\(\int \frac {-6 x+2 x \log (\frac {2}{e^3})}{9+6 x^2+x^4+(-6-2 x^2) \log (\frac {2}{e^3})+\log ^2(\frac {2}{e^3})} \, dx\) [1940]

Optimal result

Integrand size = 48, antiderivative size = 19 \[ \int \frac {-6 x+2 x \log \left (\frac {2}{e^3}\right )}{9+6 x^2+x^4+\left (-6-2 x^2\right ) \log \left (\frac {2}{e^3}\right )+\log ^2\left (\frac {2}{e^3}\right )} \, dx=\frac {x^2}{-3-x^2+\log \left (\frac {2}{e^3}\right )} \]

[Out]

x^2/(-3-x^2+ln(2/exp(3)))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {6, 12, 2014, 28, 267} \[ \int \frac {-6 x+2 x \log \left (\frac {2}{e^3}\right )}{9+6 x^2+x^4+\left (-6-2 x^2\right ) \log \left (\frac {2}{e^3}\right )+\log ^2\left (\frac {2}{e^3}\right )} \, dx=\frac {6-\log (2)}{x^2+6-\log (2)} \]

[In]

Int[(-6*x + 2*x*Log[2/E^3])/(9 + 6*x^2 + x^4 + (-6 - 2*x^2)*Log[2/E^3] + Log[2/E^3]^2),x]

[Out]

(6 - Log[2])/(6 + x^2 - Log[2])

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 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 267

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

Rule 2014

Int[(u_)^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Int[(d*x)^m*ExpandToSum[u, x]^p, x] /; FreeQ[{d, m, p}, x] &&
TrinomialQ[u, x] &&  !TrinomialMatchQ[u, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x \left (-6+2 \log \left (\frac {2}{e^3}\right )\right )}{9+6 x^2+x^4+\left (-6-2 x^2\right ) \log \left (\frac {2}{e^3}\right )+\log ^2\left (\frac {2}{e^3}\right )} \, dx \\ & = \left (-6+2 \log \left (\frac {2}{e^3}\right )\right ) \int \frac {x}{9+6 x^2+x^4+\left (-6-2 x^2\right ) \log \left (\frac {2}{e^3}\right )+\log ^2\left (\frac {2}{e^3}\right )} \, dx \\ & = \left (-6+2 \log \left (\frac {2}{e^3}\right )\right ) \int \frac {x}{x^4+2 x^2 (6-\log (2))+(-6+\log (2))^2} \, dx \\ & = \left (-6+2 \log \left (\frac {2}{e^3}\right )\right ) \int \frac {x}{\left (6+x^2-\log (2)\right )^2} \, dx \\ & = \frac {6-\log (2)}{6+x^2-\log (2)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-6 x+2 x \log \left (\frac {2}{e^3}\right )}{9+6 x^2+x^4+\left (-6-2 x^2\right ) \log \left (\frac {2}{e^3}\right )+\log ^2\left (\frac {2}{e^3}\right )} \, dx=-\frac {-6+\log (2)}{6+x^2-\log (2)} \]

[In]

Integrate[(-6*x + 2*x*Log[2/E^3])/(9 + 6*x^2 + x^4 + (-6 - 2*x^2)*Log[2/E^3] + Log[2/E^3]^2),x]

[Out]

-((-6 + Log[2])/(6 + x^2 - Log[2]))

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89

method	result	size
norman	\(\frac {\ln \left (2\right )-6}{-x^{2}+\ln \left (2\right )-6}\)	\(17\)
default	\(-\frac {\ln \left (2 \,{\mathrm e}^{-3}\right )-3}{-\ln \left (2 \,{\mathrm e}^{-3}\right )+x^{2}+3}\)	\(26\)
risch	\(\frac {\ln \left (2\right )}{-x^{2}+\ln \left (2\right )-6}-\frac {6}{-x^{2}+\ln \left (2\right )-6}\)	\(29\)
parallelrisch	\(-\frac {-\ln \left (2 \,{\mathrm e}^{-3}\right )+3}{-3-x^{2}+\ln \left (2 \,{\mathrm e}^{-3}\right )}\)	\(30\)

[In]

int((2*x*ln(2/exp(3))-6*x)/(ln(2/exp(3))^2+(-2*x^2-6)*ln(2/exp(3))+x^4+6*x^2+9),x,method=_RETURNVERBOSE)

[Out]

