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

Optimal. Leaf size=32 \[ -\frac{1}{2 (x+1)}+\frac{1}{4} \log (1-x)-\log (x)+\frac{3}{4} \log (x+1) \]

[Out]

-1/(2*(1 + x)) + Log[1 - x]/4 - Log[x] + (3*Log[1 + x])/4

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Rubi [A]  time = 0.0123629, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {72} \[ -\frac{1}{2 (x+1)}+\frac{1}{4} \log (1-x)-\log (x)+\frac{3}{4} \log (x+1) \]

Antiderivative was successfully verified.

[In]

Int[1/((-1 + x)*x*(1 + x)^2),x]

[Out]

-1/(2*(1 + x)) + Log[1 - x]/4 - Log[x] + (3*Log[1 + x])/4

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

\begin{align*} \int \frac{1}{(-1+x) x (1+x)^2} \, dx &=\int \left (\frac{1}{4 (-1+x)}-\frac{1}{x}+\frac{1}{2 (1+x)^2}+\frac{3}{4 (1+x)}\right ) \, dx\\ &=-\frac{1}{2 (1+x)}+\frac{1}{4} \log (1-x)-\log (x)+\frac{3}{4} \log (1+x)\\ \end{align*}

Mathematica [A]  time = 0.0107612, size = 28, normalized size = 0.88 \[ \frac{1}{4} \left (-\frac{2}{x+1}+\log (1-x)-4 \log (x)+3 \log (x+1)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/((-1 + x)*x*(1 + x)^2),x]

[Out]

(-2/(1 + x) + Log[1 - x] - 4*Log[x] + 3*Log[1 + x])/4

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Maple [A]  time = 0.008, size = 25, normalized size = 0.8 \begin{align*} -\ln \left ( x \right ) -{\frac{1}{2+2\,x}}+{\frac{3\,\ln \left ( 1+x \right ) }{4}}+{\frac{\ln \left ( -1+x \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(x)-1/2/(1+x)+3/4*ln(1+x)+1/4*ln(-1+x)

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Maxima [A]  time = 0.948003, size = 32, normalized size = 1. \begin{align*} -\frac{1}{2 \,{\left (x + 1\right )}} + \frac{3}{4} \, \log \left (x + 1\right ) + \frac{1}{4} \, \log \left (x - 1\right ) - \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(x + 1) + 3/4*log(x + 1) + 1/4*log(x - 1) - log(x)

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Fricas [A]  time = 1.77226, size = 108, normalized size = 3.38 \begin{align*} \frac{3 \,{\left (x + 1\right )} \log \left (x + 1\right ) +{\left (x + 1\right )} \log \left (x - 1\right ) - 4 \,{\left (x + 1\right )} \log \left (x\right ) - 2}{4 \,{\left (x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*(x + 1)*log(x + 1) + (x + 1)*log(x - 1) - 4*(x + 1)*log(x) - 2)/(x + 1)

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Sympy [A]  time = 0.134233, size = 24, normalized size = 0.75 \begin{align*} - \log{\left (x \right )} + \frac{\log{\left (x - 1 \right )}}{4} + \frac{3 \log{\left (x + 1 \right )}}{4} - \frac{1}{2 x + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x) + log(x - 1)/4 + 3*log(x + 1)/4 - 1/(2*x + 2)

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Giac [A]  time = 1.08221, size = 46, normalized size = 1.44 \begin{align*} -\frac{1}{2 \,{\left (x + 1\right )}} - \log \left ({\left | -\frac{1}{x + 1} + 1 \right |}\right ) + \frac{1}{4} \, \log \left ({\left | -\frac{2}{x + 1} + 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(x + 1) - log(abs(-1/(x + 1) + 1)) + 1/4*log(abs(-2/(x + 1) + 1))