3.3.93 \(\int \frac {\tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\) [293]

Optimal. Leaf size=178 \[ \frac {i x}{4 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\sqrt {3} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {\log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {3}{2 d \sqrt [3]{a+i a \tan (c+d x)}} \]

[Out]

1/8*I*x*2^(2/3)/a^(1/3)+1/8*ln(cos(d*x+c))*2^(2/3)/a^(1/3)/d+3/8*ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(1/3))*
2^(2/3)/a^(1/3)/d+1/4*arctan(1/3*(a^(1/3)+2^(2/3)*(a+I*a*tan(d*x+c))^(1/3))/a^(1/3)*3^(1/2))*3^(1/2)*2^(2/3)/a
^(1/3)/d-3/2/d/(a+I*a*tan(d*x+c))^(1/3)

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Rubi [A]
time = 0.09, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3607, 3562, 57, 631, 210, 31} \begin {gather*} \frac {\sqrt {3} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {3}{2 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {\log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {i x}{4 \sqrt [3]{2} \sqrt [3]{a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]/(a + I*a*Tan[c + d*x])^(1/3),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

Rule 210

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

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 3562

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[-b/d, Subst[Int[(a + x)^(n - 1)/(a - x), x]
, x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 + b^2, 0]

Rule 3607

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

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.64, size = 140, normalized size = 0.79 \begin {gather*} -\frac {3 \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+e^{2 i d x} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (\cos (c)+i \sin (c))+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )}{4 d \left (\left (1+e^{2 i d x}\right ) \cos (c)+i \left (-1+e^{2 i d x}\right ) \sin (c)\right ) \sqrt [3]{a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]/(a + I*a*Tan[c + d*x])^(1/3),x]

[Out]

(-3*(2*(1 + E^((2*I)*d*x))*Cos[c] + E^((2*I)*d*x)*Hypergeometric2F1[2/3, 1, 5/3, E^((2*I)*(c + d*x))/(1 + E^((
2*I)*(c + d*x)))]*(Cos[c] + I*Sin[c]) + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c]))/(4*d*((1 + E^((2*I)*d*x))*Cos[c] +
I*(-1 + E^((2*I)*d*x))*Sin[c])*(a + I*a*Tan[c + d*x])^(1/3))

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Maple [A]
time = 0.12, size = 146, normalized size = 0.82

method	result	size
derivativedivides	\(\frac {\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{4 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{8 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{4 a^{\frac {1}{3}}}-\frac {3}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}}{d}\)	\(146\)
default	\(\frac {\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{4 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{8 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{4 a^{\frac {1}{3}}}-\frac {3}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}}{d}\)	\(146\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)/(a+I*a*tan(d*x+c))^(1/3),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/4*2^(2/3)/a^(1/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))-1/8*2^(2/3)/a^(1/3)*ln((a+I*a*tan(d*x+c)
)^(2/3)+2^(1/3)*a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+2^(2/3)*a^(2/3))+1/4*3^(1/2)*2^(2/3)/a^(1/3)*arctan(1/3*3^(1/
2)*(2^(2/3)/a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+1))-3/2/(a+I*a*tan(d*x+c))^(1/3))

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Maxima [A]
time = 0.53, size = 154, normalized size = 0.87 \begin {gather*} \frac {2 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {5}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} a^{\frac {5}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} a^{\frac {5}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - \frac {12 \, a^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}}{8 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

1/8*(2*sqrt(3)*2^(2/3)*a^(5/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3)*a^(1/3) + 2*(I*a*tan(d*x + c) + a)^(1/3))/a
^(1/3)) - 2^(2/3)*a^(5/3)*log(2^(2/3)*a^(2/3) + 2^(1/3)*(I*a*tan(d*x + c) + a)^(1/3)*a^(1/3) + (I*a*tan(d*x +
c) + a)^(2/3)) + 2*2^(2/3)*a^(5/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)) - 12*a^2/(I*a*tan(d*x
+ c) + a)^(1/3))/(a^2*d)

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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. 327 vs. \(2 (125) = 250\).
time = 1.17, size = 327, normalized size = 1.84 \begin {gather*} \frac {{\left (2 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} a d \left (\frac {1}{a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-2 \, \left (\frac {1}{2}\right )^{\frac {2}{3}} a d^{2} \left (\frac {1}{a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} a d + a d\right )} \left (\frac {1}{a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (i \, \sqrt {3} a d^{2} - a d^{2}\right )} \left (\frac {1}{a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} a d + a d\right )} \left (\frac {1}{a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (-i \, \sqrt {3} a d^{2} - a d^{2}\right )} \left (\frac {1}{a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 3 \cdot 2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan {\left (c + d x \right )}}{\sqrt [3]{i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))**(1/3),x)

[Out]

Integral(tan(c + d*x)/(I*a*(tan(c + d*x) - I))**(1/3), 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(tan(d*x+c)/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="giac")

[Out]

integrate(tan(d*x + c)/(I*a*tan(d*x + c) + a)^(1/3), x)

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Mupad [B]
time = 0.24, size = 172, normalized size = 0.97 \begin {gather*} -\frac {3}{2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}+\frac {4^{1/3}\,\ln \left (18\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-9\,4^{2/3}\,a^{1/3}\,d\right )}{4\,a^{1/3}\,d}+\frac {4^{1/3}\,\ln \left (18\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-144\,4^{2/3}\,a^{1/3}\,d\,{\left (-\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}^2\right )\,\left (-\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}{a^{1/3}\,d}-\frac {4^{1/3}\,\ln \left (18\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-144\,4^{2/3}\,a^{1/3}\,d\,{\left (\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}^2\right )\,\left (\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}{a^{1/3}\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)/(a + a*tan(c + d*x)*1i)^(1/3),x)

[Out]

(4^(1/3)*log(18*d*(a + a*tan(c + d*x)*1i)^(1/3) - 9*4^(2/3)*a^(1/3)*d))/(4*a^(1/3)*d) - 3/(2*d*(a + a*tan(c +
d*x)*1i)^(1/3)) + (4^(1/3)*log(18*d*(a + a*tan(c + d*x)*1i)^(1/3) - 144*4^(2/3)*a^(1/3)*d*((3^(1/2)*1i)/8 - 1/
8)^2)*((3^(1/2)*1i)/8 - 1/8))/(a^(1/3)*d) - (4^(1/3)*log(18*d*(a + a*tan(c + d*x)*1i)^(1/3) - 144*4^(2/3)*a^(1
/3)*d*((3^(1/2)*1i)/8 + 1/8)^2)*((3^(1/2)*1i)/8 + 1/8))/(a^(1/3)*d)

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