\(\int \frac {1+2 x^2}{1-6 x^2+4 x^4} \, dx\) [53]

Optimal result

Integrand size = 22, antiderivative size = 44 \[ \int \frac {1+2 x^2}{1-6 x^2+4 x^4} \, dx=\frac {\text {arctanh}\left (\sqrt {5}-2 \sqrt {2} x\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\sqrt {5}+2 \sqrt {2} x\right )}{\sqrt {2}} \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1175, 632, 212} \[ \int \frac {1+2 x^2}{1-6 x^2+4 x^4} \, dx=\frac {\text {arctanh}\left (\sqrt {5}-2 \sqrt {2} x\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (2 \sqrt {2} x+\sqrt {5}\right )}{\sqrt {2}} \]

[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 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}{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} \text {Subst}\left (\int \frac {1}{\frac {1}{2}-x^2} \, dx,x,-\sqrt {\frac {5}{2}}+2 x\right )\right )-\frac {1}{2} \text {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*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \frac {1+2 x^2}{1-6 x^2+4 x^4} \, dx=\frac {\log \left (1+\sqrt {2} x-2 x^2\right )-\log \left (-1+\sqrt {2} x+2 x^2\right )}{2 \sqrt {2}} \]

[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])

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89

method	result	size
risch	\(\frac {\sqrt {2}\, \ln \left (-x \sqrt {2}+2 x^{2}-1\right )}{4}-\frac {\sqrt {2}\, \ln \left (x \sqrt {2}+2 x^{2}-1\right )}{4}\)	\(39\)
default	\(-\frac {2 \left (-5+\sqrt {5}\right ) \sqrt {5}\, \operatorname {arctanh}\left (\frac {8 x}{2 \sqrt {10}-2 \sqrt {2}}\right )}{5 \left (2 \sqrt {10}-2 \sqrt {2}\right )}-\frac {2 \sqrt {5}\, \left (5+\sqrt {5}\right ) \operatorname {arctanh}\left (\frac {8 x}{2 \sqrt {10}+2 \sqrt {2}}\right )}{5 \left (2 \sqrt {10}+2 \sqrt {2}\right )}\)	\(82\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.07 \[ \int \frac {1+2 x^2}{1-6 x^2+4 x^4} \, dx=\frac {1}{4} \, \sqrt {2} \log \left (\frac {4 \, x^{4} - 2 \, x^{2} - 2 \, \sqrt {2} {\left (2 \, x^{3} - x\right )} + 1}{4 \, x^{4} - 6 \, x^{2} + 1}\right ) \]

[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))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \frac {1+2 x^2}{1-6 x^2+4 x^4} \, dx=\frac {\sqrt {2} \log {\left (x^{2} - \frac {\sqrt {2} x}{2} - \frac {1}{2} \right )}}{4} - \frac {\sqrt {2} \log {\left (x^{2} + \frac {\sqrt {2} x}{2} - \frac {1}{2} \right )}}{4} \]

[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

Maxima [F]

\[ \int \frac {1+2 x^2}{1-6 x^2+4 x^4} \, dx=\int { \frac {2 \, x^{2} + 1}{4 \, x^{4} - 6 \, x^{2} + 1} \,d x } \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (35) = 70\).

Time = 0.31 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.75 \[ \int \frac {1+2 x^2}{1-6 x^2+4 x^4} \, dx=-\frac {1}{4} \, \sqrt {2} \log \left ({\left | x + \frac {1}{4} \, \sqrt {10} + \frac {1}{4} \, \sqrt {2} \right |}\right ) + \frac {1}{4} \, \sqrt {2} \log \left ({\left | x + \frac {1}{4} \, \sqrt {10} - \frac {1}{4} \, \sqrt {2} \right |}\right ) - \frac {1}{4} \, \sqrt {2} \log \left ({\left | x - \frac {1}{4} \, \sqrt {10} + \frac {1}{4} \, \sqrt {2} \right |}\right ) + \frac {1}{4} \, \sqrt {2} \log \left ({\left | x - \frac {1}{4} \, \sqrt {10} - \frac {1}{4} \, \sqrt {2} \right |}\right ) \]

[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)))

Mupad [B] (verification not implemented)

Time = 13.33 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.45 \[ \int \frac {1+2 x^2}{1-6 x^2+4 x^4} \, dx=-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,x}{2\,x^2-1}\right )}{2} \]

[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