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

Optimal. Leaf size=74 \[ \frac {5 x^4}{4}-\frac {17 x^2}{2}+\frac {203 \tan ^{-1}\left (\frac {x^2+1}{\sqrt {2}}\right )}{8 \sqrt {2}}+\frac {25 \left (3-x^2\right )}{8 \left (x^4+2 x^2+3\right )}+\frac {19}{4} \log \left (x^4+2 x^2+3\right ) \]

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Rubi [A]  time = 0.12, antiderivative size = 74, 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, 1660, 1657, 634, 618, 204, 628} \begin {gather*} \frac {5 x^4}{4}-\frac {17 x^2}{2}+\frac {25 \left (3-x^2\right )}{8 \left (x^4+2 x^2+3\right )}+\frac {19}{4} \log \left (x^4+2 x^2+3\right )+\frac {203 \tan ^{-1}\left (\frac {x^2+1}{\sqrt {2}}\right )}{8 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-17*x^2)/2 + (5*x^4)/4 + (25*(3 - x^2))/(8*(3 + 2*x^2 + x^4)) + (203*ArcTan[(1 + x^2)/Sqrt[2]])/(8*Sqrt[2]) +
(19*Log[3 + 2*x^2 + x^4])/4

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 1657

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

Rule 1660

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

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 {x^5 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2 \left (4+x+3 x^2+5 x^3\right )}{\left (3+2 x+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac {25 \left (3-x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac {1}{16} \operatorname {Subst}\left (\int \frac {150-56 x^2+40 x^3}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=\frac {25 \left (3-x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac {1}{16} \operatorname {Subst}\left (\int \left (-136+40 x+\frac {2 (279+76 x)}{3+2 x+x^2}\right ) \, dx,x,x^2\right )\\ &=-\frac {17 x^2}{2}+\frac {5 x^4}{4}+\frac {25 \left (3-x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac {1}{8} \operatorname {Subst}\left (\int \frac {279+76 x}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=-\frac {17 x^2}{2}+\frac {5 x^4}{4}+\frac {25 \left (3-x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac {19}{4} \operatorname {Subst}\left (\int \frac {2+2 x}{3+2 x+x^2} \, dx,x,x^2\right )+\frac {203}{8} \operatorname {Subst}\left (\int \frac {1}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=-\frac {17 x^2}{2}+\frac {5 x^4}{4}+\frac {25 \left (3-x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac {19}{4} \log \left (3+2 x^2+x^4\right )-\frac {203}{4} \operatorname {Subst}\left (\int \frac {1}{-8-x^2} \, dx,x,2 \left (1+x^2\right )\right )\\ &=-\frac {17 x^2}{2}+\frac {5 x^4}{4}+\frac {25 \left (3-x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac {203 \tan ^{-1}\left (\frac {1+x^2}{\sqrt {2}}\right )}{8 \sqrt {2}}+\frac {19}{4} \log \left (3+2 x^2+x^4\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 66, normalized size = 0.89 \begin {gather*} \frac {1}{16} \left (20 x^4-136 x^2+203 \sqrt {2} \tan ^{-1}\left (\frac {x^2+1}{\sqrt {2}}\right )-\frac {50 \left (x^2-3\right )}{x^4+2 x^2+3}+76 \log \left (x^4+2 x^2+3\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-136*x^2 + 20*x^4 - (50*(-3 + x^2))/(3 + 2*x^2 + x^4) + 203*Sqrt[2]*ArcTan[(1 + x^2)/Sqrt[2]] + 76*Log[3 + 2*
x^2 + x^4])/16

