\(\int \frac {-x^2+3 x^2 \log (x)-5 \log (4) \log ^2(x)}{-x^3 \log (x)+(-5+5 x \log (4)) \log ^2(x)} \, dx\) [7148]

Optimal result

Integrand size = 44, antiderivative size = 27 \[ \int \frac {-x^2+3 x^2 \log (x)-5 \log (4) \log ^2(x)}{-x^3 \log (x)+(-5+5 x \log (4)) \log ^2(x)} \, dx=-5+\log \left (\frac {4}{-5+\frac {x \left (-x^2+5 \log (4) \log (x)\right )}{\log (x)}}\right ) \]

[Out]

ln(4/(x*(10*ln(2)*ln(x)-x^2)/ln(x)-5))-5

Rubi [F]

\[ \int \frac {-x^2+3 x^2 \log (x)-5 \log (4) \log ^2(x)}{-x^3 \log (x)+(-5+5 x \log (4)) \log ^2(x)} \, dx=\int \frac {-x^2+3 x^2 \log (x)-5 \log (4) \log ^2(x)}{-x^3 \log (x)+(-5+5 x \log (4)) \log ^2(x)} \, dx \]

[In]

Int[(-x^2 + 3*x^2*Log[x] - 5*Log[4]*Log[x]^2)/(-(x^3*Log[x]) + (-5 + 5*x*Log[4])*Log[x]^2),x]

[Out]

-Log[1 - x*Log[4]] + Log[Log[x]] - 5*Defer[Int][1/(x*(x^3 + 5*Log[x] - 5*x*Log[4]*Log[x])), x] - 2*Defer[Int][
x^2/(x^3 + 5*Log[x] - 5*x*Log[4]*Log[x]), x] - Defer[Int][(-x^3 - 5*Log[x] + 5*x*Log[4]*Log[x])^(-1), x]/Log[4
]^2 - 5*Log[4]*Defer[Int][(-x^3 - 5*Log[x] + 5*x*Log[4]*Log[x])^(-1), x] - Defer[Int][x/(-x^3 - 5*Log[x] + 5*x
*Log[4]*Log[x]), x]/Log[4] - Defer[Int][1/((-1 + x*Log[4])*(-x^3 - 5*Log[x] + 5*x*Log[4]*Log[x])), x]/Log[4]^2

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2-3 x^2 \log (x)+5 \log (4) \log ^2(x)}{\log (x) \left (x^3+5 \log (x)-5 x \log (4) \log (x)\right )} \, dx \\ & = \int \left (-\frac {\log (4)}{-1+x \log (4)}+\frac {1}{x \log (x)}+\frac {5 (-1+x \log (4))}{x \left (x^3+5 \log (x)-5 x \log (4) \log (x)\right )}-\frac {x^2 (-3+2 x \log (4))}{(-1+x \log (4)) \left (x^3+5 \log (x)-5 x \log (4) \log (x)\right )}\right ) \, dx \\ & = -\log (1-x \log (4))+5 \int \frac {-1+x \log (4)}{x \left (x^3+5 \log (x)-5 x \log (4) \log (x)\right )} \, dx+\int \frac {1}{x \log (x)} \, dx-\int \frac {x^2 (-3+2 x \log (4))}{(-1+x \log (4)) \left (x^3+5 \log (x)-5 x \log (4) \log (x)\right )} \, dx \\ & = -\log (1-x \log (4))+5 \int \left (-\frac {1}{x \left (x^3+5 \log (x)-5 x \log (4) \log (x)\right )}-\frac {\log (4)}{-x^3-5 \log (x)+5 x \log (4) \log (x)}\right ) \, dx-\int \left (\frac {2 x^2}{x^3+5 \log (x)-5 x \log (4) \log (x)}+\frac {1}{\log ^2(4) \left (-x^3-5 \log (x)+5 x \log (4) \log (x)\right )}+\frac {x}{\log (4) \left (-x^3-5 \log (x)+5 x \log (4) \log (x)\right )}+\frac {1}{\log ^2(4) (-1+x \log (4)) \left (-x^3-5 \log (x)+5 x \log (4) \log (x)\right )}\right ) \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = -\log (1-x \log (4))+\log (\log (x))-2 \int \frac {x^2}{x^3+5 \log (x)-5 x \log (4) \log (x)} \, dx-5 \int \frac {1}{x \left (x^3+5 \log (x)-5 x \log (4) \log (x)\right )} \, dx-\frac {\int \frac {1}{-x^3-5 \log (x)+5 x \log (4) \log (x)} \, dx}{\log ^2(4)}-\frac {\int \frac {1}{(-1+x \log (4)) \left (-x^3-5 \log (x)+5 x \log (4) \log (x)\right )} \, dx}{\log ^2(4)}-\frac {\int \frac {x}{-x^3-5 \log (x)+5 x \log (4) \log (x)} \, dx}{\log (4)}-(5 \log (4)) \int \frac {1}{-x^3-5 \log (x)+5 x \log (4) \log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.99 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-x^2+3 x^2 \log (x)-5 \log (4) \log ^2(x)}{-x^3 \log (x)+(-5+5 x \log (4)) \log ^2(x)} \, dx=\log (\log (x))-\log \left (x^3+5 \log (x)-5 x \log (4) \log (x)\right ) \]

