3.1.67 \(\int \frac {1+x^2}{1-5 x^2+x^4} \, dx\)

Optimal. Leaf size=46 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {7}-2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\tanh ^{-1}\left (\frac {2 x+\sqrt {7}}{\sqrt {3}}\right )}{\sqrt {3}} \]

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Rubi [A]  time = 0.04, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1161, 618, 206} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {7}-2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\tanh ^{-1}\left (\frac {2 x+\sqrt {7}}{\sqrt {3}}\right )}{\sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^2)/(1 - 5*x^2 + x^4),x]

[Out]

ArcTanh[(Sqrt[7] - 2*x)/Sqrt[3]]/Sqrt[3] - ArcTanh[(Sqrt[7] + 2*x)/Sqrt[3]]/Sqrt[3]

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+x^2}{1-5 x^2+x^4} \, dx &=\frac {1}{2} \int \frac {1}{1-\sqrt {7} x+x^2} \, dx+\frac {1}{2} \int \frac {1}{1+\sqrt {7} x+x^2} \, dx\\ &=-\operatorname {Subst}\left (\int \frac {1}{3-x^2} \, dx,x,-\sqrt {7}+2 x\right )-\operatorname {Subst}\left (\int \frac {1}{3-x^2} \, dx,x,\sqrt {7}+2 x\right )\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {7}-2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {7}+2 x}{\sqrt {3}}\right )}{\sqrt {3}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 40, normalized size = 0.87 \begin {gather*} \frac {\log \left (-x^2+\sqrt {3} x+1\right )-\log \left (x^2+\sqrt {3} x-1\right )}{2 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^2)/(1 - 5*x^2 + x^4),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1+x^2}{1-5 x^2+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(1 + x^2)/(1 - 5*x^2 + x^4),x]

[Out]

IntegrateAlgebraic[(1 + x^2)/(1 - 5*x^2 + x^4), x]

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fricas [A]  time = 1.01, size = 39, normalized size = 0.85 \begin {gather*} \frac {1}{6} \, \sqrt {3} \log \left (\frac {x^{4} + x^{2} - 2 \, \sqrt {3} {\left (x^{3} - x\right )} + 1}{x^{4} - 5 \, x^{2} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)/(x^4-5*x^2+1),x, algorithm="fricas")

[Out]

1/6*sqrt(3)*log((x^4 + x^2 - 2*sqrt(3)*(x^3 - x) + 1)/(x^4 - 5*x^2 + 1))

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giac [A]  time = 0.24, size = 39, normalized size = 0.85 \begin {gather*} \frac {1}{6} \, \sqrt {3} \log \left (\frac {{\left | 2 \, x - 2 \, \sqrt {3} - \frac {2}{x} \right |}}{{\left | 2 \, x + 2 \, \sqrt {3} - \frac {2}{x} \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)/(x^4-5*x^2+1),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.04, size = 82, normalized size = 1.78 \begin {gather*} -\frac {2 \sqrt {21}\, \left (-7+\sqrt {21}\right ) \arctanh \left (\frac {4 x}{2 \sqrt {7}-2 \sqrt {3}}\right )}{21 \left (2 \sqrt {7}-2 \sqrt {3}\right )}-\frac {2 \left (7+\sqrt {21}\right ) \sqrt {21}\, \arctanh \left (\frac {4 x}{2 \sqrt {7}+2 \sqrt {3}}\right )}{21 \left (2 \sqrt {7}+2 \sqrt {3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+1)/(x^4-5*x^2+1),x)

[Out]

-2/21*(7+21^(1/2))*21^(1/2)/(2*7^(1/2)+2*3^(1/2))*arctanh(4/(2*7^(1/2)+2*3^(1/2))*x)-2/21*21^(1/2)*(-7+21^(1/2
))/(2*7^(1/2)-2*3^(1/2))*arctanh(4/(2*7^(1/2)-2*3^(1/2))*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} + 1}{x^{4} - 5 \, x^{2} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)/(x^4-5*x^2+1),x, algorithm="maxima")

[Out]

integrate((x^2 + 1)/(x^4 - 5*x^2 + 1), x)

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mupad [B]  time = 4.47, size = 18, normalized size = 0.39 \begin {gather*} -\frac {\sqrt {3}\,\mathrm {atanh}\left (\frac {\sqrt {3}\,x}{x^2-1}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + 1)/(x^4 - 5*x^2 + 1),x)

[Out]

-(3^(1/2)*atanh((3^(1/2)*x)/(x^2 - 1)))/3

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sympy [A]  time = 0.12, size = 39, normalized size = 0.85 \begin {gather*} \frac {\sqrt {3} \log {\left (x^{2} - \sqrt {3} x - 1 \right )}}{6} - \frac {\sqrt {3} \log {\left (x^{2} + \sqrt {3} x - 1 \right )}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+1)/(x**4-5*x**2+1),x)

[Out]

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

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