3.6.21 \(\int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx\) [521]

3.6.21.1 Optimal result

Integrand size = 36, antiderivative size = 235 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) ((2+i) A+B) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) ((2+i) A+B) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}+\frac {(A+i B) \sqrt {\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}-\frac {((3+i) A-(1+i) B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{8 \sqrt {2} a d}+\frac {((3+i) A-(1+i) B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{8 \sqrt {2} a d} \]

output

(-1/8+1/8*I)*((2+I)*A+B)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/a/d*2^(1/2)+( 
-1/8+1/8*I)*((2+I)*A+B)*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/a/d*2^(1/2)-1/1 
6*((3+I)*A-(1+I)*B)*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/a/d*2^(1/2)+ 
1/16*((3+I)*A-(1+I)*B)*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/a/d*2^(1/ 
2)+1/2*(A+I*B)*cot(d*x+c)^(1/2)/d/(I*a+a*cot(d*x+c))
 

3.6.21.2 Mathematica [A] (verified)

Time = 3.15 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.58 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left ((-i A+B) \sqrt {\tan (c+d x)}-\sqrt [4]{-1} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) (-i+\tan (c+d x))-2 \sqrt [4]{-1} A \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) (-i+\tan (c+d x))\right )}{2 a d (-i+\tan (c+d x))} \]

input

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

output

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(((-I)*A + B)*Sqrt[Tan[c + d*x]] - 
(-1)^(1/4)*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*(-I + Tan[c + d 
*x]) - 2*(-1)^(1/4)*A*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*(-I + Tan[c + 
 d*x])))/(2*a*d*(-I + Tan[c + d*x]))
 

3.6.21.3 Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.87, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.472, Rules used = {3042, 4064, 3042, 4078, 27, 3042, 4017, 25, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4064

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {1}{2} ((3+i) A-(1+i) B) \left (\frac {\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}\right )-\left (\frac {1}{2}-\frac {i}{2}\right ) (B+(2+i) A) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )}{2 a d}+\frac {(A+i B) \sqrt {\cot (c+d x)}}{2 d (a \cot (c+d x)+i a)}\)

input

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

output

((A + I*B)*Sqrt[Cot[c + d*x]])/(2*d*(I*a + a*Cot[c + d*x])) + ((-1/2 + I/2 
)*((2 + I)*A + B)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]]/Sqrt[2]) + Arc 
Tan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]/Sqrt[2]) + (((3 + I)*A - (1 + I)*B)*(- 
1/2*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]/Sqrt[2] + Log[1 + S 
qrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]/(2*Sqrt[2])))/2)/(2*a*d)
 

3.6.21.3.1 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 217

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

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 1482

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
a*c, 2]}, Simp[(d*q + a*e)/(2*a*c)   Int[(q + c*x^2)/(a + c*x^4), x], x] + 
Simp[(d*q - a*e)/(2*a*c)   Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a 
, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- 
a)*c]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4017

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[2/f   Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq 
rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & 
& NeQ[c^2 + d^2, 0]
 

rule 4064

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp 
[g^(m + n)   Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d + c 
*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&  !Integer 
Q[p] && IntegerQ[m] && IntegerQ[n]
 

rule 4078

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

3.6.21.4 Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.46

method	result	size
derivativedivides	\(\frac {\sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \sqrt {\tan \left (d x +c \right )}\, \left (i A \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}+2 i A \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right ) \tan \left (d x +c \right )-i A \arctan \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right ) \tan \left (d x +c \right )+i B \arctan \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right ) \tan \left (d x +c \right )+2 i A \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right )+i A \arctan \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right )-2 A \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right ) \tan \left (d x +c \right )-A \arctan \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right ) \tan \left (d x +c \right )-B \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}+i B \arctan \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right )-B \arctan \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right ) \tan \left (d x +c \right )+2 A \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right )-A \arctan \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right )+B \arctan \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right )\right ) \sqrt {2}}{4 a d \left (-\tan \left (d x +c \right )+i\right )}\)	\(342\)
default	\(\frac {\sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \sqrt {\tan \left (d x +c \right )}\, \left (i A \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}+2 i A \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right ) \tan \left (d x +c \right )-i A \arctan \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right ) \tan \left (d x +c \right )+i B \arctan \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right ) \tan \left (d x +c \right )+2 i A \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right )+i A \arctan \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right )-2 A \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right ) \tan \left (d x +c \right )-A \arctan \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right ) \tan \left (d x +c \right )-B \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}+i B \arctan \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right )-B \arctan \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right ) \tan \left (d x +c \right )+2 A \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right )-A \arctan \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right )+B \arctan \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan \left (d x +c \right )}\, \sqrt {2}\right )\right ) \sqrt {2}}{4 a d \left (-\tan \left (d x +c \right )+i\right )}\)	\(342\)

