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

Optimal. Leaf size=44 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {3+2 \sqrt {3}} (1-x)}{\sqrt {x^3-1}}\right )}{\sqrt {3+2 \sqrt {3}}} \]

[Out]

2*arctanh((1-x)*(3+2*3^(1/2))^(1/2)/(x^3-1)^(1/2))/(3+2*3^(1/2))^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2140, 206} \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {3+2 \sqrt {3}} (1-x)}{\sqrt {x^3-1}}\right )}{\sqrt {3+2 \sqrt {3}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {1-\sqrt {3}-x}{\left (1+\sqrt {3}-x\right ) \sqrt {-1+x^3}} \, dx &=2 \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 )\\ &=\frac {2 \tanh ^{-1}\left (\frac {\sqrt {3+2 \sqrt {3}} (1-x)}{\sqrt {-1+x^3}}\right )}{\sqrt {3+2 \sqrt {3}}}\\ \end {align*}

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Mathematica [C]  time = 0.30, size = 265, normalized size = 6.02 \[ \frac {2 \sqrt {6} \sqrt {\frac {i (x-1)}{\sqrt {3}-3 i}} \left (\sqrt {2 i x+\sqrt {3}+i} \left (\left (\sqrt {3}+(2+i)\right ) x+i \sqrt {3}+(1+2 i)\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {-2 i x+\sqrt {3}-i}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{-3 i+\sqrt {3}}\right )-4 i \sqrt {-2 i x+\sqrt {3}-i} \sqrt {x^2+x+1} \Pi \left (\frac {2 i \sqrt {3}}{3+(2+i) \sqrt {3}};\sin ^{-1}\left (\frac {\sqrt {-2 i x+\sqrt {3}-i}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{-3 i+\sqrt {3}}\right )\right )}{\left (3+(2+i) \sqrt {3}\right ) \sqrt {-2 i x+\sqrt {3}-i} \sqrt {x^3-1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Sqrt[6]*Sqrt[(I*(-1 + x))/(-3*I + Sqrt[3])]*(Sqrt[I + Sqrt[3] + (2*I)*x]*((1 + 2*I) + I*Sqrt[3] + ((2 + I)
+ Sqrt[3])*x)*EllipticF[ArcSin[Sqrt[-I + Sqrt[3] - (2*I)*x]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(-3*I + Sqrt[3])]
- (4*I)*Sqrt[-I + Sqrt[3] - (2*I)*x]*Sqrt[1 + x + x^2]*EllipticPi[((2*I)*Sqrt[3])/(3 + (2 + I)*Sqrt[3]), ArcSi
n[Sqrt[-I + Sqrt[3] - (2*I)*x]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(-3*I + Sqrt[3])]))/((3 + (2 + I)*Sqrt[3])*Sqrt
[-I + Sqrt[3] - (2*I)*x]*Sqrt[-1 + x^3])

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fricas [B]  time = 0.49, size = 204, normalized size = 4.64 \[ \frac {1}{6} \, \sqrt {3} \sqrt {2 \, \sqrt {3} - 3} \log \left (\frac {x^{8} + 16 \, x^{7} + 112 \, x^{6} + 16 \, x^{5} + 112 \, x^{4} - 224 \, x^{3} + 64 \, x^{2} - 4 \, {\left (2 \, x^{6} + 18 \, x^{5} + 42 \, x^{4} + 8 \, x^{3} + \sqrt {3} {\left (x^{6} + 12 \, x^{5} + 18 \, x^{4} + 16 \, x^{3} - 12 \, x^{2} - 8\right )} - 24 \, x + 8\right )} \sqrt {x^{3} - 1} \sqrt {2 \, \sqrt {3} - 3} + 16 \, \sqrt {3} {\left (x^{7} + 2 \, x^{6} + 6 \, x^{5} - 5 \, x^{4} + 2 \, x^{3} - 6 \, x^{2} + 4 \, x - 4\right )} - 128 \, x + 112}{x^{8} - 8 \, x^{7} + 16 \, x^{6} + 16 \, x^{5} - 56 \, x^{4} - 32 \, x^{3} + 64 \, x^{2} + 64 \, x + 16}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(3)*sqrt(2*sqrt(3) - 3)*log((x^8 + 16*x^7 + 112*x^6 + 16*x^5 + 112*x^4 - 224*x^3 + 64*x^2 - 4*(2*x^6 +
18*x^5 + 42*x^4 + 8*x^3 + sqrt(3)*(x^6 + 12*x^5 + 18*x^4 + 16*x^3 - 12*x^2 - 8) - 24*x + 8)*sqrt(x^3 - 1)*sqr
t(2*sqrt(3) - 3) + 16*sqrt(3)*(x^7 + 2*x^6 + 6*x^5 - 5*x^4 + 2*x^3 - 6*x^2 + 4*x - 4) - 128*x + 112)/(x^8 - 8*
x^7 + 16*x^6 + 16*x^5 - 56*x^4 - 32*x^3 + 64*x^2 + 64*x + 16))

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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-x-3^(1/2))/(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,-1]:[1,0,-3]%%},[2]%%%} / %%%{%%{[2,4]:[1,0,-3]%%},[2]%%%
} Error: Bad Argument Value

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maple [C]  time = 0.03, size = 245, normalized size = 5.57 \[ \frac {2 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticF \left (\sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}-1}}-\frac {4 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticPi \left (\sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, -\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{3}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}-1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-x-3^(1/2))/(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))-4*(-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)*El
lipticPi(((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 \[ \int \frac {x + \sqrt {3} - 1}{\sqrt {x^{3} - 1} {\left (x - \sqrt {3} - 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.02 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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