\(\int \frac {2-2 \log (x)}{-5 x^2+e^4 x^2} \, dx\) [7375]

Optimal result

Integrand size = 22, antiderivative size = 14 \[ \int \frac {2-2 \log (x)}{-5 x^2+e^4 x^2} \, dx=\frac {2 \log (x)}{\left (-5+e^4\right ) x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6, 12, 2340} \[ \int \frac {2-2 \log (x)}{-5 x^2+e^4 x^2} \, dx=-\frac {2 \log (x)}{\left (5-e^4\right ) x} \]

[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 2340

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {2-2 \log (x)}{-5 x^2+e^4 x^2} \, dx=\frac {2 \log (x)}{\left (-5+e^4\right ) x} \]

[In]

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

[Out]

(2*Log[x])/((-5 + E^4)*x)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {2-2 \log (x)}{-5 x^2+e^4 x^2} \, dx=\frac {2 \, \log \left (x\right )}{x e^{4} - 5 \, x} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {2-2 \log (x)}{-5 x^2+e^4 x^2} \, dx=\frac {2 \log {\left (x \right )}}{- 5 x + x e^{4}} \]

[In]

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

[Out]

2*log(x)/(-5*x + x*exp(4))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (13) = 26\).

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.93 \[ \int \frac {2-2 \log (x)}{-5 x^2+e^4 x^2} \, dx=\frac {2 \, {\left (\log \left (x\right ) + 1\right )}}{x {\left (e^{4} - 5\right )}} - \frac {2}{x {\left (e^{4} - 5\right )}} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {2-2 \log (x)}{-5 x^2+e^4 x^2} \, dx=\frac {2 \, \log \left (x\right )}{x e^{4} - 5 \, x} \]

[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)

Mupad [B] (verification not implemented)

Time = 13.38 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {2-2 \log (x)}{-5 x^2+e^4 x^2} \, dx=\frac {2\,\ln \left (x\right )}{x\,\left ({\mathrm {e}}^4-5\right )} \]

[In]

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

[Out]

(2*log(x))/(x*(exp(4) - 5))