\(\int \frac {1}{\sqrt {8+4 x+x^2}} \, dx\) [145]

Optimal result

Integrand size = 12, antiderivative size = 8 \[ \int \frac {1}{\sqrt {8+4 x+x^2}} \, dx=\text {arcsinh}\left (\frac {2+x}{2}\right ) \]

[Out]

arcsinh(1+1/2*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {633, 221} \[ \int \frac {1}{\sqrt {8+4 x+x^2}} \, dx=\text {arcsinh}\left (\frac {x+2}{2}\right ) \]

[In]

Int[1/Sqrt[8 + 4*x + x^2],x]

[Out]

ArcSinh[(2 + x)/2]

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 {1}{4} \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{16}}} \, dx,x,4+2 x\right ) \\ & = \text {arcsinh}\left (\frac {2+x}{2}\right ) \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(20\) vs. \(2(8)=16\).

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 2.50 \[ \int \frac {1}{\sqrt {8+4 x+x^2}} \, dx=-\log \left (-2-x+\sqrt {8+4 x+x^2}\right ) \]

[In]

Integrate[1/Sqrt[8 + 4*x + x^2],x]

[Out]

-Log[-2 - x + Sqrt[8 + 4*x + x^2]]

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88

method	result	size
default	\(\operatorname {arcsinh}\left (1+\frac {x}{2}\right )\)	\(7\)
trager	\(\ln \left (x +2+\sqrt {x^{2}+4 x +8}\right )\)	\(15\)

[In]

int(1/(x^2+4*x+8)^(1/2),x,method=_RETURNVERBOSE)

[Out]

arcsinh(1+1/2*x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 18 vs. \(2 (6) = 12\).

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 2.25 \[ \int \frac {1}{\sqrt {8+4 x+x^2}} \, dx=-\log \left (-x + \sqrt {x^{2} + 4 \, x + 8} - 2\right ) \]

[In]

integrate(1/(x^2+4*x+8)^(1/2),x, algorithm="fricas")

[Out]

-log(-x + sqrt(x^2 + 4*x + 8) - 2)

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.62 \[ \int \frac {1}{\sqrt {8+4 x+x^2}} \, dx=\operatorname {asinh}{\left (\frac {x}{2} + 1 \right )} \]

[In]

integrate(1/(x**2+4*x+8)**(1/2),x)

[Out]

asinh(x/2 + 1)

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\sqrt {8+4 x+x^2}} \, dx=\operatorname {arsinh}\left (\frac {1}{2} \, x + 1\right ) \]

[In]

integrate(1/(x^2+4*x+8)^(1/2),x, algorithm="maxima")

[Out]

arcsinh(1/2*x + 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (6) = 12\).

Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 4.25 \[ \int \frac {1}{\sqrt {8+4 x+x^2}} \, dx=\frac {1}{2} \, \sqrt {x^{2} + 4 \, x + 8} {\left (x + 2\right )} - 2 \, \log \left (-x + \sqrt {x^{2} + 4 \, x + 8} - 2\right ) \]

[In]

integrate(1/(x^2+4*x+8)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.75 \[ \int \frac {1}{\sqrt {8+4 x+x^2}} \, dx=\ln \left (x+\sqrt {x^2+4\,x+8}+2\right ) \]

[In]

int(1/(4*x + x^2 + 8)^(1/2),x)

[Out]

log(x + (4*x + x^2 + 8)^(1/2) + 2)