3.9.92 \(\int \frac {\sqrt {-2 x^2+x^4}}{(-1+x^2) (2+x^2)} \, dx\) [892]

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

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 83, normalized size of antiderivative = 1.00, 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 = {2081, 585, 85, 65, 209} \begin {gather*} \frac {2 \sqrt {x^4-2 x^2} \text {ArcTan}\left (\frac {\sqrt {x^2-2}}{2}\right )}{3 x \sqrt {x^2-2}}-\frac {\sqrt {x^4-2 x^2} \text {ArcTan}\left (\sqrt {x^2-2}\right )}{3 x \sqrt {x^2-2}} \end {gather*}

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 65

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 85

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 209

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

Rule 585

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 2081

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]

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} \text {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} \text {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 ) \text {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} \text {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 ) \text {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*}

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Mathematica [A]
time = 0.04, size = 56, normalized size = 0.67 \begin {gather*} \frac {x \sqrt {-2+x^2} \left (2 \tan ^{-1}\left (\frac {1}{2} \sqrt {-2+x^2}\right )-\tan ^{-1}\left (\sqrt {-2+x^2}\right )\right )}{3 \sqrt {x^2 \left (-2+x^2\right )}} \end {gather*}

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

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Maple [A]
time = 0.32, size = 63, normalized size = 0.76

method	result	size
default	\(-\frac {\sqrt {x^{4}-2 x^{2}}\, \left (\arctan \left (\frac {x -2}{\sqrt {x^{2}-2}}\right )-\arctan \left (\frac {x +2}{\sqrt {x^{2}-2}}\right )-4 \arctan \left (\frac {\sqrt {x^{2}-2}}{2}\right )\right )}{6 x \sqrt {x^{2}-2}}\)	\(63\)
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{7}-15 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{5}-6 \sqrt {x^{4}-2 x^{2}}\, x^{4}+24 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+16 \sqrt {x^{4}-2 x^{2}}\, x^{2}-12 x \RootOf \left (\textit {\_Z}^{2}+1\right )-8 \sqrt {x^{4}-2 x^{2}}}{\left (x^{2}+2\right )^{2} x \left (1+x \right ) \left (-1+x \right )}\right )}{6}\)	\(119\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.35, size = 38, normalized size = 0.46 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 [C] Result contains complex when optimal does not.
time = 3.11, size = 46, normalized size = 0.55 \begin {gather*} \frac {1}{3} \, {\left (\arctan \left (i \, \sqrt {2}\right ) - 2 \, \arctan \left (\frac {1}{2} i \, \sqrt {2}\right )\right )} \mathrm {sgn}\left (x\right ) + \frac {2}{3} \, \arctan \left (\frac {1}{2} \, \sqrt {x^{2} - 2}\right ) \mathrm {sgn}\left (x\right ) - \frac {1}{3} \, \arctan \left (\sqrt {x^{2} - 2}\right ) \mathrm {sgn}\left (x\right ) \end {gather*}

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(I*sqrt(2)) - 2*arctan(1/2*I*sqrt(2)))*sgn(x) + 2/3*arctan(1/2*sqrt(x^2 - 2))*sgn(x) - 1/3*arctan(s
qrt(x^2 - 2))*sgn(x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {x^4-2\,x^2}}{\left (x^2-1\right )\,\left (x^2+2\right )} \,d x \end {gather*}

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]

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

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