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

Optimal. Leaf size=34 \[ \frac{3}{2 (1-x)}-\frac{5}{4} \log (1-x)+2 \log (x)-\frac{3}{4} \log (x+1) \]

[Out]

3/(2*(1 - x)) - (5*Log[1 - x])/4 + 2*Log[x] - (3*Log[1 + x])/4

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Rubi [A]  time = 0.0574547, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {1612} \[ \frac{3}{2 (1-x)}-\frac{5}{4} \log (1-x)+2 \log (x)-\frac{3}{4} \log (x+1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

3/(2*(1 - x)) - (5*Log[1 - x])/4 + 2*Log[x] - (3*Log[1 + x])/4

Rule 1612

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[E
xpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && Poly
Q[Px, x] && IntegersQ[m, n]

Rubi steps

\begin{align*} \int \frac{2+x^2}{(-1+x)^2 x (1+x)} \, dx &=\int \left (\frac{3}{2 (-1+x)^2}-\frac{5}{4 (-1+x)}+\frac{2}{x}-\frac{3}{4 (1+x)}\right ) \, dx\\ &=\frac{3}{2 (1-x)}-\frac{5}{4} \log (1-x)+2 \log (x)-\frac{3}{4} \log (1+x)\\ \end{align*}

Mathematica [A]  time = 0.0166073, size = 32, normalized size = 0.94 \[ -\frac{3}{2 (x-1)}-\frac{5}{4} \log (1-x)+2 \log (x)-\frac{3}{4} \log (x+1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-3/(2*(-1 + x)) - (5*Log[1 - x])/4 + 2*Log[x] - (3*Log[1 + x])/4

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Maple [A]  time = 0.007, size = 25, normalized size = 0.7 \begin{align*} -{\frac{3}{2\,x-2}}-{\frac{5\,\ln \left ( x-1 \right ) }{4}}+2\,\ln \left ( x \right ) -{\frac{3\,\ln \left ( 1+x \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2/(x-1)-5/4*ln(x-1)+2*ln(x)-3/4*ln(1+x)

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Maxima [A]  time = 0.984737, size = 32, normalized size = 0.94 \begin{align*} -\frac{3}{2 \,{\left (x - 1\right )}} - \frac{3}{4} \, \log \left (x + 1\right ) - \frac{5}{4} \, \log \left (x - 1\right ) + 2 \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2/(x - 1) - 3/4*log(x + 1) - 5/4*log(x - 1) + 2*log(x)

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Fricas [A]  time = 1.53738, size = 112, normalized size = 3.29 \begin{align*} -\frac{3 \,{\left (x - 1\right )} \log \left (x + 1\right ) + 5 \,{\left (x - 1\right )} \log \left (x - 1\right ) - 8 \,{\left (x - 1\right )} \log \left (x\right ) + 6}{4 \,{\left (x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(3*(x - 1)*log(x + 1) + 5*(x - 1)*log(x - 1) - 8*(x - 1)*log(x) + 6)/(x - 1)

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Sympy [A]  time = 0.133397, size = 27, normalized size = 0.79 \begin{align*} 2 \log{\left (x \right )} - \frac{5 \log{\left (x - 1 \right )}}{4} - \frac{3 \log{\left (x + 1 \right )}}{4} - \frac{3}{2 x - 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(x) - 5*log(x - 1)/4 - 3*log(x + 1)/4 - 3/(2*x - 2)

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Giac [A]  time = 1.15256, size = 46, normalized size = 1.35 \begin{align*} -\frac{3}{2 \,{\left (x - 1\right )}} + 2 \, \log \left ({\left | -\frac{1}{x - 1} - 1 \right |}\right ) - \frac{3}{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((x^2+2)/(-1+x)^2/x/(1+x),x, algorithm="giac")

[Out]

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