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

Optimal. Leaf size=144 \[ -\frac {2 (-1)^{3/4} \sqrt {a} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {(1+i) \sqrt {a} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d} \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4326, 3644, 3625, 211, 3680, 65, 223, 209} \begin {gather*} -\frac {2 (-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {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} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-1)^(3/4)*Sqrt[a]*ArcTan[((-1)^(3/4)*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]])/d - ((1 + I)*Sqrt[a]*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]])/d

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

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

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 3644

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(a*c - b*d)/a, Int[Sqrt[a + b*Tan[e + f*x]]/Sqrt[c + d*Tan[e + f*x]], x], x] + Dist[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 3680

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[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 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*} \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx\\ &=-\left (\left (i \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\right )+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} (i a+a \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx}{a}\\ &=\frac {\left (i a \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{d}-\frac {\left (2 a^2 \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 {(1+i) \sqrt {a} \tanh ^{-1}\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}+\frac {\left (2 i a \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {(1+i) \sqrt {a} \tanh ^{-1}\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}+\frac {\left (2 i a \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac {2 (-1)^{3/4} \sqrt {a} \tan ^{-1}\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} \tanh ^{-1}\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}\\ \end {align*}

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Mathematica [F]
time = 43.24, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 575 vs. \(2 (114 ) = 228\).
time = 41.44, size = 576, normalized size = 4.00

method	result	size
default	\(\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \left (i \sqrt {2}\, \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+1\right )-2 i \sqrt {2}\, \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\right )-i \sqrt {2}\, \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-1\right )+\sqrt {2}\, \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+1\right )+2 \sqrt {2}\, \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\right )-\sqrt {2}\, \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-1\right )+2 i \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}+1\right )+2 i \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}-1\right )+i \ln \left (\frac {-\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \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 {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}+1\right )-2 \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}-1\right )-\ln \left (\frac {\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \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 {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}\right )\right ) \left (-1+\cos \left (d x +c \right )\right ) \sqrt {2}}{2 d \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}}\)	\(576\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)*(I*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/
2)+1)-2*I*2^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))-I*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1
)+2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)+2*2^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))-2^(1/2
)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)+2*I*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)+2*I*arctan
(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)+I*ln((-((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+co
s(d*x+c)+sin(d*x+c)-1)/(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1))-2*arct
an(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)-2*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)-ln((((
-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1)/(-((-1+cos(d*x+c))/sin(d*x+c))^(1
/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)))*(-1+cos(d*x+c))/sin(d*x+c)/(cos(d*x+c)/sin(d*x+c))^(1/2)/(I*
sin(d*x+c)+cos(d*x+c)-1)/((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] 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 = 0.76, size = 470, normalized size = 3.26 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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
*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*sq
rt(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))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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