3.79.85 \(\int -\frac {2 x}{-6-e^5+x^2+2 \log (4)} \, dx\)

Optimal. Leaf size=19 \[ 1-\log \left (6+e^5-x^2-2 \log (4)\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 0.89, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 260} \begin {gather*} -\log \left (-x^2+e^5+6-2 \log (4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x)/(-6 - E^5 + x^2 + 2*Log[4]),x]

[Out]

-Log[6 + E^5 - x^2 - 2*Log[4]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 260

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (2 \int \frac {x}{-6-e^5+x^2+2 \log (4)} \, dx\right )\\ &=-\log \left (6+e^5-x^2-2 \log (4)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 17, normalized size = 0.89 \begin {gather*} -\log \left (6+e^5-x^2-2 \log (4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x)/(-6 - E^5 + x^2 + 2*Log[4]),x]

[Out]

-Log[6 + E^5 - x^2 - 2*Log[4]]

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fricas [A]  time = 0.71, size = 16, normalized size = 0.84 \begin {gather*} -\log \left (x^{2} - e^{5} + 4 \, \log \relax (2) - 6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*x/(4*log(2)-exp(5)+x^2-6),x, algorithm="fricas")

[Out]

-log(x^2 - e^5 + 4*log(2) - 6)

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giac [A]  time = 0.21, size = 17, normalized size = 0.89 \begin {gather*} -\log \left ({\left | x^{2} - e^{5} + 4 \, \log \relax (2) - 6 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*x/(4*log(2)-exp(5)+x^2-6),x, algorithm="giac")

[Out]

-log(abs(x^2 - e^5 + 4*log(2) - 6))

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maple [A]  time = 0.25, size = 17, normalized size = 0.89

method	result	size
derivativedivides	\(-\ln \left (4 \ln \relax (2)-{\mathrm e}^{5}+x^{2}-6\right )\)	\(17\)
default	\(-\ln \left (6-x^{2}-4 \ln \relax (2)+{\mathrm e}^{5}\right )\)	\(17\)
norman	\(-\ln \left (6-x^{2}-4 \ln \relax (2)+{\mathrm e}^{5}\right )\)	\(17\)
risch	\(-\ln \left (4 \ln \relax (2)-{\mathrm e}^{5}+x^{2}-6\right )\)	\(17\)
meijerg	\(\frac {\left ({\mathrm e}^{5}-4 \ln \relax (2)+6\right ) \ln \left (1-\frac {x^{2}}{{\mathrm e}^{5}-4 \ln \relax (2)+6}\right )}{-{\mathrm e}^{5}+4 \ln \relax (2)-6}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-2*x/(4*ln(2)-exp(5)+x^2-6),x,method=_RETURNVERBOSE)

[Out]

-ln(4*ln(2)-exp(5)+x^2-6)

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maxima [A]  time = 0.36, size = 16, normalized size = 0.84 \begin {gather*} -\log \left (x^{2} - e^{5} + 4 \, \log \relax (2) - 6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*x/(4*log(2)-exp(5)+x^2-6),x, algorithm="maxima")

[Out]

-log(x^2 - e^5 + 4*log(2) - 6)

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mupad [B]  time = 0.18, size = 14, normalized size = 0.74 \begin {gather*} -\ln \left (x^2-{\mathrm {e}}^5+\ln \left (16\right )-6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x)/(exp(5) - 4*log(2) - x^2 + 6),x)

[Out]

-log(log(16) - exp(5) + x^2 - 6)

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sympy [A]  time = 0.18, size = 15, normalized size = 0.79 \begin {gather*} - \log {\left (x^{2} - e^{5} - 6 + 4 \log {\relax (2 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*x/(4*ln(2)-exp(5)+x**2-6),x)

[Out]

-log(x**2 - exp(5) - 6 + 4*log(2))

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