3.158 \(\int \frac{1-\sqrt{3}-x}{x \sqrt{-1+x^3}} \, dx\)

Optimal. Leaf size=144 \[ \frac{2}{3} \left (1-\sqrt{3}\right ) \tan ^{-1}\left (\sqrt{x^3-1}\right )+\frac{2 \sqrt{2-\sqrt{3}} (1-x) \sqrt{\frac{x^2+x+1}{\left (-x-\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{-x+\sqrt{3}+1}{-x-\sqrt{3}+1}\right )|-7+4 \sqrt{3}\right )}{\sqrt [4]{3} \sqrt{-\frac{1-x}{\left (-x-\sqrt{3}+1\right )^2}} \sqrt{x^3-1}} \]

[Out]

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

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Rubi [A]  time = 0.046942, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1832, 266, 63, 203, 219} \[ \frac{2}{3} \left (1-\sqrt{3}\right ) \tan ^{-1}\left (\sqrt{x^3-1}\right )+\frac{2 \sqrt{2-\sqrt{3}} (1-x) \sqrt{\frac{x^2+x+1}{\left (-x-\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{-x+\sqrt{3}+1}{-x-\sqrt{3}+1}\right )|-7+4 \sqrt{3}\right )}{\sqrt [4]{3} \sqrt{-\frac{1-x}{\left (-x-\sqrt{3}+1\right )^2}} \sqrt{x^3-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1832

Int[(Pq_)/((x_)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Dist[Coeff[Pq, x, 0], Int[1/(x*Sqrt[a + b*x^n]), x
], x] + Int[ExpandToSum[(Pq - Coeff[Pq, x, 0])/x, x]/Sqrt[a + b*x^n], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] &
& IGtQ[n, 0] && NeQ[Coeff[Pq, x, 0], 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 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]

Rubi steps

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

Mathematica [C]  time = 0.031346, size = 60, normalized size = 0.42 \[ \frac{2}{3} \left (1-\sqrt{3}\right ) \tan ^{-1}\left (\sqrt{x^3-1}\right )-\frac{x \sqrt{1-x^3} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};x^3\right )}{\sqrt{x^3-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(1 - Sqrt[3])*ArcTan[Sqrt[-1 + x^3]])/3 - (x*Sqrt[1 - x^3]*Hypergeometric2F1[1/3, 1/2, 4/3, x^3])/Sqrt[-1 +
x^3]

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Maple [A]  time = 0.012, size = 140, normalized size = 1. \begin{align*} -2\,{\frac{-3/2-i/2\sqrt{3}}{\sqrt{{x}^{3}-1}}\sqrt{{\frac{x-1}{-3/2-i/2\sqrt{3}}}}\sqrt{{\frac{x+1/2-i/2\sqrt{3}}{3/2-i/2\sqrt{3}}}}\sqrt{{\frac{x+1/2+i/2\sqrt{3}}{3/2+i/2\sqrt{3}}}}{\it EllipticF} \left ( \sqrt{{\frac{x-1}{-3/2-i/2\sqrt{3}}}},\sqrt{{\frac{3/2+i/2\sqrt{3}}{3/2-i/2\sqrt{3}}}} \right ) }-{\frac{2\,\sqrt{3}}{3}\arctan \left ( \sqrt{{x}^{3}-1} \right ) }+{\frac{2}{3}\arctan \left ( \sqrt{{x}^{3}-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{x + \sqrt{3} - 1}{\sqrt{x^{3} - 1} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{x^{3} - 1}{\left (x + \sqrt{3} - 1\right )}}{x^{4} - x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 8.45124, size = 94, normalized size = 0.65 \begin{align*} \frac{i x \Gamma \left (\frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{1}{2} \\ \frac{4}{3} \end{matrix}\middle |{x^{3}} \right )}}{3 \Gamma \left (\frac{4}{3}\right )} - \sqrt{3} \left (\begin{cases} \frac{2 i \operatorname{acosh}{\left (\frac{1}{x^{\frac{3}{2}}} \right )}}{3} & \text{for}\: \frac{1}{\left |{x^{3}}\right |} > 1 \\- \frac{2 \operatorname{asin}{\left (\frac{1}{x^{\frac{3}{2}}} \right )}}{3} & \text{otherwise} \end{cases}\right ) + \begin{cases} \frac{2 i \operatorname{acosh}{\left (\frac{1}{x^{\frac{3}{2}}} \right )}}{3} & \text{for}\: \frac{1}{\left |{x^{3}}\right |} > 1 \\- \frac{2 \operatorname{asin}{\left (\frac{1}{x^{\frac{3}{2}}} \right )}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*x*gamma(1/3)*hyper((1/3, 1/2), (4/3,), x**3)/(3*gamma(4/3)) - sqrt(3)*Piecewise((2*I*acosh(x**(-3/2))/3, 1/A
bs(x**3) > 1), (-2*asin(x**(-3/2))/3, True)) + Piecewise((2*I*acosh(x**(-3/2))/3, 1/Abs(x**3) > 1), (-2*asin(x
**(-3/2))/3, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x + \sqrt{3} - 1}{\sqrt{x^{3} - 1} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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