\(\int -\frac {18 x}{-1+9 x^2} \, dx\) [691]

Optimal result

Integrand size = 12, antiderivative size = 16 \[ \int -\frac {18 x}{-1+9 x^2} \, dx=\frac {1}{4}+\log \left (\frac {2}{-1+9 x^2}\right ) \]

[Out]

1/4+ln(2/(9*x^2-1))

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 266} \[ \int -\frac {18 x}{-1+9 x^2} \, dx=-\log \left (1-9 x^2\right ) \]

[In]

Int[(-18*x)/(-1 + 9*x^2),x]

[Out]

-Log[1 - 9*x^2]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 266

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

Rubi steps \begin{align*} \text {integral}& = -\left (18 \int \frac {x}{-1+9 x^2} \, dx\right ) \\ & = -\log \left (1-9 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int -\frac {18 x}{-1+9 x^2} \, dx=-\log \left (1-9 x^2\right ) \]

[In]

Integrate[(-18*x)/(-1 + 9*x^2),x]

[Out]

-Log[1 - 9*x^2]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69

method	result	size
derivativedivides	\(-\ln \left (9 x^{2}-1\right )\)	\(11\)
default	\(-\ln \left (9 x^{2}-1\right )\)	\(11\)
meijerg	\(-\ln \left (-9 x^{2}+1\right )\)	\(11\)
risch	\(-\ln \left (9 x^{2}-1\right )\)	\(11\)
parallelrisch	\(-\ln \left (x -\frac {1}{3}\right )-\ln \left (x +\frac {1}{3}\right )\)	\(14\)
norman	\(-\ln \left (-1+3 x \right )-\ln \left (1+3 x \right )\)	\(18\)

[In]

int(-18*x/(9*x^2-1),x,method=_RETURNVERBOSE)

[Out]

-ln(9*x^2-1)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int -\frac {18 x}{-1+9 x^2} \, dx=-\log \left (9 \, x^{2} - 1\right ) \]

[In]

integrate(-18*x/(9*x^2-1),x, algorithm="fricas")

[Out]

-log(9*x^2 - 1)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.50 \[ \int -\frac {18 x}{-1+9 x^2} \, dx=- \log {\left (9 x^{2} - 1 \right )} \]

[In]

integrate(-18*x/(9*x**2-1),x)

[Out]

-log(9*x**2 - 1)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int -\frac {18 x}{-1+9 x^2} \, dx=-\log \left (9 \, x^{2} - 1\right ) \]

[In]

integrate(-18*x/(9*x^2-1),x, algorithm="maxima")

[Out]

-log(9*x^2 - 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69 \[ \int -\frac {18 x}{-1+9 x^2} \, dx=-\log \left ({\left | 9 \, x^{2} - 1 \right |}\right ) \]

[In]

integrate(-18*x/(9*x^2-1),x, algorithm="giac")

[Out]

-log(abs(9*x^2 - 1))

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.50 \[ \int -\frac {18 x}{-1+9 x^2} \, dx=-\ln \left (x^2-\frac {1}{9}\right ) \]

[In]

int(-(18*x)/(9*x^2 - 1),x)

[Out]

-log(x^2 - 1/9)