3.335 \(\int \frac {-x+2 x^3+4 x^5}{(3+2 x^2+x^4)^2} \, dx\)

Optimal. Leaf size=45 \[ \frac {9 \tan ^{-1}\left (\frac {x^2+1}{\sqrt {2}}\right )}{8 \sqrt {2}}+\frac {5-7 x^2}{8 \left (x^4+2 x^2+3\right )} \]

[Out]

1/8*(-7*x^2+5)/(x^4+2*x^2+3)+9/16*arctan(1/2*(x^2+1)*2^(1/2))*2^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1594, 1663, 1660, 12, 618, 204} \[ \frac {5-7 x^2}{8 \left (x^4+2 x^2+3\right )}+\frac {9 \tan ^{-1}\left (\frac {x^2+1}{\sqrt {2}}\right )}{8 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[(-x + 2*x^3 + 4*x^5)/(3 + 2*x^2 + x^4)^2,x]

[Out]

(5 - 7*x^2)/(8*(3 + 2*x^2 + x^4)) + (9*ArcTan[(1 + x^2)/Sqrt[2]])/(8*Sqrt[2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 1663

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)
*SubstFor[x^2, Pq, x]*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2] && Inte
gerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {-x+2 x^3+4 x^5}{\left (3+2 x^2+x^4\right )^2} \, dx &=\int \frac {x \left (-1+2 x^2+4 x^4\right )}{\left (3+2 x^2+x^4\right )^2} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {-1+2 x+4 x^2}{\left (3+2 x+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac {5-7 x^2}{8 \left (3+2 x^2+x^4\right )}+\frac {1}{16} \operatorname {Subst}\left (\int \frac {18}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=\frac {5-7 x^2}{8 \left (3+2 x^2+x^4\right )}+\frac {9}{8} \operatorname {Subst}\left (\int \frac {1}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=\frac {5-7 x^2}{8 \left (3+2 x^2+x^4\right )}-\frac {9}{4} \operatorname {Subst}\left (\int \frac {1}{-8-x^2} \, dx,x,2 \left (1+x^2\right )\right )\\ &=\frac {5-7 x^2}{8 \left (3+2 x^2+x^4\right )}+\frac {9 \tan ^{-1}\left (\frac {1+x^2}{\sqrt {2}}\right )}{8 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 45, normalized size = 1.00 \[ \frac {9 \tan ^{-1}\left (\frac {x^2+1}{\sqrt {2}}\right )}{8 \sqrt {2}}+\frac {5-7 x^2}{8 \left (x^4+2 x^2+3\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(-x + 2*x^3 + 4*x^5)/(3 + 2*x^2 + x^4)^2,x]

[Out]

(5 - 7*x^2)/(8*(3 + 2*x^2 + x^4)) + (9*ArcTan[(1 + x^2)/Sqrt[2]])/(8*Sqrt[2])

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fricas [A]  time = 0.93, size = 47, normalized size = 1.04 \[ \frac {9 \, \sqrt {2} {\left (x^{4} + 2 \, x^{2} + 3\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) - 14 \, x^{2} + 10}{16 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^5+2*x^3-x)/(x^4+2*x^2+3)^2,x, algorithm="fricas")

[Out]

1/16*(9*sqrt(2)*(x^4 + 2*x^2 + 3)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 14*x^2 + 10)/(x^4 + 2*x^2 + 3)

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giac [A]  time = 1.11, size = 38, normalized size = 0.84 \[ \frac {9}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) - \frac {7 \, x^{2} - 5}{8 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^5+2*x^3-x)/(x^4+2*x^2+3)^2,x, algorithm="giac")

[Out]

9/16*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 1/8*(7*x^2 - 5)/(x^4 + 2*x^2 + 3)

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maple [A]  time = 0.01, size = 41, normalized size = 0.91 \[ \frac {9 \sqrt {2}\, \arctan \left (\frac {\left (2 x^{2}+2\right ) \sqrt {2}}{4}\right )}{16}+\frac {-\frac {7 x^{2}}{4}+\frac {5}{4}}{2 x^{4}+4 x^{2}+6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^5+2*x^3-x)/(x^4+2*x^2+3)^2,x)

[Out]

1/2*(-7/4*x^2+5/4)/(x^4+2*x^2+3)+9/16*2^(1/2)*arctan(1/4*(2*x^2+2)*2^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {7 \, x^{2} - 5}{8 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} + \frac {9}{4} \, \int \frac {x}{x^{4} + 2 \, x^{2} + 3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^5+2*x^3-x)/(x^4+2*x^2+3)^2,x, algorithm="maxima")

[Out]

-1/8*(7*x^2 - 5)/(x^4 + 2*x^2 + 3) + 9/4*integrate(x/(x^4 + 2*x^2 + 3), x)

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mupad [B]  time = 2.19, size = 42, normalized size = 0.93 \[ \frac {9\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x^2}{2}+\frac {\sqrt {2}}{2}\right )}{16}-\frac {\frac {7\,x^2}{8}-\frac {5}{8}}{x^4+2\,x^2+3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^3 - x + 4*x^5)/(2*x^2 + x^4 + 3)^2,x)

[Out]

(9*2^(1/2)*atan(2^(1/2)/2 + (2^(1/2)*x^2)/2))/16 - ((7*x^2)/8 - 5/8)/(2*x^2 + x^4 + 3)

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sympy [A]  time = 0.15, size = 44, normalized size = 0.98 \[ \frac {5 - 7 x^{2}}{8 x^{4} + 16 x^{2} + 24} + \frac {9 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x^{2}}{2} + \frac {\sqrt {2}}{2} \right )}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**5+2*x**3-x)/(x**4+2*x**2+3)**2,x)

[Out]

(5 - 7*x**2)/(8*x**4 + 16*x**2 + 24) + 9*sqrt(2)*atan(sqrt(2)*x**2/2 + sqrt(2)/2)/16

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