\(\int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log (\frac {e^{-4-x} x^3}{10 \log (x)})} \, dx\) [4705]

Optimal result

Integrand size = 39, antiderivative size = 26 \[ \int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx=\log \left (\frac {1}{2} \left (x+\log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )\right )\right ) \]

[Out]

ln(1/2*ln(1/10*x^3/exp(4)/exp(x)/ln(x))+1/2*x)

Rubi [F]

\[ \int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx=\int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx=\log \left (x+\log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )\right ) \]

[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])]]

Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85

method	result	size
default	\(\ln \left (\ln \left (\frac {x^{3} {\mathrm e}^{-4} {\mathrm e}^{-x}}{10 \ln \left (x \right )}\right )+x \right )\)	\(22\)
parallelrisch	\(\ln \left (\ln \left (\frac {x^{3} {\mathrm e}^{-4} {\mathrm e}^{-x}}{10 \ln \left (x \right )}\right )+x \right )\)	\(22\)
risch	\(\ln \left (4+\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right )}{2}+\frac {i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \left (x \right )}\right )^{3}}{2}+\frac {i \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i x^{3} {\mathrm e}^{-x}}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (i x^{3}\right )}{2}-i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-\frac {i \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i x^{3} {\mathrm e}^{-x}}{\ln \left (x \right )}\right )^{2}}{2}+\frac {i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )}{2}-\frac {i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}}{2}+\frac {i \pi \operatorname {csgn}\left (\frac {i x^{3} {\mathrm e}^{-x}}{\ln \left (x \right )}\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i x^{3}\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{2}-\frac {i \pi \operatorname {csgn}\left (\frac {i x^{3} {\mathrm e}^{-x}}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (i x^{3}\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \left (x \right )}\right )^{2}}{2}-x +\ln \left ({\mathrm e}^{x}\right )+\ln \left (5\right )+\ln \left (2\right )+\ln \left (\ln \left (x \right )\right )-3 \ln \left (x \right )\right )\)	\(366\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx=\log \left (x + \log \left (\frac {x^{3} e^{\left (-x - 4\right )}}{10 \, \log \left (x\right )}\right )\right ) \]

[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)))

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx=\log {\left (x + \log {\left (\frac {x^{3} e^{- x}}{10 e^{4} \log {\left (x \right )}} \right )} \right )} \]

[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))))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.54 \[ \int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx=\log \left (\log \left (5\right ) + \log \left (2\right ) - 3 \, \log \left (x\right ) + \log \left (\log \left (x\right )\right ) + 4\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.46 \[ \int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx=\log \left (\log \left (10\right ) - 3 \, \log \left (x\right ) + \log \left (\log \left (x\right )\right ) + 4\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 11.96 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.50 \[ \int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx=\ln \left (\ln \left (\frac {x^3}{10\,\ln \left (x\right )}\right )-4\right ) \]

[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)