3.200 \(\int \frac{2+2 x-x^2}{(2+x^2) \sqrt{-1+x^3}} \, dx\)

Optimal. Leaf size=18 \[ -2 \tanh ^{-1}\left (\frac{1-x}{\sqrt{x^3-1}}\right ) \]

[Out]

-2*ArcTanh[(1 - x)/Sqrt[-1 + x^3]]

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Rubi [A]  time = 0.0792075, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {2146, 206} \[ -2 \tanh ^{-1}\left (\frac{1-x}{\sqrt{x^3-1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*ArcTanh[(1 - x)/Sqrt[-1 + x^3]]

Rule 2146

Int[((f_) + (g_.)*(x_) + (h_.)*(x_)^2)/(((c_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> -Dist[g
/e, Subst[Int[1/(1 + a*x^2), x], x, (1 + (2*h*x)/g)/Sqrt[a + b*x^3]], x] /; FreeQ[{a, b, c, e, f, g, h}, x] &&
EqQ[b*g^3 - 8*a*h^3, 0] && EqQ[g^2 + 2*f*h, 0] && EqQ[b*c*g - 4*a*e*h, 0]

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])

Rubi steps

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

Mathematica [C]  time = 0.227572, size = 278, normalized size = 15.44 \[ \frac{2 \sqrt{\frac{1-x}{1+\sqrt [3]{-1}}} \sqrt{x^2+x+1} \left (\frac{\sqrt{3} \left (1+\sqrt [3]{-1}\right ) \left (x+\sqrt [3]{-1}\right ) F\left (\sin ^{-1}\left (\sqrt{\frac{1-(-1)^{2/3} x}{1+\sqrt [3]{-1}}}\right )|\sqrt [3]{-1}\right )}{(-1)^{2/3} x-1}+\frac{6 \left (1+i \sqrt{2}\right ) \Pi \left (\frac{2 \sqrt{3}}{-i-2 \sqrt{2}+\sqrt{3}};\sin ^{-1}\left (\sqrt{\frac{1-(-1)^{2/3} x}{1+\sqrt [3]{-1}}}\right )|\sqrt [3]{-1}\right )}{i+2 \sqrt{2}-\sqrt{3}}+\frac{3 \left (1-i \sqrt{2}\right ) \Pi \left (\frac{2 \sqrt{3}}{-i+2 \sqrt{2}+\sqrt{3}};\sin ^{-1}\left (\sqrt{\frac{1-(-1)^{2/3} x}{1+\sqrt [3]{-1}}}\right )|\sqrt [3]{-1}\right )}{(-1)^{5/6}-\sqrt{2}}\right )}{3 \sqrt{x^3-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(1 - x)/(1 + (-1)^(1/3))]*Sqrt[1 + x + x^2]*((Sqrt[3]*(1 + (-1)^(1/3))*((-1)^(1/3) + x)*EllipticF[ArcS
in[Sqrt[(1 - (-1)^(2/3)*x)/(1 + (-1)^(1/3))]], (-1)^(1/3)])/(-1 + (-1)^(2/3)*x) + (6*(1 + I*Sqrt[2])*EllipticP
i[(2*Sqrt[3])/(-I - 2*Sqrt[2] + Sqrt[3]), ArcSin[Sqrt[(1 - (-1)^(2/3)*x)/(1 + (-1)^(1/3))]], (-1)^(1/3)])/(I +
2*Sqrt[2] - Sqrt[3]) + (3*(1 - I*Sqrt[2])*EllipticPi[(2*Sqrt[3])/(-I + 2*Sqrt[2] + Sqrt[3]), ArcSin[Sqrt[(1 -
(-1)^(2/3)*x)/(1 + (-1)^(1/3))]], (-1)^(1/3)])/((-1)^(5/6) - Sqrt[2])))/(3*Sqrt[-1 + x^3])

