3.67.70 $\int \frac{-1+\log (x^2)}{x-2x^2+x\log (x^2)}dx$

Optimal. Leaf size=20 $$-\frac{5}{3}+\log (x)-\log (-1+2x-\log \left(x^2\right))$$

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Rubi [A]  time = 0.11, antiderivative size = 15, normalized size of antiderivative = 0.75, number of steps used = 3, number of rules used = 2, integrand size = 22, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.091, Rules used = {6742, 6684} $$\begin{array}{c}\log (x)-\log (\log \left(x^2\right)-2x+1)\end{array}$$

Antiderivative was successfully verified.

[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 6684

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

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{1}{x}-\frac{2(-1+x)}{x(-1+2x-\log \left(x^2\right))})dx\\ & =\log (x)-2\int \frac{-1+x}{x(-1+2x-\log \left(x^2\right))}dx\\ & =\log (x)-\log (1-2x+\log \left(x^2\right))\end{array}\end{array}$$

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Mathematica [A]  time = 0.11, size = 15, normalized size = 0.75 $$\begin{array}{c}\log (x)-\log (1-2x+\log \left(x^2\right))\end{array}$$

Antiderivative was successfully verified.

[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]]

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fricas [A]  time = 0.53, size = 19, normalized size = 0.95 $$\begin{array}{c}\frac{1}{2}\log \left(x^2\right)-\log (-2x+\log \left(x^2\right)+1)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.23, size = 15, normalized size = 0.75 $$\begin{array}{c}\log (x)-\log (-2x+\log \left(x^2\right)+1)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.02, size = 16, normalized size = 0.80

method	result	size
risch	$\ln (x)-\ln (\ln \left(x^2\right)-2x+1)$	$16$
norman	$\ln (x)-\ln (2x-1-\ln \left(x^2\right))$	$18$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.40, size = 13, normalized size = 0.65 $$\begin{array}{c}\log (x)-\log (-x+\log (x)+\frac{1}{2})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 4.21, size = 21, normalized size = 1.05 $$\begin{array}{c}\frac{\ln \left(x^2\right)}{2}-\ln (2x-\ln \left(x^2\right)-1)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.12, size = 14, normalized size = 0.70 $$\begin{array}{c}\log (x)-\log (-2x+\log \left(x^2\right)+1)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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