[In]

Integrate[(-x^2 + 3*x^2*Log[x] - 5*Log[4]*Log[x]^2)/(-(x^3*Log[x]) + (-5 + 5*x*Log[4])*Log[x]^2),x]

[Out]

Log[Log[x]] - Log[x^3 + 5*Log[x] - 5*x*Log[4]*Log[x]]

Maple [A] (verified)

Time = 1.15 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85

method	result	size
parallelrisch	\(\ln \left (\ln \left (x \right )\right )-\ln \left (-10 x \ln \left (2\right ) \ln \left (x \right )+x^{3}+5 \ln \left (x \right )\right )\)	\(23\)
default	\(\ln \left (\ln \left (x \right )\right )-\ln \left (10 x \ln \left (2\right ) \ln \left (x \right )-x^{3}-5 \ln \left (x \right )\right )\)	\(25\)
norman	\(\ln \left (\ln \left (x \right )\right )-\ln \left (10 x \ln \left (2\right ) \ln \left (x \right )-x^{3}-5 \ln \left (x \right )\right )\)	\(25\)
risch	\(-\ln \left (2 x \ln \left (2\right )-1\right )+\ln \left (\ln \left (x \right )\right )-\ln \left (\ln \left (x \right )-\frac {x^{3}}{5 \left (2 x \ln \left (2\right )-1\right )}\right )\)	\(35\)

[In]

int((-10*ln(2)*ln(x)^2+3*x^2*ln(x)-x^2)/((10*x*ln(2)-5)*ln(x)^2-x^3*ln(x)),x,method=_RETURNVERBOSE)

[Out]

ln(ln(x))-ln(-10*x*ln(2)*ln(x)+x^3+5*ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {-x^2+3 x^2 \log (x)-5 \log (4) \log ^2(x)}{-x^3 \log (x)+(-5+5 x \log (4)) \log ^2(x)} \, dx=-\log \left (2 \, x \log \left (2\right ) - 1\right ) - \log \left (-\frac {x^{3} - 5 \, {\left (2 \, x \log \left (2\right ) - 1\right )} \log \left (x\right )}{2 \, x \log \left (2\right ) - 1}\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

integrate((-10*log(2)*log(x)^2+3*x^2*log(x)-x^2)/((10*x*log(2)-5)*log(x)^2-x^3*log(x)),x, algorithm="fricas")

[Out]

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {-x^2+3 x^2 \log (x)-5 \log (4) \log ^2(x)}{-x^3 \log (x)+(-5+5 x \log (4)) \log ^2(x)} \, dx=\text {Exception raised: PolynomialError} \]

[In]

integrate((-10*ln(2)*ln(x)**2+3*x**2*ln(x)-x**2)/((10*x*ln(2)-5)*ln(x)**2-x**3*ln(x)),x)

[Out]

Exception raised: PolynomialError >> 1/(20*x**3*log(2)**2 - 20*x**2*log(2) + 5*x) contains an element of the s
et of generators.

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {-x^2+3 x^2 \log (x)-5 \log (4) \log ^2(x)}{-x^3 \log (x)+(-5+5 x \log (4)) \log ^2(x)} \, dx=-\log \left (2 \, x \log \left (2\right ) - 1\right ) - \log \left (-\frac {x^{3} - 5 \, {\left (2 \, x \log \left (2\right ) - 1\right )} \log \left (x\right )}{5 \, {\left (2 \, x \log \left (2\right ) - 1\right )}}\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

integrate((-10*log(2)*log(x)^2+3*x^2*log(x)-x^2)/((10*x*log(2)-5)*log(x)^2-x^3*log(x)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-x^2+3 x^2 \log (x)-5 \log (4) \log ^2(x)}{-x^3 \log (x)+(-5+5 x \log (4)) \log ^2(x)} \, dx=-\log \left (-x^{3} + 10 \, x \log \left (2\right ) \log \left (x\right ) - 5 \, \log \left (x\right )\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

integrate((-10*log(2)*log(x)^2+3*x^2*log(x)-x^2)/((10*x*log(2)-5)*log(x)^2-x^3*log(x)),x, algorithm="giac")

[Out]

-log(-x^3 + 10*x*log(2)*log(x) - 5*log(x)) + log(log(x))

Mupad [B] (verification not implemented)

Time = 11.96 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-x^2+3 x^2 \log (x)-5 \log (4) \log ^2(x)}{-x^3 \log (x)+(-5+5 x \log (4)) \log ^2(x)} \, dx=\ln \left (\ln \left (x\right )\right )-\ln \left (5\,\ln \left (x\right )+x^3-10\,x\,\ln \left (2\right )\,\ln \left (x\right )\right ) \]

[In]

int((10*log(2)*log(x)^2 - 3*x^2*log(x) + x^2)/(x^3*log(x) - log(x)^2*(10*x*log(2) - 5)),x)

[Out]

log(log(x)) - log(5*log(x) + x^3 - 10*x*log(2)*log(x))