\(\int \frac {1-3 x+2 x^2-x^3}{(1+x^2)^2} \, dx\) [309]

Optimal result

Integrand size = 23, antiderivative size = 33 \[ \int \frac {1-3 x+2 x^2-x^3}{\left (1+x^2\right )^2} \, dx=\frac {2-x}{2 \left (1+x^2\right )}+\frac {3 \arctan (x)}{2}-\frac {1}{2} \log \left (1+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {1828, 649, 209, 266} \[ \int \frac {1-3 x+2 x^2-x^3}{\left (1+x^2\right )^2} \, dx=\frac {3 \arctan (x)}{2}+\frac {2-x}{2 \left (x^2+1\right )}-\frac {1}{2} \log \left (x^2+1\right ) \]

[In]

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

[Out]

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

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 1828

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*
g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {1-3 x+2 x^2-x^3}{\left (1+x^2\right )^2} \, dx=\frac {1}{2} \left (\frac {2-x}{1+x^2}+3 \arctan (x)-\log \left (1+x^2\right )\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82

method	result	size
risch	\(\frac {1-\frac {x}{2}}{x^{2}+1}-\frac {\ln \left (x^{2}+1\right )}{2}+\frac {3 \arctan \left (x \right )}{2}\)	\(27\)
default	\(-\frac {\frac {x}{2}-1}{x^{2}+1}-\frac {\ln \left (x^{2}+1\right )}{2}+\frac {3 \arctan \left (x \right )}{2}\)	\(28\)
meijerg	\(-\frac {x^{2}}{x^{2}+1}-\frac {\ln \left (x^{2}+1\right )}{2}-\frac {x}{x^{2}+1}+\frac {3 \arctan \left (x \right )}{2}+\frac {x}{2 x^{2}+2}\)	\(47\)
parallelrisch	\(-\frac {3 i \ln \left (x -i\right ) x^{2}-3 i \ln \left (x +i\right ) x^{2}+2 \ln \left (x -i\right ) x^{2}+2 \ln \left (x +i\right ) x^{2}-4+3 i \ln \left (x -i\right )-3 i \ln \left (x +i\right )+2 \ln \left (x -i\right )+2 \ln \left (x +i\right )+2 x}{4 \left (x^{2}+1\right )}\)	\(87\)

[In]

int((-x^3+2*x^2-3*x+1)/(x^2+1)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {1-3 x+2 x^2-x^3}{\left (1+x^2\right )^2} \, dx=\frac {3 \, {\left (x^{2} + 1\right )} \arctan \left (x\right ) - {\left (x^{2} + 1\right )} \log \left (x^{2} + 1\right ) - x + 2}{2 \, {\left (x^{2} + 1\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {1-3 x+2 x^2-x^3}{\left (1+x^2\right )^2} \, dx=- \frac {x - 2}{2 x^{2} + 2} - \frac {\log {\left (x^{2} + 1 \right )}}{2} + \frac {3 \operatorname {atan}{\left (x \right )}}{2} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {1-3 x+2 x^2-x^3}{\left (1+x^2\right )^2} \, dx=-\frac {x - 2}{2 \, {\left (x^{2} + 1\right )}} + \frac {3}{2} \, \arctan \left (x\right ) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {1-3 x+2 x^2-x^3}{\left (1+x^2\right )^2} \, dx=-\frac {x - 2}{2 \, {\left (x^{2} + 1\right )}} + \frac {3}{2} \, \arctan \left (x\right ) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {1-3 x+2 x^2-x^3}{\left (1+x^2\right )^2} \, dx=\frac {3\,\mathrm {atan}\left (x\right )}{2}-\frac {\ln \left (x^2+1\right )}{2}-\frac {x}{2\,\left (x^2+1\right )}+\frac {1}{x^2+1} \]

[In]

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

[Out]

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