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

Optimal. Leaf size=39 \[ \frac{x+3}{2 \left (1-x^2\right )}-\frac{3}{4} \log (1-x)+2 \log (x)-\frac{5}{4} \log (x+1) \]

[Out]

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

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Rubi [A]  time = 0.0362482, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {1805, 801} \[ \frac{x+3}{2 \left (1-x^2\right )}-\frac{3}{4} \log (1-x)+2 \log (x)-\frac{5}{4} \log (x+1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

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

Mathematica [A]  time = 0.0165338, size = 47, normalized size = 1.21 \[ \frac{1}{4} \left (-\frac{4}{x^2-1}-4 \log \left (1-x^2\right )-\frac{2}{x-1}+\log (1-x)+8 \log (x)-\log (x+1)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2/(-1 + x) - 4/(-1 + x^2) + Log[1 - x] + 8*Log[x] - Log[1 + x] - 4*Log[1 - x^2])/4

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Maple [A]  time = 0.011, size = 32, normalized size = 0.8 \begin{align*} 2\,\ln \left ( x \right ) +{\frac{1}{2\,x+2}}-{\frac{5\,\ln \left ( 1+x \right ) }{4}}- \left ( -1+x \right ) ^{-1}-{\frac{3\,\ln \left ( -1+x \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.956347, size = 39, normalized size = 1. \begin{align*} -\frac{x + 3}{2 \,{\left (x^{2} - 1\right )}} - \frac{5}{4} \, \log \left (x + 1\right ) - \frac{3}{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^3+x^2+2)/x/(x^2-1)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.19858, size = 131, normalized size = 3.36 \begin{align*} -\frac{5 \,{\left (x^{2} - 1\right )} \log \left (x + 1\right ) + 3 \,{\left (x^{2} - 1\right )} \log \left (x - 1\right ) - 8 \,{\left (x^{2} - 1\right )} \log \left (x\right ) + 2 \, x + 6}{4 \,{\left (x^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(5*(x^2 - 1)*log(x + 1) + 3*(x^2 - 1)*log(x - 1) - 8*(x^2 - 1)*log(x) + 2*x + 6)/(x^2 - 1)

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Sympy [A]  time = 0.136416, size = 31, normalized size = 0.79 \begin{align*} - \frac{x + 3}{2 x^{2} - 2} + 2 \log{\left (x \right )} - \frac{3 \log{\left (x - 1 \right )}}{4} - \frac{5 \log{\left (x + 1 \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.04904, size = 47, normalized size = 1.21 \begin{align*} -\frac{x + 3}{2 \,{\left (x + 1\right )}{\left (x - 1\right )}} - \frac{5}{4} \, \log \left ({\left | x + 1 \right |}\right ) - \frac{3}{4} \, \log \left ({\left | x - 1 \right |}\right ) + 2 \, \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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