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

Optimal. Leaf size=21 \[ e^4 \left (4-(2+x) \log \left (\frac {x}{4+2 x}\right )\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.62, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {14, 2486, 31} \begin {gather*} -2 e^4 \log (x)-e^4 x \log \left (\frac {x}{2 (x+2)}\right )+2 e^4 \log (x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 31

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

Rule 2486

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.), x_Symbol] :> Simp[((
a + b*x)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/b, x] + Dist[(q*r*s*(b*c - a*d))/b, Int[Log[e*(f*(a + b*x)^p*
(c + d*x)^q)^r]^(s - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] &&
EqQ[p + q, 0] && IGtQ[s, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {2 e^4}{x}-e^4 \log \left (\frac {x}{4+2 x}\right )\right ) \, dx\\ &=-2 e^4 \log (x)-e^4 \int \log \left (\frac {x}{4+2 x}\right ) \, dx\\ &=-2 e^4 \log (x)-e^4 x \log \left (\frac {x}{2 (2+x)}\right )+\left (4 e^4\right ) \int \frac {1}{4+2 x} \, dx\\ &=-2 e^4 \log (x)-e^4 x \log \left (\frac {x}{2 (2+x)}\right )+2 e^4 \log (2+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 33, normalized size = 1.57 \begin {gather*} -2 e^4 \log (x)+2 e^4 \log (2+x)-e^4 x \log \left (\frac {x}{4+2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.70, size = 16, normalized size = 0.76 \begin {gather*} -{\left (x + 2\right )} e^{4} \log \left (\frac {x}{2 \, {\left (x + 2\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x + 2)*e^4*log(1/2*x/(x + 2))

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giac [A]  time = 0.16, size = 24, normalized size = 1.14 \begin {gather*} \frac {2 \, e^{4} \log \left (\frac {x}{2 \, {\left (x + 2\right )}}\right )}{\frac {x}{x + 2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*e^4*log(1/2*x/(x + 2))/(x/(x + 2) - 1)

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maple [A]  time = 0.13, size = 31, normalized size = 1.48

method	result	size
risch	\(-x \,{\mathrm e}^{4} \ln \left (\frac {x}{2 x +4}\right )-2 \,{\mathrm e}^{4} \ln \relax (x )+2 \,{\mathrm e}^{4} \ln \left (2+x \right )\)	\(31\)
norman	\(-2 \,{\mathrm e}^{4} \ln \left (\frac {x}{2 x +4}\right )-x \,{\mathrm e}^{4} \ln \left (\frac {x}{2 x +4}\right )\)	\(35\)
derivativedivides	\(-2 \,{\mathrm e}^{4} \ln \left (\frac {1}{2}-\frac {1}{2+x}\right ) \left (\frac {1}{2}-\frac {1}{2+x}\right ) \left (2+x \right )-2 \,{\mathrm e}^{4} \ln \left (\frac {1}{2}-\frac {1}{2+x}\right )\)	\(46\)
default	\(-2 \,{\mathrm e}^{4} \ln \left (\frac {1}{2}-\frac {1}{2+x}\right ) \left (\frac {1}{2}-\frac {1}{2+x}\right ) \left (2+x \right )-2 \,{\mathrm e}^{4} \ln \left (\frac {1}{2}-\frac {1}{2+x}\right )\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 29, normalized size = 1.38 \begin {gather*} -{\left (x \log \left (\frac {x}{2 \, {\left (x + 2\right )}}\right ) - 2 \, \log \left (x + 2\right )\right )} e^{4} - 2 \, e^{4} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x*log(1/2*x/(x + 2)) - 2*log(x + 2))*e^4 - 2*e^4*log(x)

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mupad [B]  time = 4.85, size = 17, normalized size = 0.81 \begin {gather*} -{\mathrm {e}}^4\,\ln \left (\frac {x}{2\,x+4}\right )\,\left (x+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(4)*log(x/(2*x + 4))*(x + 2)

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sympy [B]  time = 0.16, size = 31, normalized size = 1.48 \begin {gather*} - x e^{4} \log {\left (\frac {x}{2 x + 4} \right )} - 4 \left (\frac {\log {\relax (x )}}{2} - \frac {\log {\left (x + 2 \right )}}{2}\right ) e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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