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

Optimal. Leaf size=41 \[ \frac {1}{12} \log \left (3 x^4-2 x^2+1\right )-\frac {\tan ^{-1}\left (\frac {1-3 x^2}{\sqrt {2}}\right )}{6 \sqrt {2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1114, 634, 618, 204, 628} \[ \frac {1}{12} \log \left (3 x^4-2 x^2+1\right )-\frac {\tan ^{-1}\left (\frac {1-3 x^2}{\sqrt {2}}\right )}{6 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(1 - 2*x^2 + 3*x^4),x]

[Out]

-ArcTan[(1 - 3*x^2)/Sqrt[2]]/(6*Sqrt[2]) + Log[1 - 2*x^2 + 3*x^4]/12

Rule 204

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

Rule 618

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

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

Rule 1114

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

Rubi steps

\begin {align*} \int \frac {x^3}{1-2 x^2+3 x^4} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{1-2 x+3 x^2} \, dx,x,x^2\right )\\ &=\frac {1}{12} \operatorname {Subst}\left (\int \frac {-2+6 x}{1-2 x+3 x^2} \, dx,x,x^2\right )+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1-2 x+3 x^2} \, dx,x,x^2\right )\\ &=\frac {1}{12} \log \left (1-2 x^2+3 x^4\right )-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{-8-x^2} \, dx,x,2 \left (-1+3 x^2\right )\right )\\ &=-\frac {\tan ^{-1}\left (\frac {1-3 x^2}{\sqrt {2}}\right )}{6 \sqrt {2}}+\frac {1}{12} \log \left (1-2 x^2+3 x^4\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 38, normalized size = 0.93 \[ \frac {1}{12} \left (\sqrt {2} \tan ^{-1}\left (\frac {3 x^2-1}{\sqrt {2}}\right )+\log \left (3 x^4-2 x^2+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(1 - 2*x^2 + 3*x^4),x]

[Out]

(Sqrt[2]*ArcTan[(-1 + 3*x^2)/Sqrt[2]] + Log[1 - 2*x^2 + 3*x^4])/12

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IntegrateAlgebraic [A]  time = 0.02, size = 44, normalized size = 1.07 \[ \frac {1}{12} \log \left (3 x^4-2 x^2+1\right )-\frac {\tan ^{-1}\left (\frac {1}{\sqrt {2}}-\frac {3 x^2}{\sqrt {2}}\right )}{6 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*ArcTan[1/Sqrt[2] - (3*x^2)/Sqrt[2]]/Sqrt[2] + Log[1 - 2*x^2 + 3*x^4]/12

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fricas [A]  time = 1.15, size = 34, normalized size = 0.83 \[ \frac {1}{12} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x^{2} - 1\right )}\right ) + \frac {1}{12} \, \log \left (3 \, x^{4} - 2 \, x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(2)*arctan(1/2*sqrt(2)*(3*x^2 - 1)) + 1/12*log(3*x^4 - 2*x^2 + 1)

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giac [A]  time = 1.51, size = 34, normalized size = 0.83 \[ \frac {1}{12} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x^{2} - 1\right )}\right ) + \frac {1}{12} \, \log \left (3 \, x^{4} - 2 \, x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(2)*arctan(1/2*sqrt(2)*(3*x^2 - 1)) + 1/12*log(3*x^4 - 2*x^2 + 1)

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maple [A]  time = 0.03, size = 35, normalized size = 0.85

method	result	size
default	\(\frac {\ln \left (3 x^{4}-2 x^{2}+1\right )}{12}+\frac {\sqrt {2}\, \arctan \left (\frac {\left (6 x^{2}-2\right ) \sqrt {2}}{4}\right )}{12}\)	\(35\)
risch	\(\frac {\ln \left (9 x^{4}-6 x^{2}+3\right )}{12}+\frac {\sqrt {2}\, \arctan \left (\frac {\left (3 x^{2}-1\right ) \sqrt {2}}{2}\right )}{12}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*ln(3*x^4-2*x^2+1)+1/12*2^(1/2)*arctan(1/4*(6*x^2-2)*2^(1/2))

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maxima [A]  time = 0.97, size = 34, normalized size = 0.83 \[ \frac {1}{12} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x^{2} - 1\right )}\right ) + \frac {1}{12} \, \log \left (3 \, x^{4} - 2 \, x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(2)*arctan(1/2*sqrt(2)*(3*x^2 - 1)) + 1/12*log(3*x^4 - 2*x^2 + 1)

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mupad [B]  time = 0.20, size = 34, normalized size = 0.83 \[ \frac {\ln \left (x^4-\frac {2\,x^2}{3}+\frac {1}{3}\right )}{12}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}}{2}-\frac {3\,\sqrt {2}\,x^2}{2}\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(3*x^4 - 2*x^2 + 1),x)

[Out]

log(x^4 - (2*x^2)/3 + 1/3)/12 - (2^(1/2)*atan(2^(1/2)/2 - (3*2^(1/2)*x^2)/2))/12

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sympy [A]  time = 0.13, size = 42, normalized size = 1.02 \[ \frac {\log {\left (x^{4} - \frac {2 x^{2}}{3} + \frac {1}{3} \right )}}{12} + \frac {\sqrt {2} \operatorname {atan}{\left (\frac {3 \sqrt {2} x^{2}}{2} - \frac {\sqrt {2}}{2} \right )}}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x**4 - 2*x**2/3 + 1/3)/12 + sqrt(2)*atan(3*sqrt(2)*x**2/2 - sqrt(2)/2)/12

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