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

Optimal. Leaf size=71 \[ -\frac{25 \left (x^2+5\right )}{72 \left (x^4+2 x^2+3\right )}-\frac{2}{9 x^2}+\frac{13}{108} \log \left (x^4+2 x^2+3\right )-\frac{71 \tan ^{-1}\left (\frac{x^2+1}{\sqrt{2}}\right )}{216 \sqrt{2}}-\frac{13 \log (x)}{27} \]

[Out]

-2/(9*x^2) - (25*(5 + x^2))/(72*(3 + 2*x^2 + x^4)) - (71*ArcTan[(1 + x^2)/Sqrt[2]])/(216*Sqrt[2]) - (13*Log[x]
)/27 + (13*Log[3 + 2*x^2 + x^4])/108

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Rubi [A]  time = 0.133811, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {1663, 1646, 1628, 634, 618, 204, 628} \[ -\frac{25 \left (x^2+5\right )}{72 \left (x^4+2 x^2+3\right )}-\frac{2}{9 x^2}+\frac{13}{108} \log \left (x^4+2 x^2+3\right )-\frac{71 \tan ^{-1}\left (\frac{x^2+1}{\sqrt{2}}\right )}{216 \sqrt{2}}-\frac{13 \log (x)}{27} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(9*x^2) - (25*(5 + x^2))/(72*(3 + 2*x^2 + x^4)) - (71*ArcTan[(1 + x^2)/Sqrt[2]])/(216*Sqrt[2]) - (13*Log[x]
)/27 + (13*Log[3 + 2*x^2 + x^4])/108

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 1646

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

Rule 1628

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

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{4+x^2+3 x^4+5 x^6}{x^3 \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}{x^2 \left (3+2 x+x^2\right )^2} \, dx,x,x^2\right )\\ &=-\frac{25 \left (5+x^2\right )}{72 \left (3+2 x^2+x^4\right )}+\frac{1}{16} \operatorname{Subst}\left (\int \frac{\frac{32}{3}-\frac{40 x}{9}-\frac{50 x^2}{9}}{x^2 \left (3+2 x+x^2\right )} \, dx,x,x^2\right )\\ &=-\frac{25 \left (5+x^2\right )}{72 \left (3+2 x^2+x^4\right )}+\frac{1}{16} \operatorname{Subst}\left (\int \left (\frac{32}{9 x^2}-\frac{104}{27 x}+\frac{2 (-19+52 x)}{27 \left (3+2 x+x^2\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac{2}{9 x^2}-\frac{25 \left (5+x^2\right )}{72 \left (3+2 x^2+x^4\right )}-\frac{13 \log (x)}{27}+\frac{1}{216} \operatorname{Subst}\left (\int \frac{-19+52 x}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=-\frac{2}{9 x^2}-\frac{25 \left (5+x^2\right )}{72 \left (3+2 x^2+x^4\right )}-\frac{13 \log (x)}{27}+\frac{13}{108} \operatorname{Subst}\left (\int \frac{2+2 x}{3+2 x+x^2} \, dx,x,x^2\right )-\frac{71}{216} \operatorname{Subst}\left (\int \frac{1}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=-\frac{2}{9 x^2}-\frac{25 \left (5+x^2\right )}{72 \left (3+2 x^2+x^4\right )}-\frac{13 \log (x)}{27}+\frac{13}{108} \log \left (3+2 x^2+x^4\right )+\frac{71}{108} \operatorname{Subst}\left (\int \frac{1}{-8-x^2} \, dx,x,2 \left (1+x^2\right )\right )\\ &=-\frac{2}{9 x^2}-\frac{25 \left (5+x^2\right )}{72 \left (3+2 x^2+x^4\right )}-\frac{71 \tan ^{-1}\left (\frac{1+x^2}{\sqrt{2}}\right )}{216 \sqrt{2}}-\frac{13 \log (x)}{27}+\frac{13}{108} \log \left (3+2 x^2+x^4\right )\\ \end{align*}

Mathematica [C]  time = 0.0512072, size = 97, normalized size = 1.37 \[ \frac{1}{864} \left (-\frac{300 \left (x^2+5\right )}{x^4+2 x^2+3}-\frac{192}{x^2}+\sqrt{2} \left (52 \sqrt{2}+71 i\right ) \log \left (x^2-i \sqrt{2}+1\right )+\sqrt{2} \left (52 \sqrt{2}-71 i\right ) \log \left (x^2+i \sqrt{2}+1\right )-416 \log (x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-192/x^2 - (300*(5 + x^2))/(3 + 2*x^2 + x^4) - 416*Log[x] + Sqrt[2]*(71*I + 52*Sqrt[2])*Log[1 - I*Sqrt[2] + x
^2] + Sqrt[2]*(-71*I + 52*Sqrt[2])*Log[1 + I*Sqrt[2] + x^2])/864

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Maple [A]  time = 0.014, size = 63, normalized size = 0.9 \begin{align*}{\frac{1}{54\,{x}^{4}+108\,{x}^{2}+162} \left ( -{\frac{75\,{x}^{2}}{4}}-{\frac{375}{4}} \right ) }+{\frac{13\,\ln \left ({x}^{4}+2\,{x}^{2}+3 \right ) }{108}}-{\frac{71\,\sqrt{2}}{432}\arctan \left ({\frac{ \left ( 2\,{x}^{2}+2 \right ) \sqrt{2}}{4}} \right ) }-{\frac{2}{9\,{x}^{2}}}-{\frac{13\,\ln \left ( x \right ) }{27}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/54*(-75/4*x^2-375/4)/(x^4+2*x^2+3)+13/108*ln(x^4+2*x^2+3)-71/432*2^(1/2)*arctan(1/4*(2*x^2+2)*2^(1/2))-2/9/x
^2-13/27*ln(x)

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Maxima [A]  time = 1.45779, size = 89, normalized size = 1.25 \begin{align*} -\frac{71}{432} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (x^{2} + 1\right )}\right ) - \frac{41 \, x^{4} + 157 \, x^{2} + 48}{72 \,{\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )}} + \frac{13}{108} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) - \frac{13}{54} \, \log \left (x^{2}\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^3/(x^4+2*x^2+3)^2,x, algorithm="maxima")

[Out]

-71/432*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 1/72*(41*x^4 + 157*x^2 + 48)/(x^6 + 2*x^4 + 3*x^2) + 13/108*lo
g(x^4 + 2*x^2 + 3) - 13/54*log(x^2)

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Fricas [A]  time = 1.61482, size = 275, normalized size = 3.87 \begin{align*} -\frac{246 \, x^{4} + 71 \, \sqrt{2}{\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (x^{2} + 1\right )}\right ) + 942 \, x^{2} - 52 \,{\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )} \log \left (x^{4} + 2 \, x^{2} + 3\right ) + 208 \,{\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )} \log \left (x\right ) + 288}{432 \,{\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\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^3/(x^4+2*x^2+3)^2,x, algorithm="fricas")

[Out]

-1/432*(246*x^4 + 71*sqrt(2)*(x^6 + 2*x^4 + 3*x^2)*arctan(1/2*sqrt(2)*(x^2 + 1)) + 942*x^2 - 52*(x^6 + 2*x^4 +
3*x^2)*log(x^4 + 2*x^2 + 3) + 208*(x^6 + 2*x^4 + 3*x^2)*log(x) + 288)/(x^6 + 2*x^4 + 3*x^2)

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Sympy [A]  time = 0.200721, size = 75, normalized size = 1.06 \begin{align*} - \frac{41 x^{4} + 157 x^{2} + 48}{72 x^{6} + 144 x^{4} + 216 x^{2}} - \frac{13 \log{\left (x \right )}}{27} + \frac{13 \log{\left (x^{4} + 2 x^{2} + 3 \right )}}{108} - \frac{71 \sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} x^{2}}{2} + \frac{\sqrt{2}}{2} \right )}}{432} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(41*x**4 + 157*x**2 + 48)/(72*x**6 + 144*x**4 + 216*x**2) - 13*log(x)/27 + 13*log(x**4 + 2*x**2 + 3)/108 - 71
*sqrt(2)*atan(sqrt(2)*x**2/2 + sqrt(2)/2)/432

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Giac [A]  time = 1.09812, size = 89, normalized size = 1.25 \begin{align*} -\frac{71}{432} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (x^{2} + 1\right )}\right ) - \frac{41 \, x^{4} + 157 \, x^{2} + 48}{72 \,{\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )}} + \frac{13}{108} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) - \frac{13}{54} \, \log \left (x^{2}\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^3/(x^4+2*x^2+3)^2,x, algorithm="giac")

[Out]

-71/432*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 1/72*(41*x^4 + 157*x^2 + 48)/(x^6 + 2*x^4 + 3*x^2) + 13/108*lo
g(x^4 + 2*x^2 + 3) - 13/54*log(x^2)