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

[Out]

-2/9/x^2-25/72*(x^2+5)/(x^4+2*x^2+3)-13/27*ln(x)+13/108*ln(x^4+2*x^2+3)-71/432*arctan(1/2*(x^2+1)*2^(1/2))*2^(
1/2)

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Rubi [A]  time = 0.13, antiderivative size = 71, normalized size of antiderivative = 1.00, 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 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 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]

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 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 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 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 {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*}

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Mathematica [C]  time = 0.05, size = 97, normalized size = 1.37 \[ \frac {1}{864} \left (-\frac {192}{x^2}+\sqrt {2} \left (52 \sqrt {2}+71 i\right ) \log \left (x^2-i \sqrt {2}+1\right )+\sqrt {2} \left (52 \sqrt {2}-71 i\right ) \log \left (x^2+i \sqrt {2}+1\right )-\frac {300 \left (x^2+5\right )}{x^4+2 x^2+3}-416 \log (x)\right ) \]

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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fricas [A]  time = 0.58, size = 105, normalized size = 1.48 \[ -\frac {246 \, x^{4} + 71 \, \sqrt {2} {\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) + 942 \, x^{2} - 52 \, {\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )} \log \left (x^{4} + 2 \, x^{2} + 3\right ) + 208 \, {\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )} \log \relax (x) + 288}{432 \, {\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )}} \]

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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giac [A]  time = 1.07, size = 66, normalized size = 0.93 \[ -\frac {71}{432} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) - \frac {41 \, x^{4} + 157 \, x^{2} + 48}{72 \, {\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )}} + \frac {13}{108} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) - \frac {13}{54} \, \log \left (x^{2}\right ) \]

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)

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maple [A]  time = 0.01, size = 63, normalized size = 0.89 \[ -\frac {71 \sqrt {2}\, \arctan \left (\frac {\left (2 x^{2}+2\right ) \sqrt {2}}{4}\right )}{432}-\frac {13 \ln \relax (x )}{27}+\frac {13 \ln \left (x^{4}+2 x^{2}+3\right )}{108}-\frac {2}{9 x^{2}}+\frac {-\frac {75 x^{2}}{4}-\frac {375}{4}}{54 x^{4}+108 x^{2}+162} \]

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]

-2/9/x^2-13/27*ln(x)+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))

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maxima [A]  time = 1.43, size = 66, normalized size = 0.93 \[ -\frac {71}{432} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) - \frac {41 \, x^{4} + 157 \, x^{2} + 48}{72 \, {\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )}} + \frac {13}{108} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) - \frac {13}{54} \, \log \left (x^{2}\right ) \]

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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mupad [B]  time = 0.06, size = 68, normalized size = 0.96 \[ \frac {13\,\ln \left (x^4+2\,x^2+3\right )}{108}-\frac {13\,\ln \relax (x)}{27}-\frac {\frac {41\,x^4}{72}+\frac {157\,x^2}{72}+\frac {2}{3}}{x^6+2\,x^4+3\,x^2}-\frac {71\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x^2}{2}+\frac {\sqrt {2}}{2}\right )}{432} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(13*log(2*x^2 + x^4 + 3))/108 - (13*log(x))/27 - ((157*x^2)/72 + (41*x^4)/72 + 2/3)/(3*x^2 + 2*x^4 + x^6) - (7
1*2^(1/2)*atan(2^(1/2)/2 + (2^(1/2)*x^2)/2))/432

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sympy [A]  time = 0.21, size = 76, normalized size = 1.07 \[ \frac {- 41 x^{4} - 157 x^{2} - 48}{72 x^{6} + 144 x^{4} + 216 x^{2}} - \frac {13 \log {\relax (x )}}{27} + \frac {13 \log {\left (x^{4} + 2 x^{2} + 3 \right )}}{108} - \frac {71 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x^{2}}{2} + \frac {\sqrt {2}}{2} \right )}}{432} \]

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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