3.892 \(\int \frac{\sqrt{-2 x^2+x^4}}{(-1+x^2) (2+x^2)} \, dx\)

Optimal. Leaf size=83 \[ \frac{2 \sqrt{x^4-2 x^2} \tan ^{-1}\left (\frac{\sqrt{x^2-2}}{2}\right )}{3 x \sqrt{x^2-2}}-\frac{\sqrt{x^4-2 x^2} \tan ^{-1}\left (\sqrt{x^2-2}\right )}{3 x \sqrt{x^2-2}} \]

[Out]

(2*Sqrt[-2*x^2 + x^4]*ArcTan[Sqrt[-2 + x^2]/2])/(3*x*Sqrt[-2 + x^2]) - (Sqrt[-2*x^2 + x^4]*ArcTan[Sqrt[-2 + x^
2]])/(3*x*Sqrt[-2 + x^2])

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Rubi [A]  time = 0.154027, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {2056, 571, 83, 63, 203} \[ \frac{2 \sqrt{x^4-2 x^2} \tan ^{-1}\left (\frac{\sqrt{x^2-2}}{2}\right )}{3 x \sqrt{x^2-2}}-\frac{\sqrt{x^4-2 x^2} \tan ^{-1}\left (\sqrt{x^2-2}\right )}{3 x \sqrt{x^2-2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-2*x^2 + x^4]/((-1 + x^2)*(2 + x^2)),x]

[Out]

(2*Sqrt[-2*x^2 + x^4]*ArcTan[Sqrt[-2 + x^2]/2])/(3*x*Sqrt[-2 + x^2]) - (Sqrt[-2*x^2 + x^4]*ArcTan[Sqrt[-2 + x^
2]])/(3*x*Sqrt[-2 + x^2])

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 571

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x
_Symbol] :> Dist[1/n, Subst[Int[(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^n], x] /; FreeQ[{a, b, c, d, e,
f, m, n, p, q, r}, x] && EqQ[m - n + 1, 0]

Rule 83

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[(b*e - a*f)/(b*c
- a*d), Int[(e + f*x)^(p - 1)/(a + b*x), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[(e + f*x)^(p - 1)/(c + d*
x), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[0, p, 1]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

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

Rubi steps

\begin{align*} \int \frac{\sqrt{-2 x^2+x^4}}{\left (-1+x^2\right ) \left (2+x^2\right )} \, dx &=\frac{\sqrt{-2 x^2+x^4} \int \frac{x \sqrt{-2+x^2}}{\left (-1+x^2\right ) \left (2+x^2\right )} \, dx}{x \sqrt{-2+x^2}}\\ &=\frac{\sqrt{-2 x^2+x^4} \operatorname{Subst}\left (\int \frac{\sqrt{-2+x}}{(-1+x) (2+x)} \, dx,x,x^2\right )}{2 x \sqrt{-2+x^2}}\\ &=-\frac{\sqrt{-2 x^2+x^4} \operatorname{Subst}\left (\int \frac{1}{\sqrt{-2+x} (-1+x)} \, dx,x,x^2\right )}{6 x \sqrt{-2+x^2}}+\frac{\left (2 \sqrt{-2 x^2+x^4}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-2+x} (2+x)} \, dx,x,x^2\right )}{3 x \sqrt{-2+x^2}}\\ &=-\frac{\sqrt{-2 x^2+x^4} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{-2+x^2}\right )}{3 x \sqrt{-2+x^2}}+\frac{\left (4 \sqrt{-2 x^2+x^4}\right ) \operatorname{Subst}\left (\int \frac{1}{4+x^2} \, dx,x,\sqrt{-2+x^2}\right )}{3 x \sqrt{-2+x^2}}\\ &=\frac{2 \sqrt{-2 x^2+x^4} \tan ^{-1}\left (\frac{1}{2} \sqrt{-2+x^2}\right )}{3 x \sqrt{-2+x^2}}-\frac{\sqrt{-2 x^2+x^4} \tan ^{-1}\left (\sqrt{-2+x^2}\right )}{3 x \sqrt{-2+x^2}}\\ \end{align*}

Mathematica [A]  time = 0.061988, size = 52, normalized size = 0.63 \[ -\frac{x \sqrt{x^2-2} \left (2 \tan ^{-1}\left (\frac{2}{\sqrt{x^2-2}}\right )+\tan ^{-1}\left (\sqrt{x^2-2}\right )\right )}{3 \sqrt{x^2 \left (x^2-2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-2*x^2 + x^4]/((-1 + x^2)*(2 + x^2)),x]

[Out]

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

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Maple [A]  time = 0.023, size = 63, normalized size = 0.8 \begin{align*} -{\frac{1}{6\,x}\sqrt{{x}^{4}-2\,{x}^{2}} \left ( \arctan \left ({(-2+x){\frac{1}{\sqrt{{x}^{2}-2}}}} \right ) -4\,\arctan \left ( 1/2\,\sqrt{{x}^{2}-2} \right ) -\arctan \left ({(2+x){\frac{1}{\sqrt{{x}^{2}-2}}}} \right ) \right ){\frac{1}{\sqrt{{x}^{2}-2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-2*x^2)^(1/2)/(x^2-1)/(x^2+2),x)

[Out]

-1/6*(x^4-2*x^2)^(1/2)*(arctan((-2+x)/(x^2-2)^(1/2))-4*arctan(1/2*(x^2-2)^(1/2))-arctan((2+x)/(x^2-2)^(1/2)))/
x/(x^2-2)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{4} - 2 \, x^{2}}}{{\left (x^{2} + 2\right )}{\left (x^{2} - 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-2*x^2)^(1/2)/(x^2-1)/(x^2+2),x, algorithm="maxima")

[Out]

integrate(sqrt(x^4 - 2*x^2)/((x^2 + 2)*(x^2 - 1)), x)

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Fricas [A]  time = 1.48531, size = 97, normalized size = 1.17 \begin{align*} -\frac{1}{3} \, \arctan \left (\frac{\sqrt{x^{4} - 2 \, x^{2}}}{x}\right ) + \frac{2}{3} \, \arctan \left (\frac{\sqrt{x^{4} - 2 \, x^{2}}}{2 \, x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-2*x^2)^(1/2)/(x^2-1)/(x^2+2),x, algorithm="fricas")

[Out]

-1/3*arctan(sqrt(x^4 - 2*x^2)/x) + 2/3*arctan(1/2*sqrt(x^4 - 2*x^2)/x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4-2*x**2)**(1/2)/(x**2-1)/(x**2+2),x)

[Out]

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

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Giac [A]  time = 1.1604, size = 65, normalized size = 0.78 \begin{align*} \frac{1}{3} \,{\left (\arctan \left (\sqrt{2} i\right ) - 2 \, \arctan \left (\frac{1}{2} \, \sqrt{2} i\right )\right )} \mathrm{sgn}\left (x\right ) + \frac{1}{3} \,{\left (2 \, \arctan \left (\frac{1}{2} \, \sqrt{x^{2} - 2}\right ) - \arctan \left (\sqrt{x^{2} - 2}\right )\right )} \mathrm{sgn}\left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-2*x^2)^(1/2)/(x^2-1)/(x^2+2),x, algorithm="giac")

[Out]

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