3.48.5 \(\int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log (\frac {e^{-4-x} x^3}{10 \log (x)})} \, dx\)

Optimal. Leaf size=26 \[ \log \left (\frac {1}{2} \left (x+\log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )\right )\right ) \]

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Rubi [F]  time = 0.37, 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 {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1 + 3*Log[x])/(x^2*Log[x] + x*Log[x]*Log[(E^(-4 - x)*x^3)/(10*Log[x])]),x]

[Out]

3*Defer[Int][1/(x*(x + Log[(E^(-4 - x)*x^3)/(10*Log[x])])), x] - Defer[Int][1/(x*Log[x]*(x + Log[(E^(-4 - x)*x
^3)/(10*Log[x])])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1+3 \log (x)}{x \log (x) \left (x+\log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )\right )} \, dx\\ &=\int \left (\frac {3}{x \left (x+\log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )\right )}-\frac {1}{x \log (x) \left (x+\log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )\right )}\right ) \, dx\\ &=3 \int \frac {1}{x \left (x+\log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )\right )} \, dx-\int \frac {1}{x \log (x) \left (x+\log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 22, normalized size = 0.85 \begin {gather*} \log \left (x+\log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 3*Log[x])/(x^2*Log[x] + x*Log[x]*Log[(E^(-4 - x)*x^3)/(10*Log[x])]),x]

[Out]

Log[x + Log[(E^(-4 - x)*x^3)/(10*Log[x])]]

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fricas [A]  time = 0.95, size = 19, normalized size = 0.73 \begin {gather*} \log \left (x + \log \left (\frac {x^{3} e^{\left (-x - 4\right )}}{10 \, \log \relax (x)}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*log(x)-1)/(x*log(x)*log(1/10*x^3/exp(4)/exp(x)/log(x))+x^2*log(x)),x, algorithm="fricas")

[Out]

log(x + log(1/10*x^3*e^(-x - 4)/log(x)))

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giac [A]  time = 0.19, size = 12, normalized size = 0.46 \begin {gather*} \log \left (\log \left (10\right ) - 3 \, \log \relax (x) + \log \left (\log \relax (x)\right ) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*log(x)-1)/(x*log(x)*log(1/10*x^3/exp(4)/exp(x)/log(x))+x^2*log(x)),x, algorithm="giac")

[Out]

log(log(10) - 3*log(x) + log(log(x)) + 4)

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maple [A]  time = 0.35, size = 22, normalized size = 0.85

method	result	size
default	\(\ln \left (\ln \left (\frac {x^{3} {\mathrm e}^{-4} {\mathrm e}^{-x}}{10 \ln \relax (x )}\right )+x \right )\)	\(22\)
risch	\(\ln \left (\ln \left ({\mathrm e}^{x}\right )+\frac {i \left (-8 i+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+\pi \,\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i x^{3} {\mathrm e}^{-x}}{\ln \relax (x )}\right )+\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \relax (x )}\right )+6 i \ln \relax (x )+2 i x +\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \relax (x )}\right )^{3}+\pi \mathrm {csgn}\left (i x^{3}\right )^{3}+\pi \mathrm {csgn}\left (\frac {i x^{3} {\mathrm e}^{-x}}{\ln \relax (x )}\right )^{3}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \relax (x )}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \relax (x )}\right )^{2}-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}-\pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}-\pi \,\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (\frac {i x^{3} {\mathrm e}^{-x}}{\ln \relax (x )}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i x^{3} {\mathrm e}^{-x}}{\ln \relax (x )}\right )^{2}+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-2 i \ln \relax (5)-2 i \ln \relax (2)-2 i \ln \left (\ln \relax (x )\right )\right )}{2}\right )\)	\(359\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*ln(x)-1)/(x*ln(x)*ln(1/10*x^3/exp(4)/exp(x)/ln(x))+x^2*ln(x)),x,method=_RETURNVERBOSE)

[Out]

ln(ln(1/10*x^3/exp(4)/exp(x)/ln(x))+x)

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maxima [A]  time = 0.53, size = 14, normalized size = 0.54 \begin {gather*} \log \left (\log \relax (5) + \log \relax (2) - 3 \, \log \relax (x) + \log \left (\log \relax (x)\right ) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*log(x)-1)/(x*log(x)*log(1/10*x^3/exp(4)/exp(x)/log(x))+x^2*log(x)),x, algorithm="maxima")

[Out]

log(log(5) + log(2) - 3*log(x) + log(log(x)) + 4)

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mupad [B]  time = 3.58, size = 13, normalized size = 0.50 \begin {gather*} \ln \left (\ln \left (\frac {x^3}{10\,\ln \relax (x)}\right )-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*log(x) - 1)/(x^2*log(x) + x*log((x^3*exp(-x)*exp(-4))/(10*log(x)))*log(x)),x)

[Out]

log(log(x^3/(10*log(x))) - 4)

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sympy [A]  time = 0.36, size = 19, normalized size = 0.73 \begin {gather*} \log {\left (x + \log {\left (\frac {x^{3} e^{- x}}{10 e^{4} \log {\relax (x )}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*ln(x)-1)/(x*ln(x)*ln(1/10*x**3/exp(4)/exp(x)/ln(x))+x**2*ln(x)),x)

[Out]

log(x + log(x**3*exp(-4)*exp(-x)/(10*log(x))))

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