\(\int \frac {2+5 x^2+(-2+10 x) \log (x)+5 \log ^2(x)}{-2 x^2+5 x^3+(-2 x+10 x^2) \log (x)+5 x \log ^2(x)} \, dx\) [4883]

Optimal result

Integrand size = 54, antiderivative size = 20 \[ \int \frac {2+5 x^2+(-2+10 x) \log (x)+5 \log ^2(x)}{-2 x^2+5 x^3+\left (-2 x+10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx=\log \left (\frac {5}{3} \left (-x+\frac {2 x}{5 (x+\log (x))}\right )\right ) \]

[Out]

ln(2/3*x/(x+ln(x))-5/3*x)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6874, 6816} \[ \int \frac {2+5 x^2+(-2+10 x) \log (x)+5 \log ^2(x)}{-2 x^2+5 x^3+\left (-2 x+10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx=\log (x)+\log (-5 x-5 \log (x)+2)-\log (x+\log (x)) \]

[In]

Int[(2 + 5*x^2 + (-2 + 10*x)*Log[x] + 5*Log[x]^2)/(-2*x^2 + 5*x^3 + (-2*x + 10*x^2)*Log[x] + 5*x*Log[x]^2),x]

[Out]

Log[x] + Log[2 - 5*x - 5*Log[x]] - Log[x + Log[x]]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {2+5 x^2+(-2+10 x) \log (x)+5 \log ^2(x)}{-2 x^2+5 x^3+\left (-2 x+10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx=\log (x)+\log (2-5 x-5 \log (x))-\log (x+\log (x)) \]

[In]

Integrate[(2 + 5*x^2 + (-2 + 10*x)*Log[x] + 5*Log[x]^2)/(-2*x^2 + 5*x^3 + (-2*x + 10*x^2)*Log[x] + 5*x*Log[x]^
2),x]

[Out]

Log[x] + Log[2 - 5*x - 5*Log[x]] - Log[x + Log[x]]

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85

method	result	size
risch	\(\ln \left (x \right )+\ln \left (-\frac {2}{5}+x +\ln \left (x \right )\right )-\ln \left (x +\ln \left (x \right )\right )\)	\(17\)
parallelrisch	\(\ln \left (x \right )+\ln \left (-\frac {2}{5}+x +\ln \left (x \right )\right )-\ln \left (x +\ln \left (x \right )\right )\)	\(17\)
default	\(\ln \left (x \right )-\ln \left (x +\ln \left (x \right )\right )+\ln \left (5 \ln \left (x \right )+5 x -2\right )\)	\(21\)
norman	\(\ln \left (x \right )-\ln \left (x +\ln \left (x \right )\right )+\ln \left (5 \ln \left (x \right )+5 x -2\right )\)	\(21\)

[In]

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

[Out]

ln(x)+ln(-2/5+x+ln(x))-ln(x+ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {2+5 x^2+(-2+10 x) \log (x)+5 \log ^2(x)}{-2 x^2+5 x^3+\left (-2 x+10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx=\log \left (5 \, x + 5 \, \log \left (x\right ) - 2\right ) - \log \left (x + \log \left (x\right )\right ) + \log \left (x\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {2+5 x^2+(-2+10 x) \log (x)+5 \log ^2(x)}{-2 x^2+5 x^3+\left (-2 x+10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx=\log {\left (x \right )} - \log {\left (x + \log {\left (x \right )} \right )} + \log {\left (x + \log {\left (x \right )} - \frac {2}{5} \right )} \]

[In]

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

[Out]

log(x) - log(x + log(x)) + log(x + log(x) - 2/5)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {2+5 x^2+(-2+10 x) \log (x)+5 \log ^2(x)}{-2 x^2+5 x^3+\left (-2 x+10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx=-\log \left (x + \log \left (x\right )\right ) + \log \left (x + \log \left (x\right ) - \frac {2}{5}\right ) + \log \left (x\right ) \]

[In]

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

[Out]

-log(x + log(x)) + log(x + log(x) - 2/5) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {2+5 x^2+(-2+10 x) \log (x)+5 \log ^2(x)}{-2 x^2+5 x^3+\left (-2 x+10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx=\log \left (5 \, x + 5 \, \log \left (x\right ) - 2\right ) - \log \left (x + \log \left (x\right )\right ) + \log \left (x\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.09 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {2+5 x^2+(-2+10 x) \log (x)+5 \log ^2(x)}{-2 x^2+5 x^3+\left (-2 x+10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx=\ln \left (x+\ln \left (x\right )-\frac {2}{5}\right )-\ln \left (x+\ln \left (x\right )\right )+\ln \left (x\right ) \]

[In]

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

[Out]

log(x + log(x) - 2/5) - log(x + log(x)) + log(x)