\(\int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [95]

Optimal result

Integrand size = 36, antiderivative size = 167 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {(i A-2 B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {(A+i B) \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(2 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{a d} \] Output:

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

Mathematica [A] (verified)

Time = 2.09 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.78 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i (A+2 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {-2 i A+B-A \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.14 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.10, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3042, 4079, 27, 3042, 4081, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4079

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4081

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4083

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3961

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 4081

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*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 
1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + 
f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* 
m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr 
eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 
0] && 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 = 0.24 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {2 i a^{2} \left (-\frac {\left (i B -A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4 a^{\frac {5}{2}}}-\frac {i B +A}{2 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {\frac {i A \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}+\frac {\left (2 i B +A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{2}}\right )}{d}\)	\(139\)
default	\(\frac {2 i a^{2} \left (-\frac {\left (i B -A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4 a^{\frac {5}{2}}}-\frac {i B +A}{2 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {\frac {i A \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}+\frac {\left (2 i B +A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{2}}\right )}{d}\)	\(139\)

Input:

int(cot(d*x+c)^2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETUR 
NVERBOSE)
 

Output:

2*I/d*a^2*(-1/4*(-A+I*B)/a^(5/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1 
/2)*2^(1/2)/a^(1/2))-1/2/a^2*(A+I*B)/(a+I*a*tan(d*x+c))^(1/2)+1/a^2*(1/2*I 
*A*(a+I*a*tan(d*x+c))^(1/2)/a/tan(d*x+c)+1/2*(A+2*I*B)/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. 742 vs. \(2 (135) = 270\).

Time = 0.11 (sec) , antiderivative size = 742, normalized size of antiderivative = 4.44 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate(cot(d*x+c)^2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x, algori 
thm="fricas")
 

Output:

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

Sympy [F]

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

integrate(cot(d*x+c)**2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(1/2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.10 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i \, a {\left (\frac {\sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}} + \frac {2 \, {\left (A + 2 i \, B\right )} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {4 \, {\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (2 \, A + i \, B\right )} - {\left (A + i \, B\right )} a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a - \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2}}\right )}}{4 \, d} \] Input:

integrate(cot(d*x+c)^2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x, algori 
thm="maxima")
 

Output:

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

Giac [F(-2)]

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

integrate(cot(d*x+c)^2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),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 TypeDone
 

Mupad [B] (verification not implemented)

Time = 5.27 (sec) , antiderivative size = 2961, normalized size of antiderivative = 17.73 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

2*atanh((3*d^4*(a + a*tan(c + d*x)*1i)^(1/2)*((9*B^2)/(16*a*d^2) - (3*A^2) 
/(16*a*d^2) - ((A^4*a^10)/d^4 + (49*B^4*a^10)/d^4 - (114*A^2*B^2*a^10)/d^4 
 - (A*B^3*a^10*140i)/d^4 + (A^3*B*a^10*20i)/d^4)^(1/2)/(16*a^6) - (A*B*3i) 
/(8*a*d^2))^(1/2)*((A^4*a^10)/d^4 + (49*B^4*a^10)/d^4 - (114*A^2*B^2*a^10) 
/d^4 - (A*B^3*a^10*140i)/d^4 + (A^3*B*a^10*20i)/d^4)^(1/2))/((A^3*a^5*d*1i 
)/2 + (35*B^3*a^5*d)/2 + (A*d^3*((A^4*a^10)/d^4 + (49*B^4*a^10)/d^4 - (114 
*A^2*B^2*a^10)/d^4 - (A*B^3*a^10*140i)/d^4 + (A^3*B*a^10*20i)/d^4)^(1/2)*3 
i)/2 - (3*B*d^3*((A^4*a^10)/d^4 + (49*B^4*a^10)/d^4 - (114*A^2*B^2*a^10)/d 
^4 - (A*B^3*a^10*140i)/d^4 + (A^3*B*a^10*20i)/d^4)^(1/2))/2 - (A*B^2*a^5*d 
*57i)/2 - (15*A^2*B*a^5*d)/2) + (A^2*a^2*d^2*(a + a*tan(c + d*x)*1i)^(1/2) 
*((9*B^2)/(16*a*d^2) - (3*A^2)/(16*a*d^2) - ((A^4*a^10)/d^4 + (49*B^4*a^10 
)/d^4 - (114*A^2*B^2*a^10)/d^4 - (A*B^3*a^10*140i)/d^4 + (A^3*B*a^10*20i)/ 
d^4)^(1/2)/(16*a^6) - (A*B*3i)/(8*a*d^2))^(1/2))/((A^3*a^2*d*1i)/2 + (35*B 
^3*a^2*d)/2 - (A*B^2*a^2*d*57i)/2 - (15*A^2*B*a^2*d)/2 + (A*d^3*((A^4*a^10 
)/d^4 + (49*B^4*a^10)/d^4 - (114*A^2*B^2*a^10)/d^4 - (A*B^3*a^10*140i)/d^4 
 + (A^3*B*a^10*20i)/d^4)^(1/2)*3i)/(2*a^3) - (3*B*d^3*((A^4*a^10)/d^4 + (4 
9*B^4*a^10)/d^4 - (114*A^2*B^2*a^10)/d^4 - (A*B^3*a^10*140i)/d^4 + (A^3*B* 
a^10*20i)/d^4)^(1/2))/(2*a^3)) - (7*B^2*a^2*d^2*(a + a*tan(c + d*x)*1i)^(1 
/2)*((9*B^2)/(16*a*d^2) - (3*A^2)/(16*a*d^2) - ((A^4*a^10)/d^4 + (49*B^4*a 
^10)/d^4 - (114*A^2*B^2*a^10)/d^4 - (A*B^3*a^10*140i)/d^4 + (A^3*B*a^10...
 

Reduce [F]

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

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

Output:

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