\(\int \frac {1}{x+\sqrt {1+x+x^2}} \, dx\) [288]

Optimal result

Integrand size = 14, antiderivative size = 45 \[ \int \frac {1}{x+\sqrt {1+x+x^2}} \, dx=-x+\sqrt {1+x+x^2}-\frac {3}{2} \text {arcsinh}\left (\frac {1+2 x}{\sqrt {3}}\right )+2 \log \left (x+\sqrt {1+x+x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.31, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2141, 907} \[ \int \frac {1}{x+\sqrt {1+x+x^2}} \, dx=\frac {3}{2 \left (2 \left (\sqrt {x^2+x+1}+x\right )+1\right )}+2 \log \left (\sqrt {x^2+x+1}+x\right )-\frac {3}{2} \log \left (2 \left (\sqrt {x^2+x+1}+x\right )+1\right ) \]

[In]

Int[(x + Sqrt[1 + x + x^2])^(-1),x]

[Out]

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

Rule 907

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2141

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol]
:> Dist[2, Subst[Int[(g + h*x^n)^p*((d^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x + e*x^2)/(-2*d*e + b*f^2 + 2
*e*x)^2), x], x, d + e*x + f*Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e, f, g, h, n}, x] && EqQ[e^2 -
c*f^2, 0] && IntegerQ[p]

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.20 \[ \int \frac {1}{x+\sqrt {1+x+x^2}} \, dx=-x+\sqrt {1+x+x^2}+2 \log \left (-2-x+\sqrt {1+x+x^2}\right )-\frac {1}{2} \log \left (-1-2 x+2 \sqrt {1+x+x^2}\right ) \]

[In]

Integrate[(x + Sqrt[1 + x + x^2])^(-1),x]

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.16

method	result	size
default	\(\sqrt {\left (1+x \right )^{2}-x}-\frac {\operatorname {arcsinh}\left (\frac {2 \sqrt {3}\, \left (x +\frac {1}{2}\right )}{3}\right )}{2}-\operatorname {arctanh}\left (\frac {1-x}{2 \sqrt {\left (1+x \right )^{2}-x}}\right )-x +\ln \left (1+x \right )\)	\(52\)
trager	\(\sqrt {x^{2}+x +1}-x -\frac {\ln \left (\frac {2 x^{2} \sqrt {x^{2}+x +1}+2 x^{3}+8 x \sqrt {x^{2}+x +1}+9 x^{2}+14 \sqrt {x^{2}+x +1}+12 x +13}{\left (1+x \right )^{4}}\right )}{2}\)	\(71\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.40 \[ \int \frac {1}{x+\sqrt {1+x+x^2}} \, dx=-x + \sqrt {x^{2} + x + 1} + \log \left (x + 1\right ) - \log \left (-x + \sqrt {x^{2} + x + 1}\right ) + \log \left (-x + \sqrt {x^{2} + x + 1} - 2\right ) + \frac {1}{2} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} + x + 1} - 1\right ) \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {1}{x+\sqrt {1+x+x^2}} \, dx=\int \frac {1}{x + \sqrt {x^{2} + x + 1}}\, dx \]

[In]

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

[Out]

Integral(1/(x + sqrt(x**2 + x + 1)), x)

Maxima [F]

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

[In]

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

[Out]

integrate(1/(x + sqrt(x^2 + x + 1)), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.47 \[ \int \frac {1}{x+\sqrt {1+x+x^2}} \, dx=-x + \sqrt {x^{2} + x + 1} + \frac {1}{2} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} + x + 1} - 1\right ) + \log \left ({\left | x + 1 \right |}\right ) - \log \left ({\left | -x + \sqrt {x^{2} + x + 1} \right |}\right ) + \log \left ({\left | -x + \sqrt {x^{2} + x + 1} - 2 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x+\sqrt {1+x+x^2}} \, dx=\ln \left (x+1\right )-x+\int \frac {\sqrt {x^2+x+1}}{x+1} \,d x \]

[In]

int(1/(x + (x + x^2 + 1)^(1/2)),x)

[Out]

log(x + 1) - x + int((x + x^2 + 1)^(1/2)/(x + 1), x)