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

Optimal. Leaf size=70 \[ \frac{\sqrt{x^2+x-1}}{2 \left (1-x^2\right )}-\frac{1}{8} \tan ^{-1}\left (\frac{x+3}{2 \sqrt{x^2+x-1}}\right )-\frac{5}{8} \tanh ^{-1}\left (\frac{1-3 x}{2 \sqrt{x^2+x-1}}\right ) \]

[Out]

Sqrt[-1 + x + x^2]/(2*(1 - x^2)) - ArcTan[(3 + x)/(2*Sqrt[-1 + x + x^2])]/8 - (5*ArcTanh[(1 - 3*x)/(2*Sqrt[-1
+ x + x^2])])/8

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Rubi [A]  time = 0.0514388, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {976, 1033, 724, 206, 204} \[ \frac{\sqrt{x^2+x-1}}{2 \left (1-x^2\right )}-\frac{1}{8} \tan ^{-1}\left (\frac{x+3}{2 \sqrt{x^2+x-1}}\right )-\frac{5}{8} \tanh ^{-1}\left (\frac{1-3 x}{2 \sqrt{x^2+x-1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[-1 + x + x^2]/(2*(1 - x^2)) - ArcTan[(3 + x)/(2*Sqrt[-1 + x + x^2])]/8 - (5*ArcTanh[(1 - 3*x)/(2*Sqrt[-1
+ x + x^2])])/8

Rule 976

Int[((a_.) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[((2*a*c^2*e + c*(2
*c^2*d - c*(2*a*f))*x)*(a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^(q + 1))/((-4*a*c)*(a*c*e^2 + (c*d - a*f)^2)*(p +
1)), x] - Dist[1/((-4*a*c)*(a*c*e^2 + (c*d - a*f)^2)*(p + 1)), Int[(a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^q*Si
mp[2*c*((c*d - a*f)^2 - (-(a*e))*(c*e))*(p + 1) - (2*c^2*d - c*(2*a*f))*(a*f*(p + 1) - c*d*(p + 2)) - e*(-2*a*
c^2*e)*(p + q + 2) + (2*f*(2*a*c^2*e)*(p + q + 2) - (2*c^2*d - c*(2*a*f))*(-(c*e*(2*p + q + 4))))*x + c*f*(2*c
^2*d - c*(2*a*f))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, q}, x] && NeQ[e^2 - 4*d*f, 0] && Lt
Q[p, -1] && NeQ[a*c*e^2 + (c*d - a*f)^2, 0] &&  !( !IntegerQ[p] && ILtQ[q, -1]) &&  !IGtQ[q, 0]

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}{\left (-1+x^2\right )^2 \sqrt{-1+x+x^2}} \, dx &=\frac{\sqrt{-1+x+x^2}}{2 \left (1-x^2\right )}-\frac{1}{4} \int \frac{3+2 x}{\left (-1+x^2\right ) \sqrt{-1+x+x^2}} \, dx\\ &=\frac{\sqrt{-1+x+x^2}}{2 \left (1-x^2\right )}+\frac{1}{8} \int \frac{1}{(1+x) \sqrt{-1+x+x^2}} \, dx-\frac{5}{8} \int \frac{1}{(-1+x) \sqrt{-1+x+x^2}} \, dx\\ &=\frac{\sqrt{-1+x+x^2}}{2 \left (1-x^2\right )}-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{-4-x^2} \, dx,x,\frac{-3-x}{\sqrt{-1+x+x^2}}\right )+\frac{5}{4} \operatorname{Subst}\left (\int \frac{1}{4-x^2} \, dx,x,\frac{-1+3 x}{\sqrt{-1+x+x^2}}\right )\\ &=\frac{\sqrt{-1+x+x^2}}{2 \left (1-x^2\right )}-\frac{1}{8} \tan ^{-1}\left (\frac{3+x}{2 \sqrt{-1+x+x^2}}\right )-\frac{5}{8} \tanh ^{-1}\left (\frac{1-3 x}{2 \sqrt{-1+x+x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0834212, size = 66, normalized size = 0.94 \[ \frac{1}{8} \left (-\frac{4 \sqrt{x^2+x-1}}{x^2-1}-\tan ^{-1}\left (\frac{x+3}{2 \sqrt{x^2+x-1}}\right )-5 \tanh ^{-1}\left (\frac{1-3 x}{2 \sqrt{x^2+x-1}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*Sqrt[-1 + x + x^2])/(-1 + x^2) - ArcTan[(3 + x)/(2*Sqrt[-1 + x + x^2])] - 5*ArcTanh[(1 - 3*x)/(2*Sqrt[-1
+ x + x^2])])/8

