\(\int \frac {-1+\log (x^2)}{x-2 x^2+x \log (x^2)} \, dx\) [6670]

Optimal result

Integrand size = 22, antiderivative size = 20 \[ \int \frac {-1+\log \left (x^2\right )}{x-2 x^2+x \log \left (x^2\right )} \, dx=-\frac {5}{3}+\log (x)-\log \left (-1+2 x-\log \left (x^2\right )\right ) \]

[Out]

ln(x)-5/3-ln(2*x-1-ln(x^2))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6874, 6816} \[ \int \frac {-1+\log \left (x^2\right )}{x-2 x^2+x \log \left (x^2\right )} \, dx=\log (x)-\log \left (\log \left (x^2\right )-2 x+1\right ) \]

[In]

Int[(-1 + Log[x^2])/(x - 2*x^2 + x*Log[x^2]),x]

[Out]

Log[x] - Log[1 - 2*x + Log[x^2]]

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 {2 (-1+x)}{x \left (-1+2 x-\log \left (x^2\right )\right )}\right ) \, dx \\ & = \log (x)-2 \int \frac {-1+x}{x \left (-1+2 x-\log \left (x^2\right )\right )} \, dx \\ & = \log (x)-\log \left (1-2 x+\log \left (x^2\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-1+\log \left (x^2\right )}{x-2 x^2+x \log \left (x^2\right )} \, dx=\log (x)-\log \left (1-2 x+\log \left (x^2\right )\right ) \]

[In]

Integrate[(-1 + Log[x^2])/(x - 2*x^2 + x*Log[x^2]),x]

[Out]

Log[x] - Log[1 - 2*x + Log[x^2]]

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80

method	result	size
risch	\(\ln \left (x \right )-\ln \left (\ln \left (x^{2}\right )-2 x +1\right )\)	\(16\)
parallelrisch	\(\frac {\ln \left (x^{2}\right )}{2}-\ln \left (-\frac {\ln \left (x^{2}\right )}{2}+x -\frac {1}{2}\right )\)	\(20\)
norman	\(\frac {\ln \left (x^{2}\right )}{2}-\ln \left (2 x -1-\ln \left (x^{2}\right )\right )\)	\(22\)

[In]

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

[Out]

ln(x)-ln(ln(x^2)-2*x+1)

Fricas [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {-1+\log \left (x^2\right )}{x-2 x^2+x \log \left (x^2\right )} \, dx=\frac {1}{2} \, \log \left (x^{2}\right ) - \log \left (-2 \, x + \log \left (x^{2}\right ) + 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-1+\log \left (x^2\right )}{x-2 x^2+x \log \left (x^2\right )} \, dx=\log {\left (x \right )} - \log {\left (- 2 x + \log {\left (x^{2} \right )} + 1 \right )} \]

[In]

integrate((ln(x**2)-1)/(x*ln(x**2)-2*x**2+x),x)

[Out]

log(x) - log(-2*x + log(x**2) + 1)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int \frac {-1+\log \left (x^2\right )}{x-2 x^2+x \log \left (x^2\right )} \, dx=\log \left (x\right ) - \log \left (-x + \log \left (x\right ) + \frac {1}{2}\right ) \]

[In]

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

[Out]

log(x) - log(-x + log(x) + 1/2)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-1+\log \left (x^2\right )}{x-2 x^2+x \log \left (x^2\right )} \, dx=\log \left (x\right ) - \log \left (-2 \, x + \log \left (x^{2}\right ) + 1\right ) \]

[In]

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

[Out]

log(x) - log(-2*x + log(x^2) + 1)

Mupad [B] (verification not implemented)

Time = 14.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-1+\log \left (x^2\right )}{x-2 x^2+x \log \left (x^2\right )} \, dx=\frac {\ln \left (x^2\right )}{2}-\ln \left (2\,x-\ln \left (x^2\right )-1\right ) \]

[In]

int((log(x^2) - 1)/(x + x*log(x^2) - 2*x^2),x)

[Out]

log(x^2)/2 - log(2*x - log(x^2) - 1)