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

Optimal. Leaf size=49 \[ \frac {5 x^2}{2}-\frac {7}{2} \log \left (x^2+1\right )-10 \log \left (x^2+2\right )-\frac {51 x^2+50}{2 \left (x^4+3 x^2+2\right )} \]

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Rubi [A]  time = 0.09, antiderivative size = 49, 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^2}{2}-\frac {51 x^2+50}{2 \left (x^4+3 x^2+2\right )}-\frac {7}{2} \log \left (x^2+1\right )-10 \log \left (x^2+2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x^2)/2 - (50 + 51*x^2)/(2*(2 + 3*x^2 + x^4)) - (7*Log[1 + x^2])/2 - 10*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^3 \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 \left (4+x+3 x^2+5 x^3\right )}{\left (2+3 x+x^2\right )^2} \, dx,x,x^2\right )\\ &=-\frac {50+51 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {24+12 x-5 x^2}{2+3 x+x^2} \, dx,x,x^2\right )\\ &=-\frac {50+51 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac {1}{2} \operatorname {Subst}\left (\int \left (-5+\frac {34+27 x}{2+3 x+x^2}\right ) \, dx,x,x^2\right )\\ &=\frac {5 x^2}{2}-\frac {50+51 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {34+27 x}{2+3 x+x^2} \, dx,x,x^2\right )\\ &=\frac {5 x^2}{2}-\frac {50+51 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac {7}{2} \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right )-10 \operatorname {Subst}\left (\int \frac {1}{2+x} \, dx,x,x^2\right )\\ &=\frac {5 x^2}{2}-\frac {50+51 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac {7}{2} \log \left (1+x^2\right )-10 \log \left (2+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 49, normalized size = 1.00 \begin {gather*} \frac {5 x^2}{2}-\frac {7}{2} \log \left (x^2+1\right )-10 \log \left (x^2+2\right )+\frac {-51 x^2-50}{2 \left (x^4+3 x^2+2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x^2)/2 + (-50 - 51*x^2)/(2*(2 + 3*x^2 + x^4)) - (7*Log[1 + x^2])/2 - 10*Log[2 + x^2]

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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 (2+3 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))/(2 + 3*x^2 + x^4)^2,x]

[Out]

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

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fricas [A]  time = 1.18, size = 67, normalized size = 1.37 \begin {gather*} \frac {5 \, x^{6} + 15 \, x^{4} - 41 \, x^{2} - 20 \, {\left (x^{4} + 3 \, x^{2} + 2\right )} \log \left (x^{2} + 2\right ) - 7 \, {\left (x^{4} + 3 \, x^{2} + 2\right )} \log \left (x^{2} + 1\right ) - 50}{2 \, {\left (x^{4} + 3 \, x^{2} + 2\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+3*x^2+2)^2,x, algorithm="fricas")

[Out]

1/2*(5*x^6 + 15*x^4 - 41*x^2 - 20*(x^4 + 3*x^2 + 2)*log(x^2 + 2) - 7*(x^4 + 3*x^2 + 2)*log(x^2 + 1) - 50)/(x^4
+ 3*x^2 + 2)

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giac [A]  time = 0.39, size = 45, normalized size = 0.92 \begin {gather*} \frac {5}{2} \, x^{2} - \frac {51 \, x^{2} + 50}{2 \, {\left (x^{2} + 2\right )} {\left (x^{2} + 1\right )}} - 10 \, \log \left (x^{2} + 2\right ) - \frac {7}{2} \, \log \left (x^{2} + 1\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+3*x^2+2)^2,x, algorithm="giac")

[Out]

5/2*x^2 - 1/2*(51*x^2 + 50)/((x^2 + 2)*(x^2 + 1)) - 10*log(x^2 + 2) - 7/2*log(x^2 + 1)

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maple [A]  time = 0.02, size = 41, normalized size = 0.84 \begin {gather*} \frac {5 x^{2}}{2}-\frac {7 \ln \left (x^{2}+1\right )}{2}-10 \ln \left (x^{2}+2\right )+\frac {1}{2 x^{2}+2}-\frac {26}{x^{2}+2} \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+3*x^2+2)^2,x)

[Out]

5/2*x^2-7/2*ln(x^2+1)+1/2/(x^2+1)-26/(x^2+2)-10*ln(x^2+2)

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maxima [A]  time = 0.51, size = 43, normalized size = 0.88 \begin {gather*} \frac {5}{2} \, x^{2} - \frac {51 \, x^{2} + 50}{2 \, {\left (x^{4} + 3 \, x^{2} + 2\right )}} - 10 \, \log \left (x^{2} + 2\right ) - \frac {7}{2} \, \log \left (x^{2} + 1\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+3*x^2+2)^2,x, algorithm="maxima")

[Out]

5/2*x^2 - 1/2*(51*x^2 + 50)/(x^4 + 3*x^2 + 2) - 10*log(x^2 + 2) - 7/2*log(x^2 + 1)

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mupad [B]  time = 0.04, size = 43, normalized size = 0.88 \begin {gather*} \frac {5\,x^2}{2}-10\,\ln \left (x^2+2\right )-\frac {\frac {51\,x^2}{2}+25}{x^4+3\,x^2+2}-\frac {7\,\ln \left (x^2+1\right )}{2} \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))/(3*x^2 + x^4 + 2)^2,x)

[Out]

(5*x^2)/2 - 10*log(x^2 + 2) - ((51*x^2)/2 + 25)/(3*x^2 + x^4 + 2) - (7*log(x^2 + 1))/2

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sympy [A]  time = 0.17, size = 44, normalized size = 0.90 \begin {gather*} \frac {5 x^{2}}{2} + \frac {- 51 x^{2} - 50}{2 x^{4} + 6 x^{2} + 4} - \frac {7 \log {\left (x^{2} + 1 \right )}}{2} - 10 \log {\left (x^{2} + 2 \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+3*x**2+2)**2,x)

[Out]

5*x**2/2 + (-51*x**2 - 50)/(2*x**4 + 6*x**2 + 4) - 7*log(x**2 + 1)/2 - 10*log(x**2 + 2)

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