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

Optimal result

Integrand size = 14, antiderivative size = 29 \[ \int \frac {1}{\sqrt {4+12 x+9 x^2}} \, dx=\frac {(2+3 x) \log (2+3 x)}{3 \sqrt {4+12 x+9 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 29, 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 = {622, 31} \[ \int \frac {1}{\sqrt {4+12 x+9 x^2}} \, dx=\frac {(3 x+2) \log (3 x+2)}{3 \sqrt {9 x^2+12 x+4}} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\sqrt {4+12 x+9 x^2}} \, dx=\frac {(2+3 x) \log (2+3 x)}{3 \sqrt {(2+3 x)^2}} \]

[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])

Maple [A] (verified)

Time = 2.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.31

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	\(\frac {\sqrt {\left (2+3 x \right )^{2}}\, \ln \left (2+3 x \right )}{9 x +6}\)	\(25\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.28 \[ \int \frac {1}{\sqrt {4+12 x+9 x^2}} \, dx=\frac {1}{3} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

1/3*log(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\sqrt {4+12 x+9 x^2}} \, dx=\frac {\left (x + \frac {2}{3}\right ) \log {\left (x + \frac {2}{3} \right )}}{3 \sqrt {\left (x + \frac {2}{3}\right )^{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.21 \[ \int \frac {1}{\sqrt {4+12 x+9 x^2}} \, dx=\frac {1}{3} \, \log \left (x + \frac {2}{3}\right ) \]

[In]

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

[Out]

1/3*log(x + 2/3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\sqrt {4+12 x+9 x^2}} \, dx=\frac {\log \left ({\left | 3 \, x + 2 \right |} {\left | \mathrm {sgn}\left (3 \, x + 2\right ) \right |}\right )}{3 \, \mathrm {sgn}\left (3 \, x + 2\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 9.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.48 \[ \int \frac {1}{\sqrt {4+12 x+9 x^2}} \, dx=\frac {\ln \left (9\,x+6\right )\,\mathrm {sign}\left (18\,x+12\right )}{3} \]

[In]

int(1/(12*x + 9*x^2 + 4)^(1/2),x)

[Out]

(log(9*x + 6)*sign(18*x + 12))/3