3.2525 \(\int \frac{-1+x}{(1+x) \sqrt{1+x+x^2}} \, dx\)

Optimal. Leaf size=35 \[ 2 \tanh ^{-1}\left (\frac{1-x}{2 \sqrt{x^2+x+1}}\right )+\sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0303598, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {843, 619, 215, 724, 206} \[ 2 \tanh ^{-1}\left (\frac{1-x}{2 \sqrt{x^2+x+1}}\right )+\sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 215

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

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 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{-1+x}{(1+x) \sqrt{1+x+x^2}} \, dx &=-\left (2 \int \frac{1}{(1+x) \sqrt{1+x+x^2}} \, dx\right )+\int \frac{1}{\sqrt{1+x+x^2}} \, dx\\ &=4 \operatorname{Subst}\left (\int \frac{1}{4-x^2} \, dx,x,\frac{1-x}{\sqrt{1+x+x^2}}\right )+\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{3}}} \, dx,x,1+2 x\right )}{\sqrt{3}}\\ &=\sinh ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )+2 \tanh ^{-1}\left (\frac{1-x}{2 \sqrt{1+x+x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0065447, size = 35, normalized size = 1. \[ 2 \tanh ^{-1}\left (\frac{1-x}{2 \sqrt{x^2+x+1}}\right )+\sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 32, normalized size = 0.9 \begin{align*}{\it Arcsinh} \left ({\frac{2\,\sqrt{3}}{3} \left ( x+{\frac{1}{2}} \right ) } \right ) +2\,{\it Artanh} \left ( 1/2\,{\frac{1-x}{\sqrt{ \left ( 1+x \right ) ^{2}-x}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.59517, size = 55, normalized size = 1.57 \begin{align*} \operatorname{arsinh}\left (\frac{2}{3} \, \sqrt{3} x + \frac{1}{3} \, \sqrt{3}\right ) - 2 \, \operatorname{arsinh}\left (\frac{\sqrt{3} x}{3 \,{\left | x + 1 \right |}} - \frac{\sqrt{3}}{3 \,{\left | x + 1 \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsinh(2/3*sqrt(3)*x + 1/3*sqrt(3)) - 2*arcsinh(1/3*sqrt(3)*x/abs(x + 1) - 1/3*sqrt(3)/abs(x + 1))

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Fricas [A]  time = 2.09671, size = 142, normalized size = 4.06 \begin{align*} 2 \, \log \left (-x + \sqrt{x^{2} + x + 1}\right ) - 2 \, \log \left (-x + \sqrt{x^{2} + x + 1} - 2\right ) - \log \left (-2 \, x + 2 \, \sqrt{x^{2} + x + 1} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10529, size = 70, normalized size = 2. \begin{align*} -\log \left (-2 \, x + 2 \, \sqrt{x^{2} + x + 1} - 1\right ) + 2 \, \log \left ({\left | -x + \sqrt{x^{2} + x + 1} \right |}\right ) - 2 \, \log \left ({\left | -x + \sqrt{x^{2} + x + 1} - 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(-2*x + 2*sqrt(x^2 + x + 1) - 1) + 2*log(abs(-x + sqrt(x^2 + x + 1))) - 2*log(abs(-x + sqrt(x^2 + x + 1) -
2))