\(\int \frac {\sqrt {\cot (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [770]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 28, antiderivative size = 105 \[ \int \frac {\sqrt {\cot (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {a} d}+\frac {1}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {\cot (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {i a \tan (c+d x)}}+\frac {2}{\sqrt {a+i a \tan (c+d x)}}}{2 d \sqrt {\cot (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.39, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 4729, 3042, 4031, 3042, 4029, 3042, 4027, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4031

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4029

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Defintions of rubi rules used

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 4029
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/(2*b*f*m)), x] - Simp[(a*c - b*d)/(2*b^2)   Int[(a + b*Tan[e + f 
*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1), 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] && Eq 
Q[m + n, 0] && LeQ[m, -2^(-1)]
 

rule 4031
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-d)*(a + b*Tan[e + f*x])^m*((c + d*Ta 
n[e + f*x])^(n + 1)/(f*m*(c^2 + d^2))), x] + Simp[a/(a*c - b*d)   Int[(a + 
b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d 
, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^ 
2, 0] && EqQ[m + n + 1, 0] &&  !LtQ[m, -1]
 

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] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (83 ) = 166\).

Time = 1.87 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.97

method result size
default \(\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\, \left (i \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \sqrt {2}+2 i \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\right )-2 \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\right ) \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-\sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \sqrt {2}\right ) \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{d \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\) \(207\)

Input:

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

Output:

(1/2+1/2*I)/d*2^(1/2)*cot(d*x+c)^(1/2)*(I*(-csc(d*x+c)+cot(d*x+c))^(1/2)*2 
^(1/2)+2*I*arctan((1/2+1/2*I)*2^(1/2)*(-csc(d*x+c)+cot(d*x+c))^(1/2))-2*ar 
ctan((1/2+1/2*I)*2^(1/2)*(-csc(d*x+c)+cot(d*x+c))^(1/2))*(-cot(d*x+c)+csc( 
d*x+c))-(-csc(d*x+c)+cot(d*x+c))^(1/2)*2^(1/2))/(-csc(d*x+c)+cot(d*x+c))^( 
1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)/(a*(1+I*tan(d*x+c)))^(1/2)*(-cot(d* 
x+c)+csc(d*x+c))
 

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. 333 vs. \(2 (77) = 154\).

Time = 0.11 (sec) , antiderivative size = 333, normalized size of antiderivative = 3.17 \[ \int \frac {\sqrt {\cot (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {{\left (a d \sqrt {-\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-2 \, {\left (\sqrt {2} {\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {2 i}{a d^{2}}} - 2 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - a d \sqrt {-\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-2 \, {\left (\sqrt {2} {\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {2 i}{a d^{2}}} - 2 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a d} \] Input:

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

Output:

-1/4*(a*d*sqrt(-2*I/(a*d^2))*e^(I*d*x + I*c)*log(-2*(sqrt(2)*(I*a*d*e^(2*I 
*d*x + 2*I*c) - I*a*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(-2*I/(a*d^2)) - 2*I*a*e^(I 
*d*x + I*c))*e^(-I*d*x - I*c)) - a*d*sqrt(-2*I/(a*d^2))*e^(I*d*x + I*c)*lo 
g(-2*(sqrt(2)*(-I*a*d*e^(2*I*d*x + 2*I*c) + I*a*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))*sqr 
t(-2*I/(a*d^2)) - 2*I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 2*sqrt(2)*sqr 
t(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))*(I*e^(2*I*d*x + 2*I*c) - I))*e^(-I*d*x - I*c)/(a*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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