3.692 \(\int \frac{1}{\sqrt [3]{a+b \tan (c+d x)}} \, dx\)

Optimal. Leaf size=415 \[ -\frac{\sqrt{3} b \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-\sqrt{-b^2}}}+1}{\sqrt{3}}\right )}{2 \sqrt{-b^2} d \sqrt [3]{a-\sqrt{-b^2}}}+\frac{\sqrt{3} b \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+\sqrt{-b^2}}}+1}{\sqrt{3}}\right )}{2 \sqrt{-b^2} d \sqrt [3]{a+\sqrt{-b^2}}}-\frac{3 b \log \left (\sqrt [3]{a-\sqrt{-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt{-b^2} d \sqrt [3]{a-\sqrt{-b^2}}}+\frac{3 b \log \left (\sqrt [3]{a+\sqrt{-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt{-b^2} d \sqrt [3]{a+\sqrt{-b^2}}}-\frac{b \log (\cos (c+d x))}{4 \sqrt{-b^2} d \sqrt [3]{a-\sqrt{-b^2}}}+\frac{b \log (\cos (c+d x))}{4 \sqrt{-b^2} d \sqrt [3]{a+\sqrt{-b^2}}}-\frac{x}{4 \sqrt [3]{a-\sqrt{-b^2}}}-\frac{x}{4 \sqrt [3]{a+\sqrt{-b^2}}} \]

[Out]

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

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Rubi [A]  time = 0.263521, antiderivative size = 415, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3485, 712, 55, 617, 204, 31} \[ -\frac{\sqrt{3} b \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-\sqrt{-b^2}}}+1}{\sqrt{3}}\right )}{2 \sqrt{-b^2} d \sqrt [3]{a-\sqrt{-b^2}}}+\frac{\sqrt{3} b \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+\sqrt{-b^2}}}+1}{\sqrt{3}}\right )}{2 \sqrt{-b^2} d \sqrt [3]{a+\sqrt{-b^2}}}-\frac{3 b \log \left (\sqrt [3]{a-\sqrt{-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt{-b^2} d \sqrt [3]{a-\sqrt{-b^2}}}+\frac{3 b \log \left (\sqrt [3]{a+\sqrt{-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt{-b^2} d \sqrt [3]{a+\sqrt{-b^2}}}-\frac{b \log (\cos (c+d x))}{4 \sqrt{-b^2} d \sqrt [3]{a-\sqrt{-b^2}}}+\frac{b \log (\cos (c+d x))}{4 \sqrt{-b^2} d \sqrt [3]{a+\sqrt{-b^2}}}-\frac{x}{4 \sqrt [3]{a-\sqrt{-b^2}}}-\frac{x}{4 \sqrt [3]{a+\sqrt{-b^2}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(-1/3),x]

[Out]

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

Rule 3485

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

Rule 712

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[m]

Rule 55

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 617

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 204

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

Rule 31

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

Rubi steps

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

Mathematica [C]  time = 0.292508, size = 251, normalized size = 0.6 \[ \frac{i \left (\frac{2 \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )}{\sqrt [3]{a-i b}}-\frac{2 \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )}{\sqrt [3]{a+i b}}+\frac{\log (-\tan (c+d x)+i)}{\sqrt [3]{a+i b}}-\frac{\log (\tan (c+d x)+i)}{\sqrt [3]{a-i b}}+\frac{3 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{\sqrt [3]{a-i b}}-\frac{3 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{\sqrt [3]{a+i b}}\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.012, size = 58, normalized size = 0.1 \begin{align*}{\frac{b}{2\,d}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{6}-2\,{{\it \_Z}}^{3}a+{a}^{2}+{b}^{2} \right ) }{\frac{{\it \_R}}{{{\it \_R}}^{5}-{{\it \_R}}^{2}a}\ln \left ( \sqrt [3]{a+b\tan \left ( dx+c \right ) }-{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*b*sum(_R/(_R^5-_R^2*a)*ln((a+b*tan(d*x+c))^(1/3)-_R),_R=RootOf(_Z^6-2*_Z^3*a+a^2+b^2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt [3]{a + b \tan{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*tan(c + d*x))**(-1/3), x)

