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

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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4326, 3629, 3627, 3625, 211} \[ \int \frac {\sqrt {\cot (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {1}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

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

Rule 211

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

Rule 3625

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
-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] && NeQ[c^2 + d^2, 0]

Rule 3627

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*b*f*m)), x] - Dist[(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] && EqQ[m + n, 0] && LeQ[m, -2^(-1)]

Rule 3629

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-d)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*m*(c^2 + d^2))), x] + Dist[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 4326

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

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx \\ & = \frac {2}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\left (i \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx \\ & = \frac {1}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{2 a} \\ & = \frac {1}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (i a \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \\ & = \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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {\cot (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {i a \tan (c+d x)}}+\frac {2}{\sqrt {a+i a \tan (c+d x)}}}{2 d \sqrt {\cot (c+d x)}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (83 ) = 166\).

Time = 38.92 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.92

method result size
default \(\frac {\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (\sqrt {\cot }\left (d x +c \right )\right ) \left (i \tan \left (d x +c \right ) \sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \sqrt {2}\right )+i \tan \left (d x +c \right ) \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}+\sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \sqrt {2}\right )-\sec \left (d x +c \right ) \sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \sqrt {2}\right )-\tan \left (d x +c \right ) \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\right )}{d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}}\) \(202\)

[In]

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

[Out]

(-1/2-1/2*I)/d*cot(d*x+c)^(1/2)/(a*(1+I*tan(d*x+c)))^(1/2)/(cot(d*x+c)-csc(d*x+c))^(1/2)*(I*tan(d*x+c)*2^(1/2)
*arctan((1/2+1/2*I)*(cot(d*x+c)-csc(d*x+c))^(1/2)*2^(1/2))+I*tan(d*x+c)*(cot(d*x+c)-csc(d*x+c))^(1/2)+2^(1/2)*
arctan((1/2+1/2*I)*(cot(d*x+c)-csc(d*x+c))^(1/2)*2^(1/2))-sec(d*x+c)*2^(1/2)*arctan((1/2+1/2*I)*(cot(d*x+c)-cs
c(d*x+c))^(1/2)*2^(1/2))-tan(d*x+c)*(cot(d*x+c)-csc(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. 333 vs. \(2 (77) = 154\).

Time = 0.28 (sec) , antiderivative size = 333, normalized size of antiderivative = 3.17 \[ \int \frac {\sqrt {\cot (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {{\left (a d \sqrt {-\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-2 \, {\left (\sqrt {2} {\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {2 i}{a d^{2}}} - 2 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - a d \sqrt {-\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-2 \, {\left (\sqrt {2} {\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {2 i}{a d^{2}}} - 2 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a d} \]

[In]

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

[Out]

-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)*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)) + 2*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*s
qrt((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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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