\(\int \frac {1-2 x^2}{1+5 x^2+4 x^4} \, dx\) [56]

Optimal result

Integrand size = 22, antiderivative size = 9 \[ \int \frac {1-2 x^2}{1+5 x^2+4 x^4} \, dx=-\arctan (x)+\arctan (2 x) \]

[Out]

-arctan(x)+arctan(2*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1177, 209} \[ \int \frac {1-2 x^2}{1+5 x^2+4 x^4} \, dx=\arctan (2 x)-\arctan (x) \]

[In]

Int[(1 - 2*x^2)/(1 + 5*x^2 + 4*x^4),x]

[Out]

-ArcTan[x] + ArcTan[2*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 1177

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && GtQ[b^2
- 4*a*c, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.33 \[ \int \frac {1-2 x^2}{1+5 x^2+4 x^4} \, dx=\arctan \left (\frac {x}{1+2 x^2}\right ) \]

[In]

Integrate[(1 - 2*x^2)/(1 + 5*x^2 + 4*x^4),x]

[Out]

ArcTan[x/(1 + 2*x^2)]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.11

method	result	size
default	\(-\arctan \left (x \right )+\arctan \left (2 x \right )\)	\(10\)
risch	\(-\arctan \left (2 x \right )+\arctan \left (4 x^{3}+3 x \right )\)	\(18\)
parallelrisch	\(\frac {i \ln \left (x -i\right )}{2}-\frac {i \ln \left (x +i\right )}{2}-\frac {i \ln \left (x -\frac {i}{2}\right )}{2}+\frac {i \ln \left (x +\frac {i}{2}\right )}{2}\)	\(34\)

[In]

int((-2*x^2+1)/(4*x^4+5*x^2+1),x,method=_RETURNVERBOSE)

[Out]

-arctan(x)+arctan(2*x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.89 \[ \int \frac {1-2 x^2}{1+5 x^2+4 x^4} \, dx=\arctan \left (4 \, x^{3} + 3 \, x\right ) - \arctan \left (2 \, x\right ) \]

[In]

integrate((-2*x^2+1)/(4*x^4+5*x^2+1),x, algorithm="fricas")

[Out]

arctan(4*x^3 + 3*x) - arctan(2*x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.56 \[ \int \frac {1-2 x^2}{1+5 x^2+4 x^4} \, dx=- \operatorname {atan}{\left (2 x \right )} + \operatorname {atan}{\left (4 x^{3} + 3 x \right )} \]

[In]

integrate((-2*x**2+1)/(4*x**4+5*x**2+1),x)

[Out]

-atan(2*x) + atan(4*x**3 + 3*x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {1-2 x^2}{1+5 x^2+4 x^4} \, dx=\arctan \left (2 \, x\right ) - \arctan \left (x\right ) \]

[In]

integrate((-2*x^2+1)/(4*x^4+5*x^2+1),x, algorithm="maxima")

[Out]

arctan(2*x) - arctan(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {1-2 x^2}{1+5 x^2+4 x^4} \, dx=\arctan \left (2 \, x\right ) - \arctan \left (x\right ) \]

[In]

integrate((-2*x^2+1)/(4*x^4+5*x^2+1),x, algorithm="giac")

[Out]

arctan(2*x) - arctan(x)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.89 \[ \int \frac {1-2 x^2}{1+5 x^2+4 x^4} \, dx=\mathrm {atan}\left (4\,x^3+3\,x\right )-\mathrm {atan}\left (2\,x\right ) \]

[In]

int(-(2*x^2 - 1)/(5*x^2 + 4*x^4 + 1),x)

[Out]

atan(3*x + 4*x^3) - atan(2*x)