3.1.73 \(\int \frac {x^7 (4+x^2+3 x^4+5 x^6)}{(2+3 x^2+x^4)^2} \, dx\)

Optimal. Leaf size=61 \[ \frac {5 x^6}{6}-\frac {27 x^4}{4}+49 x^2-\frac {5}{2} \log \left (x^2+1\right )-144 \log \left (x^2+2\right )-\frac {207 x^2+206}{2 \left (x^4+3 x^2+2\right )} \]

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Rubi [A]  time = 0.12, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {1663, 1660, 1657, 632, 31} \begin {gather*} \frac {5 x^6}{6}-\frac {27 x^4}{4}+49 x^2-\frac {207 x^2+206}{2 \left (x^4+3 x^2+2\right )}-\frac {5}{2} \log \left (x^2+1\right )-144 \log \left (x^2+2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^7*(4 + x^2 + 3*x^4 + 5*x^6))/(2 + 3*x^2 + x^4)^2,x]

[Out]

49*x^2 - (27*x^4)/4 + (5*x^6)/6 - (206 + 207*x^2)/(2*(2 + 3*x^2 + x^4)) - (5*Log[1 + x^2])/2 - 144*Log[2 + x^2
]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), 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^7 \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^3 \left (4+x+3 x^2+5 x^3\right )}{\left (2+3 x+x^2\right )^2} \, dx,x,x^2\right )\\ &=-\frac {206+207 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {102+53 x-27 x^2+12 x^3-5 x^4}{2+3 x+x^2} \, dx,x,x^2\right )\\ &=-\frac {206+207 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac {1}{2} \operatorname {Subst}\left (\int \left (-98+27 x-5 x^2+\frac {298+293 x}{2+3 x+x^2}\right ) \, dx,x,x^2\right )\\ &=49 x^2-\frac {27 x^4}{4}+\frac {5 x^6}{6}-\frac {206+207 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {298+293 x}{2+3 x+x^2} \, dx,x,x^2\right )\\ &=49 x^2-\frac {27 x^4}{4}+\frac {5 x^6}{6}-\frac {206+207 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac {5}{2} \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right )-144 \operatorname {Subst}\left (\int \frac {1}{2+x} \, dx,x,x^2\right )\\ &=49 x^2-\frac {27 x^4}{4}+\frac {5 x^6}{6}-\frac {206+207 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac {5}{2} \log \left (1+x^2\right )-144 \log \left (2+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 61, normalized size = 1.00 \begin {gather*} \frac {5 x^6}{6}-\frac {27 x^4}{4}+49 x^2-\frac {5}{2} \log \left (x^2+1\right )-144 \log \left (x^2+2\right )+\frac {-207 x^2-206}{2 \left (x^4+3 x^2+2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^7*(4 + x^2 + 3*x^4 + 5*x^6))/(2 + 3*x^2 + x^4)^2,x]

[Out]

49*x^2 - (27*x^4)/4 + (5*x^6)/6 + (-206 - 207*x^2)/(2*(2 + 3*x^2 + x^4)) - (5*Log[1 + x^2])/2 - 144*Log[2 + x^
2]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^7 \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^7*(4 + x^2 + 3*x^4 + 5*x^6))/(2 + 3*x^2 + x^4)^2,x]

[Out]

IntegrateAlgebraic[(x^7*(4 + x^2 + 3*x^4 + 5*x^6))/(2 + 3*x^2 + x^4)^2, x]

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fricas [A]  time = 1.64, size = 77, normalized size = 1.26 \begin {gather*} \frac {10 \, x^{10} - 51 \, x^{8} + 365 \, x^{6} + 1602 \, x^{4} - 66 \, x^{2} - 1728 \, {\left (x^{4} + 3 \, x^{2} + 2\right )} \log \left (x^{2} + 2\right ) - 30 \, {\left (x^{4} + 3 \, x^{2} + 2\right )} \log \left (x^{2} + 1\right ) - 1236}{12 \, {\left (x^{4} + 3 \, x^{2} + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(5*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^2,x, algorithm="fricas")