input

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

output

1/4/a/d*(1/tan(d*x+c))^(1/2)*tan(d*x+c)^(1/2)*(I*A*tan(d*x+c)^(1/2)*2^(1/2 
)+2*I*A*arctan((1/2+1/2*I)*tan(d*x+c)^(1/2)*2^(1/2))*tan(d*x+c)-I*A*arctan 
((1/2-1/2*I)*tan(d*x+c)^(1/2)*2^(1/2))*tan(d*x+c)+I*B*arctan((1/2-1/2*I)*t 
an(d*x+c)^(1/2)*2^(1/2))*tan(d*x+c)+2*I*A*arctan((1/2+1/2*I)*tan(d*x+c)^(1 
/2)*2^(1/2))+I*A*arctan((1/2-1/2*I)*tan(d*x+c)^(1/2)*2^(1/2))-2*A*arctan(( 
1/2+1/2*I)*tan(d*x+c)^(1/2)*2^(1/2))*tan(d*x+c)-A*arctan((1/2-1/2*I)*tan(d 
*x+c)^(1/2)*2^(1/2))*tan(d*x+c)-B*tan(d*x+c)^(1/2)*2^(1/2)+I*B*arctan((1/2 
-1/2*I)*tan(d*x+c)^(1/2)*2^(1/2))-B*arctan((1/2-1/2*I)*tan(d*x+c)^(1/2)*2^ 
(1/2))*tan(d*x+c)+2*A*arctan((1/2+1/2*I)*tan(d*x+c)^(1/2)*2^(1/2))-A*arcta 
n((1/2-1/2*I)*tan(d*x+c)^(1/2)*2^(1/2))+B*arctan((1/2-1/2*I)*tan(d*x+c)^(1 
/2)*2^(1/2)))*2^(1/2)/(-tan(d*x+c)+I)
 

3.6.21.5 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. 570 vs. \(2 (178) = 356\).

Time = 0.26 (sec) , antiderivative size = 570, normalized size of antiderivative = 2.43 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {{\left (a d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{2} d^{2}}} + {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - a d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{2} d^{2}}} - {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - 2 \, a d \sqrt {\frac {i \, A^{2}}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i \, A^{2}}{a^{2} d^{2}}} + A\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) + 2 \, a d \sqrt {\frac {i \, A^{2}}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i \, A^{2}}{a^{2} d^{2}}} - A\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) + 2 \, {\left ({\left (-i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A - B\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a d} \]

input

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

output

1/8*(a*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^2*d^2))*e^(2*I*d*x + 2*I*c)*log( 
-2*((a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2 
*I*d*x + 2*I*c) - 1))*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^2*d^2)) + (A - I*B) 
*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - a*d*sqrt((-I*A^2 - 
 2*A*B + I*B^2)/(a^2*d^2))*e^(2*I*d*x + 2*I*c)*log(2*((a*d*e^(2*I*d*x + 2* 
I*c) - a*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sq 
rt((-I*A^2 - 2*A*B + I*B^2)/(a^2*d^2)) - (A - I*B)*e^(2*I*d*x + 2*I*c))*e^ 
(-2*I*d*x - 2*I*c)/(I*A + B)) - 2*a*d*sqrt(I*A^2/(a^2*d^2))*e^(2*I*d*x + 2 
*I*c)*log(-((a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + 
I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(I*A^2/(a^2*d^2)) + A)*e^(-2*I*d*x - 2*I 
*c)/(a*d)) + 2*a*d*sqrt(I*A^2/(a^2*d^2))*e^(2*I*d*x + 2*I*c)*log(((a*d*e^( 
2*I*d*x + 2*I*c) - a*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I 
*c) - 1))*sqrt(I*A^2/(a^2*d^2)) - A)*e^(-2*I*d*x - 2*I*c)/(a*d)) + 2*((-I* 
A + B)*e^(2*I*d*x + 2*I*c) + I*A - B)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^ 
(2*I*d*x + 2*I*c) - 1)))*e^(-2*I*d*x - 2*I*c)/(a*d)
 

3.6.21.6 Sympy [F]

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

input

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

output

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

3.6.21.7 Maxima [F(-2)]

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

input

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

output

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

3.6.21.8 Giac [F(-2)]

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

input

integrate(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c)),x, algorith 
m="giac")
 

output

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:Not invertible Error: Bad Argument Value
 

3.6.21.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\int \frac {\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

input

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

output

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