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

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

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Rubi [A]  time = 0.02, 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 1.31 \[ \frac {1}{2} \left (\sinh ^{-1}\left (x^2\right )-\tanh ^{-1}\left (\sqrt {x^4+1}\right )\right ) \]

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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IntegrateAlgebraic [B]  time = 0.19, size = 35, normalized size = 2.19 \[ \frac {1}{2} \log \left (\sqrt {x^4+1}+x^2\right )-\tanh ^{-1}\left (\sqrt {x^4+1}+x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.75, size = 49, normalized size = 3.06 \[ -\frac {1}{2} \, \log \left (2 \, x^{4} - x^{2} - \sqrt {x^{4} + 1} {\left (2 \, x^{2} - 1\right )} + 1\right ) + \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 1} - 1\right ) \]

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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giac [B]  time = 0.64, size = 51, normalized size = 3.19 \[ \frac {1}{2} \, \log \left (x^{2} - \sqrt {x^{4} + 1} + 1\right ) - \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 1} + 1\right ) - \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 1}\right ) \]

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]

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

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maple [A]  time = 0.30, size = 18, normalized size = 1.12




method result size



default \(-\frac {\arctanh \left (\frac {1}{\sqrt {x^{4}+1}}\right )}{2}+\frac {\arcsinh \left (x^{2}\right )}{2}\) \(18\)
trager \(\ln \left (\frac {x^{2}+\sqrt {x^{4}+1}-1}{x}\right )\) \(18\)
elliptic \(-\frac {\arctanh \left (\frac {1}{\sqrt {x^{4}+1}}\right )}{2}+\frac {\arcsinh \left (x^{2}\right )}{2}\) \(18\)
meijerg \(\frac {\arcsinh \left (x^{2}\right )}{2}+\frac {-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{4}+1}}{2}\right )+\left (-2 \ln \relax (2)+4 \ln \relax (x )\right ) \sqrt {\pi }}{4 \sqrt {\pi }}\) \(44\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.17, size = 57, normalized size = 3.56 \[ -\frac {1}{4} \, \log \left (\sqrt {x^{4} + 1} + 1\right ) + \frac {1}{4} \, \log \left (\sqrt {x^{4} + 1} - 1\right ) + \frac {1}{4} \, \log \left (\frac {\sqrt {x^{4} + 1}}{x^{2}} + 1\right ) - \frac {1}{4} \, \log \left (\frac {\sqrt {x^{4} + 1}}{x^{2}} - 1\right ) \]

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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mupad [B]  time = 0.15, size = 17, normalized size = 1.06 \[ \frac {\mathrm {asinh}\left (x^2\right )}{2}-\frac {\mathrm {atanh}\left (\sqrt {x^4+1}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + 1)/(x*(x^4 + 1)^(1/2)),x)

[Out]

asinh(x^2)/2 - atanh((x^4 + 1)^(1/2))/2

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sympy [A]  time = 9.51, size = 14, normalized size = 0.88 \[ - \frac {\operatorname {asinh}{\left (\frac {1}{x^{2}} \right )}}{2} + \frac {\operatorname {asinh}{\left (x^{2} \right )}}{2} \]

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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