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^5 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.05, size = 85, normalized size = 1.15 \begin {gather*} \frac {20 \, x^{8} - 96 \, x^{6} - 212 \, x^{4} + 203 \, \sqrt {2} {\left (x^{4} + 2 \, x^{2} + 3\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) - 458 \, x^{2} + 76 \, {\left (x^{4} + 2 \, x^{2} + 3\right )} \log \left (x^{4} + 2 \, x^{2} + 3\right ) + 150}{16 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(20*x^8 - 96*x^6 - 212*x^4 + 203*sqrt(2)*(x^4 + 2*x^2 + 3)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 458*x^2 + 76*(
x^4 + 2*x^2 + 3)*log(x^4 + 2*x^2 + 3) + 150)/(x^4 + 2*x^2 + 3)

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giac [A]  time = 1.09, size = 66, normalized size = 0.89 \begin {gather*} \frac {5}{4} \, x^{4} - \frac {17}{2} \, x^{2} + \frac {203}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) - \frac {38 \, x^{4} + 101 \, x^{2} + 39}{8 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} + \frac {19}{4} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*x^4 - 17/2*x^2 + 203/16*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 1/8*(38*x^4 + 101*x^2 + 39)/(x^4 + 2*x^2 +
3) + 19/4*log(x^4 + 2*x^2 + 3)

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maple [A]  time = 0.01, size = 64, normalized size = 0.86 \begin {gather*} \frac {5 x^{4}}{4}-\frac {17 x^{2}}{2}+\frac {203 \sqrt {2}\, \arctan \left (\frac {\left (2 x^{2}+2\right ) \sqrt {2}}{4}\right )}{16}+\frac {19 \ln \left (x^{4}+2 x^{2}+3\right )}{4}+\frac {-\frac {25 x^{2}}{4}+\frac {75}{4}}{2 x^{4}+4 x^{2}+6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*x^4-17/2*x^2+1/2*(-25/4*x^2+75/4)/(x^4+2*x^2+3)+19/4*ln(x^4+2*x^2+3)+203/16*2^(1/2)*arctan(1/4*(2*x^2+2)*2
^(1/2))

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maxima [A]  time = 1.59, size = 59, normalized size = 0.80 \begin {gather*} \frac {5}{4} \, x^{4} - \frac {17}{2} \, x^{2} + \frac {203}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) - \frac {25 \, {\left (x^{2} - 3\right )}}{8 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} + \frac {19}{4} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*x^4 - 17/2*x^2 + 203/16*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 25/8*(x^2 - 3)/(x^4 + 2*x^2 + 3) + 19/4*lo
g(x^4 + 2*x^2 + 3)

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mupad [B]  time = 0.05, size = 65, normalized size = 0.88 \begin {gather*} \frac {19\,\ln \left (x^4+2\,x^2+3\right )}{4}-\frac {\frac {25\,x^2}{8}-\frac {75}{8}}{x^4+2\,x^2+3}+\frac {203\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x^2}{2}+\frac {\sqrt {2}}{2}\right )}{16}-\frac {17\,x^2}{2}+\frac {5\,x^4}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(19*log(2*x^2 + x^4 + 3))/4 - ((25*x^2)/8 - 75/8)/(2*x^2 + x^4 + 3) + (203*2^(1/2)*atan(2^(1/2)/2 + (2^(1/2)*x
^2)/2))/16 - (17*x^2)/2 + (5*x^4)/4

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sympy [A]  time = 0.18, size = 73, normalized size = 0.99 \begin {gather*} \frac {5 x^{4}}{4} - \frac {17 x^{2}}{2} + \frac {75 - 25 x^{2}}{8 x^{4} + 16 x^{2} + 24} + \frac {19 \log {\left (x^{4} + 2 x^{2} + 3 \right )}}{4} + \frac {203 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x^{2}}{2} + \frac {\sqrt {2}}{2} \right )}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*x**4/4 - 17*x**2/2 + (75 - 25*x**2)/(8*x**4 + 16*x**2 + 24) + 19*log(x**4 + 2*x**2 + 3)/4 + 203*sqrt(2)*atan
(sqrt(2)*x**2/2 + sqrt(2)/2)/16

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