(ln(2)-6)/(-x^2+ln(2)-6)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-6 x+2 x \log \left (\frac {2}{e^3}\right )}{9+6 x^2+x^4+\left (-6-2 x^2\right ) \log \left (\frac {2}{e^3}\right )+\log ^2\left (\frac {2}{e^3}\right )} \, dx=-\frac {\log \left (2\right ) - 6}{x^{2} - \log \left (2\right ) + 6} \]

[In]

integrate((2*x*log(2/exp(3))-6*x)/(log(2/exp(3))^2+(-2*x^2-6)*log(2/exp(3))+x^4+6*x^2+9),x, algorithm="fricas"
)

[Out]

-(log(2) - 6)/(x^2 - log(2) + 6)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-6 x+2 x \log \left (\frac {2}{e^3}\right )}{9+6 x^2+x^4+\left (-6-2 x^2\right ) \log \left (\frac {2}{e^3}\right )+\log ^2\left (\frac {2}{e^3}\right )} \, dx=- \frac {-12 + 2 \log {\left (2 \right )}}{2 x^{2} - 2 \log {\left (2 \right )} + 12} \]

[In]

integrate((2*x*ln(2/exp(3))-6*x)/(ln(2/exp(3))**2+(-2*x**2-6)*ln(2/exp(3))+x**4+6*x**2+9),x)

[Out]

-(-12 + 2*log(2))/(2*x**2 - 2*log(2) + 12)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {-6 x+2 x \log \left (\frac {2}{e^3}\right )}{9+6 x^2+x^4+\left (-6-2 x^2\right ) \log \left (\frac {2}{e^3}\right )+\log ^2\left (\frac {2}{e^3}\right )} \, dx=-\frac {\log \left (2 \, e^{\left (-3\right )}\right ) - 3}{x^{2} - \log \left (2 \, e^{\left (-3\right )}\right ) + 3} \]

[In]

integrate((2*x*log(2/exp(3))-6*x)/(log(2/exp(3))^2+(-2*x^2-6)*log(2/exp(3))+x^4+6*x^2+9),x, algorithm="maxima"
)

[Out]

-(log(2*e^(-3)) - 3)/(x^2 - log(2*e^(-3)) + 3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (19) = 38\).

Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.74 \[ \int \frac {-6 x+2 x \log \left (\frac {2}{e^3}\right )}{9+6 x^2+x^4+\left (-6-2 x^2\right ) \log \left (\frac {2}{e^3}\right )+\log ^2\left (\frac {2}{e^3}\right )} \, dx=-\frac {\log \left (2 \, e^{\left (-3\right )}\right )^{2} - 6 \, \log \left (2 \, e^{\left (-3\right )}\right ) + 9}{x^{2} \log \left (2 \, e^{\left (-3\right )}\right ) - 3 \, x^{2} - \log \left (2 \, e^{\left (-3\right )}\right )^{2} + 6 \, \log \left (2 \, e^{\left (-3\right )}\right ) - 9} \]

[In]

integrate((2*x*log(2/exp(3))-6*x)/(log(2/exp(3))^2+(-2*x^2-6)*log(2/exp(3))+x^4+6*x^2+9),x, algorithm="giac")

[Out]

-(log(2*e^(-3))^2 - 6*log(2*e^(-3)) + 9)/(x^2*log(2*e^(-3)) - 3*x^2 - log(2*e^(-3))^2 + 6*log(2*e^(-3)) - 9)

Mupad [F(-1)]

Timed out. \[ \int \frac {-6 x+2 x \log \left (\frac {2}{e^3}\right )}{9+6 x^2+x^4+\left (-6-2 x^2\right ) \log \left (\frac {2}{e^3}\right )+\log ^2\left (\frac {2}{e^3}\right )} \, dx=\int -\frac {6\,x-2\,x\,\ln \left (2\,{\mathrm {e}}^{-3}\right )}{{\ln \left (2\,{\mathrm {e}}^{-3}\right )}^2-\ln \left (2\,{\mathrm {e}}^{-3}\right )\,\left (2\,x^2+6\right )+6\,x^2+x^4+9} \,d x \]

[In]

int(-(6*x - 2*x*log(2*exp(-3)))/(log(2*exp(-3))^2 - log(2*exp(-3))*(2*x^2 + 6) + 6*x^2 + x^4 + 9),x)

[Out]

int(-(6*x - 2*x*log(2*exp(-3)))/(log(2*exp(-3))^2 - log(2*exp(-3))*(2*x^2 + 6) + 6*x^2 + x^4 + 9), x)