[Out]

1/12*(10*x^10 - 51*x^8 + 365*x^6 + 1602*x^4 - 66*x^2 - 1728*(x^4 + 3*x^2 + 2)*log(x^2 + 2) - 30*(x^4 + 3*x^2 +
2)*log(x^2 + 1) - 1236)/(x^4 + 3*x^2 + 2)

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giac [A]  time = 0.32, size = 58, normalized size = 0.95 \begin {gather*} \frac {5}{6} \, x^{6} - \frac {27}{4} \, x^{4} + 49 \, x^{2} + \frac {293 \, x^{4} + 465 \, x^{2} + 174}{4 \, {\left (x^{4} + 3 \, x^{2} + 2\right )}} - 144 \, \log \left (x^{2} + 2\right ) - \frac {5}{2} \, \log \left (x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(5*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^2,x, algorithm="giac")

[Out]

5/6*x^6 - 27/4*x^4 + 49*x^2 + 1/4*(293*x^4 + 465*x^2 + 174)/(x^4 + 3*x^2 + 2) - 144*log(x^2 + 2) - 5/2*log(x^2
+ 1)

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maple [A]  time = 0.02, size = 51, normalized size = 0.84 \begin {gather*} \frac {5 x^{6}}{6}-\frac {27 x^{4}}{4}+49 x^{2}-\frac {5 \ln \left (x^{2}+1\right )}{2}-144 \ln \left (x^{2}+2\right )+\frac {1}{2 x^{2}+2}-\frac {104}{x^{2}+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(5*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^2,x)

[Out]

5/6*x^6-27/4*x^4+49*x^2-5/2*ln(x^2+1)+1/2/(x^2+1)-104/(x^2+2)-144*ln(x^2+2)

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maxima [A]  time = 0.72, size = 53, normalized size = 0.87 \begin {gather*} \frac {5}{6} \, x^{6} - \frac {27}{4} \, x^{4} + 49 \, x^{2} - \frac {207 \, x^{2} + 206}{2 \, {\left (x^{4} + 3 \, x^{2} + 2\right )}} - 144 \, \log \left (x^{2} + 2\right ) - \frac {5}{2} \, \log \left (x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(5*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^2,x, algorithm="maxima")

[Out]

5/6*x^6 - 27/4*x^4 + 49*x^2 - 1/2*(207*x^2 + 206)/(x^4 + 3*x^2 + 2) - 144*log(x^2 + 2) - 5/2*log(x^2 + 1)

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mupad [B]  time = 0.04, size = 53, normalized size = 0.87 \begin {gather*} 49\,x^2-144\,\ln \left (x^2+2\right )-\frac {\frac {207\,x^2}{2}+103}{x^4+3\,x^2+2}-\frac {5\,\ln \left (x^2+1\right )}{2}-\frac {27\,x^4}{4}+\frac {5\,x^6}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^7*(x^2 + 3*x^4 + 5*x^6 + 4))/(3*x^2 + x^4 + 2)^2,x)

[Out]

49*x^2 - 144*log(x^2 + 2) - ((207*x^2)/2 + 103)/(3*x^2 + x^4 + 2) - (5*log(x^2 + 1))/2 - (27*x^4)/4 + (5*x^6)/
6

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sympy [A]  time = 0.17, size = 56, normalized size = 0.92 \begin {gather*} \frac {5 x^{6}}{6} - \frac {27 x^{4}}{4} + 49 x^{2} + \frac {- 207 x^{2} - 206}{2 x^{4} + 6 x^{2} + 4} - \frac {5 \log {\left (x^{2} + 1 \right )}}{2} - 144 \log {\left (x^{2} + 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(5*x**6+3*x**4+x**2+4)/(x**4+3*x**2+2)**2,x)

[Out]

5*x**6/6 - 27*x**4/4 + 49*x**2 + (-207*x**2 - 206)/(2*x**4 + 6*x**2 + 4) - 5*log(x**2 + 1)/2 - 144*log(x**2 +
2)

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