3.104 \(\int \frac{x (4+x^2+3 x^4+5 x^6)}{(3+2 x^2+x^4)^2} \, dx\)

Optimal. Leaf size=58 \[ \frac{25 \left (x^2+1\right )}{8 \left (x^4+2 x^2+3\right )}+\frac{5}{4} \log \left (x^4+2 x^2+3\right )-\frac{23 \tan ^{-1}\left (\frac{x^2+1}{\sqrt{2}}\right )}{8 \sqrt{2}} \]

[Out]

(25*(1 + x^2))/(8*(3 + 2*x^2 + x^4)) - (23*ArcTan[(1 + x^2)/Sqrt[2]])/(8*Sqrt[2]) + (5*Log[3 + 2*x^2 + x^4])/4

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Rubi [A]  time = 0.0669888, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {1663, 1660, 634, 618, 204, 628} \[ \frac{25 \left (x^2+1\right )}{8 \left (x^4+2 x^2+3\right )}+\frac{5}{4} \log \left (x^4+2 x^2+3\right )-\frac{23 \tan ^{-1}\left (\frac{x^2+1}{\sqrt{2}}\right )}{8 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[(x*(4 + x^2 + 3*x^4 + 5*x^6))/(3 + 2*x^2 + x^4)^2,x]

[Out]

(25*(1 + x^2))/(8*(3 + 2*x^2 + x^4)) - (23*ArcTan[(1 + x^2)/Sqrt[2]])/(8*Sqrt[2]) + (5*Log[3 + 2*x^2 + x^4])/4

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]

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 634

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

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 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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{x \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{4+x+3 x^2+5 x^3}{\left (3+2 x+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{25 \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac{1}{16} \operatorname{Subst}\left (\int \frac{-6+40 x}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=\frac{25 \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac{5}{4} \operatorname{Subst}\left (\int \frac{2+2 x}{3+2 x+x^2} \, dx,x,x^2\right )-\frac{23}{8} \operatorname{Subst}\left (\int \frac{1}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=\frac{25 \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac{5}{4} \log \left (3+2 x^2+x^4\right )+\frac{23}{4} \operatorname{Subst}\left (\int \frac{1}{-8-x^2} \, dx,x,2 \left (1+x^2\right )\right )\\ &=\frac{25 \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac{23 \tan ^{-1}\left (\frac{1+x^2}{\sqrt{2}}\right )}{8 \sqrt{2}}+\frac{5}{4} \log \left (3+2 x^2+x^4\right )\\ \end{align*}

Mathematica [A]  time = 0.0223489, size = 58, normalized size = 1. \[ \frac{25 \left (x^2+1\right )}{8 \left (x^4+2 x^2+3\right )}+\frac{5}{4} \log \left (x^4+2 x^2+3\right )-\frac{23 \tan ^{-1}\left (\frac{x^2+1}{\sqrt{2}}\right )}{8 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(4 + x^2 + 3*x^4 + 5*x^6))/(3 + 2*x^2 + x^4)^2,x]

[Out]

(25*(1 + x^2))/(8*(3 + 2*x^2 + x^4)) - (23*ArcTan[(1 + x^2)/Sqrt[2]])/(8*Sqrt[2]) + (5*Log[3 + 2*x^2 + x^4])/4

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Maple [A]  time = 0.01, size = 54, normalized size = 0.9 \begin{align*}{\frac{1}{2\,{x}^{4}+4\,{x}^{2}+6} \left ({\frac{25\,{x}^{2}}{4}}+{\frac{25}{4}} \right ) }+{\frac{5\,\ln \left ({x}^{4}+2\,{x}^{2}+3 \right ) }{4}}-{\frac{23\,\sqrt{2}}{16}\arctan \left ({\frac{ \left ( 2\,{x}^{2}+2 \right ) \sqrt{2}}{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(5*x^6+3*x^4+x^2+4)/(x^4+2*x^2+3)^2,x)

[Out]

1/2*(25/4*x^2+25/4)/(x^4+2*x^2+3)+5/4*ln(x^4+2*x^2+3)-23/16*2^(1/2)*arctan(1/4*(2*x^2+2)*2^(1/2))

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Maxima [A]  time = 1.47022, size = 66, normalized size = 1.14 \begin{align*} -\frac{23}{16} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (x^{2} + 1\right )}\right ) + \frac{25 \,{\left (x^{2} + 1\right )}}{8 \,{\left (x^{4} + 2 \, x^{2} + 3\right )}} + \frac{5}{4} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(5*x^6+3*x^4+x^2+4)/(x^4+2*x^2+3)^2,x, algorithm="maxima")

[Out]

-23/16*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) + 25/8*(x^2 + 1)/(x^4 + 2*x^2 + 3) + 5/4*log(x^4 + 2*x^2 + 3)

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Fricas [A]  time = 1.52721, size = 194, normalized size = 3.34 \begin{align*} -\frac{23 \, \sqrt{2}{\left (x^{4} + 2 \, x^{2} + 3\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (x^{2} + 1\right )}\right ) - 50 \, x^{2} - 20 \,{\left (x^{4} + 2 \, x^{2} + 3\right )} \log \left (x^{4} + 2 \, x^{2} + 3\right ) - 50}{16 \,{\left (x^{4} + 2 \, x^{2} + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(5*x^6+3*x^4+x^2+4)/(x^4+2*x^2+3)^2,x, algorithm="fricas")

[Out]

-1/16*(23*sqrt(2)*(x^4 + 2*x^2 + 3)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 50*x^2 - 20*(x^4 + 2*x^2 + 3)*log(x^4 + 2*
x^2 + 3) - 50)/(x^4 + 2*x^2 + 3)

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Sympy [A]  time = 0.161755, size = 60, normalized size = 1.03 \begin{align*} \frac{25 x^{2} + 25}{8 x^{4} + 16 x^{2} + 24} + \frac{5 \log{\left (x^{4} + 2 x^{2} + 3 \right )}}{4} - \frac{23 \sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} x^{2}}{2} + \frac{\sqrt{2}}{2} \right )}}{16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(5*x**6+3*x**4+x**2+4)/(x**4+2*x**2+3)**2,x)

[Out]

(25*x**2 + 25)/(8*x**4 + 16*x**2 + 24) + 5*log(x**4 + 2*x**2 + 3)/4 - 23*sqrt(2)*atan(sqrt(2)*x**2/2 + sqrt(2)
/2)/16

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Giac [A]  time = 1.10919, size = 66, normalized size = 1.14 \begin{align*} -\frac{23}{16} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (x^{2} + 1\right )}\right ) + \frac{25 \,{\left (x^{2} + 1\right )}}{8 \,{\left (x^{4} + 2 \, x^{2} + 3\right )}} + \frac{5}{4} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(5*x^6+3*x^4+x^2+4)/(x^4+2*x^2+3)^2,x, algorithm="giac")

[Out]

-23/16*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) + 25/8*(x^2 + 1)/(x^4 + 2*x^2 + 3) + 5/4*log(x^4 + 2*x^2 + 3)