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

Optimal. Leaf size=66 \[ \frac {89 \tan ^{-1}\left (\frac {x^2+1}{\sqrt {2}}\right )}{72 \sqrt {2}}+\frac {25 \left (1-x^2\right )}{24 \left (x^4+2 x^2+3\right )}-\frac {1}{9} \log \left (x^4+2 x^2+3\right )+\frac {4 \log (x)}{9} \]

[Out]

25/24*(-x^2+1)/(x^4+2*x^2+3)+4/9*ln(x)-1/9*ln(x^4+2*x^2+3)+89/144*arctan(1/2*(x^2+1)*2^(1/2))*2^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 66, 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, 800, 634, 618, 204, 628} \[ \frac {25 \left (1-x^2\right )}{24 \left (x^4+2 x^2+3\right )}-\frac {1}{9} \log \left (x^4+2 x^2+3\right )+\frac {89 \tan ^{-1}\left (\frac {x^2+1}{\sqrt {2}}\right )}{72 \sqrt {2}}+\frac {4 \log (x)}{9} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(25*(1 - x^2))/(24*(3 + 2*x^2 + x^4)) + (89*ArcTan[(1 + x^2)/Sqrt[2]])/(72*Sqrt[2]) + (4*Log[x])/9 - Log[3 + 2
*x^2 + x^4]/9

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 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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 \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 \left (3+2 x+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac {25 \left (1-x^2\right )}{24 \left (3+2 x^2+x^4\right )}+\frac {1}{16} \operatorname {Subst}\left (\int \frac {\frac {32}{3}+\frac {70 x}{3}}{x \left (3+2 x+x^2\right )} \, dx,x,x^2\right )\\ &=\frac {25 \left (1-x^2\right )}{24 \left (3+2 x^2+x^4\right )}+\frac {1}{16} \operatorname {Subst}\left (\int \left (\frac {32}{9 x}-\frac {2 (-73+16 x)}{9 \left (3+2 x+x^2\right )}\right ) \, dx,x,x^2\right )\\ &=\frac {25 \left (1-x^2\right )}{24 \left (3+2 x^2+x^4\right )}+\frac {4 \log (x)}{9}-\frac {1}{72} \operatorname {Subst}\left (\int \frac {-73+16 x}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=\frac {25 \left (1-x^2\right )}{24 \left (3+2 x^2+x^4\right )}+\frac {4 \log (x)}{9}-\frac {1}{9} \operatorname {Subst}\left (\int \frac {2+2 x}{3+2 x+x^2} \, dx,x,x^2\right )+\frac {89}{72} \operatorname {Subst}\left (\int \frac {1}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=\frac {25 \left (1-x^2\right )}{24 \left (3+2 x^2+x^4\right )}+\frac {4 \log (x)}{9}-\frac {1}{9} \log \left (3+2 x^2+x^4\right )-\frac {89}{36} \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 )}{24 \left (3+2 x^2+x^4\right )}+\frac {89 \tan ^{-1}\left (\frac {1+x^2}{\sqrt {2}}\right )}{72 \sqrt {2}}+\frac {4 \log (x)}{9}-\frac {1}{9} \log \left (3+2 x^2+x^4\right )\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 93, normalized size = 1.41 \[ \frac {1}{288} \left (-\sqrt {2} \left (16 \sqrt {2}+89 i\right ) \log \left (x^2-i \sqrt {2}+1\right )+\sqrt {2} \left (-16 \sqrt {2}+89 i\right ) \log \left (x^2+i \sqrt {2}+1\right )-\frac {300 \left (x^2-1\right )}{x^4+2 x^2+3}+128 \log (x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-300*(-1 + x^2))/(3 + 2*x^2 + x^4) + 128*Log[x] - Sqrt[2]*(89*I + 16*Sqrt[2])*Log[1 - I*Sqrt[2] + x^2] + Sqr
t[2]*(89*I - 16*Sqrt[2])*Log[1 + I*Sqrt[2] + x^2])/288

