### 3.74 $$\int \frac{x^9 (4+x^2+3 x^4+5 x^6)}{(2+3 x^2+x^4)^2} \, dx$$

Optimal. Leaf size=68 $\frac{5 x^8}{8}-\frac{9 x^6}{2}+\frac{49 x^4}{2}-\frac{293 x^2}{2}+\frac{415 x^2+414}{2 \left (x^4+3 x^2+2\right )}+2 \log \left (x^2+1\right )+392 \log \left (x^2+2\right )$

[Out]

(-293*x^2)/2 + (49*x^4)/2 - (9*x^6)/2 + (5*x^8)/8 + (414 + 415*x^2)/(2*(2 + 3*x^2 + x^4)) + 2*Log[1 + x^2] + 3
92*Log[2 + x^2]

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Rubi [A]  time = 0.126049, antiderivative size = 68, 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} $\frac{5 x^8}{8}-\frac{9 x^6}{2}+\frac{49 x^4}{2}-\frac{293 x^2}{2}+\frac{415 x^2+414}{2 \left (x^4+3 x^2+2\right )}+2 \log \left (x^2+1\right )+392 \log \left (x^2+2\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[(x^9*(4 + x^2 + 3*x^4 + 5*x^6))/(2 + 3*x^2 + x^4)^2,x]

[Out]

(-293*x^2)/2 + (49*x^4)/2 - (9*x^6)/2 + (5*x^8)/8 + (414 + 415*x^2)/(2*(2 + 3*x^2 + x^4)) + 2*Log[1 + x^2] + 3
92*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^9 \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^4 \left (4+x+3 x^2+5 x^3\right )}{\left (2+3 x+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{414+415 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac{1}{2} \operatorname{Subst}\left (\int \frac{-206-105 x+53 x^2-27 x^3+12 x^4-5 x^5}{2+3 x+x^2} \, dx,x,x^2\right )\\ &=\frac{414+415 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac{1}{2} \operatorname{Subst}\left (\int \left (293-98 x+27 x^2-5 x^3-\frac{4 (198+197 x)}{2+3 x+x^2}\right ) \, dx,x,x^2\right )\\ &=-\frac{293 x^2}{2}+\frac{49 x^4}{2}-\frac{9 x^6}{2}+\frac{5 x^8}{8}+\frac{414+415 x^2}{2 \left (2+3 x^2+x^4\right )}+2 \operatorname{Subst}\left (\int \frac{198+197 x}{2+3 x+x^2} \, dx,x,x^2\right )\\ &=-\frac{293 x^2}{2}+\frac{49 x^4}{2}-\frac{9 x^6}{2}+\frac{5 x^8}{8}+\frac{414+415 x^2}{2 \left (2+3 x^2+x^4\right )}+2 \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,x^2\right )+392 \operatorname{Subst}\left (\int \frac{1}{2+x} \, dx,x,x^2\right )\\ &=-\frac{293 x^2}{2}+\frac{49 x^4}{2}-\frac{9 x^6}{2}+\frac{5 x^8}{8}+\frac{414+415 x^2}{2 \left (2+3 x^2+x^4\right )}+2 \log \left (1+x^2\right )+392 \log \left (2+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.035152, size = 62, normalized size = 0.91 $\frac{1}{8} \left (5 x^8-36 x^6+196 x^4-1172 x^2+\frac{4 \left (415 x^2+414\right )}{x^4+3 x^2+2}+16 \log \left (x^2+1\right )+3136 \log \left (x^2+2\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(x^9*(4 + x^2 + 3*x^4 + 5*x^6))/(2 + 3*x^2 + x^4)^2,x]

[Out]

(-1172*x^2 + 196*x^4 - 36*x^6 + 5*x^8 + (4*(414 + 415*x^2))/(2 + 3*x^2 + x^4) + 16*Log[1 + x^2] + 3136*Log[2 +
x^2])/8

