3.2422 \(\int \frac{1}{x \sqrt{-4-12 x-9 x^2}} \, dx\)

Optimal. Leaf size=27 \[ -\frac{(3 x+2) \tanh ^{-1}(3 x+1)}{\sqrt{-9 x^2-12 x-4}} \]

[Out]

-(((2 + 3*x)*ArcTanh[1 + 3*x])/Sqrt[-4 - 12*x - 9*x^2])

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Rubi [B]  time = 0.0137119, antiderivative size = 55, normalized size of antiderivative = 2.04, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {646, 36, 31, 29} \[ \frac{(3 x+2) \log (x)}{2 \sqrt{-9 x^2-12 x-4}}-\frac{(3 x+2) \log (3 x+2)}{2 \sqrt{-9 x^2-12 x-4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[-4 - 12*x - 9*x^2]),x]

[Out]

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

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 36

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

Rule 31

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rubi steps

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

Mathematica [A]  time = 0.0071819, size = 33, normalized size = 1.22 \[ \frac{(3 x+2) (\log (x)-\log (3 x+2))}{2 \sqrt{-(3 x+2)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sqrt[-4 - 12*x - 9*x^2]),x]

[Out]

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

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Maple [A]  time = 0.103, size = 30, normalized size = 1.1 \begin{align*}{\frac{ \left ( 2+3\,x \right ) \left ( \ln \left ( x \right ) -\ln \left ( 2+3\,x \right ) \right ) }{2}{\frac{1}{\sqrt{- \left ( 2+3\,x \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(-(2+3*x)^2)^(1/2),x)

[Out]

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

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Maxima [C]  time = 1.51444, size = 32, normalized size = 1.19 \begin{align*} -\frac{1}{2} i \, \left (-1\right )^{12 \, x + 8} \log \left (\frac{12 \, x}{{\left | x \right |}} + \frac{8}{{\left | x \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*(-1)^(12*x + 8)*log(12*x/abs(x) + 8/abs(x))

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Fricas [C]  time = 2.00307, size = 47, normalized size = 1.74 \begin{align*} \frac{1}{2} i \, \log \left (x + \frac{2}{3}\right ) - \frac{1}{2} i \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*log(x + 2/3) - 1/2*I*log(x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \sqrt{- \left (3 x + 2\right )^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x*sqrt(-(3*x + 2)**2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef