\(\int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx\) [8242]

Optimal result

Integrand size = 32, antiderivative size = 19 \[ \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx=6 x-\log (4) \log \left (\frac {5}{x}-\log (x)\right ) \]

[Out]

6*x-2*ln(2)*ln(5/x-ln(x))

Rubi [F]

\[ \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx=\int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx \]

[In]

Int[(-30*x + (-5 - x)*Log[4] + 6*x^2*Log[x])/(-5*x + x^2*Log[x]),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{x^2 \left (-\frac {5}{x}+\log (x)\right )} \, dx \\ & = \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{x (-5+x \log (x))} \, dx \\ & = \int \left (6-\frac {(5+x) \log (4)}{x (-5+x \log (x))}\right ) \, dx \\ & = 6 x-\log (4) \int \frac {5+x}{x (-5+x \log (x))} \, dx \\ & = 6 x-\log (4) \int \left (\frac {1}{-5+x \log (x)}+\frac {5}{x (-5+x \log (x))}\right ) \, dx \\ & = 6 x-\log (4) \int \frac {1}{-5+x \log (x)} \, dx-(5 \log (4)) \int \frac {1}{x (-5+x \log (x))} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx=6 x+\log (4) \log (x)-\log (4) \log (5-x \log (x)) \]

[In]

Integrate[(-30*x + (-5 - x)*Log[4] + 6*x^2*Log[x])/(-5*x + x^2*Log[x]),x]

[Out]

6*x + Log[4]*Log[x] - Log[4]*Log[5 - x*Log[x]]

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

method	result	size
risch	\(6 x -2 \ln \left (2\right ) \ln \left (\ln \left (x \right )-\frac {5}{x}\right )\)	\(18\)
default	\(6 x -2 \ln \left (2\right ) \left (-\ln \left (x \right )+\ln \left (x \ln \left (x \right )-5\right )\right )\)	\(21\)
norman	\(2 \ln \left (2\right ) \ln \left (x \right )+6 x -2 \ln \left (2\right ) \ln \left (x \ln \left (x \right )-5\right )\)	\(22\)
parallelrisch	\(2 \ln \left (2\right ) \ln \left (x \right )+6 x -2 \ln \left (2\right ) \ln \left (x \ln \left (x \right )-5\right )\)	\(22\)

[In]

int((6*x^2*ln(x)+2*(-x-5)*ln(2)-30*x)/(x^2*ln(x)-5*x),x,method=_RETURNVERBOSE)

[Out]

6*x-2*ln(2)*ln(ln(x)-5/x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx=-2 \, \log \left (2\right ) \log \left (\frac {x \log \left (x\right ) - 5}{x}\right ) + 6 \, x \]

[In]

integrate((6*x^2*log(x)+2*(-x-5)*log(2)-30*x)/(x^2*log(x)-5*x),x, algorithm="fricas")

[Out]

-2*log(2)*log((x*log(x) - 5)/x) + 6*x

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx=6 x - 2 \log {\left (2 \right )} \log {\left (\log {\left (x \right )} - \frac {5}{x} \right )} \]

[In]

integrate((6*x**2*ln(x)+2*(-x-5)*ln(2)-30*x)/(x**2*ln(x)-5*x),x)

[Out]

6*x - 2*log(2)*log(log(x) - 5/x)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx=-2 \, \log \left (2\right ) \log \left (\frac {x \log \left (x\right ) - 5}{x}\right ) + 6 \, x \]

[In]

integrate((6*x^2*log(x)+2*(-x-5)*log(2)-30*x)/(x^2*log(x)-5*x),x, algorithm="maxima")

[Out]

-2*log(2)*log((x*log(x) - 5)/x) + 6*x

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx=-2 \, \log \left (2\right ) \log \left (x \log \left (x\right ) - 5\right ) + 2 \, \log \left (2\right ) \log \left (x\right ) + 6 \, x \]

[In]

integrate((6*x^2*log(x)+2*(-x-5)*log(2)-30*x)/(x^2*log(x)-5*x),x, algorithm="giac")

[Out]

-2*log(2)*log(x*log(x) - 5) + 2*log(2)*log(x) + 6*x

Mupad [B] (verification not implemented)

Time = 12.83 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {-30 x+(-5-x) \log (4)+6 x^2 \log (x)}{-5 x+x^2 \log (x)} \, dx=6\,x-2\,\ln \left (2\right )\,\ln \left (x\,\ln \left (x\right )-5\right )+2\,\ln \left (2\right )\,\ln \left (x\right ) \]

[In]

int((30*x - 6*x^2*log(x) + 2*log(2)*(x + 5))/(5*x - x^2*log(x)),x)

[Out]

6*x - 2*log(2)*log(x*log(x) - 5) + 2*log(2)*log(x)