\(\int \frac {\cot (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [114]

Optimal result

Integrand size = 24, antiderivative size = 99 \[ \int \frac {\cot (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {1}{d \sqrt {a+i a \tan (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \frac {\cot (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {1}{d \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

(-2*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(Sqrt[a]*d) + ArcTanh[Sqr 
t[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(Sqrt[2]*Sqrt[a]*d) + 1/(d*Sqrt 
[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.08, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {3042, 4042, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (c+d x) \sqrt {a+i a \tan (c+d x)}}dx\)

\(\Big \downarrow \) 4042

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4083

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3961

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 4042

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(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))   In 
t[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m 
 + n + 1) + b*d*(m + n + 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, 
e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] 
 && LtQ[m, 0] && (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 [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. 409 vs. \(2 (80 ) = 160\).

Time = 13.99 (sec) , antiderivative size = 410, normalized size of antiderivative = 4.14

method	result	size
default	\(\frac {i \left (\cos \left (d x +c \right )+1\right ) \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )+i\right )}{2 \sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+i \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+i \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {2}\, \sin \left (d x +c \right )-\operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )+i\right )}{2 \sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right ) \sin \left (d x +c \right )-\ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {2}\, \sin \left (d x +c \right )+\sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}+2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )}{2 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\)	\(410\)

Input:

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

Output:

1/2/d*(I*(cos(d*x+c)+1)*arctanh(1/2*2^(1/2)*(cot(d*x+c)-csc(d*x+c)+I)/(cot 
(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2))+I*2^(1/2)*(cos(d* 
x+c)+1)*ln((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))+I*a 
rctan(1/2*2^(1/2)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)*sin(d*x+c)-a 
rctanh(1/2*2^(1/2)*(cot(d*x+c)-csc(d*x+c)+I)/(cot(d*x+c)^2-2*cot(d*x+c)*cs 
c(d*x+c)+csc(d*x+c)^2-1)^(1/2))*sin(d*x+c)-ln((-2*cos(d*x+c)/(cos(d*x+c)+1 
))^(1/2)-cot(d*x+c)+csc(d*x+c))*2^(1/2)*sin(d*x+c)+2^(1/2)*(cos(d*x+c)+1)* 
arctan(1/2*2^(1/2)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+(-2*cos(d*x+c)/(cos 
(d*x+c)+1))^(1/2)*2^(1/2)+2*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)) 
/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/(a*(1+I*tan(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. 459 vs. \(2 (77) = 154\).

Time = 0.10 (sec) , antiderivative size = 459, normalized size of antiderivative = 4.64 \[ \int \frac {\cot (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {{\left (\sqrt {2} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (4 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt {2} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-4 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 2 \, a d \sqrt {\frac {1}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, \sqrt {2} {\left (a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{2} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + 2 \, a d \sqrt {\frac {1}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, \sqrt {2} {\left (a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{2} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + 2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.23 \[ \int \frac {\cot (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {\frac {\sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} - \frac {4 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{\sqrt {a}} - \frac {4}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}}{4 \, d} \] Input:

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

Output:

-1/4*(sqrt(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)))/sqrt(a) - 4*log((sqrt(I*a*tan(d*x 
+ c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))/sqrt(a) - 4/s 
qrt(I*a*tan(d*x + c) + a))/d
 

Giac [F(-2)]

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

integrate(cot(d*x+c)/(a+I*a*tan(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 [B] (verification not implemented)

Time = 0.99 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.81 \[ \int \frac {\cot (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {1}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}-\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {a}}\right )}{\sqrt {a}\,d}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{2\,\sqrt {a}\,d} \] Input:

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

Output:

1/(d*(a + a*tan(c + d*x)*1i)^(1/2)) - (2*atanh((a + a*tan(c + d*x)*1i)^(1/ 
2)/a^(1/2)))/(a^(1/2)*d) + (2^(1/2)*atanh((2^(1/2)*(a + a*tan(c + d*x)*1i) 
^(1/2))/(2*a^(1/2))))/(2*a^(1/2)*d)
 

Reduce [F]

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

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