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

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

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Rubi [A]  time = 0.04, antiderivative size = 39, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.02, 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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IntegrateAlgebraic [A]  time = 0.02, size = 35, normalized size = 0.90 \[ \frac {-x-3}{2 \left (x^2-1\right )}-\log \left (x^2-1\right )+2 \log (x)-\frac {1}{2} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.89, size = 45, normalized size = 1.15 \[ -\frac {5 \, {\left (x^{2} - 1\right )} \log \left (x + 1\right ) + 3 \, {\left (x^{2} - 1\right )} \log \left (x - 1\right ) - 8 \, {\left (x^{2} - 1\right )} \log \relax (x) + 2 \, x + 6}{4 \, {\left (x^{2} - 1\right )}} \]

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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giac [A]  time = 1.11, size = 35, normalized size = 0.90 \[ -\frac {x + 3}{2 \, {\left (x + 1\right )} {\left (x - 1\right )}} - \frac {5}{4} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {3}{4} \, \log \left ({\left | x - 1 \right |}\right ) + 2 \, \log \left ({\left | x \right |}\right ) \]

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))

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maple [A]  time = 0.35, size = 31, normalized size = 0.79

method	result	size
norman	\(\frac {-\frac {x}{2}-\frac {3}{2}}{x^{2}-1}+2 \ln \relax (x )-\frac {3 \ln \left (-1+x \right )}{4}-\frac {5 \ln \left (1+x \right )}{4}\)	\(31\)
risch	\(\frac {-\frac {x}{2}-\frac {3}{2}}{x^{2}-1}+2 \ln \relax (x )-\frac {3 \ln \left (-1+x \right )}{4}-\frac {5 \ln \left (1+x \right )}{4}\)	\(31\)
default	\(2 \ln \relax (x )-\frac {1}{-1+x}-\frac {3 \ln \left (-1+x \right )}{4}+\frac {1}{2 x +2}-\frac {5 \ln \left (1+x \right )}{4}\)	\(32\)
meijerg	\(\frac {i \left (-\frac {i x}{-x^{2}+1}+i \arctanh \relax (x )\right )}{2}+\frac {3 x^{2}}{-2 x^{2}+2}-\ln \left (-x^{2}+1\right )+1+2 \ln \relax (x )+i \pi \)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 29, normalized size = 0.74 \[ -\frac {x + 3}{2 \, {\left (x^{2} - 1\right )}} - \frac {5}{4} \, \log \left (x + 1\right ) - \frac {3}{4} \, \log \left (x - 1\right ) + 2 \, \log \relax (x) \]

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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mupad [B]  time = 0.05, size = 31, normalized size = 0.79 \[ 2\,\ln \relax (x)-\frac {5\,\ln \left (x+1\right )}{4}-\frac {3\,\ln \left (x-1\right )}{4}-\frac {\frac {x}{2}+\frac {3}{2}}{x^2-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.16, size = 32, normalized size = 0.82 \[ \frac {- x - 3}{2 x^{2} - 2} + 2 \log {\relax (x )} - \frac {3 \log {\left (x - 1 \right )}}{4} - \frac {5 \log {\left (x + 1 \right )}}{4} \]

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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