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fricas [A]  time = 0.78, size = 84, normalized size = 1.27 \[ \frac {89 \, \sqrt {2} {\left (x^{4} + 2 \, x^{2} + 3\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) - 150 \, x^{2} - 16 \, {\left (x^{4} + 2 \, x^{2} + 3\right )} \log \left (x^{4} + 2 \, x^{2} + 3\right ) + 64 \, {\left (x^{4} + 2 \, x^{2} + 3\right )} \log \relax (x) + 150}{144 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/144*(89*sqrt(2)*(x^4 + 2*x^2 + 3)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 150*x^2 - 16*(x^4 + 2*x^2 + 3)*log(x^4 + 2
*x^2 + 3) + 64*(x^4 + 2*x^2 + 3)*log(x) + 150)/(x^4 + 2*x^2 + 3)

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giac [A]  time = 1.10, size = 62, normalized size = 0.94 \[ \frac {89}{144} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) + \frac {8 \, x^{4} - 59 \, x^{2} + 99}{72 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} - \frac {1}{9} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) + \frac {2}{9} \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

89/144*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) + 1/72*(8*x^4 - 59*x^2 + 99)/(x^4 + 2*x^2 + 3) - 1/9*log(x^4 + 2*
x^2 + 3) + 2/9*log(x^2)

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maple [A]  time = 0.01, size = 58, normalized size = 0.88 \[ \frac {89 \sqrt {2}\, \arctan \left (\frac {\left (2 x^{2}+2\right ) \sqrt {2}}{4}\right )}{144}+\frac {4 \ln \relax (x )}{9}-\frac {\ln \left (x^{4}+2 x^{2}+3\right )}{9}-\frac {\frac {75 x^{2}}{4}-\frac {75}{4}}{18 \left (x^{4}+2 x^{2}+3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/9*ln(x)-1/18*(75/4*x^2-75/4)/(x^4+2*x^2+3)-1/9*ln(x^4+2*x^2+3)+89/144*2^(1/2)*arctan(1/4*(2*x^2+2)*2^(1/2))

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maxima [A]  time = 1.68, size = 55, normalized size = 0.83 \[ \frac {89}{144} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) - \frac {25 \, {\left (x^{2} - 1\right )}}{24 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} - \frac {1}{9} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) + \frac {2}{9} \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

89/144*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 25/24*(x^2 - 1)/(x^4 + 2*x^2 + 3) - 1/9*log(x^4 + 2*x^2 + 3) +
2/9*log(x^2)

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mupad [B]  time = 0.91, size = 59, normalized size = 0.89 \[ \frac {4\,\ln \relax (x)}{9}-\frac {\ln \left (x^4+2\,x^2+3\right )}{9}-\frac {\frac {25\,x^2}{24}-\frac {25}{24}}{x^4+2\,x^2+3}+\frac {89\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x^2}{2}+\frac {\sqrt {2}}{2}\right )}{144} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*log(x))/9 - log(2*x^2 + x^4 + 3)/9 - ((25*x^2)/24 - 25/24)/(2*x^2 + x^4 + 3) + (89*2^(1/2)*atan(2^(1/2)/2 +
(2^(1/2)*x^2)/2))/144

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sympy [A]  time = 0.20, size = 65, normalized size = 0.98 \[ \frac {25 - 25 x^{2}}{24 x^{4} + 48 x^{2} + 72} + \frac {4 \log {\relax (x )}}{9} - \frac {\log {\left (x^{4} + 2 x^{2} + 3 \right )}}{9} + \frac {89 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x^{2}}{2} + \frac {\sqrt {2}}{2} \right )}}{144} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(25 - 25*x**2)/(24*x**4 + 48*x**2 + 72) + 4*log(x)/9 - log(x**4 + 2*x**2 + 3)/9 + 89*sqrt(2)*atan(sqrt(2)*x**2
/2 + sqrt(2)/2)/144

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