3.1.66 \(\int \frac {1}{\sqrt {4+12 x+9 x^2}} \, dx\) [66]

Optimal. Leaf size=29 \[ \frac {(2+3 x) \log (2+3 x)}{3 \sqrt {4+12 x+9 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {622, 31} \begin {gather*} \frac {(3 x+2) \log (3 x+2)}{3 \sqrt {9 x^2+12 x+4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[4 + 12*x + 9*x^2],x]

[Out]

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

Rule 31

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

Rule 622

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]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 26, normalized size = 0.90 \begin {gather*} \frac {(2+3 x) \log (2+3 x)}{3 \sqrt {(2+3 x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[4 + 12*x + 9*x^2],x]

[Out]

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

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Maple [A]
time = 0.45, size = 23, normalized size = 0.79

method	result	size
meijerg	\(\frac {\ln \left (1+\frac {3 x}{2}\right )}{3}\)	\(9\)
default	\(\frac {\left (2+3 x \right ) \ln \left (2+3 x \right )}{3 \sqrt {\left (2+3 x \right )^{2}}}\)	\(23\)
risch	\(\frac {\sqrt {\left (2+3 x \right )^{2}}\, \ln \left (2+3 x \right )}{6+9 x}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 6, normalized size = 0.21 \begin {gather*} \frac {1}{3} \, \log \left (x + \frac {2}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*log(x + 2/3)

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Fricas [A]
time = 1.50, size = 8, normalized size = 0.28 \begin {gather*} \frac {1}{3} \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*log(3*x + 2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {9 x^{2} + 12 x + 4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.42, size = 25, normalized size = 0.86 \begin {gather*} \frac {\log \left ({\left | 3 \, x + 2 \right |} {\left | \mathrm {sgn}\left (3 \, x + 2\right ) \right |}\right )}{3 \, \mathrm {sgn}\left (3 \, x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*log(abs(3*x + 2)*abs(sgn(3*x + 2)))/sgn(3*x + 2)

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Mupad [B]
time = 0.28, size = 14, normalized size = 0.48 \begin {gather*} \frac {\ln \left (9\,x+6\right )\,\mathrm {sign}\left (18\,x+12\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(12*x + 9*x^2 + 4)^(1/2),x)

[Out]

(log(9*x + 6)*sign(18*x + 12))/3

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