\(\int \frac {x^4}{4+5 x^2+x^4} \, dx\) [121]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 16, antiderivative size = 18 \[ \int \frac {x^4}{4+5 x^2+x^4} \, dx=x-\frac {8}{3} \arctan \left (\frac {x}{2}\right )+\frac {\arctan (x)}{3} \]

[Out]

x-8/3*arctan(1/2*x)+1/3*arctan(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1136, 1180, 209} \[ \int \frac {x^4}{4+5 x^2+x^4} \, dx=-\frac {8}{3} \arctan \left (\frac {x}{2}\right )+\frac {\arctan (x)}{3}+x \]

[In]

Int[x^4/(4 + 5*x^2 + x^4),x]

[Out]

x - (8*ArcTan[x/2])/3 + ArcTan[x]/3

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 1136

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[d^3*(d*x)^(m - 3)*((a + b*
x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Dist[d^4/(c*(m + 4*p + 1)), Int[(d*x)^(m - 4)*Simp[a*(m - 3) + b
*(m + 2*p - 1)*x^2, x]*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && Gt
Q[m, 3] && NeQ[m + 4*p + 1, 0] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1180

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] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rubi steps \begin{align*} \text {integral}& = x-\int \frac {4+5 x^2}{4+5 x^2+x^4} \, dx \\ & = x+\frac {1}{3} \int \frac {1}{1+x^2} \, dx-\frac {16}{3} \int \frac {1}{4+x^2} \, dx \\ & = x-\frac {8}{3} \arctan \left (\frac {x}{2}\right )+\frac {\arctan (x)}{3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {x^4}{4+5 x^2+x^4} \, dx=x+\frac {8}{3} \arctan \left (\frac {2}{x}\right )+\frac {\arctan (x)}{3} \]

[In]

Integrate[x^4/(4 + 5*x^2 + x^4),x]

[Out]

x + (8*ArcTan[2/x])/3 + ArcTan[x]/3

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72

method result size
default \(x -\frac {8 \arctan \left (\frac {x}{2}\right )}{3}+\frac {\arctan \left (x \right )}{3}\) \(13\)
risch \(x -\frac {8 \arctan \left (\frac {x}{2}\right )}{3}+\frac {\arctan \left (x \right )}{3}\) \(13\)
parallelrisch \(x +\frac {i \ln \left (x +i\right )}{6}-\frac {i \ln \left (x -i\right )}{6}+\frac {4 i \ln \left (x -2 i\right )}{3}-\frac {4 i \ln \left (x +2 i\right )}{3}\) \(35\)

[In]

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

[Out]

x-8/3*arctan(1/2*x)+1/3*arctan(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {x^4}{4+5 x^2+x^4} \, dx=x - \frac {8}{3} \, \arctan \left (\frac {1}{2} \, x\right ) + \frac {1}{3} \, \arctan \left (x\right ) \]

[In]

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

[Out]

x - 8/3*arctan(1/2*x) + 1/3*arctan(x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {x^4}{4+5 x^2+x^4} \, dx=x - \frac {8 \operatorname {atan}{\left (\frac {x}{2} \right )}}{3} + \frac {\operatorname {atan}{\left (x \right )}}{3} \]

[In]

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

[Out]

x - 8*atan(x/2)/3 + atan(x)/3

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {x^4}{4+5 x^2+x^4} \, dx=x - \frac {8}{3} \, \arctan \left (\frac {1}{2} \, x\right ) + \frac {1}{3} \, \arctan \left (x\right ) \]

[In]

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

[Out]

x - 8/3*arctan(1/2*x) + 1/3*arctan(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {x^4}{4+5 x^2+x^4} \, dx=x - \frac {8}{3} \, \arctan \left (\frac {1}{2} \, x\right ) + \frac {1}{3} \, \arctan \left (x\right ) \]

[In]

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

[Out]

x - 8/3*arctan(1/2*x) + 1/3*arctan(x)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {x^4}{4+5 x^2+x^4} \, dx=x-\frac {8\,\mathrm {atan}\left (\frac {x}{2}\right )}{3}+\frac {\mathrm {atan}\left (x\right )}{3} \]

[In]

int(x^4/(5*x^2 + x^4 + 4),x)

[Out]

x - (8*atan(x/2))/3 + atan(x)/3