\(\int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [104]

Optimal result

Integrand size = 26, antiderivative size = 114 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {5 i a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d} \] Output:

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

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.96 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {-5 i a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )+4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )-a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d} \] Input:

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

Output:

((-5*I)*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]] + (4*I)*Sqrt[2 
]*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] - a^2*Cot[ 
c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {3042, 4036, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4036

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4083

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3961

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (\frac {10 a^2 \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}+\frac {8 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}\right )-\frac {a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} \left (\frac {8 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {10 i a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}\right )-\frac {a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\)

Input:

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

Output:

(((-10*I)*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + ((8*I)* 
Sqrt[2]*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d)/ 
2 - (a^2*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d
 

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 4036

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x] 
)^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] + Si 
mp[a/(d*(b*c + a*d)*(n + 1))   Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[ 
e + f*x])^(n + 1)*Simp[b*(b*c*(m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) 
 + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /; FreeQ[{a, 
 b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + 
d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

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 = 1.57 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {2 i a^{3} \left (\frac {2 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{\sqrt {a}}+\frac {i \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}-\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )}{d}\)	\(94\)
default	\(\frac {2 i a^{3} \left (\frac {2 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{\sqrt {a}}+\frac {i \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}-\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )}{d}\)	\(94\)

Input:

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

Output:

2*I/d*a^3*(2*2^(1/2)/a^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/ 
a^(1/2))+1/2*I*(a+I*a*tan(d*x+c))^(1/2)/a/tan(d*x+c)-5/2/a^(1/2)*arctanh(( 
a+I*a*tan(d*x+c))^(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. \(2 (87) = 174\).

Time = 0.11 (sec) , antiderivative size = 511, normalized size of antiderivative = 4.48 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {8 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{5}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 8 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{5}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) + 5 \, \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {16 \, {\left (3 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} - 2 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (i \, d e^{\left (3 i \, d x + 3 i \, c\right )} + i \, d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) - 5 \, \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {16 \, {\left (3 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} - 2 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (-i \, d e^{\left (3 i \, d x + 3 i \, c\right )} - i \, d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) + 4 \, \sqrt {2} {\left (i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \] Input:

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

Output:

-1/4*(8*sqrt(2)*sqrt(-a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(4*(a^3*e^(I 
*d*x + I*c) + sqrt(-a^5/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2* 
I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - 8*sqrt(2)*sqrt(-a^5/d^2)*(d* 
e^(2*I*d*x + 2*I*c) - d)*log(4*(a^3*e^(I*d*x + I*c) + sqrt(-a^5/d^2)*(-I*d 
*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - 
 I*c)/a^2) + 5*sqrt(-a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(16*(3*a^3*e^ 
(2*I*d*x + 2*I*c) + a^3 - 2*sqrt(2)*sqrt(-a^5/d^2)*(I*d*e^(3*I*d*x + 3*I*c 
) + I*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 
2*I*c)/a) - 5*sqrt(-a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(16*(3*a^3*e^( 
2*I*d*x + 2*I*c) + a^3 - 2*sqrt(2)*sqrt(-a^5/d^2)*(-I*d*e^(3*I*d*x + 3*I*c 
) - I*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 
2*I*c)/a) + 4*sqrt(2)*(I*a^2*e^(3*I*d*x + 3*I*c) + I*a^2*e^(I*d*x + I*c))* 
sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(2*I*d*x + 2*I*c) - d)
 

Sympy [F(-1)]

Timed out. \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.17 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {i \, {\left (4 \, \sqrt {2} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 5 \, a^{\frac {3}{2}} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right ) - \frac {2 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a}{\tan \left (d x + c\right )}\right )} a}{2 \, d} \] Input:

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

Output:

-1/2*I*(4*sqrt(2)*a^(3/2)*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))) - 5*a^(3/2)*log((sqrt( 
I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a))) - 
 2*I*sqrt(I*a*tan(d*x + c) + a)*a/tan(d*x + c))*a/d
 

Giac [F(-2)]

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

integrate(cot(d*x+c)^2*(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 = 1.13 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {-a^5}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^3}\right )\,\sqrt {-a^5}\,5{}\mathrm {i}}{d}-\frac {a^2\,\mathrm {cot}\left (c+d\,x\right )\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {-a^5}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,a^3}\right )\,\sqrt {-a^5}\,4{}\mathrm {i}}{d} \] Input:

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

Output:

(atan(((-a^5)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/a^3)*(-a^5)^(1/2)*5i)/d 
 - (a^2*cot(c + d*x)*(a + a*tan(c + d*x)*1i)^(1/2))/d - (2^(1/2)*atan((2^( 
1/2)*(-a^5)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*a^3))*(-a^5)^(1/2)*4i) 
/d
 

Reduce [F]

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

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

Output:

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