3.1.11 \(\int \frac {1}{(-1+x)^2 (1+x^2)^2} \, dx\) [11]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 49, normalized size of antiderivative = 1.00, 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 = {755, 815, 649, 209, 266} \begin {gather*} -\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) \end {gather*}

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 209

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

Rule 266

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]

Rule 649

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 755

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 815

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 {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*}

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Mathematica [A]
time = 0.02, size = 35, normalized size = 0.71 \begin {gather*} \frac {1}{4} \left (\frac {1}{1-x}-\frac {1}{1+x^2}+\tan ^{-1}(x)-2 \log (-1+x)+\log \left (1+x^2\right )\right ) \end {gather*}

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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Mathics [A]
time = 2.00, size = 49, normalized size = 1.00 \begin {gather*} \frac {-\frac {x \left (1+x\right )}{4}+\frac {\left (-1+x-x^2+x^3\right ) \left (\text {ArcTan}\left [x\right ]+\text {Log}\left [1+x^2\right ]-2 \text {Log}\left [-1+x\right ]\right )}{4}}{-1+x-x^2+x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.04, size = 36, normalized size = 0.73

method	result	size
default	\(-\frac {1}{4 \left (-1+x \right )}-\frac {\ln \left (-1+x \right )}{2}-\frac {1}{4 \left (x^{2}+1\right )}+\frac {\ln \left (x^{2}+1\right )}{4}+\frac {\arctan \left (x \right )}{4}\)	\(36\)
risch	\(\frac {-\frac {1}{4} x^{2}-\frac {1}{4} x}{\left (x^{2}+1\right ) \left (-1+x \right )}-\frac {\ln \left (-1+x \right )}{2}+\frac {\ln \left (x^{2}+1\right )}{4}+\frac {\arctan \left (x \right )}{4}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.35, size = 39, normalized size = 0.80 \begin {gather*} -\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 {gather*}

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 = 0.42, size = 71, normalized size = 1.45 \begin {gather*} -\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 {gather*}

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 = 0.09, size = 41, normalized size = 0.84 \begin {gather*} \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 {gather*}

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 [A]
time = 0.00, size = 45, normalized size = 0.92 \begin {gather*} -\frac {\ln \left |x-1\right |}{2}+\frac {\ln \left (x^{2}+1\right )}{4}+\frac {\arctan x}{4}+\frac {-x^{2}-x}{4 \left (x^{3}-x^{2}+x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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(abs(x - 1))

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Mupad [B]
time = 0.13, size = 49, normalized size = 1.00 \begin {gather*} -\frac {\ln \left (x-1\right )}{2}-\frac {\frac {x^2}{4}+\frac {x}{4}}{x^3-x^2+x-1}+\ln \left (x-\mathrm {i}\right )\,\left (\frac {1}{4}-\frac {1}{8}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (\frac {1}{4}+\frac {1}{8}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x^2 + 1)^2*(x - 1)^2),x)

[Out]

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

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