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

Optimal. Leaf size=80 \[ -\frac {1}{9 x^4}+\frac {13}{54 x^2}+\frac {125 \tan ^{-1}\left (\frac {x^2+1}{\sqrt {2}}\right )}{216 \sqrt {2}}+\frac {25 \left (5 x^2+7\right )}{216 \left (x^4+2 x^2+3\right )}-\frac {13}{108} \log \left (x^4+2 x^2+3\right )+\frac {13 \log (x)}{27} \]

[Out]

-1/9/x^4+13/54/x^2+25/216*(5*x^2+7)/(x^4+2*x^2+3)+13/27*ln(x)-13/108*ln(x^4+2*x^2+3)+125/432*arctan(1/2*(x^2+1
)*2^(1/2))*2^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 80, 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} \[ \frac {25 \left (5 x^2+7\right )}{216 \left (x^4+2 x^2+3\right )}+\frac {13}{54 x^2}-\frac {1}{9 x^4}-\frac {13}{108} \log \left (x^4+2 x^2+3\right )+\frac {125 \tan ^{-1}\left (\frac {x^2+1}{\sqrt {2}}\right )}{216 \sqrt {2}}+\frac {13 \log (x)}{27} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(9*x^4) + 13/(54*x^2) + (25*(7 + 5*x^2))/(216*(3 + 2*x^2 + x^4)) + (125*ArcTan[(1 + x^2)/Sqrt[2]])/(216*Sqr
t[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^5 \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^3 \left (3+2 x+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac {25 \left (7+5 x^2\right )}{216 \left (3+2 x^2+x^4\right )}+\frac {1}{16} \operatorname {Subst}\left (\int \frac {\frac {32}{3}-\frac {40 x}{9}+\frac {200 x^2}{27}+\frac {250 x^3}{27}}{x^3 \left (3+2 x+x^2\right )} \, dx,x,x^2\right )\\ &=\frac {25 \left (7+5 x^2\right )}{216 \left (3+2 x^2+x^4\right )}+\frac {1}{16} \operatorname {Subst}\left (\int \left (\frac {32}{9 x^3}-\frac {104}{27 x^2}+\frac {104}{27 x}-\frac {2 (-73+52 x)}{27 \left (3+2 x+x^2\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac {1}{9 x^4}+\frac {13}{54 x^2}+\frac {25 \left (7+5 x^2\right )}{216 \left (3+2 x^2+x^4\right )}+\frac {13 \log (x)}{27}-\frac {1}{216} \operatorname {Subst}\left (\int \frac {-73+52 x}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=-\frac {1}{9 x^4}+\frac {13}{54 x^2}+\frac {25 \left (7+5 x^2\right )}{216 \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 {125}{216} \operatorname {Subst}\left (\int \frac {1}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=-\frac {1}{9 x^4}+\frac {13}{54 x^2}+\frac {25 \left (7+5 x^2\right )}{216 \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 {125}{108} \operatorname {Subst}\left (\int \frac {1}{-8-x^2} \, dx,x,2 \left (1+x^2\right )\right )\\ &=-\frac {1}{9 x^4}+\frac {13}{54 x^2}+\frac {25 \left (7+5 x^2\right )}{216 \left (3+2 x^2+x^4\right )}+\frac {125 \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 = 0.06, size = 105, normalized size = 1.31 \[ \frac {1}{864} \left (-\frac {96}{x^4}+\frac {208}{x^2}-\sqrt {2} \left (52 \sqrt {2}+125 i\right ) \log \left (x^2-i \sqrt {2}+1\right )+\sqrt {2} \left (-52 \sqrt {2}+125 i\right ) \log \left (x^2+i \sqrt {2}+1\right )+\frac {100 \left (5 x^2+7\right )}{x^4+2 x^2+3}+416 \log (x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-96/x^4 + 208/x^2 + (100*(7 + 5*x^2))/(3 + 2*x^2 + x^4) + 416*Log[x] - Sqrt[2]*(125*I + 52*Sqrt[2])*Log[1 - I
*Sqrt[2] + x^2] + Sqrt[2]*(125*I - 52*Sqrt[2])*Log[1 + I*Sqrt[2] + x^2])/864