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Giac [C]  time = 5.38487, size = 609, normalized size = 1.47 \begin{align*} -\frac{3}{2} \,{\left ({\left (i \, \sqrt{3} + 1\right )} \left (-\frac{1}{216 i \, a b^{3} d^{3} - 216 \, b^{4} d^{3}}\right )^{\frac{1}{3}} \log \left (-b d{\left (\sqrt{3} + i\right )} + a d{\left (i \, \sqrt{3} - 1\right )} - 2 \,{\left (a^{2} + 2 i \, a b - b^{2}\right )}^{\frac{1}{3}}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}} d\right ) +{\left (-i \, \sqrt{3} + 1\right )} \left (-\frac{1}{216 i \, a b^{3} d^{3} - 216 \, b^{4} d^{3}}\right )^{\frac{1}{3}} \log \left (b d{\left (\sqrt{3} - i\right )} + a d{\left (-i \, \sqrt{3} - 1\right )} - 2 \,{\left (a^{2} + 2 i \, a b - b^{2}\right )}^{\frac{1}{3}}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}} d\right ) + \frac{{\left (i \, \sqrt{3} + 1\right )} \log \left (b d{\left (\sqrt{3} + i\right )} + a d{\left (i \, \sqrt{3} - 1\right )} + 2 \,{\left (-a^{2} + 2 i \, a b + b^{2}\right )}^{\frac{1}{3}}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}} d\right )}{{\left (216 i \, a b^{3} d^{3} + 216 \, b^{4} d^{3}\right )}^{\frac{1}{3}}} + \frac{{\left (-i \, \sqrt{3} + 1\right )} \log \left (-b d{\left (\sqrt{3} - i\right )} + a d{\left (-i \, \sqrt{3} - 1\right )} + 2 \,{\left (-a^{2} + 2 i \, a b + b^{2}\right )}^{\frac{1}{3}}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}} d\right )}{{\left (216 i \, a b^{3} d^{3} + 216 \, b^{4} d^{3}\right )}^{\frac{1}{3}}} - 2 \, \left (-\frac{1}{216 i \, a b^{3} d^{3} - 216 \, b^{4} d^{3}}\right )^{\frac{1}{3}} \log \left (-a d - i \, b d +{\left (a^{2} + 2 i \, a b - b^{2}\right )}^{\frac{1}{3}}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}} d\right ) - \frac{2 \, \log \left (a d - i \, b d +{\left (-a^{2} + 2 i \, a b + b^{2}\right )}^{\frac{1}{3}}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}} d\right )}{{\left (216 i \, a b^{3} d^{3} + 216 \, b^{4} d^{3}\right )}^{\frac{1}{3}}}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*((I*sqrt(3) + 1)*(-1/(216*I*a*b^3*d^3 - 216*b^4*d^3))^(1/3)*log(-b*d*(sqrt(3) + I) + a*d*(I*sqrt(3) - 1)
- 2*(a^2 + 2*I*a*b - b^2)^(1/3)*(b*tan(d*x + c) + a)^(1/3)*d) + (-I*sqrt(3) + 1)*(-1/(216*I*a*b^3*d^3 - 216*b^
4*d^3))^(1/3)*log(b*d*(sqrt(3) - I) + a*d*(-I*sqrt(3) - 1) - 2*(a^2 + 2*I*a*b - b^2)^(1/3)*(b*tan(d*x + c) + a
)^(1/3)*d) + (I*sqrt(3) + 1)*log(b*d*(sqrt(3) + I) + a*d*(I*sqrt(3) - 1) + 2*(-a^2 + 2*I*a*b + b^2)^(1/3)*(b*t
an(d*x + c) + a)^(1/3)*d)/(216*I*a*b^3*d^3 + 216*b^4*d^3)^(1/3) + (-I*sqrt(3) + 1)*log(-b*d*(sqrt(3) - I) + a*
d*(-I*sqrt(3) - 1) + 2*(-a^2 + 2*I*a*b + b^2)^(1/3)*(b*tan(d*x + c) + a)^(1/3)*d)/(216*I*a*b^3*d^3 + 216*b^4*d
^3)^(1/3) - 2*(-1/(216*I*a*b^3*d^3 - 216*b^4*d^3))^(1/3)*log(-a*d - I*b*d + (a^2 + 2*I*a*b - b^2)^(1/3)*(b*tan
(d*x + c) + a)^(1/3)*d) - 2*log(a*d - I*b*d + (-a^2 + 2*I*a*b + b^2)^(1/3)*(b*tan(d*x + c) + a)^(1/3)*d)/(216*
I*a*b^3*d^3 + 216*b^4*d^3)^(1/3))*b