3.11.47 \(\int \frac {8 e^{6+2 e^3} x^4-48 e^{6+2 e^3} x^3 \log (x)+8 e^{6+2 e^3} x^4 \log (x) \log (\frac {3}{\log (x)})}{-27 \log (x)+27 x \log (x) \log (\frac {3}{\log (x)})-9 x^2 \log (x) \log ^2(\frac {3}{\log (x)})+x^3 \log (x) \log ^3(\frac {3}{\log (x)})} \, dx\)

Optimal. Leaf size=31 \[ 4+\frac {4 e^{6+2 e^3} x^2}{\left (-\frac {3}{x}+\log \left (\frac {3}{\log (x)}\right )\right )^2} \]

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Rubi [F]  time = 1.30, 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 {8 e^{6+2 e^3} x^4-48 e^{6+2 e^3} x^3 \log (x)+8 e^{6+2 e^3} x^4 \log (x) \log \left (\frac {3}{\log (x)}\right )}{-27 \log (x)+27 x \log (x) \log \left (\frac {3}{\log (x)}\right )-9 x^2 \log (x) \log ^2\left (\frac {3}{\log (x)}\right )+x^3 \log (x) \log ^3\left (\frac {3}{\log (x)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(8*E^(6 + 2*E^3)*x^4 - 48*E^(6 + 2*E^3)*x^3*Log[x] + 8*E^(6 + 2*E^3)*x^4*Log[x]*Log[3/Log[x]])/(-27*Log[x]
 + 27*x*Log[x]*Log[3/Log[x]] - 9*x^2*Log[x]*Log[3/Log[x]]^2 + x^3*Log[x]*Log[3/Log[x]]^3),x]

[Out]

-24*E^(6 + 2*E^3)*Defer[Int][x^3/(-3 + x*Log[3/Log[x]])^3, x] + 8*E^(6 + 2*E^3)*Defer[Int][x^4/(Log[x]*(-3 + x
*Log[3/Log[x]])^3), x] + 8*E^(6 + 2*E^3)*Defer[Int][x^3/(-3 + x*Log[3/Log[x]])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8 e^{6+2 e^3} x^3 \left (-x-\log (x) \left (-6+x \log \left (\frac {3}{\log (x)}\right )\right )\right )}{\log (x) \left (3-x \log \left (\frac {3}{\log (x)}\right )\right )^3} \, dx\\ &=\left (8 e^{6+2 e^3}\right ) \int \frac {x^3 \left (-x-\log (x) \left (-6+x \log \left (\frac {3}{\log (x)}\right )\right )\right )}{\log (x) \left (3-x \log \left (\frac {3}{\log (x)}\right )\right )^3} \, dx\\ &=\left (8 e^{6+2 e^3}\right ) \int \left (\frac {x^3 (x-3 \log (x))}{\log (x) \left (-3+x \log \left (\frac {3}{\log (x)}\right )\right )^3}+\frac {x^3}{\left (-3+x \log \left (\frac {3}{\log (x)}\right )\right )^2}\right ) \, dx\\ &=\left (8 e^{6+2 e^3}\right ) \int \frac {x^3 (x-3 \log (x))}{\log (x) \left (-3+x \log \left (\frac {3}{\log (x)}\right )\right )^3} \, dx+\left (8 e^{6+2 e^3}\right ) \int \frac {x^3}{\left (-3+x \log \left (\frac {3}{\log (x)}\right )\right )^2} \, dx\\ &=\left (8 e^{6+2 e^3}\right ) \int \frac {x^3}{\left (-3+x \log \left (\frac {3}{\log (x)}\right )\right )^2} \, dx+\left (8 e^{6+2 e^3}\right ) \int \left (-\frac {3 x^3}{\left (-3+x \log \left (\frac {3}{\log (x)}\right )\right )^3}+\frac {x^4}{\log (x) \left (-3+x \log \left (\frac {3}{\log (x)}\right )\right )^3}\right ) \, dx\\ &=\left (8 e^{6+2 e^3}\right ) \int \frac {x^4}{\log (x) \left (-3+x \log \left (\frac {3}{\log (x)}\right )\right )^3} \, dx+\left (8 e^{6+2 e^3}\right ) \int \frac {x^3}{\left (-3+x \log \left (\frac {3}{\log (x)}\right )\right )^2} \, dx-\left (24 e^{6+2 e^3}\right ) \int \frac {x^3}{\left (-3+x \log \left (\frac {3}{\log (x)}\right )\right )^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.36, size = 27, normalized size = 0.87 \begin {gather*} \frac {4 e^{6+2 e^3} x^4}{\left (-3+x \log \left (\frac {3}{\log (x)}\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*E^(6 + 2*E^3)*x^4 - 48*E^(6 + 2*E^3)*x^3*Log[x] + 8*E^(6 + 2*E^3)*x^4*Log[x]*Log[3/Log[x]])/(-27*
Log[x] + 27*x*Log[x]*Log[3/Log[x]] - 9*x^2*Log[x]*Log[3/Log[x]]^2 + x^3*Log[x]*Log[3/Log[x]]^3),x]

[Out]

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

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fricas [A]  time = 0.61, size = 39, normalized size = 1.26 \begin {gather*} \frac {4 \, x^{4} e^{\left (2 \, e^{3} + 6\right )}}{x^{2} \log \left (\frac {3}{\log \relax (x)}\right )^{2} - 6 \, x \log \left (\frac {3}{\log \relax (x)}\right ) + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^4*exp(exp(3)+3)^2*log(x)*log(3/log(x))-48*x^3*exp(exp(3)+3)^2*log(x)+8*x^4*exp(exp(3)+3)^2)/(x^
3*log(x)*log(3/log(x))^3-9*x^2*log(x)*log(3/log(x))^2+27*x*log(x)*log(3/log(x))-27*log(x)),x, algorithm="frica
s")

