3.305 \(\int \frac {-1+2 x+x^2}{-2 x+3 x^2+2 x^3} \, dx\)

Optimal. Leaf size=25 \[ \frac {1}{10} \log (1-2 x)+\frac {\log (x)}{2}-\frac {1}{10} \log (x+2) \]

[Out]

1/10*ln(1-2*x)+1/2*ln(x)-1/10*ln(2+x)

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Rubi [A]  time = 0.04, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1594, 1628} \[ \frac {1}{10} \log (1-2 x)+\frac {\log (x)}{2}-\frac {1}{10} \log (x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[1 - 2*x]/10 + Log[x]/2 - Log[2 + x]/10

Rule 1594

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

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {align*} \int \frac {-1+2 x+x^2}{-2 x+3 x^2+2 x^3} \, dx &=\int \frac {-1+2 x+x^2}{x \left (-2+3 x+2 x^2\right )} \, dx\\ &=\int \left (\frac {1}{2 x}-\frac {1}{10 (2+x)}+\frac {1}{5 (-1+2 x)}\right ) \, dx\\ &=\frac {1}{10} \log (1-2 x)+\frac {\log (x)}{2}-\frac {1}{10} \log (2+x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.00 \[ \frac {1}{10} \log (1-2 x)+\frac {\log (x)}{2}-\frac {1}{10} \log (x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[1 - 2*x]/10 + Log[x]/2 - Log[2 + x]/10

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fricas [A]  time = 0.80, size = 19, normalized size = 0.76 \[ \frac {1}{10} \, \log \left (2 \, x - 1\right ) - \frac {1}{10} \, \log \left (x + 2\right ) + \frac {1}{2} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*log(2*x - 1) - 1/10*log(x + 2) + 1/2*log(x)

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giac [A]  time = 0.29, size = 22, normalized size = 0.88 \[ \frac {1}{10} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) - \frac {1}{10} \, \log \left ({\left | x + 2 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*log(abs(2*x - 1)) - 1/10*log(abs(x + 2)) + 1/2*log(abs(x))

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maple [A]  time = 0.01, size = 20, normalized size = 0.80 \[ \frac {\ln \relax (x )}{2}+\frac {\ln \left (2 x -1\right )}{10}-\frac {\ln \left (x +2\right )}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*ln(2*x-1)-1/10*ln(x+2)+1/2*ln(x)

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maxima [A]  time = 1.09, size = 19, normalized size = 0.76 \[ \frac {1}{10} \, \log \left (2 \, x - 1\right ) - \frac {1}{10} \, \log \left (x + 2\right ) + \frac {1}{2} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*log(2*x - 1) - 1/10*log(x + 2) + 1/2*log(x)

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mupad [B]  time = 0.06, size = 19, normalized size = 0.76 \[ \frac {\mathrm {atanh}\left (\frac {24}{145\,\left (\frac {29\,x}{100}-\frac {11}{50}\right )}+\frac {35}{29}\right )}{5}+\frac {\ln \relax (x)}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atanh(24/(145*((29*x)/100 - 11/50)) + 35/29)/5 + log(x)/2

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sympy [A]  time = 0.14, size = 19, normalized size = 0.76 \[ \frac {\log {\relax (x )}}{2} + \frac {\log {\left (x - \frac {1}{2} \right )}}{10} - \frac {\log {\left (x + 2 \right )}}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)/2 + log(x - 1/2)/10 - log(x + 2)/10

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