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Maple [C]  time = 0.032, size = 1656, normalized size = 92. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2+2*x+2)/(x^2+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))-3*(1/(-3/2-1/2*I*3^(1/2))*x-1/(-3/2-1/2*I*3^(1/2)))^(1/2)*(1/(3
/2-1/2*I*3^(1/2))*x+1/2/(3/2-1/2*I*3^(1/2))-1/2*I/(3/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*(1/(3/2+1/2*I*3^(1/2))*x+
1/2/(3/2+1/2*I*3^(1/2))+1/2*I/(3/2+1/2*I*3^(1/2))*3^(1/2))^(1/2)/(x^3-1)^(1/2)/(-I*2^(1/2)+1)*EllipticPi(((x-1
)/(-3/2-1/2*I*3^(1/2)))^(1/2),(3/2+1/2*I*3^(1/2))/(-I*2^(1/2)+1),((3/2+1/2*I*3^(1/2))/(3/2-1/2*I*3^(1/2)))^(1/
2))-I*(1/(-3/2-1/2*I*3^(1/2))*x-1/(-3/2-1/2*I*3^(1/2)))^(1/2)*(1/(3/2-1/2*I*3^(1/2))*x+1/2/(3/2-1/2*I*3^(1/2))
-1/2*I/(3/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*(1/(3/2+1/2*I*3^(1/2))*x+1/2/(3/2+1/2*I*3^(1/2))+1/2*I/(3/2+1/2*I*3^
(1/2))*3^(1/2))^(1/2)/(x^3-1)^(1/2)/(-I*2^(1/2)+1)*EllipticPi(((x-1)/(-3/2-1/2*I*3^(1/2)))^(1/2),(3/2+1/2*I*3^
(1/2))/(-I*2^(1/2)+1),((3/2+1/2*I*3^(1/2))/(3/2-1/2*I*3^(1/2)))^(1/2))*3^(1/2)+3*I*2^(1/2)*(1/(-3/2-1/2*I*3^(1
/2))*x-1/(-3/2-1/2*I*3^(1/2)))^(1/2)*(1/(3/2-1/2*I*3^(1/2))*x+1/2/(3/2-1/2*I*3^(1/2))-1/2*I/(3/2-1/2*I*3^(1/2)
)*3^(1/2))^(1/2)*(1/(3/2+1/2*I*3^(1/2))*x+1/2/(3/2+1/2*I*3^(1/2))+1/2*I/(3/2+1/2*I*3^(1/2))*3^(1/2))^(1/2)/(x^
3-1)^(1/2)/(-I*2^(1/2)+1)*EllipticPi(((x-1)/(-3/2-1/2*I*3^(1/2)))^(1/2),(3/2+1/2*I*3^(1/2))/(-I*2^(1/2)+1),((3
/2+1/2*I*3^(1/2))/(3/2-1/2*I*3^(1/2)))^(1/2))-2^(1/2)*(1/(-3/2-1/2*I*3^(1/2))*x-1/(-3/2-1/2*I*3^(1/2)))^(1/2)*
(1/(3/2-1/2*I*3^(1/2))*x+1/2/(3/2-1/2*I*3^(1/2))-1/2*I/(3/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*(1/(3/2+1/2*I*3^(1/2
))*x+1/2/(3/2+1/2*I*3^(1/2))+1/2*I/(3/2+1/2*I*3^(1/2))*3^(1/2))^(1/2)/(x^3-1)^(1/2)/(-I*2^(1/2)+1)*EllipticPi(
((x-1)/(-3/2-1/2*I*3^(1/2)))^(1/2),(3/2+1/2*I*3^(1/2))/(-I*2^(1/2)+1),((3/2+1/2*I*3^(1/2))/(3/2-1/2*I*3^(1/2))
)^(1/2))*3^(1/2)-3*(1/(-3/2-1/2*I*3^(1/2))*x-1/(-3/2-1/2*I*3^(1/2)))^(1/2)*(1/(3/2-1/2*I*3^(1/2))*x+1/2/(3/2-1
/2*I*3^(1/2))-1/2*I/(3/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*(1/(3/2+1/2*I*3^(1/2))*x+1/2/(3/2+1/2*I*3^(1/2))+1/2*I/
(3/2+1/2*I*3^(1/2))*3^(1/2))^(1/2)/(x^3-1)^(1/2)/(I*2^(1/2)+1)*EllipticPi(((x-1)/(-3/2-1/2*I*3^(1/2)))^(1/2),(
