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

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Rubi [A]  time = 0.0143797, antiderivative size = 24, 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 = {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*}

Mathematica [A]  time = 0.0083749, 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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Maple [A]  time = 0.007, size = 19, normalized size = 0.8 \begin{align*} \left ( 1+x \right ) ^{-1}-{\frac{\ln \left ( 1+x \right ) }{2}}+{\frac{3\,\ln \left ( -1+x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.947586, size = 24, normalized size = 1. \begin{align*} \frac{1}{x + 1} - \frac{1}{2} \, \log \left (x + 1\right ) + \frac{3}{2} \, \log \left (x - 1\right ) \end{align*}

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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Fricas [A]  time = 0.666378, size = 84, normalized size = 3.5 \begin{align*} -\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 )}} \end{align*}

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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Sympy [A]  time = 0.104982, size = 19, normalized size = 0.79 \begin{align*} \frac{3 \log{\left (x - 1 \right )}}{2} - \frac{\log{\left (x + 1 \right )}}{2} + \frac{1}{x + 1} \end{align*}

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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Giac [A]  time = 1.09003, size = 32, normalized size = 1.33 \begin{align*} \frac{1}{x + 1} + \log \left ({\left | x + 1 \right |}\right ) + \frac{3}{2} \, \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*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))