a^5}}+\frac {\frac {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{2\,a}+\frac {7\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4\,a^2}+\frac {1}{5}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a^5}}\right )}{8\,d\,\sqrt {a^5}} \] Input:

int(cot(c + d*x)/(a + a*tan(c + d*x)*1i)^(5/2),x)
 

Output:

((a + a*tan(c + d*x)*1i)/(2*a) + (7*(a + a*tan(c + d*x)*1i)^2)/(4*a^2) + 1 
/5)/(d*(a + a*tan(c + d*x)*1i)^(5/2)) - (2*atanh((a^2*(a + a*tan(c + d*x)* 
1i)^(1/2))/(a^5)^(1/2)))/(d*(a^5)^(1/2)) + (2^(1/2)*atanh((2^(1/2)*a^2*(a 
+ a*tan(c + d*x)*1i)^(1/2))/(2*(a^5)^(1/2))))/(8*d*(a^5)^(1/2))
 

Reduce [F]

\[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\int \frac {\cot \left (d x +c \right )}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}}d x}{\sqrt {a}\, a^{2}} \] Input:

int(cot(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x)
 

Output:

( - int(cot(c + d*x)/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2*sqrt(ta 
n(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)),x))/(sqrt(a)* 
a**2)