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

Optimal. Leaf size=18 \[ \frac {1}{2} \sinh ^{-1}\left (\frac {2 x^2+1}{\sqrt {3}}\right ) \]

[Out]

1/2*arcsinh(1/3*(2*x^2+1)*3^(1/2))

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Rubi [A]  time = 0.02, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1107, 619, 215} \[ \frac {1}{2} \sinh ^{-1}\left (\frac {2 x^2+1}{\sqrt {3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[1 + x^2 + x^4],x]

[Out]

ArcSinh[(1 + 2*x^2)/Sqrt[3]]/2

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 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 1107

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
x, x^2], x] /; FreeQ[{a, b, c, p}, x]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 18, normalized size = 1.00 \[ \frac {1}{2} \sinh ^{-1}\left (\frac {2 x^2+1}{\sqrt {3}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[1 + x^2 + x^4],x]

[Out]

ArcSinh[(1 + 2*x^2)/Sqrt[3]]/2

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fricas [A]  time = 0.41, size = 22, normalized size = 1.22 \[ -\frac {1}{2} \, \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} + x^{2} + 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.94, size = 22, normalized size = 1.22 \[ -\frac {1}{2} \, \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} + x^{2} + 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 14, normalized size = 0.78 \[ \frac {\arcsinh \left (\frac {2 \sqrt {3}\, \left (x^{2}+\frac {1}{2}\right )}{3}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arcsinh(2/3*3^(1/2)*(x^2+1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {x^{4} + x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/sqrt(x^4 + x^2 + 1), x)

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mupad [B]  time = 0.36, size = 18, normalized size = 1.00 \[ \frac {\ln \left (\sqrt {x^4+x^2+1}+x^2+\frac {1}{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^2 + x^4 + 1)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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