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

Optimal. Leaf size=126 \[ \frac{\sqrt [4]{3} \sqrt{2-\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} E\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{\sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^3+1}}-\frac{2 \sqrt{x^3+1}}{x+\sqrt{3}+1} \]

[Out]

(-2*Sqrt[1 + x^3])/(1 + Sqrt[3] + x) + (3^(1/4)*Sqrt[2 - Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)
^2]*EllipticE[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]
*Sqrt[1 + x^3])

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Rubi [A]  time = 0.0246693, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {1877} \[ \frac{\sqrt [4]{3} \sqrt{2-\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} E\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{\sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^3+1}}-\frac{2 \sqrt{x^3+1}}{x+\sqrt{3}+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[1 + x^3])/(1 + Sqrt[3] + x) + (3^(1/4)*Sqrt[2 - Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)
^2]*EllipticE[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]
*Sqrt[1 + x^3])

Rule 1877

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Simplify[((1 - Sqrt[3])*d)/c]]
, s = Denom[Simplify[((1 - Sqrt[3])*d)/c]]}, Simp[(2*d*s^3*Sqrt[a + b*x^3])/(a*r^2*((1 + Sqrt[3])*s + r*x)), x
] - Simp[(3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*Elli
pticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(r^2*Sqrt[a + b*x^3]*Sqrt[(s*(
s + r*x))/((1 + Sqrt[3])*s + r*x)^2]), x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && EqQ[b*c^3 - 2*(5 - 3*Sqrt[3
])*a*d^3, 0]

Rubi steps

\begin{align*} \int \frac{-1+\sqrt{3}-x}{\sqrt{1+x^3}} \, dx &=-\frac{2 \sqrt{1+x^3}}{1+\sqrt{3}+x}+\frac{\sqrt [4]{3} \sqrt{2-\sqrt{3}} (1+x) \sqrt{\frac{1-x+x^2}{\left (1+\sqrt{3}+x\right )^2}} E\left (\sin ^{-1}\left (\frac{1-\sqrt{3}+x}{1+\sqrt{3}+x}\right )|-7-4 \sqrt{3}\right )}{\sqrt{\frac{1+x}{\left (1+\sqrt{3}+x\right )^2}} \sqrt{1+x^3}}\\ \end{align*}

Mathematica [C]  time = 0.0153862, size = 47, normalized size = 0.37 \[ \left (\sqrt{3}-1\right ) x \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};-x^3\right )-\frac{1}{2} x^2 \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};-x^3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1 + Sqrt[3])*x*Hypergeometric2F1[1/3, 1/2, 4/3, -x^3] - (x^2*Hypergeometric2F1[1/2, 2/3, 5/3, -x^3])/2

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Maple [B]  time = 0.01, size = 407, normalized size = 3.2 \begin{align*} -2\,{\frac{3/2-i/2\sqrt{3}}{\sqrt{{x}^{3}+1}}\sqrt{{\frac{1+x}{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}}}} \left ( \left ( -3/2-i/2\sqrt{3} \right ){\it EllipticE} \left ( \sqrt{{\frac{1+x}{3/2-i/2\sqrt{3}}}},\sqrt{{\frac{-3/2+i/2\sqrt{3}}{-3/2-i/2\sqrt{3}}}} \right ) + \left ( 1/2+i/2\sqrt{3} \right ){\it EllipticF} \left ( \sqrt{{\frac{1+x}{3/2-i/2\sqrt{3}}}},\sqrt{{\frac{-3/2+i/2\sqrt{3}}{-3/2-i/2\sqrt{3}}}} \right ) \right ) }-2\,{\frac{3/2-i/2\sqrt{3}}{\sqrt{{x}^{3}+1}}\sqrt{{\frac{1+x}{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{1+x}{3/2-i/2\sqrt{3}}}},\sqrt{{\frac{-3/2+i/2\sqrt{3}}{-3/2-i/2\sqrt{3}}}} \right ) }+2\,{\frac{\sqrt{3} \left ( 3/2-i/2\sqrt{3} \right ) }{\sqrt{{x}^{3}+1}}\sqrt{{\frac{1+x}{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{1+x}{3/2-i/2\sqrt{3}}}},\sqrt{{\frac{-3/2+i/2\sqrt{3}}{-3/2-i/2\sqrt{3}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)*((-3/2-1/2*I*3^(1/2))*EllipticE(((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))+(1/2+1/2*I*3^(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)))-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))+2*3^(1/2)*(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))

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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}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.30229, size = 92, normalized size = 0.73 \begin{align*} - \frac{x^{2} \Gamma \left (\frac{2}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{2}{3} \\ \frac{5}{3} \end{matrix}\middle |{x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac{5}{3}\right )} - \frac{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} e^{i \pi }} \right )}}{3 \Gamma \left (\frac{4}{3}\right )} + \frac{\sqrt{3} 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} e^{i \pi }} \right )}}{3 \Gamma \left (\frac{4}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2*gamma(2/3)*hyper((1/2, 2/3), (5/3,), x**3*exp_polar(I*pi))/(3*gamma(5/3)) - x*gamma(1/3)*hyper((1/3, 1/2
), (4/3,), x**3*exp_polar(I*pi))/(3*gamma(4/3)) + sqrt(3)*x*gamma(1/3)*hyper((1/3, 1/2), (4/3,), x**3*exp_pola
r(I*pi))/(3*gamma(4/3))

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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}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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