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

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Rubi [A]  time = 0.06, antiderivative size = 34, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.60, size = 34, normalized size = 1.00 \[ -\frac {3 \, {\left (x - 1\right )} \log \left (x + 1\right ) + 5 \, {\left (x - 1\right )} \log \left (x - 1\right ) - 8 \, {\left (x - 1\right )} \log \relax (x) + 6}{4 \, {\left (x - 1\right )}} \]

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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giac [A]  time = 0.39, size = 34, normalized size = 1.00 \[ -\frac {3}{2 \, {\left (x - 1\right )}} + 2 \, \log \left ({\left | -\frac {1}{x - 1} - 1 \right |}\right ) - \frac {3}{4} \, \log \left ({\left | -\frac {2}{x - 1} - 1 \right |}\right ) \]

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))

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maple [A]  time = 0.01, size = 25, normalized size = 0.74 \[ 2 \ln \relax (x )-\frac {5 \ln \left (x -1\right )}{4}-\frac {3 \ln \left (x +1\right )}{4}-\frac {3}{2 \left (x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.13, size = 24, normalized size = 0.71 \[ -\frac {3}{2 \, {\left (x - 1\right )}} - \frac {3}{4} \, \log \left (x + 1\right ) - \frac {5}{4} \, \log \left (x - 1\right ) + 2 \, \log \relax (x) \]

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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mupad [B]  time = 2.11, size = 26, normalized size = 0.76 \[ 2\,\ln \relax (x)-\frac {3\,\ln \left (x+1\right )}{4}-\frac {5\,\ln \left (x-1\right )}{4}-\frac {3}{2\,\left (x-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.14, size = 27, normalized size = 0.79 \[ 2 \log {\relax (x )} - \frac {5 \log {\left (x - 1 \right )}}{4} - \frac {3 \log {\left (x + 1 \right )}}{4} - \frac {3}{2 x - 2} \]

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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