3.330 \(\int \frac{-1+x+4 x^3}{(-1+x) x^2 (1+x^2)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.1668, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6725, 635, 203, 260} \[ -\log \left (x^2+1\right )-\frac{1}{x}+2 \log (1-x)+\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(-1 + x + 4*x^3)/((-1 + x)*x^2*(1 + x^2)),x]

[Out]

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

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 0.0085351, size = 24, normalized size = 1. \[ -\log \left (x^2+1\right )-\frac{1}{x}+2 \log (1-x)+\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x + 4*x^3)/((-1 + x)*x^2*(1 + x^2)),x]

[Out]

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

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Maple [A]  time = 0.007, size = 23, normalized size = 1. \begin{align*} -\ln \left ({x}^{2}+1 \right ) +\arctan \left ( x \right ) +2\,\ln \left ( x-1 \right ) -{x}^{-1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^3+x-1)/(x-1)/x^2/(x^2+1),x)

[Out]

-ln(x^2+1)+arctan(x)+2*ln(x-1)-1/x

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Maxima [A]  time = 1.48614, size = 30, normalized size = 1.25 \begin{align*} -\frac{1}{x} + \arctan \left (x\right ) - \log \left (x^{2} + 1\right ) + 2 \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3+x-1)/(-1+x)/x^2/(x^2+1),x, algorithm="maxima")

[Out]

-1/x + arctan(x) - log(x^2 + 1) + 2*log(x - 1)

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Fricas [A]  time = 1.46124, size = 74, normalized size = 3.08 \begin{align*} \frac{x \arctan \left (x\right ) - x \log \left (x^{2} + 1\right ) + 2 \, x \log \left (x - 1\right ) - 1}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3+x-1)/(-1+x)/x^2/(x^2+1),x, algorithm="fricas")

[Out]

(x*arctan(x) - x*log(x^2 + 1) + 2*x*log(x - 1) - 1)/x

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Sympy [A]  time = 0.142021, size = 19, normalized size = 0.79 \begin{align*} 2 \log{\left (x - 1 \right )} - \log{\left (x^{2} + 1 \right )} + \operatorname{atan}{\left (x \right )} - \frac{1}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**3+x-1)/(-1+x)/x**2/(x**2+1),x)

[Out]

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

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Giac [A]  time = 1.13051, size = 31, normalized size = 1.29 \begin{align*} -\frac{1}{x} + \arctan \left (x\right ) - \log \left (x^{2} + 1\right ) + 2 \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3+x-1)/(-1+x)/x^2/(x^2+1),x, algorithm="giac")

[Out]

-1/x + arctan(x) - log(x^2 + 1) + 2*log(abs(x - 1))