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

Optimal. Leaf size=65 \[ \frac {5 x^2}{2}-\frac {17 \tan ^{-1}\left (\frac {x^2+1}{\sqrt {2}}\right )}{8 \sqrt {2}}-\frac {25 \left (x^2+3\right )}{8 \left (x^4+2 x^2+3\right )}-\frac {17}{4} \log \left (x^4+2 x^2+3\right ) \]

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Rubi [A]  time = 0.11, antiderivative size = 65, 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^2}{2}-\frac {25 \left (x^2+3\right )}{8 \left (x^4+2 x^2+3\right )}-\frac {17}{4} \log \left (x^4+2 x^2+3\right )-\frac {17 \tan ^{-1}\left (\frac {x^2+1}{\sqrt {2}}\right )}{8 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x^2)/2 - (25*(3 + x^2))/(8*(3 + 2*x^2 + x^4)) - (17*ArcTan[(1 + x^2)/Sqrt[2]])/(8*Sqrt[2]) - (17*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^3 \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 \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 {-50-56 x+40 x^2}{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 (40-\frac {34 (5+4 x)}{3+2 x+x^2}\right ) \, dx,x,x^2\right )\\ &=\frac {5 x^2}{2}-\frac {25 \left (3+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac {17}{8} \operatorname {Subst}\left (\int \frac {5+4 x}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=\frac {5 x^2}{2}-\frac {25 \left (3+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac {17}{8} \operatorname {Subst}\left (\int \frac {1}{3+2 x+x^2} \, dx,x,x^2\right )-\frac {17}{4} \operatorname {Subst}\left (\int \frac {2+2 x}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=\frac {5 x^2}{2}-\frac {25 \left (3+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac {17}{4} \log \left (3+2 x^2+x^4\right )+\frac {17}{4} \operatorname {Subst}\left (\int \frac {1}{-8-x^2} \, dx,x,2 \left (1+x^2\right )\right )\\ &=\frac {5 x^2}{2}-\frac {25 \left (3+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac {17 \tan ^{-1}\left (\frac {1+x^2}{\sqrt {2}}\right )}{8 \sqrt {2}}-\frac {17}{4} \log \left (3+2 x^2+x^4\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 61, normalized size = 0.94 \begin {gather*} \frac {1}{16} \left (40 x^2-17 \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}-68 \log \left (x^4+2 x^2+3\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(40*x^2 - (50*(3 + x^2))/(3 + 2*x^2 + x^4) - 17*Sqrt[2]*ArcTan[(1 + x^2)/Sqrt[2]] - 68*Log[3 + 2*x^2 + x^4])/1
6

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

[Out]

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

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fricas [A]  time = 1.00, size = 80, normalized size = 1.23 \begin {gather*} \frac {40 \, x^{6} + 80 \, x^{4} - 17 \, \sqrt {2} {\left (x^{4} + 2 \, x^{2} + 3\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) + 70 \, x^{2} - 68 \, {\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^3*(5*x^6+3*x^4+x^2+4)/(x^4+2*x^2+3)^2,x, algorithm="fricas")

[Out]

1/16*(40*x^6 + 80*x^4 - 17*sqrt(2)*(x^4 + 2*x^2 + 3)*arctan(1/2*sqrt(2)*(x^2 + 1)) + 70*x^2 - 68*(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 = 1.18, size = 54, normalized size = 0.83 \begin {gather*} \frac {5}{2} \, x^{2} - \frac {17}{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 {17}{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^3*(5*x^6+3*x^4+x^2+4)/(x^4+2*x^2+3)^2,x, algorithm="giac")

[Out]

5/2*x^2 - 17/16*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 25/8*(x^2 + 3)/(x^4 + 2*x^2 + 3) - 17/4*log(x^4 + 2*x^
2 + 3)

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maple [A]  time = 0.01, size = 59, normalized size = 0.91 \begin {gather*} \frac {5 x^{2}}{2}-\frac {17 \sqrt {2}\, \arctan \left (\frac {\left (2 x^{2}+2\right ) \sqrt {2}}{4}\right )}{16}-\frac {17 \ln \left (x^{4}+2 x^{2}+3\right )}{4}-\frac {\frac {25 x^{2}}{4}+\frac {75}{4}}{2 \left (x^{4}+2 x^{2}+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.72, size = 54, normalized size = 0.83 \begin {gather*} \frac {5}{2} \, x^{2} - \frac {17}{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 {17}{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^3*(5*x^6+3*x^4+x^2+4)/(x^4+2*x^2+3)^2,x, algorithm="maxima")

[Out]

5/2*x^2 - 17/16*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 25/8*(x^2 + 3)/(x^4 + 2*x^2 + 3) - 17/4*log(x^4 + 2*x^
2 + 3)

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mupad [B]  time = 0.92, size = 60, normalized size = 0.92 \begin {gather*} \frac {5\,x^2}{2}-\frac {\frac {25\,x^2}{8}+\frac {75}{8}}{x^4+2\,x^2+3}-\frac {17\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x^2}{2}+\frac {\sqrt {2}}{2}\right )}{16}-\frac {17\,\ln \left (x^4+2\,x^2+3\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*x^2)/2 - ((25*x^2)/8 + 75/8)/(2*x^2 + x^4 + 3) - (17*2^(1/2)*atan(2^(1/2)/2 + (2^(1/2)*x^2)/2))/16 - (17*lo
g(2*x^2 + x^4 + 3))/4

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sympy [A]  time = 0.18, size = 68, normalized size = 1.05 \begin {gather*} \frac {5 x^{2}}{2} + \frac {- 25 x^{2} - 75}{8 x^{4} + 16 x^{2} + 24} - \frac {17 \log {\left (x^{4} + 2 x^{2} + 3 \right )}}{4} - \frac {17 \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**3*(5*x**6+3*x**4+x**2+4)/(x**4+2*x**2+3)**2,x)

[Out]

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

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