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

Optimal. Leaf size=19 \[ \frac {x^2}{-3-x^2+\log \left (\frac {2}{e^3}\right )} \]

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Rubi [A]  time = 0.03, antiderivative size = 18, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 5, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {6, 12, 1989, 28, 261} \begin {gather*} \frac {6-\log (2)}{x^2+6-\log (2)} \end {gather*}

Antiderivative was successfully verified.

[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 261

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 1989

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.89 \begin {gather*} -\frac {-6+\log (2)}{6+x^2-\log (2)} \end {gather*}

Antiderivative was successfully verified.

[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]))

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fricas [A]  time = 0.85, size = 17, normalized size = 0.89 \begin {gather*} -\frac {\log \relax (2) - 6}{x^{2} - \log \relax (2) + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.20, size = 52, normalized size = 2.74 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.05, size = 17, normalized size = 0.89

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.40, size = 23, normalized size = 1.21 \begin {gather*} -\frac {\log \left (2 \, e^{\left (-3\right )}\right ) - 3}{x^{2} - \log \left (2 \, e^{\left (-3\right )}\right ) + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.17, size = 19, normalized size = 1.00 \begin {gather*} - \frac {-12 + 2 \log {\relax (2 )}}{2 x^{2} - 2 \log {\relax (2 )} + 12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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