### 3.201 $$\int \frac{5-4 x+3 x^2}{(-1+x) (1+x^2)} \, dx$$

Optimal. Leaf size=23 $\frac{1}{2} \log \left (x^2+1\right )+2 \log (1-x)-3 \tan ^{-1}(x)$

[Out]

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

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Rubi [A]  time = 0.0346737, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.174, Rules used = {1629, 635, 203, 260} $\frac{1}{2} \log \left (x^2+1\right )+2 \log (1-x)-3 \tan ^{-1}(x)$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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

Mathematica [A]  time = 0.006862, size = 28, normalized size = 1.22 $\frac{1}{2} \log \left ((x-1)^2+2 (x-1)+2\right )+2 \log (x-1)-3 \tan ^{-1}(x)$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 20, normalized size = 0.9 \begin{align*}{\frac{\ln \left ({x}^{2}+1 \right ) }{2}}-3\,\arctan \left ( x \right ) +2\,\ln \left ( -1+x \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.41652, size = 26, normalized size = 1.13 \begin{align*} -3 \, \arctan \left (x\right ) + \frac{1}{2} \, \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((3*x^2-4*x+5)/(-1+x)/(x^2+1),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.75993, size = 65, normalized size = 2.83 \begin{align*} -3 \, \arctan \left (x\right ) + \frac{1}{2} \, \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((3*x^2-4*x+5)/(-1+x)/(x^2+1),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.11887, size = 19, normalized size = 0.83 \begin{align*} 2 \log{\left (x - 1 \right )} + \frac{\log{\left (x^{2} + 1 \right )}}{2} - 3 \operatorname{atan}{\left (x \right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.06216, size = 27, normalized size = 1.17 \begin{align*} -3 \, \arctan \left (x\right ) + \frac{1}{2} \, \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((3*x^2-4*x+5)/(-1+x)/(x^2+1),x, algorithm="giac")

[Out]

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