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

Optimal. Leaf size=49 \[ -\frac{1}{4 \left (x^2+1\right )}+\frac{1}{4} \log \left (x^2+1\right )+\frac{1}{4 (1-x)}-\frac{1}{2} \log (1-x)+\frac{1}{4} \tan ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.0288985, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {741, 801, 635, 203, 260} \[ -\frac{1}{4 \left (x^2+1\right )}+\frac{1}{4} \log \left (x^2+1\right )+\frac{1}{4 (1-x)}-\frac{1}{2} \log (1-x)+\frac{1}{4} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[1/((-1 + x)^2*(1 + x^2)^2),x]

[Out]

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

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

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

Mathematica [A]  time = 0.0195791, size = 35, normalized size = 0.71 \[ \frac{1}{4} \left (-\frac{1}{x^2+1}+\log \left (x^2+1\right )+\frac{1}{1-x}-2 \log (x-1)+\tan ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.011, size = 36, normalized size = 0.7 \begin{align*} -{\frac{1}{4\,{x}^{2}+4}}+{\frac{\ln \left ({x}^{2}+1 \right ) }{4}}+{\frac{\arctan \left ( x \right ) }{4}}-{\frac{1}{-4+4\,x}}-{\frac{\ln \left ( -1+x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.43306, size = 53, normalized size = 1.08 \begin{align*} -\frac{x^{2} + x}{4 \,{\left (x^{3} - x^{2} + x - 1\right )}} + \frac{1}{4} \, \arctan \left (x\right ) + \frac{1}{4} \, \log \left (x^{2} + 1\right ) - \frac{1}{2} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.93164, size = 186, normalized size = 3.8 \begin{align*} -\frac{x^{2} -{\left (x^{3} - x^{2} + x - 1\right )} \arctan \left (x\right ) -{\left (x^{3} - x^{2} + x - 1\right )} \log \left (x^{2} + 1\right ) + 2 \,{\left (x^{3} - x^{2} + x - 1\right )} \log \left (x - 1\right ) + x}{4 \,{\left (x^{3} - x^{2} + x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.161398, size = 39, normalized size = 0.8 \begin{align*} - \frac{x^{2} + x}{4 x^{3} - 4 x^{2} + 4 x - 4} - \frac{\log{\left (x - 1 \right )}}{2} + \frac{\log{\left (x^{2} + 1 \right )}}{4} + \frac{\operatorname{atan}{\left (x \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.11861, size = 108, normalized size = 2.2 \begin{align*} \frac{1}{16} \, \pi - \frac{1}{4} \, \pi \left \lfloor \frac{\pi + 4 \, \arctan \left (x\right )}{4 \, \pi } + \frac{1}{2} \right \rfloor + \frac{\frac{2}{x - 1} + 1}{8 \,{\left (\frac{2}{x - 1} + \frac{2}{{\left (x - 1\right )}^{2}} + 1\right )}} - \frac{1}{4 \,{\left (x - 1\right )}} + \frac{1}{4} \, \arctan \left (x\right ) + \frac{1}{4} \, \log \left (\frac{2}{x - 1} + \frac{2}{{\left (x - 1\right )}^{2}} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*pi - 1/4*pi*floor(1/4*(pi + 4*arctan(x))/pi + 1/2) + 1/8*(2/(x - 1) + 1)/(2/(x - 1) + 2/(x - 1)^2 + 1) -
1/4/(x - 1) + 1/4*arctan(x) + 1/4*log(2/(x - 1) + 2/(x - 1)^2 + 1)