3.41.71 $\int \frac{-1+3x+2x^2}{-x+2x^2}dx$

Optimal. Leaf size=15 $$-5+x+\log \left(\frac{6x(-1+2x)}{e^5}\right)$$

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Rubi [A]  time = 0.02, antiderivative size = 10, normalized size of antiderivative = 0.67, 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 = {1593, 893} $$\begin{array}{c}x+\log (1-2x)+\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-1 + 3*x + 2*x^2)/(-x + 2*x^2),x]

[Out]

x + Log[1 - 2*x] + Log[x]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-1+3x+2x^2}{x(-1+2x)}dx\\ & =\int (1+\frac{1}{x}+\frac{2}{-1+2x})dx\\ & =x+\log (1-2x)+\log (x)\end{array}\end{array}$$

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Mathematica [A]  time = 0.00, size = 10, normalized size = 0.67 $$\begin{array}{c}x+\log (1-2x)+\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 3*x + 2*x^2)/(-x + 2*x^2),x]

[Out]

x + Log[1 - 2*x] + Log[x]

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fricas [A]  time = 0.57, size = 12, normalized size = 0.80 $$\begin{array}{c}x+\log (2x^2-x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+3*x-1)/(2*x^2-x),x, algorithm="fricas")

[Out]

x + log(2*x^2 - x)

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giac [A]  time = 0.21, size = 12, normalized size = 0.80 $$\begin{array}{c}x+\log \left(|2x-1|\right)+\log \left(\left|x\right|\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+3*x-1)/(2*x^2-x),x, algorithm="giac")

[Out]

x + log(abs(2*x - 1)) + log(abs(x))

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maple [A]  time = 0.07, size = 11, normalized size = 0.73

method	result	size
default	$x+\ln (2x-1)+\ln (x)$	$11$
norman	$x+\ln (2x-1)+\ln (x)$	$11$
risch	$x+\ln (2x^2-x)$	$13$
meijerg	$\ln (x)+\ln (2)+i\pi +\ln (1-2x)+x$	$17$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2+3*x-1)/(2*x^2-x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.37, size = 10, normalized size = 0.67 $$\begin{array}{c}x+\log (2x-1)+\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+3*x-1)/(2*x^2-x),x, algorithm="maxima")

[Out]

x + log(2*x - 1) + log(x)

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mupad [B]  time = 3.08, size = 10, normalized size = 0.67 $$\begin{array}{c}x+\ln \left(x(2x-1)\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x + 2*x^2 - 1)/(x - 2*x^2),x)

[Out]

x + log(x*(2*x - 1))

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sympy [A]  time = 0.08, size = 8, normalized size = 0.53 $$\begin{array}{c}x+\log (2x^2-x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2+3*x-1)/(2*x**2-x),x)

[Out]

x + log(2*x**2 - x)

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