### 3.707 $$\int \frac{(2+3 i x)^2}{\sqrt [3]{4-27 x^2}} \, dx$$

Optimal. Leaf size=557 $\frac{8\ 2^{5/6} \left (2^{2/3}-\sqrt [3]{4-27 x^2}\right ) \sqrt{\frac{\left (4-27 x^2\right )^{2/3}+2^{2/3} \sqrt [3]{4-27 x^2}+2 \sqrt [3]{2}}{\left (2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{2^{2/3} \left (1+\sqrt{3}\right )-\sqrt [3]{4-27 x^2}}{2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}}\right ),4 \sqrt{3}-7\right )}{21 \sqrt [4]{3} x \sqrt{-\frac{2^{2/3}-\sqrt [3]{4-27 x^2}}{\left (2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}\right )^2}}}-\frac{1}{21} i \left (4-27 x^2\right )^{2/3} (2+3 i x)-\frac{5}{21} i \left (4-27 x^2\right )^{2/3}-\frac{72 x}{7 \left (2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}\right )}-\frac{4 \sqrt [3]{2} \sqrt{2+\sqrt{3}} \left (2^{2/3}-\sqrt [3]{4-27 x^2}\right ) \sqrt{\frac{\left (4-27 x^2\right )^{2/3}+2^{2/3} \sqrt [3]{4-27 x^2}+2 \sqrt [3]{2}}{\left (2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}\right )^2}} E\left (\sin ^{-1}\left (\frac{2^{2/3} \left (1+\sqrt{3}\right )-\sqrt [3]{4-27 x^2}}{2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}}\right )|-7+4 \sqrt{3}\right )}{7\ 3^{3/4} x \sqrt{-\frac{2^{2/3}-\sqrt [3]{4-27 x^2}}{\left (2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}\right )^2}}}$

[Out]

((-5*I)/21)*(4 - 27*x^2)^(2/3) - (I/21)*(2 + (3*I)*x)*(4 - 27*x^2)^(2/3) - (72*x)/(7*(2^(2/3)*(1 - Sqrt[3]) -
(4 - 27*x^2)^(1/3))) - (4*2^(1/3)*Sqrt[2 + Sqrt[3]]*(2^(2/3) - (4 - 27*x^2)^(1/3))*Sqrt[(2*2^(1/3) + 2^(2/3)*(
4 - 27*x^2)^(1/3) + (4 - 27*x^2)^(2/3))/(2^(2/3)*(1 - Sqrt[3]) - (4 - 27*x^2)^(1/3))^2]*EllipticE[ArcSin[(2^(2
/3)*(1 + Sqrt[3]) - (4 - 27*x^2)^(1/3))/(2^(2/3)*(1 - Sqrt[3]) - (4 - 27*x^2)^(1/3))], -7 + 4*Sqrt[3]])/(7*3^(
3/4)*x*Sqrt[-((2^(2/3) - (4 - 27*x^2)^(1/3))/(2^(2/3)*(1 - Sqrt[3]) - (4 - 27*x^2)^(1/3))^2)]) + (8*2^(5/6)*(2
^(2/3) - (4 - 27*x^2)^(1/3))*Sqrt[(2*2^(1/3) + 2^(2/3)*(4 - 27*x^2)^(1/3) + (4 - 27*x^2)^(2/3))/(2^(2/3)*(1 -
Sqrt[3]) - (4 - 27*x^2)^(1/3))^2]*EllipticF[ArcSin[(2^(2/3)*(1 + Sqrt[3]) - (4 - 27*x^2)^(1/3))/(2^(2/3)*(1 -
Sqrt[3]) - (4 - 27*x^2)^(1/3))], -7 + 4*Sqrt[3]])/(21*3^(1/4)*x*Sqrt[-((2^(2/3) - (4 - 27*x^2)^(1/3))/(2^(2/3)
*(1 - Sqrt[3]) - (4 - 27*x^2)^(1/3))^2)])

