\(\int \frac {1-3 x+2 x^2-4 x^3+x^4}{1+3 x^2+3 x^4+x^6} \, dx\) [332]

Optimal result

Integrand size = 36, antiderivative size = 23 \[ \int \frac {1-3 x+2 x^2-4 x^3+x^4}{1+3 x^2+3 x^4+x^6} \, dx=-\frac {1}{4 \left (1+x^2\right )^2}+\frac {2}{1+x^2}+\arctan (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2098, 267, 209} \[ \int \frac {1-3 x+2 x^2-4 x^3+x^4}{1+3 x^2+3 x^4+x^6} \, dx=\arctan (x)+\frac {2}{x^2+1}-\frac {1}{4 \left (x^2+1\right )^2} \]

[In]

Int[(1 - 3*x + 2*x^2 - 4*x^3 + x^4)/(1 + 3*x^2 + 3*x^4 + x^6),x]

[Out]

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

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 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 2098

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P /. x -> Sqrt[x]]}, Int[ExpandIntegrand[(PP /. x ->
x^2)^p*Q^q, x], x] /;  !SumQ[NonfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x^2] && PolyQ[Q, x] && ILtQ[p,
0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {x}{\left (1+x^2\right )^3}-\frac {4 x}{\left (1+x^2\right )^2}+\frac {1}{1+x^2}\right ) \, dx \\ & = -\left (4 \int \frac {x}{\left (1+x^2\right )^2} \, dx\right )+\int \frac {x}{\left (1+x^2\right )^3} \, dx+\int \frac {1}{1+x^2} \, dx \\ & = -\frac {1}{4 \left (1+x^2\right )^2}+\frac {2}{1+x^2}+\tan ^{-1}(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1-3 x+2 x^2-4 x^3+x^4}{1+3 x^2+3 x^4+x^6} \, dx=-\frac {1}{4 \left (1+x^2\right )^2}+\frac {2}{1+x^2}+\arctan (x) \]

[In]

Integrate[(1 - 3*x + 2*x^2 - 4*x^3 + x^4)/(1 + 3*x^2 + 3*x^4 + x^6),x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {2 x^{2}+\frac {7}{4}}{\left (x^{2}+1\right )^{2}}+\arctan \left (x \right )\)	\(19\)
risch	\(\frac {2 x^{2}+\frac {7}{4}}{x^{4}+2 x^{2}+1}+\arctan \left (x \right )\)	\(24\)
parallelrisch	\(-\frac {2 i \ln \left (x -i\right ) x^{4}-2 i \ln \left (x +i\right ) x^{4}-3+4 i \ln \left (x -i\right ) x^{2}-4 i \ln \left (x +i\right ) x^{2}+4 x^{4}+2 i \ln \left (x -i\right )-2 i \ln \left (x +i\right )}{4 \left (x^{4}+2 x^{2}+1\right )}\)	\(82\)

[In]

int((x^4-4*x^3+2*x^2-3*x+1)/(x^6+3*x^4+3*x^2+1),x,method=_RETURNVERBOSE)

[Out]

(2*x^2+7/4)/(x^2+1)^2+arctan(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {1-3 x+2 x^2-4 x^3+x^4}{1+3 x^2+3 x^4+x^6} \, dx=\frac {8 \, x^{2} + 4 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (x\right ) + 7}{4 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \]

[In]

integrate((x^4-4*x^3+2*x^2-3*x+1)/(x^6+3*x^4+3*x^2+1),x, algorithm="fricas")

[Out]

1/4*(8*x^2 + 4*(x^4 + 2*x^2 + 1)*arctan(x) + 7)/(x^4 + 2*x^2 + 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {1-3 x+2 x^2-4 x^3+x^4}{1+3 x^2+3 x^4+x^6} \, dx=\frac {8 x^{2} + 7}{4 x^{4} + 8 x^{2} + 4} + \operatorname {atan}{\left (x \right )} \]

[In]

integrate((x**4-4*x**3+2*x**2-3*x+1)/(x**6+3*x**4+3*x**2+1),x)

[Out]

(8*x**2 + 7)/(4*x**4 + 8*x**2 + 4) + atan(x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1-3 x+2 x^2-4 x^3+x^4}{1+3 x^2+3 x^4+x^6} \, dx=\frac {8 \, x^{2} + 7}{4 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} + \arctan \left (x\right ) \]

[In]

integrate((x^4-4*x^3+2*x^2-3*x+1)/(x^6+3*x^4+3*x^2+1),x, algorithm="maxima")

[Out]

1/4*(8*x^2 + 7)/(x^4 + 2*x^2 + 1) + arctan(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1-3 x+2 x^2-4 x^3+x^4}{1+3 x^2+3 x^4+x^6} \, dx=\frac {8 \, x^{2} + 7}{4 \, {\left (x^{2} + 1\right )}^{2}} + \arctan \left (x\right ) \]

[In]

integrate((x^4-4*x^3+2*x^2-3*x+1)/(x^6+3*x^4+3*x^2+1),x, algorithm="giac")

[Out]

1/4*(8*x^2 + 7)/(x^2 + 1)^2 + arctan(x)

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1-3 x+2 x^2-4 x^3+x^4}{1+3 x^2+3 x^4+x^6} \, dx=\mathrm {atan}\left (x\right )+\frac {2\,x^2+\frac {7}{4}}{x^4+2\,x^2+1} \]

[In]

int((2*x^2 - 3*x - 4*x^3 + x^4 + 1)/(3*x^2 + 3*x^4 + x^6 + 1),x)

[Out]

atan(x) + (2*x^2 + 7/4)/(2*x^2 + x^4 + 1)