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

Optimal. Leaf size=30 \[ \frac {1}{10} \log \left (\frac {x-x \log \left (e^{\frac {10}{5+\frac {5}{3 x}}}\right )}{x}\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 8, normalized size of antiderivative = 0.27, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 207} \begin {gather*} -\frac {1}{5} \tanh ^{-1}(3 x) \end {gather*}

Antiderivative was successfully verified.

[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 207

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

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 23, normalized size = 0.77 \begin {gather*} 3 \left (\frac {1}{30} \log (1-3 x)-\frac {1}{30} \log (1+3 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[3/(-5 + 45*x^2),x]

[Out]

3*(Log[1 - 3*x]/30 - Log[1 + 3*x]/30)

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fricas [A]  time = 0.67, size = 17, normalized size = 0.57 \begin {gather*} -\frac {1}{10} \, \log \left (3 \, x + 1\right ) + \frac {1}{10} \, \log \left (3 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.13, size = 15, normalized size = 0.50 \begin {gather*} -\frac {1}{10} \, \log \left ({\left | x + \frac {1}{3} \right |}\right ) + \frac {1}{10} \, \log \left ({\left | x - \frac {1}{3} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.08, size = 7, normalized size = 0.23

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*arctanh(3*x)

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maxima [A]  time = 0.36, size = 17, normalized size = 0.57 \begin {gather*} -\frac {1}{10} \, \log \left (3 \, x + 1\right ) + \frac {1}{10} \, \log \left (3 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.10, size = 6, normalized size = 0.20 \begin {gather*} -\frac {\mathrm {atanh}\left (3\,x\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(3/(45*x^2 - 5),x)

[Out]

-atanh(3*x)/5

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sympy [A]  time = 0.09, size = 15, normalized size = 0.50 \begin {gather*} \frac {\log {\left (x - \frac {1}{3} \right )}}{10} - \frac {\log {\left (x + \frac {1}{3} \right )}}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x - 1/3)/10 - log(x + 1/3)/10

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