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

Optimal result

Integrand size = 34, antiderivative size = 80 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {2 \sqrt [4]{-1} a (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 a (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x)}{3 d} \]

[Out]

2*(-1)^(1/4)*a*(I*A+B)*arctan((-1)^(3/4)*tan(d*x+c)^(1/2))/d+2*a*(I*A+B)*tan(d*x+c)^(1/2)/d+2/3*I*a*B*tan(d*x+
c)^(3/2)/d

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3673, 3609, 3614, 211} \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {2 \sqrt [4]{-1} a (B+i A) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 a (B+i A) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x)}{3 d} \]

[In]

Int[Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])*(A + B*Tan[c + d*x]),x]

[Out]

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

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 3609

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

Rule 3614

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2*(c^2/f), S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 3673

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 i a B \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\int \sqrt {\tan (c+d x)} (a (A-i B)+a (i A+B) \tan (c+d x)) \, dx \\ & = \frac {2 a (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\int \frac {-a (i A+B)+a (A-i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {2 a (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {\left (2 a^2 (i A+B)^2\right ) \text {Subst}\left (\int \frac {1}{-a (i A+B)-a (A-i B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = \frac {2 \sqrt [4]{-1} a (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 a (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {2 a \left (3 \sqrt [4]{-1} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\sqrt {\tan (c+d x)} (3 i A+3 B+i B \tan (c+d x))\right )}{3 d} \]

[In]

Integrate[Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])*(A + B*Tan[c + d*x]),x]

[Out]

(2*a*(3*(-1)^(1/4)*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + Sqrt[Tan[c + d*x]]*((3*I)*A + 3*B + I*B*T
an[c + d*x])))/(3*d)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (65 ) = 130\).

Time = 0.03 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.82

method	result	size
derivativedivides	\(\frac {a \left (\frac {2 i B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+2 i A \left (\sqrt {\tan }\left (d x +c \right )\right )+2 B \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (-i A -B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-i B +A \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(226\)
default	\(\frac {a \left (\frac {2 i B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+2 i A \left (\sqrt {\tan }\left (d x +c \right )\right )+2 B \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (-i A -B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-i B +A \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(226\)
parts	\(\frac {\left (i a A +B a \right ) \left (2 \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {a A \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {i a B \left (\frac {2 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(305\)

[In]

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

[Out]

1/d*a*(2/3*I*B*tan(d*x+c)^(3/2)+2*I*A*tan(d*x+c)^(1/2)+2*B*tan(d*x+c)^(1/2)+1/4*(-I*A-B)*2^(1/2)*(ln((1+2^(1/2
)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2
*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*(A-I*B)*2^(1/2)*(ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/
2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*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. 372 vs. \(2 (62) = 124\).

Time = 0.25 (sec) , antiderivative size = 372, normalized size of antiderivative = 4.65 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) - 3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) + 4 \, {\left ({\left (-3 i \, A - 4 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-3 i \, A - 2 \, B\right )} a\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{6 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

-1/6*(3*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^2/d^2)*log(2*((A - I*B)*a*e^(2*I*d*x + 2*
I*c) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^2/d^2)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(
e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((I*A + B)*a)) - 3*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(-I*A^2 -
2*A*B + I*B^2)*a^2/d^2)*log(2*((A - I*B)*a*e^(2*I*d*x + 2*I*c) - (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(-I*A^2 -
2*A*B + I*B^2)*a^2/d^2)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((I
*A + B)*a)) + 4*((-3*I*A - 4*B)*a*e^(2*I*d*x + 2*I*c) + (-3*I*A - 2*B)*a)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e
^(2*I*d*x + 2*I*c) + 1)))/(d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F]

\[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=i a \left (\int A \tan ^{\frac {3}{2}}{\left (c + d x \right )}\, dx + \int B \tan ^{\frac {5}{2}}{\left (c + d x \right )}\, dx + \int \left (- i A \sqrt {\tan {\left (c + d x \right )}}\right )\, dx + \int \left (- i B \tan ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx\right ) \]

[In]

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

[Out]

I*a*(Integral(A*tan(c + d*x)**(3/2), x) + Integral(B*tan(c + d*x)**(5/2), x) + Integral(-I*A*sqrt(tan(c + d*x)
), x) + Integral(-I*B*tan(c + d*x)**(3/2), x))

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (62) = 124\).

Time = 0.33 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.12 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {-8 i \, B a \tan \left (d x + c\right )^{\frac {3}{2}} + 24 \, {\left (-i \, A - B\right )} a \sqrt {\tan \left (d x + c\right )} + 3 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a}{12 \, d} \]

[In]

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

[Out]

-1/12*(-8*I*B*a*tan(d*x + c)^(3/2) + 24*(-I*A - B)*a*sqrt(tan(d*x + c)) + 3*(2*sqrt(2)*((I - 1)*A + (I + 1)*B)
*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*((I - 1)*A + (I + 1)*B)*arctan(-1/2*sqrt(2)*
(sqrt(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*(-(I + 1)*A + (I - 1)*B)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x
+ c) + 1) + sqrt(2)*(-(I + 1)*A + (I - 1)*B)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))*a)/d

Giac [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.02 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {\left (i - 1\right ) \, \sqrt {2} {\left (A a - i \, B a\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{d} - \frac {2 \, {\left (-i \, B a d^{2} \tan \left (d x + c\right )^{\frac {3}{2}} - 3 i \, A a d^{2} \sqrt {\tan \left (d x + c\right )} - 3 \, B a d^{2} \sqrt {\tan \left (d x + c\right )}\right )}}{3 \, d^{3}} \]

[In]

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

[Out]

-(I - 1)*sqrt(2)*(A*a - I*B*a)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d - 2/3*(-I*B*a*d^2*tan(d*x +
c)^(3/2) - 3*I*A*a*d^2*sqrt(tan(d*x + c)) - 3*B*a*d^2*sqrt(tan(d*x + c)))/d^3

Mupad [B] (verification not implemented)

Time = 9.75 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.24 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {A\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{d}+\frac {2\,B\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}+\frac {B\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,2{}\mathrm {i}}{3\,d}-\frac {2\,{\left (-1\right )}^{1/4}\,A\,a\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-1-\mathrm {i}\right )}{d} \]

[In]

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

[Out]

(A*a*tan(c + d*x)^(1/2)*2i)/d + (2*B*a*tan(c + d*x)^(1/2))/d + (B*a*tan(c + d*x)^(3/2)*2i)/(3*d) - (2*(-1)^(1/
4)*A*a*atanh((-1)^(1/4)*tan(c + d*x)^(1/2)))/d - (2^(1/2)*B*a*atan(2^(1/2)*tan(c + d*x)^(1/2)*(1/2 - 1i/2))*(1
+ 1i))/d