3.92.79 \(\int \frac {-6 x+2 x \log (3)}{e^4+e^{e^5}} \, dx\)

Optimal. Leaf size=25 \[ \frac {\left (-\frac {1}{e^3}+x^2\right ) (-3+\log (3))}{e^4+e^{e^5}} \]

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 0.88, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6, 12, 30} \begin {gather*} -\frac {x^2 (3-\log (3))}{e^4+e^{e^5}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6*x + 2*x*Log[3])/(E^4 + E^E^5),x]

[Out]

-((x^2*(3 - Log[3]))/(E^4 + E^E^5))

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x (-6+2 \log (3))}{e^4+e^{e^5}} \, dx\\ &=\frac {(-6+2 \log (3)) \int x \, dx}{e^4+e^{e^5}}\\ &=-\frac {x^2 (3-\log (3))}{e^4+e^{e^5}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 19, normalized size = 0.76 \begin {gather*} \frac {x^2 (-3+\log (3))}{e^4+e^{e^5}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6*x + 2*x*Log[3])/(E^4 + E^E^5),x]

[Out]

(x^2*(-3 + Log[3]))/(E^4 + E^E^5)

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fricas [A]  time = 0.52, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^{2} \log \relax (3) - 3 \, x^{2}}{e^{4} + e^{\left (e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(3)-6*x)/(exp(exp(5))+exp(4)),x, algorithm="fricas")

[Out]

(x^2*log(3) - 3*x^2)/(e^4 + e^(e^5))

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giac [A]  time = 0.21, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^{2} \log \relax (3) - 3 \, x^{2}}{e^{4} + e^{\left (e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(3)-6*x)/(exp(exp(5))+exp(4)),x, algorithm="giac")

[Out]

(x^2*log(3) - 3*x^2)/(e^4 + e^(e^5))

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maple [A]  time = 0.05, size = 17, normalized size = 0.68

method	result	size
gosper	\(\frac {x^{2} \left (\ln \relax (3)-3\right )}{{\mathrm e}^{{\mathrm e}^{5}}+{\mathrm e}^{4}}\)	\(17\)
norman	\(\frac {x^{2} \left (\ln \relax (3)-3\right )}{{\mathrm e}^{{\mathrm e}^{5}}+{\mathrm e}^{4}}\)	\(17\)
default	\(\frac {x^{2} \ln \relax (3)-3 x^{2}}{{\mathrm e}^{{\mathrm e}^{5}}+{\mathrm e}^{4}}\)	\(22\)
risch	\(\frac {x^{2} \ln \relax (3)}{{\mathrm e}^{{\mathrm e}^{5}}+{\mathrm e}^{4}}-\frac {3 x^{2}}{{\mathrm e}^{{\mathrm e}^{5}}+{\mathrm e}^{4}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*ln(3)-6*x)/(exp(exp(5))+exp(4)),x,method=_RETURNVERBOSE)

[Out]

x^2*(ln(3)-3)/(exp(exp(5))+exp(4))

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maxima [A]  time = 0.35, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^{2} \log \relax (3) - 3 \, x^{2}}{e^{4} + e^{\left (e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(3)-6*x)/(exp(exp(5))+exp(4)),x, algorithm="maxima")

[Out]

(x^2*log(3) - 3*x^2)/(e^4 + e^(e^5))

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mupad [B]  time = 7.14, size = 16, normalized size = 0.64 \begin {gather*} \frac {x^2\,\left (\ln \relax (3)-3\right )}{{\mathrm {e}}^4+{\mathrm {e}}^{{\mathrm {e}}^5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(6*x - 2*x*log(3))/(exp(4) + exp(exp(5))),x)

[Out]

(x^2*(log(3) - 3))/(exp(4) + exp(exp(5)))

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sympy [A]  time = 0.06, size = 15, normalized size = 0.60 \begin {gather*} \frac {x^{2} \left (-3 + \log {\relax (3 )}\right )}{e^{4} + e^{e^{5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*ln(3)-6*x)/(exp(exp(5))+exp(4)),x)

[Out]

x**2*(-3 + log(3))/(exp(4) + exp(exp(5)))

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