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Maple [A]  time = 0.059, size = 84, normalized size = 1.2 \begin{align*}{\frac{1}{8}\arctan \left ({\frac{-3-x}{2}{\frac{1}{\sqrt{ \left ( 1+x \right ) ^{2}-2-x}}}} \right ) }-{\frac{1}{-4+4\,x}\sqrt{ \left ( -1+x \right ) ^{2}-2+3\,x}}+{\frac{5}{8}{\it Artanh} \left ({\frac{3\,x-1}{2}{\frac{1}{\sqrt{ \left ( -1+x \right ) ^{2}-2+3\,x}}}} \right ) }+{\frac{1}{4+4\,x}\sqrt{ \left ( 1+x \right ) ^{2}-2-x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*arctan(1/2*(-3-x)/((1+x)^2-2-x)^(1/2))-1/4/(-1+x)*((-1+x)^2-2+3*x)^(1/2)+5/8*arctanh(1/2*(3*x-1)/((-1+x)^2
-2+3*x)^(1/2))+1/4/(1+x)*((1+x)^2-2-x)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.808353, size = 235, normalized size = 3.36 \begin{align*} \frac{2 \,{\left (x^{2} - 1\right )} \arctan \left (-x + \sqrt{x^{2} + x - 1} - 1\right ) + 5 \,{\left (x^{2} - 1\right )} \log \left (-x + \sqrt{x^{2} + x - 1} + 2\right ) - 5 \,{\left (x^{2} - 1\right )} \log \left (-x + \sqrt{x^{2} + x - 1}\right ) - 4 \, \sqrt{x^{2} + x - 1}}{8 \,{\left (x^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*(x^2 - 1)*arctan(-x + sqrt(x^2 + x - 1) - 1) + 5*(x^2 - 1)*log(-x + sqrt(x^2 + x - 1) + 2) - 5*(x^2 - 1
)*log(-x + sqrt(x^2 + x - 1)) - 4*sqrt(x^2 + x - 1))/(x^2 - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.16836, size = 193, normalized size = 2.76 \begin{align*} \frac{2 \,{\left (x - \sqrt{x^{2} + x - 1}\right )}^{3} + 3 \,{\left (x - \sqrt{x^{2} + x - 1}\right )}^{2} - x + \sqrt{x^{2} + x - 1} - 1}{2 \,{\left ({\left (x - \sqrt{x^{2} + x - 1}\right )}^{4} - 2 \,{\left (x - \sqrt{x^{2} + x - 1}\right )}^{2} - 4 \, x + 4 \, \sqrt{x^{2} + x - 1}\right )}} + \frac{1}{4} \, \arctan \left (-x + \sqrt{x^{2} + x - 1} - 1\right ) + \frac{5}{8} \, \log \left ({\left | -x + \sqrt{x^{2} + x - 1} + 2 \right |}\right ) - \frac{5}{8} \, \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/(x^2-1)^2/(x^2+x-1)^(1/2),x, algorithm="giac")

[Out]

1/2*(2*(x - sqrt(x^2 + x - 1))^3 + 3*(x - sqrt(x^2 + x - 1))^2 - x + sqrt(x^2 + x - 1) - 1)/((x - sqrt(x^2 + x
- 1))^4 - 2*(x - sqrt(x^2 + x - 1))^2 - 4*x + 4*sqrt(x^2 + x - 1)) + 1/4*arctan(-x + sqrt(x^2 + x - 1) - 1) +
5/8*log(abs(-x + sqrt(x^2 + x - 1) + 2)) - 5/8*log(abs(-x + sqrt(x^2 + x - 1)))