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

Optimal. Leaf size=24 \[ \frac {1}{x+1}+\frac {3}{2} \log (1-x)-\frac {1}{2} \log (x+1) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 24, 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 = {893} \[ \frac {1}{x+1}+\frac {3}{2} \log (1-x)-\frac {1}{2} \log (x+1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.92 \[ \frac {1}{x+1}+\frac {3}{2} \log (x-1)-\frac {1}{2} \log (x+1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 26, normalized size = 1.08 \[ -\frac {{\left (x + 1\right )} \log \left (x + 1\right ) - 3 \, {\left (x + 1\right )} \log \left (x - 1\right ) - 2}{2 \, {\left (x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.01, size = 24, normalized size = 1.00 \[ \frac {1}{x + 1} + \log \left ({\left | x + 1 \right |}\right ) + \frac {3}{2} \, \log \left ({\left | -\frac {2}{x + 1} + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 19, normalized size = 0.79 \[ \frac {3 \ln \left (x -1\right )}{2}-\frac {\ln \left (x +1\right )}{2}+\frac {1}{x +1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 18, normalized size = 0.75 \[ \frac {1}{x + 1} - \frac {1}{2} \, \log \left (x + 1\right ) + \frac {3}{2} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 18, normalized size = 0.75 \[ \frac {3\,\ln \left (x-1\right )}{2}-\frac {\ln \left (x+1\right )}{2}+\frac {1}{x+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.12, size = 19, normalized size = 0.79 \[ \frac {3 \log {\left (x - 1 \right )}}{2} - \frac {\log {\left (x + 1 \right )}}{2} + \frac {1}{x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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