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

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

[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 = 0.09, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2140, 203} \[ -\frac {2 \tan ^{-1}\left (\frac {\sqrt {2 \sqrt {3}-3} (x+1)}{\sqrt {-x^3-1}}\right )}{\sqrt {2 \sqrt {3}-3}} \]

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 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 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 &=-\left (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 )\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 [C]  time = 0.38, size = 269, normalized size = 6.11 \[ -\frac {2 \sqrt {6} \sqrt {\frac {i (x+1)}{\sqrt {3}+3 i}} \left (4 i \sqrt {-2 i x+\sqrt {3}+i} \sqrt {x^2-x+1} \Pi \left (\frac {2 i \sqrt {3}}{-3+(2+i) \sqrt {3}};\sin ^{-1}\left (\frac {\sqrt {-2 i x+\sqrt {3}+i}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )+\sqrt {2 i x+\sqrt {3}-i} \left (\left (\sqrt {3}+(-2-i)\right ) x-i \sqrt {3}+(1+2 i)\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {-2 i x+\sqrt {3}+i}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )\right )}{\left (-3+(2+i) \sqrt {3}\right ) \sqrt {-2 i x+\sqrt {3}+i} \sqrt {-x^3-1}} \]

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]), ArcSin
[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 [A]  time = 0.49, size = 59, normalized size = 1.34 \[ \frac {1}{3} \, \sqrt {3} \sqrt {2 \, \sqrt {3} + 3} \arctan \left (\frac {\sqrt {-x^{3} - 1} {\left (\sqrt {3} {\left (x^{2} - 4 \, x - 2\right )} + 6 \, x + 6\right )} \sqrt {2 \, \sqrt {3} + 3}}{6 \, {\left (x^{3} + 1\right )}}\right ) \]

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]

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

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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 = 0.07, size = 247, normalized size = 5.61 \[ -\frac {2 i \sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x +1}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x -\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \EllipticF \left (\frac {\sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}-1}}-\frac {4 i \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x +1}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x -\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \frac {i \sqrt {3}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}-\sqrt {3}}, \sqrt {\frac {i \sqrt {3}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {-x^{3}-1}\, \left (\frac {3}{2}+\frac {i \sqrt {3}}{2}-\sqrt {3}\right )} \]

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