3.2 \(\int \frac {1}{(2^{2/3}-x) \sqrt {1-x^3}} \, dx\)

Optimal. Leaf size=160 \[ -\frac {2 \tan ^{-1}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} x\right )}{\sqrt {1-x^3}}\right )}{3 \sqrt {3}}-\frac {2 \sqrt [3]{2} \sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x+\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {-x-\sqrt {3}+1}{-x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {1-x}{\left (-x+\sqrt {3}+1\right )^2}} \sqrt {1-x^3}} \]

[Out]

-2/9*arctan((1-2^(1/3)*x)*3^(1/2)/(-x^3+1)^(1/2))*3^(1/2)-2/9*2^(1/3)*(1-x)*EllipticF((1-x-3^(1/2))/(1-x+3^(1/
2)),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/2))*((x^2+x+1)/(1-x+3^(1/2))^2)^(1/2)*3^(3/4)/(-x^3+1)^(1/2)/((1-x)/(
1-x+3^(1/2))^2)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2134, 218, 2137, 203} \[ -\frac {2 \sqrt [3]{2} \sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x+\sqrt {3}+1\right )^2}} \text {EllipticF}\left (\sin ^{-1}\left (\frac {-x-\sqrt {3}+1}{-x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {1-x}{\left (-x+\sqrt {3}+1\right )^2}} \sqrt {1-x^3}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} x\right )}{\sqrt {1-x^3}}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[1/((2^(2/3) - x)*Sqrt[1 - x^3]),x]

[Out]

(-2*ArcTan[(Sqrt[3]*(1 - 2^(1/3)*x))/Sqrt[1 - x^3]])/(3*Sqrt[3]) - (2*2^(1/3)*Sqrt[2 + Sqrt[3]]*(1 - x)*Sqrt[(
1 + x + x^2)/(1 + Sqrt[3] - x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] - x)/(1 + Sqrt[3] - x)], -7 - 4*Sqrt[3]])/(3*3
^(1/4)*Sqrt[(1 - x)/(1 + Sqrt[3] - x)^2]*Sqrt[1 - x^3])

Rule 203

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

Rule 218

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 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 2134

Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Dist[2/(3*c), Int[1/Sqrt[a + b*x^3], x], x
] + Dist[1/(3*c), Int[(c - 2*d*x)/((c + d*x)*Sqrt[a + b*x^3]), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^3 -
4*a*d^3, 0]

Rule 2137

Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Dist[(2*e)/d, Subst[Int[
1/(1 + 3*a*x^2), x], x, (1 + (2*d*x)/c)/Sqrt[a + b*x^3]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f,
0] && EqQ[b*c^3 - 4*a*d^3, 0] && EqQ[2*d*e + c*f, 0]

Rubi steps

\begin {align*} \int \frac {1}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx &=\frac {\int \frac {2^{2/3}+2 x}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx}{3\ 2^{2/3}}+\frac {1}{3} \sqrt [3]{2} \int \frac {1}{\sqrt {1-x^3}} \, dx\\ &=-\frac {2 \sqrt [3]{2} \sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {1+x+x^2}{\left (1+\sqrt {3}-x\right )^2}} F\left (\sin ^{-1}\left (\frac {1-\sqrt {3}-x}{1+\sqrt {3}-x}\right )|-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {1-x}{\left (1+\sqrt {3}-x\right )^2}} \sqrt {1-x^3}}-\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{1+3 x^2} \, dx,x,\frac {1-\sqrt [3]{2} x}{\sqrt {1-x^3}}\right )\\ &=-\frac {2 \tan ^{-1}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} x\right )}{\sqrt {1-x^3}}\right )}{3 \sqrt {3}}-\frac {2 \sqrt [3]{2} \sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {1+x+x^2}{\left (1+\sqrt {3}-x\right )^2}} F\left (\sin ^{-1}\left (\frac {1-\sqrt {3}-x}{1+\sqrt {3}-x}\right )|-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {1-x}{\left (1+\sqrt {3}-x\right )^2}} \sqrt {1-x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.12, size = 148, normalized size = 0.92 \[ -\frac {4 i \sqrt {2} \sqrt {-\frac {i (x-1)}{\sqrt {3}+3 i}} \sqrt {x^2+x+1} \Pi \left (\frac {2 \sqrt {3}}{i+2 i 2^{2/3}+\sqrt {3}};\sin ^{-1}\left (\frac {\sqrt {2 i x+\sqrt {3}+i}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )}{\left (1+2\ 2^{2/3}-i \sqrt {3}\right ) \sqrt {1-x^3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((2^(2/3) - x)*Sqrt[1 - x^3]),x]

[Out]

((-4*I)*Sqrt[2]*Sqrt[((-I)*(-1 + x))/(3*I + Sqrt[3])]*Sqrt[1 + x + x^2]*EllipticPi[(2*Sqrt[3])/(I + (2*I)*2^(2
/3) + Sqrt[3]), ArcSin[Sqrt[I + Sqrt[3] + (2*I)*x]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(3*I + Sqrt[3])])/((1 + 2*2
^(2/3) - I*Sqrt[3])*Sqrt[1 - x^3])

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-x^{3} + 1} {\left (x^{2} + 2^{\frac {2}{3}} x + 2 \cdot 2^{\frac {1}{3}}\right )}}{x^{6} - 5 \, x^{3} + 4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-x^3 + 1)*(x^2 + 2^(2/3)*x + 2*2^(1/3))/(x^6 - 5*x^3 + 4), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to divide, perhaps due to rounding error%%%{1,[2]%%%} / %%%{%%{[2,0]:[1,0,0,-2]%%},[2]%%%} Error: Bad Argum
ent Value

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maple [A]  time = 0.09, size = 143, normalized size = 0.89 \[ \frac {2 i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \frac {i \sqrt {3}}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}-2^{\frac {2}{3}}}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}+1}\, \left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}-2^{\frac {2}{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2^(2/3)-x)/(-x^3+1)^(1/2),x)

[Out]

2/3*I*3^(1/2)*(I*(x+1/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*((x-1)/(-3/2+1/2*I*3^(1/2)))^(1/2)*(-I*(x+1/2+1/2*I*3^(1
/2))*3^(1/2))^(1/2)/(-x^3+1)^(1/2)/(-1/2+1/2*I*3^(1/2)-2^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2-1/2*I*3^(1/2)
)*3^(1/2))^(1/2),I*3^(1/2)/(-1/2+1/2*I*3^(1/2)-2^(2/3)),(I*3^(1/2)/(-3/2+1/2*I*3^(1/2)))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {1}{\sqrt {-x^{3} + 1} {\left (x - 2^{\frac {2}{3}}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate(1/(sqrt(-x^3 + 1)*(x - 2^(2/3))), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {1}{\sqrt {1-x^3}\,\left (x-2^{2/3}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/((1 - x^3)^(1/2)*(x - 2^(2/3))),x)

[Out]

-int(1/((1 - x^3)^(1/2)*(x - 2^(2/3))), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {1}{x \sqrt {1 - x^{3}} - 2^{\frac {2}{3}} \sqrt {1 - x^{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(1/(x*sqrt(1 - x**3) - 2**(2/3)*sqrt(1 - x**3)), x)

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