\(\int \frac {1}{\sqrt {1+x+x^2}} \, dx\) [266]

Optimal result

Integrand size = 10, antiderivative size = 12 \[ \int \frac {1}{\sqrt {1+x+x^2}} \, dx=\text {arcsinh}\left (\frac {1+2 x}{\sqrt {3}}\right ) \]

[Out]

arcsinh(1/3*(1+2*x)*3^(1/2))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {633, 221} \[ \int \frac {1}{\sqrt {1+x+x^2}} \, dx=\text {arcsinh}\left (\frac {2 x+1}{\sqrt {3}}\right ) \]

[In]

Int[1/Sqrt[1 + x + x^2],x]

[Out]

ArcSinh[(1 + 2*x)/Sqrt[3]]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{3}}} \, dx,x,1+2 x\right )}{\sqrt {3}} \\ & = \text {arcsinh}\left (\frac {1+2 x}{\sqrt {3}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.67 \[ \int \frac {1}{\sqrt {1+x+x^2}} \, dx=-\log \left (-1-2 x+2 \sqrt {1+x+x^2}\right ) \]

[In]

Integrate[1/Sqrt[1 + x + x^2],x]

[Out]

-Log[-1 - 2*x + 2*Sqrt[1 + x + x^2]]

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83

method	result	size
default	\(\operatorname {arcsinh}\left (\frac {2 \sqrt {3}\, \left (x +\frac {1}{2}\right )}{3}\right )\)	\(10\)
trager	\(-\ln \left (2 \sqrt {x^{2}+x +1}-1-2 x \right )\)	\(19\)

[In]

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

[Out]

arcsinh(2/3*3^(1/2)*(x+1/2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.50 \[ \int \frac {1}{\sqrt {1+x+x^2}} \, dx=-\log \left (-2 \, x + 2 \, \sqrt {x^{2} + x + 1} - 1\right ) \]

[In]

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

[Out]

-log(-2*x + 2*sqrt(x^2 + x + 1) - 1)

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \frac {1}{\sqrt {1+x+x^2}} \, dx=\operatorname {asinh}{\left (\frac {2 \sqrt {3} \left (x + \frac {1}{2}\right )}{3} \right )} \]

[In]

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

[Out]

asinh(2*sqrt(3)*(x + 1/2)/3)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\sqrt {1+x+x^2}} \, dx=\operatorname {arsinh}\left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) \]

[In]

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

[Out]

arcsinh(1/3*sqrt(3)*(2*x + 1))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (11) = 22\).

Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.83 \[ \int \frac {1}{\sqrt {1+x+x^2}} \, dx=\frac {1}{4} \, \sqrt {x^{2} + x + 1} {\left (2 \, x + 1\right )} - \frac {3}{8} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} + x + 1} - 1\right ) \]

[In]

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

[Out]

1/4*sqrt(x^2 + x + 1)*(2*x + 1) - 3/8*log(-2*x + 2*sqrt(x^2 + x + 1) - 1)

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {1+x+x^2}} \, dx=\ln \left (x+\sqrt {x^2+x+1}+\frac {1}{2}\right ) \]

[In]

int(1/(x + x^2 + 1)^(1/2),x)

[Out]

log(x + (x + x^2 + 1)^(1/2) + 1/2)