3.211 \(\int \frac{x}{\sqrt{9+12 x+4 x^2}} \, dx\)

Optimal. Leaf size=48 \[ \frac{1}{4} \sqrt{4 x^2+12 x+9}-\frac{3 (2 x+3) \log (2 x+3)}{4 \sqrt{4 x^2+12 x+9}} \]

[Out]

Sqrt[9 + 12*x + 4*x^2]/4 - (3*(3 + 2*x)*Log[3 + 2*x])/(4*Sqrt[9 + 12*x + 4*x^2])

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Rubi [A]  time = 0.0108877, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {640, 608, 31} \[ \frac{1}{4} \sqrt{4 x^2+12 x+9}-\frac{3 (2 x+3) \log (2 x+3)}{4 \sqrt{4 x^2+12 x+9}} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[9 + 12*x + 4*x^2],x]

[Out]

Sqrt[9 + 12*x + 4*x^2]/4 - (3*(3 + 2*x)*Log[3 + 2*x])/(4*Sqrt[9 + 12*x + 4*x^2])

Rule 640

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]

Rule 608

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{x}{\sqrt{9+12 x+4 x^2}} \, dx &=\frac{1}{4} \sqrt{9+12 x+4 x^2}-\frac{3}{2} \int \frac{1}{\sqrt{9+12 x+4 x^2}} \, dx\\ &=\frac{1}{4} \sqrt{9+12 x+4 x^2}-\frac{(3 (6+4 x)) \int \frac{1}{6+4 x} \, dx}{2 \sqrt{9+12 x+4 x^2}}\\ &=\frac{1}{4} \sqrt{9+12 x+4 x^2}-\frac{3 (3+2 x) \log (3+2 x)}{4 \sqrt{9+12 x+4 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0094597, size = 33, normalized size = 0.69 \[ \frac{(2 x+3) (2 x-3 \log (2 x+3)+3)}{4 \sqrt{(2 x+3)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[9 + 12*x + 4*x^2],x]

[Out]

((3 + 2*x)*(3 + 2*x - 3*Log[3 + 2*x]))/(4*Sqrt[(3 + 2*x)^2])

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Maple [A]  time = 0.158, size = 29, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 3+2\,x \right ) \left ( -2\,x+3\,\ln \left ( 3+2\,x \right ) \right ) }{4}{\frac{1}{\sqrt{ \left ( 3+2\,x \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(4*x^2+12*x+9)^(1/2),x)

[Out]

-1/4*(3+2*x)*(-2*x+3*ln(3+2*x))/((3+2*x)^2)^(1/2)

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Maxima [A]  time = 1.68325, size = 28, normalized size = 0.58 \begin{align*} \frac{1}{4} \, \sqrt{4 \, x^{2} + 12 \, x + 9} - \frac{3}{4} \, \log \left (x + \frac{3}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(4*x^2+12*x+9)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.73834, size = 35, normalized size = 0.73 \begin{align*} \frac{1}{2} \, x - \frac{3}{4} \, \log \left (2 \, x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(4*x^2+12*x+9)^(1/2),x, algorithm="fricas")

[Out]

1/2*x - 3/4*log(2*x + 3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{\left (2 x + 3\right )^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(4*x**2+12*x+9)**(1/2),x)

[Out]

Integral(x/sqrt((2*x + 3)**2), x)

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Giac [A]  time = 1.31121, size = 49, normalized size = 1.02 \begin{align*} \frac{1}{4} \, \sqrt{4 \, x^{2} + 12 \, x + 9} + \frac{3}{4} \, \log \left ({\left | -2 \, x + \sqrt{4 \, x^{2} + 12 \, x + 9} - 3 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(4*x^2+12*x+9)^(1/2),x, algorithm="giac")

[Out]

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