3.2.3 \(\int \frac {1+\sqrt {3}-x}{(1-\sqrt {3}-x) \sqrt {-1+x^3}} \, dx\) [103]

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

[Out]

2*arctan((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 = 0.07, 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 = {2165, 209} \begin {gather*} \frac {2 \text {ArcTan}\left (\frac {\sqrt {2 \sqrt {3}-3} (1-x)}{\sqrt {x^3-1}}\right )}{\sqrt {2 \sqrt {3}-3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 209

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

Rule 2165

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 \text {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 \tan ^{-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 [A]
time = 1.62, size = 47, normalized size = 1.07 \begin {gather*} -2 \sqrt {1+\frac {2}{\sqrt {3}}} \tan ^{-1}\left (\frac {\sqrt {-3+2 \sqrt {3}} \sqrt {-1+x^3}}{1+x+x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.96, size = 245, normalized size = 5.57

method	result	size
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}+24 \sqrt {3}+36\right ) \ln \left (-\frac {6 \RootOf \left (\textit {\_Z}^{2}+24 \sqrt {3}+36\right ) x^{2}+4 \RootOf \left (\textit {\_Z}^{2}+24 \sqrt {3}+36\right ) \sqrt {3}\, x^{2}+4 \sqrt {3}\, \RootOf \left (\textit {\_Z}^{2}+24 \sqrt {3}+36\right ) x +48 \sqrt {x^{3}-1}\, \sqrt {3}+4 \RootOf \left (\textit {\_Z}^{2}+24 \sqrt {3}+36\right ) \sqrt {3}+72 \sqrt {x^{3}-1}+12 \RootOf \left (\textit {\_Z}^{2}+24 \sqrt {3}+36\right )}{\left (x \sqrt {3}+x +2\right )^{2}}\right )}{6}\)	\(130\)
default	\(\frac {2 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticF \left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}-1}}-\frac {4 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticPi \left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{3}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}-1}}\)	\(245\)
elliptic	\(\frac {2 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticF \left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}-1}}-\frac {4 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticPi \left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{3}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}-1}}\)	\(245\)

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,method=_RETURNVERBOSE)

[Out]

2*(-3/2-1/2*I*3^(1/2))*((-1+x)/(-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(((-1+x)/(-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))*((-1+x)/(-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)
*EllipticPi(((-1+x)/(-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [A]
time = 0.37, size = 50, normalized size = 1.14 \begin {gather*} \frac {1}{3} \, \sqrt {3} \sqrt {2 \, \sqrt {3} + 3} \arctan \left (\frac {{\left (\sqrt {3} {\left (x^{2} + 4 \, x - 2\right )} - 6 \, x + 6\right )} \sqrt {2 \, \sqrt {3} + 3}}{6 \, \sqrt {x^{3} - 1}}\right ) \end {gather*}

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]

1/3*sqrt(3)*sqrt(2*sqrt(3) + 3)*arctan(1/6*(sqrt(3)*(x^2 + 4*x - 2) - 6*x + 6)*sqrt(2*sqrt(3) + 3)/sqrt(x^3 -
1))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x - \sqrt {3} - 1}{\sqrt {\left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x - 1 + \sqrt {3}\right )}\, dx \end {gather*}

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{%%{[1,1]:[1,0,-3]%%},[2]%%%} / %%%{%%{[-2,4]:[1,0,-3]%%}
,[2]%%%} Er

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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