3.74.75 \(\int \frac {2-2 \log (x)}{-5 x^2+e^4 x^2} \, dx\)

Optimal. Leaf size=14 \[ \frac {2 \log (x)}{\left (-5+e^4\right ) x} \]

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.14, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6, 12, 2303} \begin {gather*} -\frac {2 \log (x)}{\left (5-e^4\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - 2*Log[x])/(-5*x^2 + E^4*x^2),x]

[Out]

(-2*Log[x])/((5 - E^4)*x)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 2303

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(b*(d*x)^(m + 1)*Log[c*x^n])/(
d*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && EqQ[a*(m + 1) - b*n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2-2 \log (x)}{\left (-5+e^4\right ) x^2} \, dx\\ &=\frac {\int \frac {2-2 \log (x)}{x^2} \, dx}{-5+e^4}\\ &=-\frac {2 \log (x)}{\left (5-e^4\right ) x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 1.00 \begin {gather*} \frac {2 \log (x)}{\left (-5+e^4\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - 2*Log[x])/(-5*x^2 + E^4*x^2),x]

[Out]

(2*Log[x])/((-5 + E^4)*x)

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fricas [A]  time = 0.60, size = 14, normalized size = 1.00 \begin {gather*} \frac {2 \, \log \relax (x)}{x e^{4} - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(x)+2)/(x^2*exp(4)-5*x^2),x, algorithm="fricas")

[Out]

2*log(x)/(x*e^4 - 5*x)

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giac [A]  time = 0.15, size = 14, normalized size = 1.00 \begin {gather*} \frac {2 \, \log \relax (x)}{x e^{4} - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(x)+2)/(x^2*exp(4)-5*x^2),x, algorithm="giac")

[Out]

2*log(x)/(x*e^4 - 5*x)

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maple [A]  time = 0.16, size = 14, normalized size = 1.00

method	result	size
norman	\(\frac {2 \ln \relax (x )}{\left ({\mathrm e}^{4}-5\right ) x}\)	\(14\)
risch	\(\frac {2 \ln \relax (x )}{\left ({\mathrm e}^{4}-5\right ) x}\)	\(14\)
default	\(-\frac {2 \left (-\frac {\ln \relax (x )}{x}-\frac {1}{x}\right )}{{\mathrm e}^{4}-5}-\frac {2}{\left ({\mathrm e}^{4}-5\right ) x}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*ln(x)+2)/(x^2*exp(4)-5*x^2),x,method=_RETURNVERBOSE)

[Out]

2/(exp(4)-5)/x*ln(x)

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maxima [B]  time = 0.39, size = 27, normalized size = 1.93 \begin {gather*} \frac {2 \, {\left (\log \relax (x) + 1\right )}}{x {\left (e^{4} - 5\right )}} - \frac {2}{x {\left (e^{4} - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(x)+2)/(x^2*exp(4)-5*x^2),x, algorithm="maxima")

[Out]

2*(log(x) + 1)/(x*(e^4 - 5)) - 2/(x*(e^4 - 5))

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mupad [B]  time = 5.33, size = 13, normalized size = 0.93 \begin {gather*} \frac {2\,\ln \relax (x)}{x\,\left ({\mathrm {e}}^4-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(x) - 2)/(x^2*exp(4) - 5*x^2),x)

[Out]

(2*log(x))/(x*(exp(4) - 5))

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sympy [A]  time = 0.14, size = 12, normalized size = 0.86 \begin {gather*} \frac {2 \log {\relax (x )}}{- 5 x + x e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*ln(x)+2)/(x**2*exp(4)-5*x**2),x)

[Out]

2*log(x)/(-5*x + x*exp(4))

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