\(\int \frac {1}{\sqrt {2+4 x+3 x^2}} \, dx\) [117]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 14, antiderivative size = 18 \[ \int \frac {1}{\sqrt {2+4 x+3 x^2}} \, dx=\frac {\text {arcsinh}\left (\frac {2+3 x}{\sqrt {2}}\right )}{\sqrt {3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {633, 221} \[ \int \frac {1}{\sqrt {2+4 x+3 x^2}} \, dx=\frac {\text {arcsinh}\left (\frac {3 x+2}{\sqrt {2}}\right )}{\sqrt {3}} \]

[In]

Int[1/Sqrt[2 + 4*x + 3*x^2],x]

[Out]

ArcSinh[(2 + 3*x)/Sqrt[2]]/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}{8}}} \, dx,x,4+6 x\right )}{2 \sqrt {6}} \\ & = \frac {\sinh ^{-1}\left (\frac {2+3 x}{\sqrt {2}}\right )}{\sqrt {3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \frac {1}{\sqrt {2+4 x+3 x^2}} \, dx=-\frac {\log \left (-2-3 x+\sqrt {6+12 x+9 x^2}\right )}{\sqrt {3}} \]

[In]

Integrate[1/Sqrt[2 + 4*x + 3*x^2],x]

[Out]

-(Log[-2 - 3*x + Sqrt[6 + 12*x + 9*x^2]]/Sqrt[3])

Maple [A] (verified)

Time = 2.41 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method result size
default \(\frac {\sqrt {3}\, \operatorname {arcsinh}\left (\frac {3 \sqrt {2}\, \left (\frac {2}{3}+x \right )}{2}\right )}{3}\) \(15\)
trager \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )+3 \sqrt {3 x^{2}+4 x +2}\right )}{3}\) \(42\)

[In]

int(1/(3*x^2+4*x+2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (16) = 32\).

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 2.11 \[ \int \frac {1}{\sqrt {2+4 x+3 x^2}} \, dx=\frac {1}{6} \, \sqrt {3} \log \left (-\sqrt {3} \sqrt {3 \, x^{2} + 4 \, x + 2} {\left (3 \, x + 2\right )} - 9 \, x^{2} - 12 \, x - 5\right ) \]

[In]

integrate(1/(3*x^2+4*x+2)^(1/2),x, algorithm="fricas")

[Out]

1/6*sqrt(3)*log(-sqrt(3)*sqrt(3*x^2 + 4*x + 2)*(3*x + 2) - 9*x^2 - 12*x - 5)

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {1}{\sqrt {2+4 x+3 x^2}} \, dx=\frac {\sqrt {3} \operatorname {asinh}{\left (\frac {3 \sqrt {2} \left (x + \frac {2}{3}\right )}{2} \right )}}{3} \]

[In]

integrate(1/(3*x**2+4*x+2)**(1/2),x)

[Out]

sqrt(3)*asinh(3*sqrt(2)*(x + 2/3)/2)/3

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\sqrt {2+4 x+3 x^2}} \, dx=\frac {1}{3} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x + 2\right )}\right ) \]

[In]

integrate(1/(3*x^2+4*x+2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (16) = 32\).

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.94 \[ \int \frac {1}{\sqrt {2+4 x+3 x^2}} \, dx=\frac {1}{6} \, \sqrt {3 \, x^{2} + 4 \, x + 2} {\left (3 \, x + 2\right )} - \frac {1}{9} \, \sqrt {3} \log \left (-\sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 4 \, x + 2}\right )} - 2\right ) \]

[In]

integrate(1/(3*x^2+4*x+2)^(1/2),x, algorithm="giac")

[Out]

1/6*sqrt(3*x^2 + 4*x + 2)*(3*x + 2) - 1/9*sqrt(3)*log(-sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 4*x + 2)) - 2)

Mupad [B] (verification not implemented)

Time = 9.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {1}{\sqrt {2+4 x+3 x^2}} \, dx=\frac {\sqrt {3}\,\ln \left (\sqrt {3}\,\left (x+\frac {2}{3}\right )+\sqrt {3\,x^2+4\,x+2}\right )}{3} \]

[In]

int(1/(4*x + 3*x^2 + 2)^(1/2),x)

[Out]

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