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Rubi [A]  time = 0.315223, antiderivative size = 557, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.286, Rules used = {743, 641, 235, 304, 219, 1879} $-\frac{1}{21} i \left (4-27 x^2\right )^{2/3} (2+3 i x)-\frac{5}{21} i \left (4-27 x^2\right )^{2/3}-\frac{72 x}{7 \left (2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}\right )}+\frac{8\ 2^{5/6} \left (2^{2/3}-\sqrt [3]{4-27 x^2}\right ) \sqrt{\frac{\left (4-27 x^2\right )^{2/3}+2^{2/3} \sqrt [3]{4-27 x^2}+2 \sqrt [3]{2}}{\left (2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac{2^{2/3} \left (1+\sqrt{3}\right )-\sqrt [3]{4-27 x^2}}{2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}}\right )|-7+4 \sqrt{3}\right )}{21 \sqrt [4]{3} x \sqrt{-\frac{2^{2/3}-\sqrt [3]{4-27 x^2}}{\left (2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}\right )^2}}}-\frac{4 \sqrt [3]{2} \sqrt{2+\sqrt{3}} \left (2^{2/3}-\sqrt [3]{4-27 x^2}\right ) \sqrt{\frac{\left (4-27 x^2\right )^{2/3}+2^{2/3} \sqrt [3]{4-27 x^2}+2 \sqrt [3]{2}}{\left (2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}\right )^2}} E\left (\sin ^{-1}\left (\frac{2^{2/3} \left (1+\sqrt{3}\right )-\sqrt [3]{4-27 x^2}}{2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}}\right )|-7+4 \sqrt{3}\right )}{7\ 3^{3/4} x \sqrt{-\frac{2^{2/3}-\sqrt [3]{4-27 x^2}}{\left (2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}\right )^2}}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(2 + (3*I)*x)^2/(4 - 27*x^2)^(1/3),x]

[Out]

((-5*I)/21)*(4 - 27*x^2)^(2/3) - (I/21)*(2 + (3*I)*x)*(4 - 27*x^2)^(2/3) - (72*x)/(7*(2^(2/3)*(1 - Sqrt[3]) -
(4 - 27*x^2)^(1/3))) - (4*2^(1/3)*Sqrt[2 + Sqrt[3]]*(2^(2/3) - (4 - 27*x^2)^(1/3))*Sqrt[(2*2^(1/3) + 2^(2/3)*(
4 - 27*x^2)^(1/3) + (4 - 27*x^2)^(2/3))/(2^(2/3)*(1 - Sqrt[3]) - (4 - 27*x^2)^(1/3))^2]*EllipticE[ArcSin[(2^(2
/3)*(1 + Sqrt[3]) - (4 - 27*x^2)^(1/3))/(2^(2/3)*(1 - Sqrt[3]) - (4 - 27*x^2)^(1/3))], -7 + 4*Sqrt[3]])/(7*3^(
3/4)*x*Sqrt[-((2^(2/3) - (4 - 27*x^2)^(1/3))/(2^(2/3)*(1 - Sqrt[3]) - (4 - 27*x^2)^(1/3))^2)]) + (8*2^(5/6)*(2
^(2/3) - (4 - 27*x^2)^(1/3))*Sqrt[(2*2^(1/3) + 2^(2/3)*(4 - 27*x^2)^(1/3) + (4 - 27*x^2)^(2/3))/(2^(2/3)*(1 -
Sqrt[3]) - (4 - 27*x^2)^(1/3))^2]*EllipticF[ArcSin[(2^(2/3)*(1 + Sqrt[3]) - (4 - 27*x^2)^(1/3))/(2^(2/3)*(1 -
Sqrt[3]) - (4 - 27*x^2)^(1/3))], -7 + 4*Sqrt[3]])/(21*3^(1/4)*x*Sqrt[-((2^(2/3) - (4 - 27*x^2)^(1/3))/(2^(2/3)
*(1 - Sqrt[3]) - (4 - 27*x^2)^(1/3))^2)])

Rule 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 235

Int[((a_) + (b_.)*(x_)^2)^(-1/3), x_Symbol] :> Dist[(3*Sqrt[b*x^2])/(2*b*x), Subst[Int[x/Sqrt[-a + x^3], x], x
, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b}, x]

Rule 304

Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, -Dist[(S
qrt[2]*s)/(Sqrt[2 - Sqrt[3]]*r), Int[1/Sqrt[a + b*x^3], x], x] + Dist[1/r, Int[((1 + Sqrt[3])*s + r*x)/Sqrt[a
+ b*x^3], x], x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rule 219

