3.53 \(\int \frac {1+2 x^2}{1-6 x^2+4 x^4} \, dx\)

Optimal. Leaf size=44 \[ \frac {\tanh ^{-1}\left (\sqrt {5}-2 \sqrt {2} x\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (2 \sqrt {2} x+\sqrt {5}\right )}{\sqrt {2}} \]

[Out]

-1/2*arctanh(2*x*2^(1/2)-5^(1/2))*2^(1/2)-1/2*arctanh(2*x*2^(1/2)+5^(1/2))*2^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1161, 618, 206} \[ \frac {\tanh ^{-1}\left (\sqrt {5}-2 \sqrt {2} x\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (2 \sqrt {2} x+\sqrt {5}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x^2)/(1 - 6*x^2 + 4*x^4),x]

[Out]

ArcTanh[Sqrt[5] - 2*Sqrt[2]*x]/Sqrt[2] - ArcTanh[Sqrt[5] + 2*Sqrt[2]*x]/Sqrt[2]

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 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 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rubi steps

\begin {align*} \int \frac {1+2 x^2}{1-6 x^2+4 x^4} \, dx &=\frac {1}{4} \int \frac {1}{\frac {1}{2}-\sqrt {\frac {5}{2}} x+x^2} \, dx+\frac {1}{4} \int \frac {1}{\frac {1}{2}+\sqrt {\frac {5}{2}} x+x^2} \, dx\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\frac {1}{2}-x^2} \, dx,x,-\sqrt {\frac {5}{2}}+2 x\right )\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\frac {1}{2}-x^2} \, dx,x,\sqrt {\frac {5}{2}}+2 x\right )\\ &=\frac {\tanh ^{-1}\left (\sqrt {5}-2 \sqrt {2} x\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (\sqrt {5}+2 \sqrt {2} x\right )}{\sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 42, normalized size = 0.95 \[ \frac {\log \left (-2 x^2+\sqrt {2} x+1\right )-\log \left (2 x^2+\sqrt {2} x-1\right )}{2 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x^2)/(1 - 6*x^2 + 4*x^4),x]

[Out]

(Log[1 + Sqrt[2]*x - 2*x^2] - Log[-1 + Sqrt[2]*x + 2*x^2])/(2*Sqrt[2])

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fricas [A]  time = 0.38, size = 47, normalized size = 1.07 \[ \frac {1}{4} \, \sqrt {2} \log \left (\frac {4 \, x^{4} - 2 \, x^{2} - 2 \, \sqrt {2} {\left (2 \, x^{3} - x\right )} + 1}{4 \, x^{4} - 6 \, x^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+1)/(4*x^4-6*x^2+1),x, algorithm="fricas")

[Out]

1/4*sqrt(2)*log((4*x^4 - 2*x^2 - 2*sqrt(2)*(2*x^3 - x) + 1)/(4*x^4 - 6*x^2 + 1))

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giac [B]  time = 0.34, size = 77, normalized size = 1.75 \[ -\frac {1}{4} \, \sqrt {2} \log \left ({\left | x + \frac {1}{4} \, \sqrt {10} + \frac {1}{4} \, \sqrt {2} \right |}\right ) + \frac {1}{4} \, \sqrt {2} \log \left ({\left | x + \frac {1}{4} \, \sqrt {10} - \frac {1}{4} \, \sqrt {2} \right |}\right ) - \frac {1}{4} \, \sqrt {2} \log \left ({\left | x - \frac {1}{4} \, \sqrt {10} + \frac {1}{4} \, \sqrt {2} \right |}\right ) + \frac {1}{4} \, \sqrt {2} \log \left ({\left | x - \frac {1}{4} \, \sqrt {10} - \frac {1}{4} \, \sqrt {2} \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+1)/(4*x^4-6*x^2+1),x, algorithm="giac")

[Out]

-1/4*sqrt(2)*log(abs(x + 1/4*sqrt(10) + 1/4*sqrt(2))) + 1/4*sqrt(2)*log(abs(x + 1/4*sqrt(10) - 1/4*sqrt(2))) -
1/4*sqrt(2)*log(abs(x - 1/4*sqrt(10) + 1/4*sqrt(2))) + 1/4*sqrt(2)*log(abs(x - 1/4*sqrt(10) - 1/4*sqrt(2)))

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maple [B]  time = 0.04, size = 82, normalized size = 1.86 \[ -\frac {2 \left (-5+\sqrt {5}\right ) \sqrt {5}\, \arctanh \left (\frac {8 x}{2 \sqrt {10}-2 \sqrt {2}}\right )}{5 \left (2 \sqrt {10}-2 \sqrt {2}\right )}-\frac {2 \left (5+\sqrt {5}\right ) \sqrt {5}\, \arctanh \left (\frac {8 x}{2 \sqrt {10}+2 \sqrt {2}}\right )}{5 \left (2 \sqrt {10}+2 \sqrt {2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2+1)/(4*x^4-6*x^2+1),x)

[Out]

-2/5*(-5+5^(1/2))*5^(1/2)/(2*10^(1/2)-2*2^(1/2))*arctanh(8/(2*10^(1/2)-2*2^(1/2))*x)-2/5*(5+5^(1/2))*5^(1/2)/(
2*10^(1/2)+2*2^(1/2))*arctanh(8/(2*10^(1/2)+2*2^(1/2))*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {2 \, x^{2} + 1}{4 \, x^{4} - 6 \, x^{2} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+1)/(4*x^4-6*x^2+1),x, algorithm="maxima")

[Out]

integrate((2*x^2 + 1)/(4*x^4 - 6*x^2 + 1), x)

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mupad [B]  time = 0.22, size = 20, normalized size = 0.45 \[ -\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,x}{2\,x^2-1}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2 + 1)/(4*x^4 - 6*x^2 + 1),x)

[Out]

-(2^(1/2)*atanh((2^(1/2)*x)/(2*x^2 - 1)))/2

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sympy [A]  time = 0.12, size = 46, normalized size = 1.05 \[ \frac {\sqrt {2} \log {\left (x^{2} - \frac {\sqrt {2} x}{2} - \frac {1}{2} \right )}}{4} - \frac {\sqrt {2} \log {\left (x^{2} + \frac {\sqrt {2} x}{2} - \frac {1}{2} \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2+1)/(4*x**4-6*x**2+1),x)

[Out]

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

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