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

Optimal. Leaf size=164 \[ \frac {2 \sqrt {\frac {7}{6}-\frac {2}{\sqrt {3}}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {3+2 \sqrt {3}} (1-x)}{\sqrt {x^3-1}}\right )}{3^{3/4}} \]

[Out]

-1/3*arctanh((1-x)*(3+2*3^(1/2))^(1/2)/(x^3-1)^(1/2))*2^(1/2)*3^(1/4)+2/3*(1-x)*EllipticF((1-x+3^(1/2))/(1-x-3
^(1/2)),2*I-I*3^(1/2))*(1/3*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 = 0.22, antiderivative size = 164, 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 = {2141, 219, 2140, 206} \[ \frac {2 \sqrt {\frac {7}{6}-\frac {2}{\sqrt {3}}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {3+2 \sqrt {3}} (1-x)}{\sqrt {x^3-1}}\right )}{3^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*ArcTanh[(Sqrt[3 + 2*Sqrt[3]]*(1 - x))/Sqrt[-1 + x^3]])/3^(3/4)) + (2*Sqrt[7/6 - 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 206

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

Rule 2140

Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> With[{k = Simplify[(d*e
+ 2*c*f)/(c*f)]}, Dist[((1 + k)*e)/d, Subst[Int[1/(1 + (3 + 2*k)*a*x^2), x], x, (1 + ((1 + k)*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^2*c^6 - 20*a*b*c^3*d^3 - 8*a^2*d^6
, 0] && EqQ[6*a*d^4*e - c*f*(b*c^3 - 22*a*d^3), 0]

Rule 2141

Int[((e_.) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> -Dist[(6*a*d^4*e - c*f*
(b*c^3 - 22*a*d^3))/(c*d*(b*c^3 - 28*a*d^3)), Int[1/Sqrt[a + b*x^3], x], x] + Dist[(d*e - c*f)/(c*d*(b*c^3 - 2
8*a*d^3)), Int[(c*(b*c^3 - 22*a*d^3) + 6*a*d^4*x)/((c + d*x)*Sqrt[a + b*x^3]), x], x] /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[d*e - c*f, 0] && EqQ[b^2*c^6 - 20*a*b*c^3*d^3 - 8*a^2*d^6, 0] && NeQ[6*a*d^4*e - c*f*(b*c^3 - 2
2*a*d^3), 0]

Rubi steps

\begin {align*} \int \frac {x}{\left (1+\sqrt {3}-x\right ) \sqrt {-1+x^3}} \, dx &=-\frac {\left (-1-\sqrt {3}\right ) \int \frac {\left (1+\sqrt {3}\right ) \left (-22+\left (1+\sqrt {3}\right )^3\right )-6 x}{\left (1+\sqrt {3}-x\right ) \sqrt {-1+x^3}} \, dx}{\left (1+\sqrt {3}\right ) \left (-28+\left (1+\sqrt {3}\right )^3\right )}-\frac {\left (-22+\left (1+\sqrt {3}\right )^3\right ) \int \frac {1}{\sqrt {-1+x^3}} \, dx}{-28+\left (1+\sqrt {3}\right )^3}\\ &=\frac {2 \sqrt {\frac {7}{6}-\frac {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 {\left (12 \left (-1-\sqrt {3}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-\left (3+2 \sqrt {3}\right ) x^2} \, dx,x,\frac {1-x}{\sqrt {-1+x^3}}\right )}{\left (1+\sqrt {3}\right ) \left (-28+\left (1+\sqrt {3}\right )^3\right )}\\ &=-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {3+2 \sqrt {3}} (1-x)}{\sqrt {-1+x^3}}\right )}{3^{3/4}}+\frac {2 \sqrt {\frac {7}{6}-\frac {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*}

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Mathematica [C]  time = 0.27, size = 230, normalized size = 1.40 \[ \frac {2 i \sqrt {\frac {1-x}{1+\sqrt [3]{-1}}} \left (2 \left (1+\sqrt {3}\right ) \sqrt {x^2+x+1} \Pi \left (\frac {2 i \sqrt {3}}{3+(2+i) \sqrt {3}};\sin ^{-1}\left (\sqrt {\frac {1-(-1)^{2/3} x}{1+\sqrt [3]{-1}}}\right )|\sqrt [3]{-1}\right )+\frac {i \sqrt {\frac {(-1)^{2/3} x+\sqrt [3]{-1}}{1+\sqrt [3]{-1}}} \left (\left (3+(2+i) \sqrt {3}\right ) x+(1+2 i) \sqrt {3}+3 i\right ) F\left (\sin ^{-1}\left (\sqrt {\frac {1-(-1)^{2/3} x}{1+\sqrt [3]{-1}}}\right )|\sqrt [3]{-1}\right )}{\sqrt {\frac {1-(-1)^{2/3} x}{1+\sqrt [3]{-1}}}}\right )}{\left (3+(2+i) \sqrt {3}\right ) \sqrt {x^3-1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {x^{3} - 1} {\left (x^{2} + \sqrt {3} x - x\right )}}{x^{5} - 2 \, x^{4} - 2 \, x^{3} - x^{2} + 2 \, x + 2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(x^3 - 1)*(x^2 + sqrt(3)*x - x)/(x^5 - 2*x^4 - 2*x^3 - x^2 + 2*x + 2), 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(x/(1-x+3^(1/2))/(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]%%%} / %%%{%%{[2,4]:[1,0,-3]%%},[2]%%%} Error: Bad Argumen
t Value

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maple [A]  time = 0.03, size = 255, normalized size = 1.55 \[ -\frac {2 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticF \left (\sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}-1}}-\frac {2 \left (-1-\sqrt {3}\right ) \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {3}\, \EllipticPi \left (\sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, -\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{3}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {x^{3}-1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(1-x+3^(1/2))/(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*(-1-3^(1/2))*(-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)*3^(1/2)*EllipticPi(((x-1)/(-3/2-1/2*I*3^(1/2)))^(1/2),-1/3*(3/2+1/2*I*3^(1/2))*3^(1/2),((3/2+1/2
*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 \[ -\int \frac {x}{\sqrt {x^{3} - 1} {\left (x - \sqrt {3} - 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.01 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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