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

Optimal. Leaf size=47 \[ \frac{3}{2} \tanh ^{-1}\left (\frac{1-3 x}{2 \sqrt{x^2+x-1}}\right )-\frac{1}{2} \tan ^{-1}\left (\frac{x+3}{2 \sqrt{x^2+x-1}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0342737, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {1033, 724, 206, 204} \[ \frac{3}{2} \tanh ^{-1}\left (\frac{1-3 x}{2 \sqrt{x^2+x-1}}\right )-\frac{1}{2} \tan ^{-1}\left (\frac{x+3}{2 \sqrt{x^2+x-1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1033

Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[-(a*c), 2]}, Dist[h/2 + (c*g)/(2*q), Int[1/((-q + c*x)*Sqrt[d + e*x + f*x^2]), x], x] + Dist[h/2 - (c*g)
/(2*q), Int[1/((q + c*x)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d, e, f, g, h}, x] && NeQ[e^2 - 4*d*f
, 0] && PosQ[-(a*c)]

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])

Rule 204

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

Rubi steps

\begin{align*} \int \frac{1+2 x}{\left (-1+x^2\right ) \sqrt{-1+x+x^2}} \, dx &=\frac{1}{2} \int \frac{1}{(1+x) \sqrt{-1+x+x^2}} \, dx+\frac{3}{2} \int \frac{1}{(-1+x) \sqrt{-1+x+x^2}} \, dx\\ &=-\left (3 \operatorname{Subst}\left (\int \frac{1}{4-x^2} \, dx,x,\frac{-1+3 x}{\sqrt{-1+x+x^2}}\right )\right )-\operatorname{Subst}\left (\int \frac{1}{-4-x^2} \, dx,x,\frac{-3-x}{\sqrt{-1+x+x^2}}\right )\\ &=\frac{1}{2} \tan ^{-1}\left (\frac{-3-x}{2 \sqrt{-1+x+x^2}}\right )+\frac{3}{2} \tanh ^{-1}\left (\frac{1-3 x}{2 \sqrt{-1+x+x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0077584, size = 49, normalized size = 1.04 \[ \frac{1}{2} \tan ^{-1}\left (\frac{-x-3}{2 \sqrt{x^2+x-1}}\right )-\frac{3}{2} \tanh ^{-1}\left (\frac{3 x-1}{2 \sqrt{x^2+x-1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.05, size = 46, normalized size = 1. \begin{align*}{\frac{1}{2}\arctan \left ({\frac{-3-x}{2}{\frac{1}{\sqrt{ \left ( 1+x \right ) ^{2}-2-x}}}} \right ) }-{\frac{3}{2}{\it Artanh} \left ({\frac{3\,x-1}{2}{\frac{1}{\sqrt{ \left ( -1+x \right ) ^{2}-2+3\,x}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.53576, size = 88, normalized size = 1.87 \begin{align*} -\frac{1}{2} \, \arcsin \left (\frac{2 \, \sqrt{5} x}{5 \,{\left | 2 \, x + 2 \right |}} + \frac{6 \, \sqrt{5}}{5 \,{\left | 2 \, x + 2 \right |}}\right ) - \frac{3}{2} \, \log \left (\frac{2 \, \sqrt{x^{2} + x - 1}}{{\left | 2 \, x - 2 \right |}} + \frac{2}{{\left | 2 \, x - 2 \right |}} + \frac{3}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*arcsin(2/5*sqrt(5)*x/abs(2*x + 2) + 6/5*sqrt(5)/abs(2*x + 2)) - 3/2*log(2*sqrt(x^2 + x - 1)/abs(2*x - 2)
+ 2/abs(2*x - 2) + 3/2)

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Fricas [A]  time = 1.52567, size = 146, normalized size = 3.11 \begin{align*} \arctan \left (-x + \sqrt{x^{2} + x - 1} - 1\right ) - \frac{3}{2} \, \log \left (-x + \sqrt{x^{2} + x - 1} + 2\right ) + \frac{3}{2} \, \log \left (-x + \sqrt{x^{2} + x - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.21779, size = 65, normalized size = 1.38 \begin{align*} \arctan \left (-x + \sqrt{x^{2} + x - 1} - 1\right ) - \frac{3}{2} \, \log \left ({\left | -x + \sqrt{x^{2} + x - 1} + 2 \right |}\right ) + \frac{3}{2} \, \log \left ({\left | -x + \sqrt{x^{2} + x - 1} \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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