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Maple [A]  time = 0.015, size = 56, normalized size = 0.8 \begin{align*}{\frac{5\,{x}^{8}}{8}}-{\frac{9\,{x}^{6}}{2}}+{\frac{49\,{x}^{4}}{2}}-{\frac{293\,{x}^{2}}{2}}+392\,\ln \left ({x}^{2}+2 \right ) +208\, \left ({x}^{2}+2 \right ) ^{-1}+2\,\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^9*(5*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^2,x)

[Out]

5/8*x^8-9/2*x^6+49/2*x^4-293/2*x^2+392*ln(x^2+2)+208/(x^2+2)+2*ln(x^2+1)-1/2/(x^2+1)

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Maxima [A]  time = 0.972301, size = 78, normalized size = 1.15 \begin{align*} \frac{5}{8} \, x^{8} - \frac{9}{2} \, x^{6} + \frac{49}{2} \, x^{4} - \frac{293}{2} \, x^{2} + \frac{415 \, x^{2} + 414}{2 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} + 392 \, \log \left (x^{2} + 2\right ) + 2 \, \log \left (x^{2} + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(5*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^2,x, algorithm="maxima")

[Out]

5/8*x^8 - 9/2*x^6 + 49/2*x^4 - 293/2*x^2 + 1/2*(415*x^2 + 414)/(x^4 + 3*x^2 + 2) + 392*log(x^2 + 2) + 2*log(x^
2 + 1)

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Fricas [A]  time = 1.9619, size = 220, normalized size = 3.24 \begin{align*} \frac{5 \, x^{12} - 21 \, x^{10} + 98 \, x^{8} - 656 \, x^{6} - 3124 \, x^{4} - 684 \, x^{2} + 3136 \,{\left (x^{4} + 3 \, x^{2} + 2\right )} \log \left (x^{2} + 2\right ) + 16 \,{\left (x^{4} + 3 \, x^{2} + 2\right )} \log \left (x^{2} + 1\right ) + 1656}{8 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(5*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^2,x, algorithm="fricas")

[Out]

1/8*(5*x^12 - 21*x^10 + 98*x^8 - 656*x^6 - 3124*x^4 - 684*x^2 + 3136*(x^4 + 3*x^2 + 2)*log(x^2 + 2) + 16*(x^4
+ 3*x^2 + 2)*log(x^2 + 1) + 1656)/(x^4 + 3*x^2 + 2)

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Sympy [A]  time = 0.162926, size = 61, normalized size = 0.9 \begin{align*} \frac{5 x^{8}}{8} - \frac{9 x^{6}}{2} + \frac{49 x^{4}}{2} - \frac{293 x^{2}}{2} + \frac{415 x^{2} + 414}{2 x^{4} + 6 x^{2} + 4} + 2 \log{\left (x^{2} + 1 \right )} + 392 \log{\left (x^{2} + 2 \right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9*(5*x**6+3*x**4+x**2+4)/(x**4+3*x**2+2)**2,x)

[Out]

5*x**8/8 - 9*x**6/2 + 49*x**4/2 - 293*x**2/2 + (415*x**2 + 414)/(2*x**4 + 6*x**2 + 4) + 2*log(x**2 + 1) + 392*
log(x**2 + 2)

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Giac [A]  time = 1.11876, size = 85, normalized size = 1.25 \begin{align*} \frac{5}{8} \, x^{8} - \frac{9}{2} \, x^{6} + \frac{49}{2} \, x^{4} - \frac{293}{2} \, x^{2} - \frac{394 \, x^{4} + 767 \, x^{2} + 374}{2 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} + 392 \, \log \left (x^{2} + 2\right ) + 2 \, \log \left (x^{2} + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(5*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^2,x, algorithm="giac")

[Out]

5/8*x^8 - 9/2*x^6 + 49/2*x^4 - 293/2*x^2 - 1/2*(394*x^4 + 767*x^2 + 374)/(x^4 + 3*x^2 + 2) + 392*log(x^2 + 2)
+ 2*log(x^2 + 1)