3.91.35 \(\int \frac {-e^{-8+4 x}-x-4 e^{-8+4 x} x \log (x)}{x^2+e^{-8+4 x} x \log (x)} \, dx\)

Optimal. Leaf size=17 \[ \log \left (\frac {4}{x+e^{4 (-2+x)} \log (x)}\right ) \]

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Rubi [F]  time = 0.77, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-e^{-8+4 x}-x-4 e^{-8+4 x} x \log (x)}{x^2+e^{-8+4 x} x \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-E^(-8 + 4*x) - x - 4*E^(-8 + 4*x)*x*Log[x])/(x^2 + E^(-8 + 4*x)*x*Log[x]),x]

[Out]

-4*x - Log[Log[x]] - E^8*Defer[Int][(E^8*x + E^(4*x)*Log[x])^(-1), x] + 4*E^8*Defer[Int][x/(E^8*x + E^(4*x)*Lo
g[x]), x] + E^8*Defer[Int][1/(Log[x]*(E^8*x + E^(4*x)*Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^{4 x}-e^8 x-4 e^{4 x} x \log (x)}{e^8 \left (x^2+e^{-8+4 x} x \log (x)\right )} \, dx\\ &=\frac {\int \frac {-e^{4 x}-e^8 x-4 e^{4 x} x \log (x)}{x^2+e^{-8+4 x} x \log (x)} \, dx}{e^8}\\ &=\frac {\int \left (-\frac {e^8 (1+4 x \log (x))}{x \log (x)}+\frac {e^{16} (1-\log (x)+4 x \log (x))}{\log (x) \left (e^8 x+e^{4 x} \log (x)\right )}\right ) \, dx}{e^8}\\ &=e^8 \int \frac {1-\log (x)+4 x \log (x)}{\log (x) \left (e^8 x+e^{4 x} \log (x)\right )} \, dx-\int \frac {1+4 x \log (x)}{x \log (x)} \, dx\\ &=e^8 \int \left (-\frac {1}{e^8 x+e^{4 x} \log (x)}+\frac {4 x}{e^8 x+e^{4 x} \log (x)}+\frac {1}{\log (x) \left (e^8 x+e^{4 x} \log (x)\right )}\right ) \, dx-\int \left (4+\frac {1}{x \log (x)}\right ) \, dx\\ &=-4 x-e^8 \int \frac {1}{e^8 x+e^{4 x} \log (x)} \, dx+e^8 \int \frac {1}{\log (x) \left (e^8 x+e^{4 x} \log (x)\right )} \, dx+\left (4 e^8\right ) \int \frac {x}{e^8 x+e^{4 x} \log (x)} \, dx-\int \frac {1}{x \log (x)} \, dx\\ &=-4 x-e^8 \int \frac {1}{e^8 x+e^{4 x} \log (x)} \, dx+e^8 \int \frac {1}{\log (x) \left (e^8 x+e^{4 x} \log (x)\right )} \, dx+\left (4 e^8\right ) \int \frac {x}{e^8 x+e^{4 x} \log (x)} \, dx-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=-4 x-\log (\log (x))-e^8 \int \frac {1}{e^8 x+e^{4 x} \log (x)} \, dx+e^8 \int \frac {1}{\log (x) \left (e^8 x+e^{4 x} \log (x)\right )} \, dx+\left (4 e^8\right ) \int \frac {x}{e^8 x+e^{4 x} \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 17, normalized size = 1.00 \begin {gather*} -\log \left (e^8 x+e^{4 x} \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^(-8 + 4*x) - x - 4*E^(-8 + 4*x)*x*Log[x])/(x^2 + E^(-8 + 4*x)*x*Log[x]),x]

[Out]

-Log[E^8*x + E^(4*x)*Log[x]]

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fricas [A]  time = 0.43, size = 25, normalized size = 1.47 \begin {gather*} -4 \, x - \log \left ({\left (e^{\left (4 \, x - 8\right )} \log \relax (x) + x\right )} e^{\left (-4 \, x + 8\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*exp(4*x-8)*log(x)-exp(4*x-8)-x)/(x*exp(4*x-8)*log(x)+x^2),x, algorithm="fricas")

[Out]

-4*x - log((e^(4*x - 8)*log(x) + x)*e^(-4*x + 8))

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giac [A]  time = 0.17, size = 15, normalized size = 0.88 \begin {gather*} -\log \left (x e^{8} + e^{\left (4 \, x\right )} \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*exp(4*x-8)*log(x)-exp(4*x-8)-x)/(x*exp(4*x-8)*log(x)+x^2),x, algorithm="giac")

[Out]

-log(x*e^8 + e^(4*x)*log(x))

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maple [A]  time = 0.04, size = 15, normalized size = 0.88

method	result	size
norman	\(-\ln \left (x +\ln \relax (x ) {\mathrm e}^{4 x -8}\right )\)	\(15\)
risch	\(-4 x -\ln \left (\ln \relax (x )+x \,{\mathrm e}^{-4 x +8}\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x*exp(4*x-8)*ln(x)-exp(4*x-8)-x)/(x*exp(4*x-8)*ln(x)+x^2),x,method=_RETURNVERBOSE)

[Out]

-ln(x+ln(x)*exp(4*x-8))

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maxima [A]  time = 0.37, size = 26, normalized size = 1.53 \begin {gather*} -\log \left (\frac {x e^{8} + e^{\left (4 \, x\right )} \log \relax (x)}{\log \relax (x)}\right ) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*exp(4*x-8)*log(x)-exp(4*x-8)-x)/(x*exp(4*x-8)*log(x)+x^2),x, algorithm="maxima")

[Out]

-log((x*e^8 + e^(4*x)*log(x))/log(x)) - log(log(x))

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mupad [B]  time = 7.27, size = 14, normalized size = 0.82 \begin {gather*} -\ln \left (x+{\mathrm {e}}^{4\,x-8}\,\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x + exp(4*x - 8) + 4*x*exp(4*x - 8)*log(x))/(x^2 + x*exp(4*x - 8)*log(x)),x)

[Out]

-log(x + exp(4*x - 8)*log(x))

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sympy [A]  time = 0.36, size = 19, normalized size = 1.12 \begin {gather*} - \log {\left (\frac {x}{\log {\relax (x )}} + e^{4 x - 8} \right )} - \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*exp(4*x-8)*ln(x)-exp(4*x-8)-x)/(x*exp(4*x-8)*ln(x)+x**2),x)

[Out]

-log(x/log(x) + exp(4*x - 8)) - log(log(x))

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