3/2+1/2*I*3^(1/2))/(I*2^(1/2)+1),((3/2+1/2*I*3^(1/2))/(3/2-1/2*I*3^(1/2)))^(1/2))-I*(1/(-3/2-1/2*I*3^(1/2))*x-
1/(-3/2-1/2*I*3^(1/2)))^(1/2)*(1/(3/2-1/2*I*3^(1/2))*x+1/2/(3/2-1/2*I*3^(1/2))-1/2*I/(3/2-1/2*I*3^(1/2))*3^(1/
2))^(1/2)*(1/(3/2+1/2*I*3^(1/2))*x+1/2/(3/2+1/2*I*3^(1/2))+1/2*I/(3/2+1/2*I*3^(1/2))*3^(1/2))^(1/2)/(x^3-1)^(1
/2)/(I*2^(1/2)+1)*EllipticPi(((x-1)/(-3/2-1/2*I*3^(1/2)))^(1/2),(3/2+1/2*I*3^(1/2))/(I*2^(1/2)+1),((3/2+1/2*I*
3^(1/2))/(3/2-1/2*I*3^(1/2)))^(1/2))*3^(1/2)-3*I*2^(1/2)*(1/(-3/2-1/2*I*3^(1/2))*x-1/(-3/2-1/2*I*3^(1/2)))^(1/
2)*(1/(3/2-1/2*I*3^(1/2))*x+1/2/(3/2-1/2*I*3^(1/2))-1/2*I/(3/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*(1/(3/2+1/2*I*3^(
1/2))*x+1/2/(3/2+1/2*I*3^(1/2))+1/2*I/(3/2+1/2*I*3^(1/2))*3^(1/2))^(1/2)/(x^3-1)^(1/2)/(I*2^(1/2)+1)*EllipticP
i(((x-1)/(-3/2-1/2*I*3^(1/2)))^(1/2),(3/2+1/2*I*3^(1/2))/(I*2^(1/2)+1),((3/2+1/2*I*3^(1/2))/(3/2-1/2*I*3^(1/2)
))^(1/2))+2^(1/2)*(1/(-3/2-1/2*I*3^(1/2))*x-1/(-3/2-1/2*I*3^(1/2)))^(1/2)*(1/(3/2-1/2*I*3^(1/2))*x+1/2/(3/2-1/
2*I*3^(1/2))-1/2*I/(3/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*(1/(3/2+1/2*I*3^(1/2))*x+1/2/(3/2+1/2*I*3^(1/2))+1/2*I/(
3/2+1/2*I*3^(1/2))*3^(1/2))^(1/2)/(x^3-1)^(1/2)/(I*2^(1/2)+1)*EllipticPi(((x-1)/(-3/2-1/2*I*3^(1/2)))^(1/2),(3
/2+1/2*I*3^(1/2))/(I*2^(1/2)+1),((3/2+1/2*I*3^(1/2))/(3/2-1/2*I*3^(1/2)))^(1/2))*3^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{x^{2} - 2 \, x - 2}{\sqrt{x^{3} - 1}{\left (x^{2} + 2\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.37915, size = 62, normalized size = 3.44 \begin{align*} \log \left (\frac{x^{2} + 2 \, x + 2 \, \sqrt{x^{3} - 1}}{x^{2} + 2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((x^2 + 2*x + 2*sqrt(x^3 - 1))/(x^2 + 2))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int - \frac{2 x}{x^{2} \sqrt{x^{3} - 1} + 2 \sqrt{x^{3} - 1}}\, dx - \int \frac{x^{2}}{x^{2} \sqrt{x^{3} - 1} + 2 \sqrt{x^{3} - 1}}\, dx - \int - \frac{2}{x^{2} \sqrt{x^{3} - 1} + 2 \sqrt{x^{3} - 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x^{2} - 2 \, x - 2}{\sqrt{x^{3} - 1}{\left (x^{2} + 2\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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