\(\int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx\) [755]

Optimal result

Integrand size = 28, antiderivative size = 144 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx=-\frac {2 (-1)^{3/4} \sqrt {a} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {(1+i) \sqrt {a} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d} \] Output:

-2*(-1)^(3/4)*a^(1/2)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*ta 
n(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d-(1+I)*a^(1/2)*arctanh 
((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2) 
*tan(d*x+c)^(1/2)/d
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {i a \tan (c+d x)} \left (2 \sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+i \tan (c+d x)}-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {a+i a \tan (c+d x)}\right )}{d \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[Cot[c + d*x]],x]
 

Output:

(Sqrt[Cot[c + d*x]]*Sqrt[I*a*Tan[c + d*x]]*(2*Sqrt[a]*ArcSinh[Sqrt[I*a*Tan 
[c + d*x]]/Sqrt[a]]*Sqrt[1 + I*Tan[c + d*x]] - Sqrt[2]*ArcTanh[(Sqrt[2]*Sq 
rt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[a + I*a*Tan[c + d*x 
]]))/(d*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.87, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 4729, 3042, 4046, 3042, 4027, 218, 4082, 65, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4046

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4082

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {i a \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {(1+i) \sqrt {a} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\right )\)

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 216

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

Input:

Int[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[Cot[c + d*x]],x]
 

Output:

((-2*(-1)^(3/4)*Sqrt[a]*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqr 
t[a + I*a*Tan[c + d*x]]])/d - ((1 + I)*Sqrt[a]*ArcTanh[((1 + I)*Sqrt[a]*Sq 
rt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d)*Sqrt[Cot[c + d*x]]*Sqrt[ 
Tan[c + d*x]]
 

Defintions of rubi rules used

rule 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 0]
 

rule 216

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

rule 218

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

rule 3042

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

rule 4027

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

rule 4046

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*tan[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[(a*c - b*d)/a   Int[Sqrt[a + b*Tan[e + f 
*x]]/Sqrt[c + d*Tan[e + f*x]], x], x] + Simp[d/a   Int[Sqrt[a + b*Tan[e + f 
*x]]*((b + a*Tan[e + f*x])/Sqrt[c + d*Tan[e + f*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]
 

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 4729

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a 
+ b*x])^m*(c*Tan[a + b*x])^m   Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ[u, 
x]
 

Maple [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 519 vs. \(2 (114 ) = 228\).

Time = 0.83 (sec) , antiderivative size = 520, normalized size of antiderivative = 3.61

method	result	size
default	\(-\frac {\sqrt {2}\, \cos \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (-2 i \sqrt {2}\, \arctan \left (\sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\right )-i \sqrt {2}\, \ln \left (\sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}+1\right )+i \sqrt {2}\, \ln \left (\sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}-1\right )+i \ln \left (-\frac {\sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \sqrt {2}\, \sin \left (d x +c \right )+\sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \csc \left (d x +c \right )}\, \sqrt {2}\, \sin \left (d x +c \right )-\sin \left (d x +c \right )-\cos \left (d x +c \right )+1}\right )+2 i \arctan \left (\sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \sqrt {2}+1\right )+2 i \arctan \left (\sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \sqrt {2}-1\right )-2 \sqrt {2}\, \arctan \left (\sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\right )+\sqrt {2}\, \ln \left (\sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}+1\right )-\sqrt {2}\, \ln \left (\sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}-1\right )+\ln \left (\frac {-\sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \sqrt {2}\, \sin \left (d x +c \right )+\sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \csc \left (d x +c \right )}\, \sqrt {2}\, \sin \left (d x +c \right )+\sin \left (d x +c \right )+\cos \left (d x +c \right )-1}\right )+2 \arctan \left (\sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \sqrt {2}+1\right )+2 \arctan \left (\sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \sqrt {2}-1\right )\right )}{d \left (2 i \sin \left (d x +c \right )+2 \cos \left (d x +c \right )+2\right ) \sqrt {\cot \left (d x +c \right )}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}}\)	\(520\)

Input:

int((a+I*a*tan(d*x+c))^(1/2)/cot(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/d*2^(1/2)*cos(d*x+c)/(2*I*sin(d*x+c)+2*cos(d*x+c)+2)*(a*(1+I*tan(d*x+c) 
))^(1/2)*(-2*I*2^(1/2)*arctan((-csc(d*x+c)+cot(d*x+c))^(1/2))-I*2^(1/2)*ln 
((-csc(d*x+c)+cot(d*x+c))^(1/2)+1)+I*2^(1/2)*ln((-csc(d*x+c)+cot(d*x+c))^( 
1/2)-1)+I*ln(-((-csc(d*x+c)+cot(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c 
)+cos(d*x+c)-1)/((-2*sin(1/2*d*x+1/2*c)^2*csc(d*x+c))^(1/2)*2^(1/2)*sin(d* 
x+c)-sin(d*x+c)-cos(d*x+c)+1))+2*I*arctan((-csc(d*x+c)+cot(d*x+c))^(1/2)*2 
^(1/2)+1)+2*I*arctan((-csc(d*x+c)+cot(d*x+c))^(1/2)*2^(1/2)-1)-2*2^(1/2)*a 
rctan((-csc(d*x+c)+cot(d*x+c))^(1/2))+2^(1/2)*ln((-csc(d*x+c)+cot(d*x+c))^ 
(1/2)+1)-2^(1/2)*ln((-csc(d*x+c)+cot(d*x+c))^(1/2)-1)+ln((-(-csc(d*x+c)+co 
t(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1)/((-2*sin(1/2*d 
*x+1/2*c)^2*csc(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1)) 
+2*arctan((-csc(d*x+c)+cot(d*x+c))^(1/2)*2^(1/2)+1)+2*arctan((-csc(d*x+c)+ 
cot(d*x+c))^(1/2)*2^(1/2)-1))/cot(d*x+c)^(1/2)/(-csc(d*x+c)+cot(d*x+c))^(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. 470 vs. \(2 (108) = 216\).

Time = 0.12 (sec) , antiderivative size = 470, normalized size of antiderivative = 3.26 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx=-\frac {1}{4} \, \sqrt {\frac {8 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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 {8 i \, a}{d^{2}}} + 4 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \frac {1}{4} \, \sqrt {\frac {8 i \, a}{d^{2}}} \log \left (-{\left (\sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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 {8 i \, a}{d^{2}}} - 4 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \frac {1}{4} \, \sqrt {\frac {4 i \, a}{d^{2}}} \log \left (-16 \, {\left (\sqrt {2} {\left (i \, a d e^{\left (3 i \, d x + 3 i \, c\right )} - i \, a 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 {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 {4 i \, a}{d^{2}}} + 3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + \frac {1}{4} \, \sqrt {\frac {4 i \, a}{d^{2}}} \log \left (-16 \, {\left (\sqrt {2} {\left (-i \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + i \, a 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 {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 {4 i \, a}{d^{2}}} + 3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

integrate((a+I*a*tan(d*x+c))^(1/2)/cot(d*x+c)^(1/2),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 [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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