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

Optimal. Leaf size=16 \[ \frac {1}{x}+2 \log (1-x)-2 \log (x) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {77} \[ \frac {1}{x}+2 \log (1-x)-2 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)/((-1 + x)*x^2),x]

[Out]

x^(-1) + 2*Log[1 - x] - 2*Log[x]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 16, normalized size = 1.00 \[ \frac {1}{x}+2 \log (1-x)-2 \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)/((-1 + x)*x^2),x]

[Out]

x^(-1) + 2*Log[1 - x] - 2*Log[x]

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fricas [A]  time = 0.62, size = 18, normalized size = 1.12 \[ \frac {2 \, x \log \left (x - 1\right ) - 2 \, x \log \relax (x) + 1}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x*log(x - 1) - 2*x*log(x) + 1)/x

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giac [A]  time = 0.36, size = 16, normalized size = 1.00 \[ \frac {1}{x} + 2 \, \log \left ({\left | x - 1 \right |}\right ) - 2 \, \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 15, normalized size = 0.94 \[ -2 \ln \relax (x )+2 \ln \left (x -1\right )+\frac {1}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 14, normalized size = 0.88 \[ \frac {1}{x} + 2 \, \log \left (x - 1\right ) - 2 \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/x + 2*log(x - 1) - 2*log(x)

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mupad [B]  time = 0.03, size = 12, normalized size = 0.75 \[ \frac {1}{x}-4\,\mathrm {atanh}\left (2\,x-1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 1)/(x^2*(x - 1)),x)

[Out]

1/x - 4*atanh(2*x - 1)

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sympy [A]  time = 0.10, size = 14, normalized size = 0.88 \[ - 2 \log {\relax (x )} + 2 \log {\left (x - 1 \right )} + \frac {1}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*log(x) + 2*log(x - 1) + 1/x

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