\(\int \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) [79]

Optimal result

Integrand size = 36, antiderivative size = 125 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {a^{3/2} (3 i A+2 B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 \sqrt {2} a^{3/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a A \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d} \] Output:

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

Mathematica [A] (verified)

Time = 2.15 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.98 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {-i a^{3/2} (3 A-2 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )+2 \sqrt {2} a^{3/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )-a A \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.306, Rules used = {3042, 4076, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4076

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4083

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3961

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} \left (\frac {4 i \sqrt {2} a^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 a^{3/2} (2 B+3 i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}\right )-\frac {a A \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\)

Input:

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

Output:

((-2*a^(3/2)*((3*I)*A + 2*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/ 
d + ((4*I)*Sqrt[2]*a^(3/2)*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(S 
qrt[2]*Sqrt[a])])/d)/2 - (a*A*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 4076

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^2)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n 
+ 1)/(d*f*(b*c + a*d)*(n + 1))), x] - Simp[a/(d*(b*c + a*d)*(n + 1))   Int[ 
(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n 
 - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m - 1) + b 
*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] 
 && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]
 

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 = 0.21 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {2 i a^{2} \left (-\frac {\left (2 i B -2 A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}+\frac {i A \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}-\frac {\left (-2 i B +3 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )}{d}\)	\(111\)
default	\(\frac {2 i a^{2} \left (-\frac {\left (2 i B -2 A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}+\frac {i A \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}-\frac {\left (-2 i B +3 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )}{d}\)	\(111\)

Input:

int(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x,method=_RETUR 
NVERBOSE)
 

Output:

2*I/d*a^2*(-1/2*(2*I*B-2*A)*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*(a+I*a*tan(d*x+c))^(1/2)/a/tan(d*x+c)-1/2* 
(-2*I*B+3*A)/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. 685 vs. \(2 (98) = 196\).

Time = 0.09 (sec) , antiderivative size = 685, normalized size of antiderivative = 5.48 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx =\text {Too large to display} \] Input:

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algori 
thm="fricas")
 

Output:

-1/4*(4*sqrt(2)*sqrt(-(A^2 - 2*I*A*B - B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c 
) - d)*log(4*((-I*A - B)*a^2*e^(I*d*x + I*c) + sqrt(-(A^2 - 2*I*A*B - B^2) 
*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e 
^(-I*d*x - I*c)/((-I*A - B)*a)) - 4*sqrt(2)*sqrt(-(A^2 - 2*I*A*B - B^2)*a^ 
3/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(4*((-I*A - B)*a^2*e^(I*d*x + I*c) - 
 sqrt(-(A^2 - 2*I*A*B - B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/( 
e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a)) - sqrt(-(9*A^2 
 - 12*I*A*B - 4*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(-16*(3*(-3*I 
*A - 2*B)*a^2*e^(2*I*d*x + 2*I*c) + (-3*I*A - 2*B)*a^2 + 2*sqrt(2)*sqrt(-( 
9*A^2 - 12*I*A*B - 4*B^2)*a^3/d^2)*(d*e^(3*I*d*x + 3*I*c) + 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)/(3*I*A + 2*B) 
) + sqrt(-(9*A^2 - 12*I*A*B - 4*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)* 
log(-16*(3*(-3*I*A - 2*B)*a^2*e^(2*I*d*x + 2*I*c) + (-3*I*A - 2*B)*a^2 - 2 
*sqrt(2)*sqrt(-(9*A^2 - 12*I*A*B - 4*B^2)*a^3/d^2)*(d*e^(3*I*d*x + 3*I*c) 
+ 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)/(3*I*A + 2*B)) + 4*sqrt(2)*(I*A*a*e^(3*I*d*x + 3*I*c) + I*A*a*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]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.16 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {i \, {\left (2 \, \sqrt {2} {\left (A - i \, B\right )} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - {\left (3 \, A - 2 i \, B\right )} \sqrt {a} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right ) - \frac {2 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} A}{\tan \left (d x + c\right )}\right )} a}{2 \, d} \] Input:

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algori 
thm="maxima")
 

Output:

-1/2*I*(2*sqrt(2)*(A - I*B)*sqrt(a)*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))) - (3*A - 2*I 
*B)*sqrt(a)*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. \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algori 
thm="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 TypeError: Bad Argument TypeError: Bad Argument TypeError: Bad Ar 
gument Ty
 

Mupad [B] (verification not implemented)

Time = 5.38 (sec) , antiderivative size = 2338, normalized size of antiderivative = 18.70 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

int(cot(c + d*x)^2*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2),x)
 

Output:

- 2*atanh((6*d^4*(a + a*tan(c + d*x)*1i)^(1/2)*((3*B^2*a^3)/(2*d^2) - (17* 
A^2*a^3)/(8*d^2) - ((A^4*a^18)/d^4 + (16*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/ 
d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2)/(8*a^6) + (A*B*a^3 
*7i)/(2*d^2))^(1/2)*((A^4*a^18)/d^4 + (16*B^4*a^18)/d^4 - (8*A^2*B^2*a^18) 
/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2))/(A^3*a^11*d*10i 
+ 32*B^3*a^11*d + A*B^2*a^11*d*72i - 32*A^2*B*a^11*d + A*a^2*d^3*((A^4*a^1 
8)/d^4 + (16*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + 
 (A^3*B*a^18*8i)/d^4)^(1/2)*2i) + (2*A^2*a^6*d^2*(a + a*tan(c + d*x)*1i)^( 
1/2)*((3*B^2*a^3)/(2*d^2) - (17*A^2*a^3)/(8*d^2) - ((A^4*a^18)/d^4 + (16*B 
^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8 
i)/d^4)^(1/2)/(8*a^6) + (A*B*a^3*7i)/(2*d^2))^(1/2))/(A^3*a^8*d*10i + 32*B 
^3*a^8*d + A*B^2*a^8*d*72i - 32*A^2*B*a^8*d + (A*d^3*((A^4*a^18)/d^4 + (16 
*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18 
*8i)/d^4)^(1/2)*2i)/a) + (8*B^2*a^6*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((3* 
B^2*a^3)/(2*d^2) - (17*A^2*a^3)/(8*d^2) - ((A^4*a^18)/d^4 + (16*B^4*a^18)/ 
d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^( 
1/2)/(8*a^6) + (A*B*a^3*7i)/(2*d^2))^(1/2))/(A^3*a^8*d*10i + 32*B^3*a^8*d 
+ A*B^2*a^8*d*72i - 32*A^2*B*a^8*d + (A*d^3*((A^4*a^18)/d^4 + (16*B^4*a^18 
)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4) 
^(1/2)*2i)/a) + (A*B*a^6*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((3*B^2*a^3)...
 

Reduce [F]

\[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cot \left (d x +c \right )^{2} \tan \left (d x +c \right )^{2}d x \right ) b i +\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cot \left (d x +c \right )^{2} \tan \left (d x +c \right )d x \right ) a i +\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cot \left (d x +c \right )^{2} \tan \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cot \left (d x +c \right )^{2}d x \right ) a \right ) \] Input:

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

Output:

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