3.311 \(\int \frac {-1+x^2}{x \sqrt {1+3 x^2+x^4}} \, dx\)

Optimal. Leaf size=21 \[ \tanh ^{-1}\left (\frac {x^2+1}{\sqrt {x^4+3 x^2+1}}\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1251, 838, 206} \[ \tanh ^{-1}\left (\frac {x^2+1}{\sqrt {x^4+3 x^2+1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(-1 + x^2)/(x*Sqrt[1 + 3*x^2 + x^4]),x]

[Out]

ArcTanh[(1 + x^2)/Sqrt[1 + 3*x^2 + x^4]]

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 838

Int[((f_) + (g_.)*(x_))/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[(4*f
*(a - d))/(b*d - a*e), Subst[Int[1/(4*(a - d) - x^2), x], x, (2*(a - d) + (b - e)*x)/Sqrt[a + b*x + c*x^2]], x
] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[4*c*(a - d) - (b - e)^2, 0] && EqQ[e*f*(b - e) - 2*g*(b*d - a*e),
0] && NeQ[b*d - a*e, 0]

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [B]  time = 0.02, size = 57, normalized size = 2.71 \[ \frac {1}{2} \left (\tanh ^{-1}\left (\frac {2 x^2+3}{2 \sqrt {x^4+3 x^2+1}}\right )+\tanh ^{-1}\left (\frac {3 x^2+2}{2 \sqrt {x^4+3 x^2+1}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x^2)/(x*Sqrt[1 + 3*x^2 + x^4]),x]

[Out]

(ArcTanh[(3 + 2*x^2)/(2*Sqrt[1 + 3*x^2 + x^4])] + ArcTanh[(2 + 3*x^2)/(2*Sqrt[1 + 3*x^2 + x^4])])/2

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IntegrateAlgebraic [B]  time = 0.23, size = 52, normalized size = 2.48 \[ -\frac {1}{2} \log \left (-2 x^2+2 \sqrt {x^4+3 x^2+1}-3\right )-\tanh ^{-1}\left (x^2-\sqrt {x^4+3 x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-1 + x^2)/(x*Sqrt[1 + 3*x^2 + x^4]),x]

[Out]

-ArcTanh[x^2 - Sqrt[1 + 3*x^2 + x^4]] - Log[-3 - 2*x^2 + 2*Sqrt[1 + 3*x^2 + x^4]]/2

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fricas [B]  time = 0.61, size = 59, normalized size = 2.81 \[ -\frac {1}{2} \, \log \left (4 \, x^{4} + 11 \, x^{2} - \sqrt {x^{4} + 3 \, x^{2} + 1} {\left (4 \, x^{2} + 5\right )} + 5\right ) + \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 3 \, x^{2} + 1} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)/x/(x^4+3*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

-1/2*log(4*x^4 + 11*x^2 - sqrt(x^4 + 3*x^2 + 1)*(4*x^2 + 5) + 5) + 1/2*log(-x^2 + sqrt(x^4 + 3*x^2 + 1) + 1)

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giac [B]  time = 0.63, size = 69, normalized size = 3.29 \[ -\frac {1}{2} \, \log \left (2 \, x^{2} - 2 \, \sqrt {x^{4} + 3 \, x^{2} + 1} + 3\right ) + \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 3 \, x^{2} + 1} + 1\right ) - \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 3 \, x^{2} + 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)/x/(x^4+3*x^2+1)^(1/2),x, algorithm="giac")

[Out]

-1/2*log(2*x^2 - 2*sqrt(x^4 + 3*x^2 + 1) + 3) + 1/2*log(-x^2 + sqrt(x^4 + 3*x^2 + 1) + 1) - 1/2*log(-x^2 + sqr
t(x^4 + 3*x^2 + 1) - 1)

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maple [A]  time = 0.20, size = 27, normalized size = 1.29

method	result	size
trager	\(-\ln \left (\frac {-x^{2}+\sqrt {x^{4}+3 x^{2}+1}-1}{x}\right )\)	\(27\)
default	\(\frac {\ln \left (\frac {3}{2}+x^{2}+\sqrt {x^{4}+3 x^{2}+1}\right )}{2}+\frac {\arctanh \left (\frac {3 x^{2}+2}{2 \sqrt {x^{4}+3 x^{2}+1}}\right )}{2}\)	\(46\)
elliptic	\(\frac {\ln \left (\frac {3}{2}+x^{2}+\sqrt {x^{4}+3 x^{2}+1}\right )}{2}+\frac {\arctanh \left (\frac {3 x^{2}+2}{2 \sqrt {x^{4}+3 x^{2}+1}}\right )}{2}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-1)/x/(x^4+3*x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.48, size = 52, normalized size = 2.48 \[ \frac {1}{2} \, \log \left (2 \, x^{2} + 2 \, \sqrt {x^{4} + 3 \, x^{2} + 1} + 3\right ) + \frac {1}{2} \, \log \left (\frac {2 \, \sqrt {x^{4} + 3 \, x^{2} + 1}}{x^{2}} + \frac {2}{x^{2}} + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)/x/(x^4+3*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.81, size = 49, normalized size = 2.33 \[ \frac {\ln \left (\frac {1}{x^2}\right )}{2}+\frac {\ln \left (\sqrt {x^4+3\,x^2+1}+x^2+\frac {3}{2}\right )}{2}+\frac {\ln \left (\frac {2\,\sqrt {x^4+3\,x^2+1}}{3}+x^2+\frac {2}{3}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - 1)/(x*(3*x^2 + x^4 + 1)^(1/2)),x)

[Out]

log(1/x^2)/2 + log((3*x^2 + x^4 + 1)^(1/2) + x^2 + 3/2)/2 + log((2*(3*x^2 + x^4 + 1)^(1/2))/3 + x^2 + 2/3)/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x - 1\right ) \left (x + 1\right )}{x \sqrt {x^{4} + 3 x^{2} + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-1)/x/(x**4+3*x**2+1)**(1/2),x)

[Out]

Integral((x - 1)*(x + 1)/(x*sqrt(x**4 + 3*x**2 + 1)), x)

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