\(\int \frac {3}{-5+45 x^2} \, dx\) [4657]

Optimal result

Integrand size = 11, antiderivative size = 30 \[ \int \frac {3}{-5+45 x^2} \, dx=\frac {1}{10} \log \left (\frac {x-x \log \left (e^{\frac {10}{5+\frac {5}{3 x}}}\right )}{x}\right ) \]

[Out]

1/10*ln((x-x*ln(exp(10/(5/3/x+5))))/x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.27, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 213} \[ \int \frac {3}{-5+45 x^2} \, dx=-\frac {1}{5} \text {arctanh}(3 x) \]

[In]

Int[3/(-5 + 45*x^2),x]

[Out]

-1/5*ArcTanh[3*x]

Rule 12

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

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps \begin{align*} \text {integral}& = 3 \int \frac {1}{-5+45 x^2} \, dx \\ & = -\frac {1}{5} \tanh ^{-1}(3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {3}{-5+45 x^2} \, dx=3 \left (\frac {1}{30} \log (1-3 x)-\frac {1}{30} \log (1+3 x)\right ) \]

[In]

Integrate[3/(-5 + 45*x^2),x]

[Out]

3*(Log[1 - 3*x]/30 - Log[1 + 3*x]/30)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.23

method	result	size
meijerg	\(-\frac {\operatorname {arctanh}\left (3 x \right )}{5}\)	\(7\)
parallelrisch	\(\frac {\ln \left (x -\frac {1}{3}\right )}{10}-\frac {\ln \left (x +\frac {1}{3}\right )}{10}\)	\(14\)
default	\(-\frac {\ln \left (1+3 x \right )}{10}+\frac {\ln \left (-1+3 x \right )}{10}\)	\(18\)
norman	\(-\frac {\ln \left (1+3 x \right )}{10}+\frac {\ln \left (-1+3 x \right )}{10}\)	\(18\)
risch	\(-\frac {\ln \left (1+3 x \right )}{10}+\frac {\ln \left (-1+3 x \right )}{10}\)	\(18\)

[In]

int(3/(45*x^2-5),x,method=_RETURNVERBOSE)

[Out]

-1/5*arctanh(3*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.57 \[ \int \frac {3}{-5+45 x^2} \, dx=-\frac {1}{10} \, \log \left (3 \, x + 1\right ) + \frac {1}{10} \, \log \left (3 \, x - 1\right ) \]

[In]

integrate(3/(45*x^2-5),x, algorithm="fricas")

[Out]

-1/10*log(3*x + 1) + 1/10*log(3*x - 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.50 \[ \int \frac {3}{-5+45 x^2} \, dx=\frac {\log {\left (x - \frac {1}{3} \right )}}{10} - \frac {\log {\left (x + \frac {1}{3} \right )}}{10} \]

[In]

integrate(3/(45*x**2-5),x)

[Out]

log(x - 1/3)/10 - log(x + 1/3)/10

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.57 \[ \int \frac {3}{-5+45 x^2} \, dx=-\frac {1}{10} \, \log \left (3 \, x + 1\right ) + \frac {1}{10} \, \log \left (3 \, x - 1\right ) \]

[In]

integrate(3/(45*x^2-5),x, algorithm="maxima")

[Out]

-1/10*log(3*x + 1) + 1/10*log(3*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.50 \[ \int \frac {3}{-5+45 x^2} \, dx=-\frac {1}{10} \, \log \left ({\left | x + \frac {1}{3} \right |}\right ) + \frac {1}{10} \, \log \left ({\left | x - \frac {1}{3} \right |}\right ) \]

[In]

integrate(3/(45*x^2-5),x, algorithm="giac")

[Out]

-1/10*log(abs(x + 1/3)) + 1/10*log(abs(x - 1/3))

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.20 \[ \int \frac {3}{-5+45 x^2} \, dx=-\frac {\mathrm {atanh}\left (3\,x\right )}{5} \]

[In]

int(3/(45*x^2 - 5),x)

[Out]

-atanh(3*x)/5