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 - Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 + Sqrt[3
])*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[-((s*(s + r*x))/((1 - S
qrt[3])*s + r*x)^2)]), x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rule 1879

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Simplify[((1 + Sqrt[3])*d)/c]]
, s = Denom[Simplify[((1 + Sqrt[3])*d)/c]]}, Simp[(2*d*s^3*Sqrt[a + b*x^3])/(a*r^2*((1 - Sqrt[3])*s + r*x)), x
] + Simp[(3^(1/4)*Sqrt[2 + Sqrt[3]]*d*s*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]*Elli
pticE[ArcSin[((1 + Sqrt[3])*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]])/(r^2*Sqrt[a + b*x^3]*Sqrt[-((s
*(s + r*x))/((1 - Sqrt[3])*s + r*x)^2)]), x]] /; FreeQ[{a, b, c, d}, x] && NegQ[a] && EqQ[b*c^3 - 2*(5 + 3*Sqr
t[3])*a*d^3, 0]

Rubi steps

\begin{align*} \int \frac{(2+3 i x)^2}{\sqrt [3]{4-27 x^2}} \, dx &=-\frac{1}{21} i (2+3 i x) \left (4-27 x^2\right )^{2/3}-\frac{1}{63} \int \frac{-216-540 i x}{\sqrt [3]{4-27 x^2}} \, dx\\ &=-\frac{5}{21} i \left (4-27 x^2\right )^{2/3}-\frac{1}{21} i (2+3 i x) \left (4-27 x^2\right )^{2/3}+\frac{24}{7} \int \frac{1}{\sqrt [3]{4-27 x^2}} \, dx\\ &=-\frac{5}{21} i \left (4-27 x^2\right )^{2/3}-\frac{1}{21} i (2+3 i x) \left (4-27 x^2\right )^{2/3}-\frac{\left (4 \sqrt{3} \sqrt{-x^2}\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{-4+x^3}} \, dx,x,\sqrt [3]{4-27 x^2}\right )}{7 x}\\ &=-\frac{5}{21} i \left (4-27 x^2\right )^{2/3}-\frac{1}{21} i (2+3 i x) \left (4-27 x^2\right )^{2/3}+\frac{\left (4 \sqrt{3} \sqrt{-x^2}\right ) \operatorname{Subst}\left (\int \frac{2^{2/3} \left (1+\sqrt{3}\right )-x}{\sqrt{-4+x^3}} \, dx,x,\sqrt [3]{4-27 x^2}\right )}{7 x}-\frac{\left (8 \sqrt [6]{2} \sqrt{3} \sqrt{-x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-4+x^3}} \, dx,x,\sqrt [3]{4-27 x^2}\right )}{7 \sqrt{2-\sqrt{3}} x}\\ &=-\frac{5}{21} i \left (4-27 x^2\right )^{2/3}-\frac{1}{21} i (2+3 i x) \left (4-27 x^2\right )^{2/3}-\frac{72 x}{7 \left (2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}\right )}-\frac{4 \sqrt [3]{2} \sqrt{2+\sqrt{3}} \left (2^{2/3}-\sqrt [3]{4-27 x^2}\right ) \sqrt{\frac{2 \sqrt [3]{2}+2^{2/3} \sqrt [3]{4-27 x^2}+\left (4-27 x^2\right )^{2/3}}{\left (2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}\right )^2}} E\left (\sin ^{-1}\left (\frac{2^{2/3} \left (1+\sqrt{3}\right )-\sqrt [3]{4-27 x^2}}{2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}}\right )|-7+4 \sqrt{3}\right )}{7\ 3^{3/4} x \sqrt{-\frac{2^{2/3}-\sqrt [3]{4-27 x^2}}{\left (2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}\right )^2}}}+\frac{8\ 2^{5/6} \left (2^{2/3}-\sqrt [3]{4-27 x^2}\right ) \sqrt{\frac{2 \sqrt [3]{2}+2^{2/3} \sqrt [3]{4-27 x^2}+\left (4-27 x^2\right )^{2/3}}{\left (2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac{2^{2/3} \left (1+\sqrt{3}\right )-\sqrt [3]{4-27 x^2}}{2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}}\right )|-7+4 \sqrt{3}\right )}{21 \sqrt [4]{3} x \sqrt{-\frac{2^{2/3}-\sqrt [3]{4-27 x^2}}{\left (2^{2/3} \left (1-\sqrt{3}\right )-\sqrt [3]{4-27 x^2}\right )^2}}}\\ \end{align*}

