### 3.76 $$\int \frac{x^5 (4+x^2+3 x^4+5 x^6)}{(2+3 x^2+x^4)^2} \, dx$$

Optimal. Leaf size=54 $\frac{5 x^4}{4}-\frac{27 x^2}{2}+\frac{103 x^2+102}{2 \left (x^4+3 x^2+2\right )}+3 \log \left (x^2+1\right )+46 \log \left (x^2+2\right )$

[Out]

(-27*x^2)/2 + (5*x^4)/4 + (102 + 103*x^2)/(2*(2 + 3*x^2 + x^4)) + 3*Log[1 + x^2] + 46*Log[2 + x^2]

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Rubi [A]  time = 0.108365, antiderivative size = 54, normalized size of antiderivative = 1., 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} $\frac{5 x^4}{4}-\frac{27 x^2}{2}+\frac{103 x^2+102}{2 \left (x^4+3 x^2+2\right )}+3 \log \left (x^2+1\right )+46 \log \left (x^2+2\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-27*x^2)/2 + (5*x^4)/4 + (102 + 103*x^2)/(2*(2 + 3*x^2 + x^4)) + 3*Log[1 + x^2] + 46*Log[2 + x^2]

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]

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 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 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 31

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

Rubi steps

\begin{align*} \int \frac{x^5 \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^2 \left (4+x+3 x^2+5 x^3\right )}{\left (2+3 x+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{102+103 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac{1}{2} \operatorname{Subst}\left (\int \frac{-50-27 x+12 x^2-5 x^3}{2+3 x+x^2} \, dx,x,x^2\right )\\ &=\frac{102+103 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac{1}{2} \operatorname{Subst}\left (\int \left (27-5 x-\frac{2 (52+49 x)}{2+3 x+x^2}\right ) \, dx,x,x^2\right )\\ &=-\frac{27 x^2}{2}+\frac{5 x^4}{4}+\frac{102+103 x^2}{2 \left (2+3 x^2+x^4\right )}+\operatorname{Subst}\left (\int \frac{52+49 x}{2+3 x+x^2} \, dx,x,x^2\right )\\ &=-\frac{27 x^2}{2}+\frac{5 x^4}{4}+\frac{102+103 x^2}{2 \left (2+3 x^2+x^4\right )}+3 \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,x^2\right )+46 \operatorname{Subst}\left (\int \frac{1}{2+x} \, dx,x,x^2\right )\\ &=-\frac{27 x^2}{2}+\frac{5 x^4}{4}+\frac{102+103 x^2}{2 \left (2+3 x^2+x^4\right )}+3 \log \left (1+x^2\right )+46 \log \left (2+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0244659, size = 54, normalized size = 1. $\frac{5 x^4}{4}-\frac{27 x^2}{2}+\frac{103 x^2+102}{2 \left (x^4+3 x^2+2\right )}+3 \log \left (x^2+1\right )+46 \log \left (x^2+2\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-27*x^2)/2 + (5*x^4)/4 + (102 + 103*x^2)/(2*(2 + 3*x^2 + x^4)) + 3*Log[1 + x^2] + 46*Log[2 + x^2]

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Maple [A]  time = 0.014, size = 46, normalized size = 0.9 \begin{align*}{\frac{5\,{x}^{4}}{4}}-{\frac{27\,{x}^{2}}{2}}+46\,\ln \left ({x}^{2}+2 \right ) +52\, \left ({x}^{2}+2 \right ) ^{-1}+3\,\ln \left ({x}^{2}+1 \right ) -{\frac{1}{2\,{x}^{2}+2}} \end{align*}

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+3*x^2+2)^2,x)

[Out]

5/4*x^4-27/2*x^2+46*ln(x^2+2)+52/(x^2+2)+3*ln(x^2+1)-1/2/(x^2+1)

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Maxima [A]  time = 0.996348, size = 65, normalized size = 1.2 \begin{align*} \frac{5}{4} \, x^{4} - \frac{27}{2} \, x^{2} + \frac{103 \, x^{2} + 102}{2 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} + 46 \, \log \left (x^{2} + 2\right ) + 3 \, \log \left (x^{2} + 1\right ) \end{align*}

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+3*x^2+2)^2,x, algorithm="maxima")

[Out]

5/4*x^4 - 27/2*x^2 + 1/2*(103*x^2 + 102)/(x^4 + 3*x^2 + 2) + 46*log(x^2 + 2) + 3*log(x^2 + 1)

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Fricas [A]  time = 1.96442, size = 186, normalized size = 3.44 \begin{align*} \frac{5 \, x^{8} - 39 \, x^{6} - 152 \, x^{4} + 98 \, x^{2} + 184 \,{\left (x^{4} + 3 \, x^{2} + 2\right )} \log \left (x^{2} + 2\right ) + 12 \,{\left (x^{4} + 3 \, x^{2} + 2\right )} \log \left (x^{2} + 1\right ) + 204}{4 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} \end{align*}

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+3*x^2+2)^2,x, algorithm="fricas")

[Out]

1/4*(5*x^8 - 39*x^6 - 152*x^4 + 98*x^2 + 184*(x^4 + 3*x^2 + 2)*log(x^2 + 2) + 12*(x^4 + 3*x^2 + 2)*log(x^2 + 1
) + 204)/(x^4 + 3*x^2 + 2)

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Sympy [A]  time = 0.154507, size = 48, normalized size = 0.89 \begin{align*} \frac{5 x^{4}}{4} - \frac{27 x^{2}}{2} + \frac{103 x^{2} + 102}{2 x^{4} + 6 x^{2} + 4} + 3 \log{\left (x^{2} + 1 \right )} + 46 \log{\left (x^{2} + 2 \right )} \end{align*}

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+3*x**2+2)**2,x)

[Out]

5*x**4/4 - 27*x**2/2 + (103*x**2 + 102)/(2*x**4 + 6*x**2 + 4) + 3*log(x**2 + 1) + 46*log(x**2 + 2)

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Giac [A]  time = 1.13079, size = 72, normalized size = 1.33 \begin{align*} \frac{5}{4} \, x^{4} - \frac{27}{2} \, x^{2} - \frac{49 \, x^{4} + 44 \, x^{2} - 4}{2 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} + 46 \, \log \left (x^{2} + 2\right ) + 3 \, \log \left (x^{2} + 1\right ) \end{align*}

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+3*x^2+2)^2,x, algorithm="giac")

[Out]

5/4*x^4 - 27/2*x^2 - 1/2*(49*x^4 + 44*x^2 - 4)/(x^4 + 3*x^2 + 2) + 46*log(x^2 + 2) + 3*log(x^2 + 1)