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

Optimal. Leaf size=123 \[ \frac{\left (1-x^2\right ) \sqrt{-\frac{2 x^2-x^4}{\left (1-x^2\right )^2}} \tan ^{-1}\left (\sqrt{x^2-2}\right )}{3 x \sqrt{x^2-2}}-\frac{2 \left (1-x^2\right ) \sqrt{-\frac{2 x^2-x^4}{\left (1-x^2\right )^2}} \tan ^{-1}\left (\frac{\sqrt{x^2-2}}{2}\right )}{3 x \sqrt{x^2-2}} \]

[Out]

(-2*(1 - x^2)*Sqrt[-((2*x^2 - x^4)/(1 - x^2)^2)]*ArcTan[Sqrt[-2 + x^2]/2])/(3*x*Sqrt[-2 + x^2]) + ((1 - x^2)*S
qrt[-((2*x^2 - x^4)/(1 - x^2)^2)]*ArcTan[Sqrt[-2 + x^2]])/(3*x*Sqrt[-2 + x^2])

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Rubi [A]  time = 0.285672, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {6719, 2056, 571, 83, 63, 203} \[ \frac{\left (1-x^2\right ) \sqrt{-\frac{2 x^2-x^4}{\left (1-x^2\right )^2}} \tan ^{-1}\left (\sqrt{x^2-2}\right )}{3 x \sqrt{x^2-2}}-\frac{2 \left (1-x^2\right ) \sqrt{-\frac{2 x^2-x^4}{\left (1-x^2\right )^2}} \tan ^{-1}\left (\frac{\sqrt{x^2-2}}{2}\right )}{3 x \sqrt{x^2-2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 - x^2)*Sqrt[-((2*x^2 - x^4)/(1 - x^2)^2)]*ArcTan[Sqrt[-2 + x^2]/2])/(3*x*Sqrt[-2 + x^2]) + ((1 - x^2)*S
qrt[-((2*x^2 - x^4)/(1 - x^2)^2)]*ArcTan[Sqrt[-2 + x^2]])/(3*x*Sqrt[-2 + x^2])

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

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

Mathematica [A]  time = 0.0202175, size = 70, normalized size = 0.57 \[ \frac{\sqrt{\frac{x^2 \left (x^2-2\right )}{\left (x^2-1\right )^2}} \left (x^2-1\right ) \left (2 \tan ^{-1}\left (\frac{\sqrt{x^2-2}}{2}\right )-\tan ^{-1}\left (\sqrt{x^2-2}\right )\right )}{3 x \sqrt{x^2-2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 75, normalized size = 0.6 \begin{align*} -{\frac{{x}^{2}-1}{6\,x}\sqrt{{\frac{{x}^{2} \left ({x}^{2}-2 \right ) }{ \left ({x}^{2}-1 \right ) ^{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)/(x^2-1)^2)^(1/2)/(x^2+2),x)

[Out]

-1/6*(x^2*(x^2-2)/(x^2-1)^2)^(1/2)*(x^2-1)*(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{\frac{x^{4} - 2 \, x^{2}}{{\left (x^{2} - 1\right )}^{2}}}}{x^{2} + 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.52951, size = 178, normalized size = 1.45 \begin{align*} -\frac{1}{3} \, \arctan \left (\frac{{\left (x^{2} - 1\right )} \sqrt{\frac{x^{4} - 2 \, x^{2}}{x^{4} - 2 \, x^{2} + 1}}}{x}\right ) + \frac{2}{3} \, \arctan \left (\frac{{\left (x^{2} - 1\right )} \sqrt{\frac{x^{4} - 2 \, x^{2}}{x^{4} - 2 \, x^{2} + 1}}}{2 \, x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.11271, size = 45, normalized size = 0.37 \begin{align*} \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^{3} - x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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