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

Optimal. Leaf size=25 \[ \log \left (5-\frac {e^{\frac {e^3}{x}}+x}{x \log (6 x)}\right ) \]

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Rubi [F]  time = 2.70, 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^{\frac {e^3}{x}} x+x^2+e^{\frac {e^3}{x}} \left (e^3+x\right ) \log (6 x)}{\left (-e^{\frac {e^3}{x}} x^2-x^3\right ) \log (6 x)+5 x^3 \log ^2(6 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

E^3/x - Log[x] - Log[Log[6*x]] - 4*Defer[Int][(E^(E^3/x) + x - 5*x*Log[6*x])^(-1), x] - E^3*Defer[Int][1/(x*(-
E^(E^3/x) - x + 5*x*Log[6*x])), x] + 5*Defer[Int][Log[6*x]/(-E^(E^3/x) - x + 5*x*Log[6*x]), x] + 5*E^3*Defer[I
nt][Log[6*x]/(x*(-E^(E^3/x) - x + 5*x*Log[6*x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^{\frac {e^3}{x}} x-x^2-e^{\frac {e^3}{x}} \left (e^3+x\right ) \log (6 x)}{x^2 \log (6 x) \left (e^{\frac {e^3}{x}}+x-5 x \log (6 x)\right )} \, dx\\ &=\int \left (\frac {-x-e^3 \log (6 x)-x \log (6 x)}{x^2 \log (6 x)}+\frac {-e^3+4 x+5 e^3 \log (6 x)+5 x \log (6 x)}{x \left (-e^{\frac {e^3}{x}}-x+5 x \log (6 x)\right )}\right ) \, dx\\ &=\int \frac {-x-e^3 \log (6 x)-x \log (6 x)}{x^2 \log (6 x)} \, dx+\int \frac {-e^3+4 x+5 e^3 \log (6 x)+5 x \log (6 x)}{x \left (-e^{\frac {e^3}{x}}-x+5 x \log (6 x)\right )} \, dx\\ &=\int \left (-\frac {e^3+x}{x^2}-\frac {1}{x \log (6 x)}\right ) \, dx+\int \left (-\frac {4}{e^{\frac {e^3}{x}}+x-5 x \log (6 x)}-\frac {e^3}{x \left (-e^{\frac {e^3}{x}}-x+5 x \log (6 x)\right )}+\frac {5 \log (6 x)}{-e^{\frac {e^3}{x}}-x+5 x \log (6 x)}+\frac {5 e^3 \log (6 x)}{x \left (-e^{\frac {e^3}{x}}-x+5 x \log (6 x)\right )}\right ) \, dx\\ &=-\left (4 \int \frac {1}{e^{\frac {e^3}{x}}+x-5 x \log (6 x)} \, dx\right )+5 \int \frac {\log (6 x)}{-e^{\frac {e^3}{x}}-x+5 x \log (6 x)} \, dx-e^3 \int \frac {1}{x \left (-e^{\frac {e^3}{x}}-x+5 x \log (6 x)\right )} \, dx+\left (5 e^3\right ) \int \frac {\log (6 x)}{x \left (-e^{\frac {e^3}{x}}-x+5 x \log (6 x)\right )} \, dx-\int \frac {e^3+x}{x^2} \, dx-\int \frac {1}{x \log (6 x)} \, dx\\ &=-\left (4 \int \frac {1}{e^{\frac {e^3}{x}}+x-5 x \log (6 x)} \, dx\right )+5 \int \frac {\log (6 x)}{-e^{\frac {e^3}{x}}-x+5 x \log (6 x)} \, dx-e^3 \int \frac {1}{x \left (-e^{\frac {e^3}{x}}-x+5 x \log (6 x)\right )} \, dx+\left (5 e^3\right ) \int \frac {\log (6 x)}{x \left (-e^{\frac {e^3}{x}}-x+5 x \log (6 x)\right )} \, dx-\int \left (\frac {e^3}{x^2}+\frac {1}{x}\right ) \, dx-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (6 x)\right )\\ &=\frac {e^3}{x}-\log (x)-\log (\log (6 x))-4 \int \frac {1}{e^{\frac {e^3}{x}}+x-5 x \log (6 x)} \, dx+5 \int \frac {\log (6 x)}{-e^{\frac {e^3}{x}}-x+5 x \log (6 x)} \, dx-e^3 \int \frac {1}{x \left (-e^{\frac {e^3}{x}}-x+5 x \log (6 x)\right )} \, dx+\left (5 e^3\right ) \int \frac {\log (6 x)}{x \left (-e^{\frac {e^3}{x}}-x+5 x \log (6 x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.47, size = 31, normalized size = 1.24 \begin {gather*} -\log (x)-\log (\log (6 x))+\log \left (e^{\frac {e^3}{x}}+x-5 x \log (6 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[x] - Log[Log[6*x]] + Log[E^(E^3/x) + x - 5*x*Log[6*x]]

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fricas [A]  time = 0.63, size = 33, normalized size = 1.32 \begin {gather*} \log \left (\frac {5 \, x \log \left (6 \, x\right ) - x - e^{\left (\frac {e^{3}}{x}\right )}}{x}\right ) - \log \left (\log \left (6 \, x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((5*x*log(6*x) - x - e^(e^3/x))/x) - log(log(6*x))

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giac [A]  time = 0.18, size = 41, normalized size = 1.64 \begin {gather*} \log \left (5 \, x e^{3} \log \left (6 \, x\right ) - x e^{3} - e^{\left (\frac {3 \, x + e^{3}}{x}\right )}\right ) - \log \relax (x) - \log \left (\log \left (6 \, x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(5*x*e^3*log(6*x) - x*e^3 - e^((3*x + e^3)/x)) - log(x) - log(log(6*x))

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maple [A]  time = 0.04, size = 29, normalized size = 1.16

method	result	size
risch	\(-\ln \left (\ln \left (6 x \right )\right )+\ln \left (\ln \left (6 x \right )-\frac {x +{\mathrm e}^{\frac {{\mathrm e}^{3}}{x}}}{5 x}\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(ln(6*x))+ln(ln(6*x)-1/5*(x+exp(exp(3)/x))/x)

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maxima [A]  time = 0.49, size = 42, normalized size = 1.68 \begin {gather*} \log \left (-x {\left (5 \, \log \relax (3) + 5 \, \log \relax (2) - 1\right )} - 5 \, x \log \relax (x) + e^{\left (\frac {e^{3}}{x}\right )}\right ) - \log \relax (x) - \log \left (\log \relax (3) + \log \relax (2) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-x*(5*log(3) + 5*log(2) - 1) - 5*x*log(x) + e^(e^3/x)) - log(x) - log(log(3) + log(2) + log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {x^2+x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^3}{x}}+\ln \left (6\,x\right )\,{\mathrm {e}}^{\frac {{\mathrm {e}}^3}{x}}\,\left (x+{\mathrm {e}}^3\right )}{\ln \left (6\,x\right )\,\left (x^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^3}{x}}+x^3\right )-5\,x^3\,{\ln \left (6\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.38, size = 27, normalized size = 1.08 \begin {gather*} - \log {\relax (x )} + \log {\left (- 5 x \log {\left (6 x \right )} + x + e^{\frac {e^{3}}{x}} \right )} - \log {\left (\log {\left (6 x \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x) + log(-5*x*log(6*x) + x + exp(exp(3)/x)) - log(log(6*x))

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