3.63.48 \(\int -\frac {4 e^{224/9}}{-x+3 e^{224/9} x+2 e^{224/9} x \log (\frac {e^4}{x^2})} \, dx\)

Optimal. Leaf size=18 \[ \log \left (-3+\frac {1}{e^{224/9}}-2 \log \left (\frac {e^4}{x^2}\right )\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.39, number of steps used = 4, number of rules used = 3, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {6, 12, 31} \begin {gather*} \log \left (-2 e^{224/9} \log \left (\frac {e^4}{x^2}\right )-3 e^{224/9}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4*E^(224/9))/(-x + 3*E^(224/9)*x + 2*E^(224/9)*x*Log[E^4/x^2]),x]

[Out]

Log[1 - 3*E^(224/9) - 2*E^(224/9)*Log[E^4/x^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 31

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int -\frac {4 e^{224/9}}{\left (-1+3 e^{224/9}\right ) x+2 e^{224/9} x \log \left (\frac {e^4}{x^2}\right )} \, dx\\ &=-\left (\left (4 e^{224/9}\right ) \int \frac {1}{\left (-1+3 e^{224/9}\right ) x+2 e^{224/9} x \log \left (\frac {e^4}{x^2}\right )} \, dx\right )\\ &=\left (2 e^{224/9}\right ) \operatorname {Subst}\left (\int \frac {1}{-1+3 e^{224/9}+2 e^{224/9} x} \, dx,x,\log \left (\frac {e^4}{x^2}\right )\right )\\ &=\log \left (1-3 e^{224/9}-2 e^{224/9} \log \left (\frac {e^4}{x^2}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 21, normalized size = 1.17 \begin {gather*} \log \left (-1+11 e^{224/9}+2 e^{224/9} \log \left (\frac {1}{x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*E^(224/9))/(-x + 3*E^(224/9)*x + 2*E^(224/9)*x*Log[E^4/x^2]),x]

[Out]

Log[-1 + 11*E^(224/9) + 2*E^(224/9)*Log[x^(-2)]]

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fricas [A]  time = 0.55, size = 18, normalized size = 1.00 \begin {gather*} \log \left (2 \, e^{\frac {224}{9}} \log \left (\frac {e^{4}}{x^{2}}\right ) + 3 \, e^{\frac {224}{9}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*exp(224/9)/(2*x*exp(224/9)*log(exp(4)/x^2)+3*x*exp(224/9)-x),x, algorithm="fricas")

[Out]

log(2*e^(224/9)*log(e^4/x^2) + 3*e^(224/9) - 1)

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giac [A]  time = 0.12, size = 15, normalized size = 0.83 \begin {gather*} \log \left (-2 \, e^{\frac {224}{9}} \log \left (x^{2}\right ) + 11 \, e^{\frac {224}{9}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*exp(224/9)/(2*x*exp(224/9)*log(exp(4)/x^2)+3*x*exp(224/9)-x),x, algorithm="giac")

[Out]

log(-2*e^(224/9)*log(x^2) + 11*e^(224/9) - 1)

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maple [A]  time = 0.17, size = 16, normalized size = 0.89

method	result	size
derivativedivides	\(\ln \left (11 \,{\mathrm e}^{\frac {224}{9}}+2 \,{\mathrm e}^{\frac {224}{9}} \ln \left (\frac {1}{x^{2}}\right )-1\right )\)	\(16\)
default	\(\ln \left (11 \,{\mathrm e}^{\frac {224}{9}}+2 \,{\mathrm e}^{\frac {224}{9}} \ln \left (\frac {1}{x^{2}}\right )-1\right )\)	\(16\)
norman	\(\ln \left (2 \,{\mathrm e}^{\frac {224}{9}} \ln \left (\frac {{\mathrm e}^{4}}{x^{2}}\right )+3 \,{\mathrm e}^{\frac {224}{9}}-1\right )\)	\(19\)
risch	\(\ln \left (\ln \left (\frac {{\mathrm e}^{4}}{x^{2}}\right )+\frac {\left (3 \,{\mathrm e}^{\frac {224}{9}}-1\right ) {\mathrm e}^{-\frac {224}{9}}}{2}\right )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-4*exp(224/9)/(2*x*exp(224/9)*ln(exp(4)/x^2)+3*x*exp(224/9)-x),x,method=_RETURNVERBOSE)

[Out]

ln(11*exp(224/9)+2*exp(224/9)*ln(1/x^2)-1)

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maxima [A]  time = 0.38, size = 17, normalized size = 0.94 \begin {gather*} \log \left (\frac {1}{4} \, {\left (4 \, e^{\frac {224}{9}} \log \relax (x) - 11 \, e^{\frac {224}{9}} + 1\right )} e^{\left (-\frac {224}{9}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*exp(224/9)/(2*x*exp(224/9)*log(exp(4)/x^2)+3*x*exp(224/9)-x),x, algorithm="maxima")

[Out]

log(1/4*(4*e^(224/9)*log(x) - 11*e^(224/9) + 1)*e^(-224/9))

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mupad [B]  time = 4.49, size = 15, normalized size = 0.83 \begin {gather*} \ln \left (11\,{\mathrm {e}}^{224/9}+2\,\ln \left (\frac {1}{x^2}\right )\,{\mathrm {e}}^{224/9}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp(224/9))/(3*x*exp(224/9) - x + 2*x*log(exp(4)/x^2)*exp(224/9)),x)

[Out]

log(11*exp(224/9) + 2*log(1/x^2)*exp(224/9) - 1)

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sympy [A]  time = 0.12, size = 24, normalized size = 1.33 \begin {gather*} \log {\left (\log {\left (\frac {e^{4}}{x^{2}} \right )} + \frac {-1 + 3 e^{\frac {224}{9}}}{2 e^{\frac {224}{9}}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*exp(224/9)/(2*x*exp(224/9)*ln(exp(4)/x**2)+3*x*exp(224/9)-x),x)

[Out]

log(log(exp(4)/x**2) + (-1 + 3*exp(224/9))*exp(-224/9)/2)

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