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fricas [A]  time = 0.80, size = 110, normalized size = 1.38 \[ \frac {354 \, x^{6} + 510 \, x^{4} + 125 \, \sqrt {2} {\left (x^{8} + 2 \, x^{6} + 3 \, x^{4}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) + 216 \, x^{2} - 52 \, {\left (x^{8} + 2 \, x^{6} + 3 \, x^{4}\right )} \log \left (x^{4} + 2 \, x^{2} + 3\right ) + 208 \, {\left (x^{8} + 2 \, x^{6} + 3 \, x^{4}\right )} \log \relax (x) - 144}{432 \, {\left (x^{8} + 2 \, x^{6} + 3 \, x^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/432*(354*x^6 + 510*x^4 + 125*sqrt(2)*(x^8 + 2*x^6 + 3*x^4)*arctan(1/2*sqrt(2)*(x^2 + 1)) + 216*x^2 - 52*(x^8
+ 2*x^6 + 3*x^4)*log(x^4 + 2*x^2 + 3) + 208*(x^8 + 2*x^6 + 3*x^4)*log(x) - 144)/(x^8 + 2*x^6 + 3*x^4)

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giac [A]  time = 1.16, size = 79, normalized size = 0.99 \[ \frac {125}{432} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) + \frac {26 \, x^{4} + 177 \, x^{2} + 253}{216 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} - \frac {39 \, x^{4} - 26 \, x^{2} + 12}{108 \, x^{4}} - \frac {13}{108} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) + \frac {13}{54} \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/432*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) + 1/216*(26*x^4 + 177*x^2 + 253)/(x^4 + 2*x^2 + 3) - 1/108*(39*x
^4 - 26*x^2 + 12)/x^4 - 13/108*log(x^4 + 2*x^2 + 3) + 13/54*log(x^2)

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maple [A]  time = 0.01, size = 68, normalized size = 0.85 \[ \frac {125 \sqrt {2}\, \arctan \left (\frac {\left (2 x^{2}+2\right ) \sqrt {2}}{4}\right )}{432}+\frac {13 \ln \relax (x )}{27}-\frac {13 \ln \left (x^{4}+2 x^{2}+3\right )}{108}+\frac {13}{54 x^{2}}-\frac {1}{9 x^{4}}-\frac {-\frac {125 x^{2}}{4}-\frac {175}{4}}{54 \left (x^{4}+2 x^{2}+3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9/x^4+13/54/x^2+13/27*ln(x)-1/54*(-125/4*x^2-175/4)/(x^4+2*x^2+3)-13/108*ln(x^4+2*x^2+3)+125/432*2^(1/2)*ar
ctan(1/4*(2*x^2+2)*2^(1/2))

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maxima [A]  time = 2.38, size = 71, normalized size = 0.89 \[ \frac {125}{432} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) + \frac {59 \, x^{6} + 85 \, x^{4} + 36 \, x^{2} - 24}{72 \, {\left (x^{8} + 2 \, x^{6} + 3 \, x^{4}\right )}} - \frac {13}{108} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) + \frac {13}{54} \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/432*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) + 1/72*(59*x^6 + 85*x^4 + 36*x^2 - 24)/(x^8 + 2*x^6 + 3*x^4) - 1
3/108*log(x^4 + 2*x^2 + 3) + 13/54*log(x^2)

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mupad [B]  time = 0.06, size = 72, normalized size = 0.90 \[ \frac {13\,\ln \relax (x)}{27}-\frac {13\,\ln \left (x^4+2\,x^2+3\right )}{108}+\frac {\frac {59\,x^6}{72}+\frac {85\,x^4}{72}+\frac {x^2}{2}-\frac {1}{3}}{x^8+2\,x^6+3\,x^4}+\frac {125\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x^2}{2}+\frac {\sqrt {2}}{2}\right )}{432} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(13*log(x))/27 - (13*log(2*x^2 + x^4 + 3))/108 + (x^2/2 + (85*x^4)/72 + (59*x^6)/72 - 1/3)/(3*x^4 + 2*x^6 + x^
8) + (125*2^(1/2)*atan(2^(1/2)/2 + (2^(1/2)*x^2)/2))/432

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sympy [A]  time = 0.23, size = 80, normalized size = 1.00 \[ \frac {13 \log {\relax (x )}}{27} - \frac {13 \log {\left (x^{4} + 2 x^{2} + 3 \right )}}{108} + \frac {125 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x^{2}}{2} + \frac {\sqrt {2}}{2} \right )}}{432} + \frac {59 x^{6} + 85 x^{4} + 36 x^{2} - 24}{72 x^{8} + 144 x^{6} + 216 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

13*log(x)/27 - 13*log(x**4 + 2*x**2 + 3)/108 + 125*sqrt(2)*atan(sqrt(2)*x**2/2 + sqrt(2)/2)/432 + (59*x**6 + 8
5*x**4 + 36*x**2 - 24)/(72*x**8 + 144*x**6 + 216*x**4)

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