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

Optimal. Leaf size=43 \[ -\frac{1-2 x}{5 \left (x^2+1\right )}-\frac{14}{25} \log \left (x^2+1\right )-\frac{47}{25} \log (2-x)-\frac{46}{25} \tan ^{-1}(x) \]

[Out]

-(1 - 2*x)/(5*(1 + x^2)) - (46*ArcTan[x])/25 - (47*Log[2 - x])/25 - (14*Log[1 + x^2])/25

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Rubi [A]  time = 0.0669825, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1647, 1629, 635, 203, 260} \[ -\frac{1-2 x}{5 \left (x^2+1\right )}-\frac{14}{25} \log \left (x^2+1\right )-\frac{47}{25} \log (2-x)-\frac{46}{25} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 - 2*x)/(5*(1 + x^2)) - (46*ArcTan[x])/25 - (47*Log[2 - x])/25 - (14*Log[1 + x^2])/25

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

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{1-3 x^4}{(-2+x) \left (1+x^2\right )^2} \, dx &=-\frac{1-2 x}{5 \left (1+x^2\right )}-\frac{1}{2} \int \frac{-\frac{18}{5}-\frac{4 x}{5}+6 x^2}{(-2+x) \left (1+x^2\right )} \, dx\\ &=-\frac{1-2 x}{5 \left (1+x^2\right )}-\frac{1}{2} \int \left (\frac{94}{25 (-2+x)}+\frac{4 (23+14 x)}{25 \left (1+x^2\right )}\right ) \, dx\\ &=-\frac{1-2 x}{5 \left (1+x^2\right )}-\frac{47}{25} \log (2-x)-\frac{2}{25} \int \frac{23+14 x}{1+x^2} \, dx\\ &=-\frac{1-2 x}{5 \left (1+x^2\right )}-\frac{47}{25} \log (2-x)-\frac{28}{25} \int \frac{x}{1+x^2} \, dx-\frac{46}{25} \int \frac{1}{1+x^2} \, dx\\ &=-\frac{1-2 x}{5 \left (1+x^2\right )}-\frac{46}{25} \tan ^{-1}(x)-\frac{47}{25} \log (2-x)-\frac{14}{25} \log \left (1+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0226428, size = 57, normalized size = 1.33 \[ \frac{2 (x-2)+3}{5 \left ((x-2)^2+4 (x-2)+5\right )}-\frac{14}{25} \log \left ((x-2)^2+4 (x-2)+5\right )-\frac{47}{25} \log (x-2)-\frac{46}{25} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3 + 2*(-2 + x))/(5*(5 + 4*(-2 + x) + (-2 + x)^2)) - (46*ArcTan[x])/25 - (14*Log[5 + 4*(-2 + x) + (-2 + x)^2])
/25 - (47*Log[-2 + x])/25

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Maple [A]  time = 0.01, size = 34, normalized size = 0.8 \begin{align*} -{\frac{2}{25\,{x}^{2}+25} \left ( -5\,x+{\frac{5}{2}} \right ) }-{\frac{14\,\ln \left ({x}^{2}+1 \right ) }{25}}-{\frac{46\,\arctan \left ( x \right ) }{25}}-{\frac{47\,\ln \left ( -2+x \right ) }{25}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/25*(-5*x+5/2)/(x^2+1)-14/25*ln(x^2+1)-46/25*arctan(x)-47/25*ln(-2+x)

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Maxima [A]  time = 1.53747, size = 45, normalized size = 1.05 \begin{align*} \frac{2 \, x - 1}{5 \,{\left (x^{2} + 1\right )}} - \frac{46}{25} \, \arctan \left (x\right ) - \frac{14}{25} \, \log \left (x^{2} + 1\right ) - \frac{47}{25} \, \log \left (x - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(2*x - 1)/(x^2 + 1) - 46/25*arctan(x) - 14/25*log(x^2 + 1) - 47/25*log(x - 2)

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Fricas [A]  time = 1.4701, size = 144, normalized size = 3.35 \begin{align*} -\frac{46 \,{\left (x^{2} + 1\right )} \arctan \left (x\right ) + 14 \,{\left (x^{2} + 1\right )} \log \left (x^{2} + 1\right ) + 47 \,{\left (x^{2} + 1\right )} \log \left (x - 2\right ) - 10 \, x + 5}{25 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/25*(46*(x^2 + 1)*arctan(x) + 14*(x^2 + 1)*log(x^2 + 1) + 47*(x^2 + 1)*log(x - 2) - 10*x + 5)/(x^2 + 1)

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Sympy [A]  time = 0.154323, size = 36, normalized size = 0.84 \begin{align*} \frac{2 x - 1}{5 x^{2} + 5} - \frac{47 \log{\left (x - 2 \right )}}{25} - \frac{14 \log{\left (x^{2} + 1 \right )}}{25} - \frac{46 \operatorname{atan}{\left (x \right )}}{25} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x - 1)/(5*x**2 + 5) - 47*log(x - 2)/25 - 14*log(x**2 + 1)/25 - 46*atan(x)/25

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Giac [A]  time = 1.14189, size = 46, normalized size = 1.07 \begin{align*} \frac{2 \, x - 1}{5 \,{\left (x^{2} + 1\right )}} - \frac{46}{25} \, \arctan \left (x\right ) - \frac{14}{25} \, \log \left (x^{2} + 1\right ) - \frac{47}{25} \, \log \left ({\left | x - 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(2*x - 1)/(x^2 + 1) - 46/25*arctan(x) - 14/25*log(x^2 + 1) - 47/25*log(abs(x - 2))