\(\int \frac {1+x^2}{1-5 x^2+x^4} \, dx\) [79]

Optimal result

Integrand size = 18, antiderivative size = 46 \[ \int \frac {1+x^2}{1-5 x^2+x^4} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {7}-2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\text {arctanh}\left (\frac {\sqrt {7}+2 x}{\sqrt {3}}\right )}{\sqrt {3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, 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} \[ \int \frac {1+x^2}{1-5 x^2+x^4} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {7}-2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\text {arctanh}\left (\frac {2 x+\sqrt {7}}{\sqrt {3}}\right )}{\sqrt {3}} \]

[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 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 {7} x+x^2} \, dx+\frac {1}{2} \int \frac {1}{1+\sqrt {7} x+x^2} \, dx \\ & = -\text {Subst}\left (\int \frac {1}{3-x^2} \, dx,x,-\sqrt {7}+2 x\right )-\text {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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.87 \[ \int \frac {1+x^2}{1-5 x^2+x^4} \, dx=\frac {\log \left (1+\sqrt {3} x-x^2\right )-\log \left (-1+\sqrt {3} x+x^2\right )}{2 \sqrt {3}} \]

[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])

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {\sqrt {3}\, \ln \left (x^{2}-x \sqrt {3}-1\right )}{6}-\frac {\sqrt {3}\, \ln \left (x^{2}+x \sqrt {3}-1\right )}{6}\)	\(35\)
default	\(-\frac {2 \sqrt {21}\, \left (7+\sqrt {21}\right ) \operatorname {arctanh}\left (\frac {4 x}{2 \sqrt {7}+2 \sqrt {3}}\right )}{21 \left (2 \sqrt {7}+2 \sqrt {3}\right )}-\frac {2 \left (-7+\sqrt {21}\right ) \sqrt {21}\, \operatorname {arctanh}\left (\frac {4 x}{2 \sqrt {7}-2 \sqrt {3}}\right )}{21 \left (2 \sqrt {7}-2 \sqrt {3}\right )}\)	\(82\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int \frac {1+x^2}{1-5 x^2+x^4} \, dx=\frac {1}{6} \, \sqrt {3} \log \left (\frac {x^{4} + x^{2} - 2 \, \sqrt {3} {\left (x^{3} - x\right )} + 1}{x^{4} - 5 \, x^{2} + 1}\right ) \]

[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))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int \frac {1+x^2}{1-5 x^2+x^4} \, dx=\frac {\sqrt {3} \log {\left (x^{2} - \sqrt {3} x - 1 \right )}}{6} - \frac {\sqrt {3} \log {\left (x^{2} + \sqrt {3} x - 1 \right )}}{6} \]

[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

Maxima [F]

\[ \int \frac {1+x^2}{1-5 x^2+x^4} \, dx=\int { \frac {x^{2} + 1}{x^{4} - 5 \, x^{2} + 1} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int \frac {1+x^2}{1-5 x^2+x^4} \, dx=\frac {1}{6} \, \sqrt {3} \log \left (\frac {{\left | 2 \, x - 2 \, \sqrt {3} - \frac {2}{x} \right |}}{{\left | 2 \, x + 2 \, \sqrt {3} - \frac {2}{x} \right |}}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.39 \[ \int \frac {1+x^2}{1-5 x^2+x^4} \, dx=-\frac {\sqrt {3}\,\mathrm {atanh}\left (\frac {\sqrt {3}\,x}{x^2-1}\right )}{3} \]

[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