\(\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) [89]

Optimal result

Integrand size = 24, antiderivative size = 78 \[ \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {\sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.91 \[ \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {\sqrt {a} \left (-2 \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )+\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )\right )}{d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4045, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4045

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3961

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

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 4045

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)/((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[a/(a*c - b*d)   Int[(a + b*Tan[e + f*x])^m, 
 x], x] - Simp[d/(a*c - b*d)   Int[(a + b*Tan[e + f*x])^m*((b + a*Tan[e + f 
*x])/(c + d*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && Ne 
Q[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]
 

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. 209 vs. \(2 (61 ) = 122\).

Time = 9.15 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.69

method	result	size
default	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}}{2}\right )+\arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-i \sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {2}}{\sqrt {\csc \left (d x +c \right )^{2} \left (1-\cos \left (d x +c \right )\right )^{2}-1}}\right )+i \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{d \left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+1\right )}\)	\(210\)

Input:

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

Output:

-1/d/(I*(csc(d*x+c)-cot(d*x+c))+1)*(a*(1+I*tan(d*x+c)))^(1/2)*(-2*cos(d*x+ 
c)/(cos(d*x+c)+1))^(1/2)*(2^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(cos(d*x+c)+1) 
)^(1/2)*2^(1/2))+arctan(1/2*2^(1/2)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-I* 
2^(1/2)*arctanh(1/(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)^(1/2)*(csc(d*x+c)-cot( 
d*x+c))*2^(1/2))+I*ln((-2*cos(d*x+c)/(cos(d*x+c)+1))^(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. 336 vs. \(2 (59) = 118\).

Time = 0.08 (sec) , antiderivative size = 336, normalized size of antiderivative = 4.31 \[ \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {1}{2} \, \sqrt {2} \sqrt {\frac {a}{d^{2}}} \log \left (4 \, {\left ({\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {a}{d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \frac {1}{2} \, \sqrt {2} \sqrt {\frac {a}{d^{2}}} \log \left (-4 \, {\left ({\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {a}{d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \frac {1}{2} \, \sqrt {\frac {a}{d^{2}}} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {a}{d^{2}}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + \frac {1}{2} \, \sqrt {\frac {a}{d^{2}}} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {a}{d^{2}}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.37 \[ \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {\sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 2 \, \sqrt {a} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{2 \, d} \] Input:

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

Output:

-1/2*(sqrt(2)*sqrt(a)*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))) - 2*sqrt(a)*log((sqrt(I*a* 
tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a))))/d
 

Giac [F(-2)]

Exception generated. \[ \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\text {Exception raised: TypeError} \] 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 = 0.19 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.78 \[ \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {a}}\right )}{d}+\frac {\sqrt {2}\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{d} \] Input:

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

Output:

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

Reduce [F]

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