\(\int \frac {x}{\sqrt {2+5 x+3 x^2}} \, dx\) [2413]

Optimal result

Integrand size = 16, antiderivative size = 57 \[ \int \frac {x}{\sqrt {2+5 x+3 x^2}} \, dx=\frac {1}{3} \sqrt {2+5 x+3 x^2}-\frac {5 \text {arctanh}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{6 \sqrt {3}} \]

[Out]

-5/18*arctanh(1/6*(5+6*x)*3^(1/2)/(3*x^2+5*x+2)^(1/2))*3^(1/2)+1/3*(3*x^2+5*x+2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {654, 635, 212} \[ \int \frac {x}{\sqrt {2+5 x+3 x^2}} \, dx=\frac {1}{3} \sqrt {3 x^2+5 x+2}-\frac {5 \text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{6 \sqrt {3}} \]

[In]

Int[x/Sqrt[2 + 5*x + 3*x^2],x]

[Out]

Sqrt[2 + 5*x + 3*x^2]/3 - (5*ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])])/(6*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 635

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

Rule 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \sqrt {2+5 x+3 x^2}-\frac {5}{6} \int \frac {1}{\sqrt {2+5 x+3 x^2}} \, dx \\ & = \frac {1}{3} \sqrt {2+5 x+3 x^2}-\frac {5}{3} \text {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {5+6 x}{\sqrt {2+5 x+3 x^2}}\right ) \\ & = \frac {1}{3} \sqrt {2+5 x+3 x^2}-\frac {5 \tanh ^{-1}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{6 \sqrt {3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.89 \[ \int \frac {x}{\sqrt {2+5 x+3 x^2}} \, dx=\frac {1}{9} \left (3 \sqrt {2+5 x+3 x^2}-5 \sqrt {3} \text {arctanh}\left (\frac {\sqrt {\frac {2}{3}+\frac {5 x}{3}+x^2}}{1+x}\right )\right ) \]

[In]

Integrate[x/Sqrt[2 + 5*x + 3*x^2],x]

[Out]

(3*Sqrt[2 + 5*x + 3*x^2] - 5*Sqrt[3]*ArcTanh[Sqrt[2/3 + (5*x)/3 + x^2]/(1 + x)])/9

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {\sqrt {3 x^{2}+5 x +2}}{3}-\frac {5 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{18}\)	\(45\)
risch	\(\frac {\sqrt {3 x^{2}+5 x +2}}{3}-\frac {5 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{18}\)	\(45\)
trager	\(\frac {\sqrt {3 x^{2}+5 x +2}}{3}-\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +6 \sqrt {3 x^{2}+5 x +2}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )\right )}{18}\)	\(57\)

[In]

int(x/(3*x^2+5*x+2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/3*(3*x^2+5*x+2)^(1/2)-5/18*ln(1/3*(5/2+3*x)*3^(1/2)+(3*x^2+5*x+2)^(1/2))*3^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93 \[ \int \frac {x}{\sqrt {2+5 x+3 x^2}} \, dx=\frac {5}{36} \, \sqrt {3} \log \left (-4 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x + 5\right )} + 72 \, x^{2} + 120 \, x + 49\right ) + \frac {1}{3} \, \sqrt {3 \, x^{2} + 5 \, x + 2} \]

[In]

integrate(x/(3*x^2+5*x+2)^(1/2),x, algorithm="fricas")

[Out]

5/36*sqrt(3)*log(-4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)*(6*x + 5) + 72*x^2 + 120*x + 49) + 1/3*sqrt(3*x^2 + 5*x + 2)

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int \frac {x}{\sqrt {2+5 x+3 x^2}} \, dx=\frac {\sqrt {3 x^{2} + 5 x + 2}}{3} - \frac {5 \sqrt {3} \log {\left (6 x + 2 \sqrt {3} \sqrt {3 x^{2} + 5 x + 2} + 5 \right )}}{18} \]

[In]

integrate(x/(3*x**2+5*x+2)**(1/2),x)

[Out]

sqrt(3*x**2 + 5*x + 2)/3 - 5*sqrt(3)*log(6*x + 2*sqrt(3)*sqrt(3*x**2 + 5*x + 2) + 5)/18

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.75 \[ \int \frac {x}{\sqrt {2+5 x+3 x^2}} \, dx=-\frac {5}{18} \, \sqrt {3} \log \left (2 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 6 \, x + 5\right ) + \frac {1}{3} \, \sqrt {3 \, x^{2} + 5 \, x + 2} \]

[In]

integrate(x/(3*x^2+5*x+2)^(1/2),x, algorithm="maxima")

[Out]

-5/18*sqrt(3)*log(2*sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 6*x + 5) + 1/3*sqrt(3*x^2 + 5*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int \frac {x}{\sqrt {2+5 x+3 x^2}} \, dx=\frac {5}{18} \, \sqrt {3} \log \left ({\left | -2 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )} - 5 \right |}\right ) + \frac {1}{3} \, \sqrt {3 \, x^{2} + 5 \, x + 2} \]

[In]

integrate(x/(3*x^2+5*x+2)^(1/2),x, algorithm="giac")

[Out]

5/18*sqrt(3)*log(abs(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2)) - 5)) + 1/3*sqrt(3*x^2 + 5*x + 2)

Mupad [B] (verification not implemented)

Time = 10.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77 \[ \int \frac {x}{\sqrt {2+5 x+3 x^2}} \, dx=\frac {\sqrt {3\,x^2+5\,x+2}}{3}-\frac {5\,\sqrt {3}\,\ln \left (\sqrt {3\,x^2+5\,x+2}+\frac {\sqrt {3}\,\left (3\,x+\frac {5}{2}\right )}{3}\right )}{18} \]

[In]

int(x/(5*x + 3*x^2 + 2)^(1/2),x)

[Out]

(5*x + 3*x^2 + 2)^(1/2)/3 - (5*3^(1/2)*log((5*x + 3*x^2 + 2)^(1/2) + (3^(1/2)*(3*x + 5/2))/3))/18