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

Optimal. Leaf size=16 \[ \tanh ^{-1}\left (\frac{x^2+1}{\sqrt{x^4+1}}\right ) \]

[Out]

ArcTanh[(1 + x^2)/Sqrt[1 + x^4]]

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Rubi [A]  time = 0.024671, antiderivative size = 23, normalized size of antiderivative = 1.44, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {1252, 844, 215, 266, 63, 207} \[ \frac{1}{2} \sinh ^{-1}\left (x^2\right )+\frac{1}{2} \tanh ^{-1}\left (\sqrt{x^4+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(-1 + x^2)/(x*Sqrt[1 + x^4]),x]

[Out]

ArcSinh[x^2]/2 + ArcTanh[Sqrt[1 + x^4]]/2

Rule 1252

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

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 215

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 207

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

Rubi steps

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

Mathematica [A]  time = 0.0143672, size = 19, normalized size = 1.19 \[ \frac{1}{2} \left (\sinh ^{-1}\left (x^2\right )+\tanh ^{-1}\left (\sqrt{x^4+1}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 18, normalized size = 1.1 \begin{align*}{\frac{1}{2}{\it Artanh} \left ({\frac{1}{\sqrt{{x}^{4}+1}}} \right ) }+{\frac{{\it Arcsinh} \left ({x}^{2} \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctanh(1/(x^4+1)^(1/2))+1/2*arcsinh(x^2)

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Maxima [B]  time = 1.45403, size = 77, normalized size = 4.81 \begin{align*} \frac{1}{4} \, \log \left (\sqrt{x^{4} + 1} + 1\right ) - \frac{1}{4} \, \log \left (\sqrt{x^{4} + 1} - 1\right ) + \frac{1}{4} \, \log \left (\frac{\sqrt{x^{4} + 1}}{x^{2}} + 1\right ) - \frac{1}{4} \, \log \left (\frac{\sqrt{x^{4} + 1}}{x^{2}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.4965, size = 123, normalized size = 7.69 \begin{align*} -\frac{1}{2} \, \log \left (2 \, x^{4} + x^{2} - \sqrt{x^{4} + 1}{\left (2 \, x^{2} + 1\right )} + 1\right ) + \frac{1}{2} \, \log \left (-x^{2} + \sqrt{x^{4} + 1} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.09624, size = 14, normalized size = 0.88 \begin{align*} \frac{\operatorname{asinh}{\left (\frac{1}{x^{2}} \right )}}{2} + \frac{\operatorname{asinh}{\left (x^{2} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

asinh(x**(-2))/2 + asinh(x**2)/2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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