\(\int \frac {1+x^2}{1-4 x^2+x^4} \, dx\) [78]

Optimal result

Integrand size = 18, antiderivative size = 43 \[ \int \frac {1+x^2}{1-4 x^2+x^4} \, dx=\frac {\text {arctanh}\left (\sqrt {3}-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\sqrt {3}+\sqrt {2} x\right )}{\sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1175, 632, 212} \[ \int \frac {1+x^2}{1-4 x^2+x^4} \, dx=\frac {\text {arctanh}\left (\sqrt {3}-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\sqrt {2} x+\sqrt {3}\right )}{\sqrt {2}} \]

[In]

Int[(1 + x^2)/(1 - 4*x^2 + x^4),x]

[Out]

ArcTanh[Sqrt[3] - Sqrt[2]*x]/Sqrt[2] - ArcTanh[Sqrt[3] + Sqrt[2]*x]/Sqrt[2]

Rule 212

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

Rule 632

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 1175

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*} \text {integral}& = \frac {1}{2} \int \frac {1}{1-\sqrt {6} x+x^2} \, dx+\frac {1}{2} \int \frac {1}{1+\sqrt {6} x+x^2} \, dx \\ & = -\text {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,-\sqrt {6}+2 x\right )-\text {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\sqrt {6}+2 x\right ) \\ & = \frac {\tanh ^{-1}\left (\sqrt {3}-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (\sqrt {3}+\sqrt {2} x\right )}{\sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.93 \[ \int \frac {1+x^2}{1-4 x^2+x^4} \, dx=\frac {\log \left (1+\sqrt {2} x-x^2\right )-\log \left (-1+\sqrt {2} x+x^2\right )}{2 \sqrt {2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {\sqrt {2}\, \ln \left (x^{2}-x \sqrt {2}-1\right )}{4}-\frac {\sqrt {2}\, \ln \left (x^{2}+x \sqrt {2}-1\right )}{4}\)	\(35\)
default	\(-\frac {\sqrt {3}\, \left (\sqrt {3}+3\right ) \operatorname {arctanh}\left (\frac {2 x}{\sqrt {6}+\sqrt {2}}\right )}{3 \left (\sqrt {6}+\sqrt {2}\right )}-\frac {\left (-3+\sqrt {3}\right ) \sqrt {3}\, \operatorname {arctanh}\left (\frac {2 x}{\sqrt {6}-\sqrt {2}}\right )}{3 \left (\sqrt {6}-\sqrt {2}\right )}\)	\(70\)

[In]

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

[Out]

1/4*2^(1/2)*ln(x^2-x*2^(1/2)-1)-1/4*2^(1/2)*ln(x^2+x*2^(1/2)-1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {1+x^2}{1-4 x^2+x^4} \, dx=\frac {1}{4} \, \sqrt {2} \log \left (\frac {x^{4} - 2 \, \sqrt {2} {\left (x^{3} - x\right )} + 1}{x^{4} - 4 \, x^{2} + 1}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \frac {1+x^2}{1-4 x^2+x^4} \, dx=\frac {\sqrt {2} \log {\left (x^{2} - \sqrt {2} x - 1 \right )}}{4} - \frac {\sqrt {2} \log {\left (x^{2} + \sqrt {2} x - 1 \right )}}{4} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1+x^2}{1-4 x^2+x^4} \, dx=\int { \frac {x^{2} + 1}{x^{4} - 4 \, x^{2} + 1} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \frac {1+x^2}{1-4 x^2+x^4} \, dx=\frac {1}{4} \, \sqrt {2} \log \left (\frac {{\left | 2 \, x - 2 \, \sqrt {2} - \frac {2}{x} \right |}}{{\left | 2 \, x + 2 \, \sqrt {2} - \frac {2}{x} \right |}}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.53 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.42 \[ \int \frac {1+x^2}{1-4 x^2+x^4} \, dx=-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,x}{x^2-1}\right )}{2} \]

[In]

int((x^2 + 1)/(x^4 - 4*x^2 + 1),x)

[Out]

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