[Out]

4*x^4*e^(2*e^3 + 6)/(x^2*log(3/log(x))^2 - 6*x*log(3/log(x)) + 9)

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giac [A]  time = 0.64, size = 54, normalized size = 1.74 \begin {gather*} \frac {4 \, x^{4} e^{\left (2 \, e^{3} + 6\right )}}{x^{2} \log \relax (3)^{2} - 2 \, x^{2} \log \relax (3) \log \left (\log \relax (x)\right ) + x^{2} \log \left (\log \relax (x)\right )^{2} - 6 \, x \log \relax (3) + 6 \, x \log \left (\log \relax (x)\right ) + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^4*exp(exp(3)+3)^2*log(x)*log(3/log(x))-48*x^3*exp(exp(3)+3)^2*log(x)+8*x^4*exp(exp(3)+3)^2)/(x^
3*log(x)*log(3/log(x))^3-9*x^2*log(x)*log(3/log(x))^2+27*x*log(x)*log(3/log(x))-27*log(x)),x, algorithm="giac"
)

[Out]

4*x^4*e^(2*e^3 + 6)/(x^2*log(3)^2 - 2*x^2*log(3)*log(log(x)) + x^2*log(log(x))^2 - 6*x*log(3) + 6*x*log(log(x)
) + 9)

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maple [C]  time = 0.46, size = 31, normalized size = 1.00




method result size



risch \(-\frac {16 x^{4} {\mathrm e}^{2 \,{\mathrm e}^{3}+6}}{\left (2 i x \ln \relax (3)-2 i x \ln \left (\ln \relax (x )\right )-6 i\right )^{2}}\) \(31\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^4*exp(exp(3)+3)^2*ln(x)*ln(3/ln(x))-48*x^3*exp(exp(3)+3)^2*ln(x)+8*x^4*exp(exp(3)+3)^2)/(x^3*ln(x)*ln
(3/ln(x))^3-9*x^2*ln(x)*ln(3/ln(x))^2+27*x*ln(x)*ln(3/ln(x))-27*ln(x)),x,method=_RETURNVERBOSE)

[Out]

-16/(2*I*x*ln(3)-2*I*x*ln(ln(x))-6*I)^2*x^4*exp(2*exp(3)+6)

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maxima [A]  time = 0.59, size = 53, normalized size = 1.71 \begin {gather*} \frac {4 \, x^{4} e^{\left (2 \, e^{3} + 6\right )}}{x^{2} \log \relax (3)^{2} + x^{2} \log \left (\log \relax (x)\right )^{2} - 6 \, x \log \relax (3) - 2 \, {\left (x^{2} \log \relax (3) - 3 \, x\right )} \log \left (\log \relax (x)\right ) + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^4*exp(exp(3)+3)^2*log(x)*log(3/log(x))-48*x^3*exp(exp(3)+3)^2*log(x)+8*x^4*exp(exp(3)+3)^2)/(x^
3*log(x)*log(3/log(x))^3-9*x^2*log(x)*log(3/log(x))^2+27*x*log(x)*log(3/log(x))-27*log(x)),x, algorithm="maxim
a")

[Out]

4*x^4*e^(2*e^3 + 6)/(x^2*log(3)^2 + x^2*log(log(x))^2 - 6*x*log(3) - 2*(x^2*log(3) - 3*x)*log(log(x)) + 9)

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mupad [B]  time = 1.18, size = 74, normalized size = 2.39 \begin {gather*} \frac {4\,\left (3\,x^4\,{\mathrm {e}}^{2\,{\mathrm {e}}^3+6}\,{\ln \relax (x)}^2-x^5\,{\mathrm {e}}^{2\,{\mathrm {e}}^3+6}\,\ln \relax (x)\right )}{\left (3\,{\ln \relax (x)}^2-x\,\ln \relax (x)\right )\,\left (x^2\,{\ln \left (\frac {3}{\ln \relax (x)}\right )}^2-6\,x\,\ln \left (\frac {3}{\ln \relax (x)}\right )+9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x^4*exp(2*exp(3) + 6) - 48*x^3*exp(2*exp(3) + 6)*log(x) + 8*x^4*log(3/log(x))*exp(2*exp(3) + 6)*log(x)
)/(27*log(x) + 9*x^2*log(3/log(x))^2*log(x) - x^3*log(3/log(x))^3*log(x) - 27*x*log(3/log(x))*log(x)),x)

[Out]

(4*(3*x^4*exp(2*exp(3) + 6)*log(x)^2 - x^5*exp(2*exp(3) + 6)*log(x)))/((3*log(x)^2 - x*log(x))*(x^2*log(3/log(
x))^2 - 6*x*log(3/log(x)) + 9))

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sympy [A]  time = 0.32, size = 37, normalized size = 1.19 \begin {gather*} \frac {4 x^{4} e^{6} e^{2 e^{3}}}{x^{2} \log {\left (\frac {3}{\log {\relax (x )}} \right )}^{2} - 6 x \log {\left (\frac {3}{\log {\relax (x )}} \right )} + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x**4*exp(exp(3)+3)**2*ln(x)*ln(3/ln(x))-48*x**3*exp(exp(3)+3)**2*ln(x)+8*x**4*exp(exp(3)+3)**2)/(
x**3*ln(x)*ln(3/ln(x))**3-9*x**2*ln(x)*ln(3/ln(x))**2+27*x*ln(x)*ln(3/ln(x))-27*ln(x)),x)

[Out]

4*x**4*exp(6)*exp(2*exp(3))/(x**2*log(3/log(x))**2 - 6*x*log(3/log(x)) + 9)

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