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

Optimal. Leaf size=53 \[ \frac{x \left (11 x^2+9\right )}{4 \left (x^4+3 x^2+2\right )}-\frac{1}{x}-\frac{19}{2} \tan ^{-1}(x)+\frac{45 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{4 \sqrt{2}} \]

[Out]

-x^(-1) + (x*(9 + 11*x^2))/(4*(2 + 3*x^2 + x^4)) - (19*ArcTan[x])/2 + (45*ArcTan[x/Sqrt[2]])/(4*Sqrt[2])

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Rubi [A]  time = 0.0729781, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {1669, 1664, 203} \[ \frac{x \left (11 x^2+9\right )}{4 \left (x^4+3 x^2+2\right )}-\frac{1}{x}-\frac{19}{2} \tan ^{-1}(x)+\frac{45 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{4 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^(-1) + (x*(9 + 11*x^2))/(4*(2 + 3*x^2 + x^4)) - (19*ArcTan[x])/2 + (45*ArcTan[x/Sqrt[2]])/(4*Sqrt[2])

Rule 1669

Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainde
r[x^m*Pq, a + b*x^2 + c*x^4, x], x, 0], e = Coeff[PolynomialRemainder[x^m*Pq, a + b*x^2 + c*x^4, x], x, 2]}, S
imp[(x*(a + b*x^2 + c*x^4)^(p + 1)*(a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2))/(2*a*(p + 1)*(b^2 - 4*a*c)
), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[x^m*(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[(2*a*(p + 1)*(b^2
- 4*a*c)*PolynomialQuotient[x^m*Pq, a + b*x^2 + c*x^4, x])/x^m + (b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e)
/x^m + c*(4*p + 7)*(b*d - 2*a*e)*x^(2 - m), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && GtQ[Expon[
Pq, x^2], 1] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && ILtQ[m/2, 0]

Rule 1664

Int[(Pq_)*((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d*x
)^m*Pq*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && PolyQ[Pq, x^2] && IGtQ[p, -2]

Rule 203

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

Rubi steps

\begin{align*} \int \frac{4+x^2+3 x^4+5 x^6}{x^2 \left (2+3 x^2+x^4\right )^2} \, dx &=\frac{x \left (9+11 x^2\right )}{4 \left (2+3 x^2+x^4\right )}-\frac{1}{4} \int \frac{-8+19 x^2-11 x^4}{x^2 \left (2+3 x^2+x^4\right )} \, dx\\ &=\frac{x \left (9+11 x^2\right )}{4 \left (2+3 x^2+x^4\right )}-\frac{1}{4} \int \left (-\frac{4}{x^2}+\frac{38}{1+x^2}-\frac{45}{2+x^2}\right ) \, dx\\ &=-\frac{1}{x}+\frac{x \left (9+11 x^2\right )}{4 \left (2+3 x^2+x^4\right )}-\frac{19}{2} \int \frac{1}{1+x^2} \, dx+\frac{45}{4} \int \frac{1}{2+x^2} \, dx\\ &=-\frac{1}{x}+\frac{x \left (9+11 x^2\right )}{4 \left (2+3 x^2+x^4\right )}-\frac{19}{2} \tan ^{-1}(x)+\frac{45 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{4 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0493476, size = 51, normalized size = 0.96 \[ \frac{1}{8} \left (\frac{2 x \left (11 x^2+9\right )}{x^4+3 x^2+2}-\frac{8}{x}-76 \tan ^{-1}(x)+45 \sqrt{2} \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8/x + (2*x*(9 + 11*x^2))/(2 + 3*x^2 + x^4) - 76*ArcTan[x] + 45*Sqrt[2]*ArcTan[x/Sqrt[2]])/8

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Maple [A]  time = 0.012, size = 43, normalized size = 0.8 \begin{align*}{\frac{13\,x}{4\,{x}^{2}+8}}+{\frac{45\,\sqrt{2}}{8}\arctan \left ({\frac{x\sqrt{2}}{2}} \right ) }-{\frac{x}{2\,{x}^{2}+2}}-{\frac{19\,\arctan \left ( x \right ) }{2}}-{x}^{-1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

13/4*x/(x^2+2)+45/8*arctan(1/2*x*2^(1/2))*2^(1/2)-1/2*x/(x^2+1)-19/2*arctan(x)-1/x

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Maxima [A]  time = 1.45781, size = 61, normalized size = 1.15 \begin{align*} \frac{45}{8} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + \frac{7 \, x^{4} - 3 \, x^{2} - 8}{4 \,{\left (x^{5} + 3 \, x^{3} + 2 \, x\right )}} - \frac{19}{2} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

45/8*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/4*(7*x^4 - 3*x^2 - 8)/(x^5 + 3*x^3 + 2*x) - 19/2*arctan(x)

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Fricas [A]  time = 2.17958, size = 185, normalized size = 3.49 \begin{align*} \frac{14 \, x^{4} + 45 \, \sqrt{2}{\left (x^{5} + 3 \, x^{3} + 2 \, x\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) - 6 \, x^{2} - 76 \,{\left (x^{5} + 3 \, x^{3} + 2 \, x\right )} \arctan \left (x\right ) - 16}{8 \,{\left (x^{5} + 3 \, x^{3} + 2 \, x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(14*x^4 + 45*sqrt(2)*(x^5 + 3*x^3 + 2*x)*arctan(1/2*sqrt(2)*x) - 6*x^2 - 76*(x^5 + 3*x^3 + 2*x)*arctan(x)
- 16)/(x^5 + 3*x^3 + 2*x)

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Sympy [A]  time = 0.201047, size = 49, normalized size = 0.92 \begin{align*} \frac{7 x^{4} - 3 x^{2} - 8}{4 x^{5} + 12 x^{3} + 8 x} - \frac{19 \operatorname{atan}{\left (x \right )}}{2} + \frac{45 \sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} x}{2} \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7*x**4 - 3*x**2 - 8)/(4*x**5 + 12*x**3 + 8*x) - 19*atan(x)/2 + 45*sqrt(2)*atan(sqrt(2)*x/2)/8

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Giac [A]  time = 1.12253, size = 61, normalized size = 1.15 \begin{align*} \frac{45}{8} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + \frac{7 \, x^{4} - 3 \, x^{2} - 8}{4 \,{\left (x^{5} + 3 \, x^{3} + 2 \, x\right )}} - \frac{19}{2} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

45/8*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/4*(7*x^4 - 3*x^2 - 8)/(x^5 + 3*x^3 + 2*x) - 19/2*arctan(x)