Mathematica [C]  time = 0.0313351, size = 51, normalized size = 0.09 $\frac{12}{7} \sqrt [3]{2} x \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{3}{2};\frac{27 x^2}{4}\right )+\left (4-27 x^2\right )^{2/3} \left (\frac{x}{7}-\frac{i}{3}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(2 + (3*I)*x)^2/(4 - 27*x^2)^(1/3),x]

[Out]

(-I/3 + x/7)*(4 - 27*x^2)^(2/3) + (12*2^(1/3)*x*Hypergeometric2F1[1/3, 1/2, 3/2, (27*x^2)/4])/7

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Maple [C]  time = 0.273, size = 43, normalized size = 0.1 \begin{align*} -{\frac{ \left ( -7\,i+3\,x \right ) \left ( 27\,{x}^{2}-4 \right ) }{21}{\frac{1}{\sqrt [3]{-27\,{x}^{2}+4}}}}+{\frac{12\,\sqrt [3]{2}x}{7}{\mbox{$_2$F$_1$}({\frac{1}{3}},{\frac{1}{2}};\,{\frac{3}{2}};\,{\frac{27\,{x}^{2}}{4}})}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((2+3*I*x)^2/(-27*x^2+4)^(1/3),x)

[Out]

-1/21*(-7*I+3*x)*(27*x^2-4)/(-27*x^2+4)^(1/3)+12/7*2^(1/3)*x*hypergeom([1/3,1/2],[3/2],27/4*x^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (3 i \, x + 2\right )}^{2}}{{\left (-27 \, x^{2} + 4\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*I*x)^2/(-27*x^2+4)^(1/3),x, algorithm="maxima")

[Out]

integrate((3*I*x + 2)^2/(-27*x^2 + 4)^(1/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{21 \, x{\rm integral}\left (\frac{32 \,{\left (-27 \, x^{2} + 4\right )}^{\frac{2}{3}}}{21 \,{\left (27 \, x^{4} - 4 \, x^{2}\right )}}, x\right ) +{\left (3 \, x^{2} - 7 i \, x - 8\right )}{\left (-27 \, x^{2} + 4\right )}^{\frac{2}{3}}}{21 \, x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*I*x)^2/(-27*x^2+4)^(1/3),x, algorithm="fricas")

[Out]

1/21*(21*x*integral(32/21*(-27*x^2 + 4)^(2/3)/(27*x^4 - 4*x^2), x) + (3*x^2 - 7*I*x - 8)*(-27*x^2 + 4)^(2/3))/
x

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Sympy [A]  time = 3.47145, size = 73, normalized size = 0.13 \begin{align*} - \frac{3 \sqrt [3]{2} x^{3}{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{3}{2} \\ \frac{5}{2} \end{matrix}\middle |{\frac{27 x^{2} e^{2 i \pi }}{4}} \right )}}{2} + 2 \sqrt [3]{2} x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{1}{2} \\ \frac{3}{2} \end{matrix}\middle |{\frac{27 x^{2} e^{2 i \pi }}{4}} \right )} - \frac{i \left (4 - 27 x^{2}\right )^{\frac{2}{3}}}{3} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*I*x)**2/(-27*x**2+4)**(1/3),x)

[Out]

-3*2**(1/3)*x**3*hyper((1/3, 3/2), (5/2,), 27*x**2*exp_polar(2*I*pi)/4)/2 + 2*2**(1/3)*x*hyper((1/3, 1/2), (3/
2,), 27*x**2*exp_polar(2*I*pi)/4) - I*(4 - 27*x**2)**(2/3)/3

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (3 i \, x + 2\right )}^{2}}{{\left (-27 \, x^{2} + 4\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*I*x)^2/(-27*x^2+4)^(1/3),x, algorithm="giac")

[Out]

integrate((3*I*x + 2)^2/(-27*x^2 + 4)^(1/3), x)