### 3.78 $$\int \frac{x (4+x^2+3 x^4+5 x^6)}{(2+3 x^2+x^4)^2} \, dx$$

Optimal. Leaf size=42 $\frac{25 x^2+24}{2 \left (x^4+3 x^2+2\right )}+4 \log \left (x^2+1\right )-\frac{3}{2} \log \left (x^2+2\right )$

[Out]

(24 + 25*x^2)/(2*(2 + 3*x^2 + x^4)) + 4*Log[1 + x^2] - (3*Log[2 + x^2])/2

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Rubi [A]  time = 0.0491766, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 29, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.138, Rules used = {1663, 1660, 632, 31} $\frac{25 x^2+24}{2 \left (x^4+3 x^2+2\right )}+4 \log \left (x^2+1\right )-\frac{3}{2} \log \left (x^2+2\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(24 + 25*x^2)/(2*(2 + 3*x^2 + x^4)) + 4*Log[1 + x^2] - (3*Log[2 + x^2])/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 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 \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{4+x+3 x^2+5 x^3}{\left (2+3 x+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{24+25 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac{1}{2} \operatorname{Subst}\left (\int \frac{-13-5 x}{2+3 x+x^2} \, dx,x,x^2\right )\\ &=\frac{24+25 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac{3}{2} \operatorname{Subst}\left (\int \frac{1}{2+x} \, dx,x,x^2\right )+4 \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,x^2\right )\\ &=\frac{24+25 x^2}{2 \left (2+3 x^2+x^4\right )}+4 \log \left (1+x^2\right )-\frac{3}{2} \log \left (2+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0170239, size = 42, normalized size = 1. $\frac{25 x^2+24}{2 \left (x^4+3 x^2+2\right )}+4 \log \left (x^2+1\right )-\frac{3}{2} \log \left (x^2+2\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(24 + 25*x^2)/(2*(2 + 3*x^2 + x^4)) + 4*Log[1 + x^2] - (3*Log[2 + x^2])/2

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Maple [A]  time = 0.015, size = 36, normalized size = 0.9 \begin{align*} -{\frac{3\,\ln \left ({x}^{2}+2 \right ) }{2}}+13\, \left ({x}^{2}+2 \right ) ^{-1}+4\,\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*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^2,x)

[Out]

-3/2*ln(x^2+2)+13/(x^2+2)+4*ln(x^2+1)-1/2/(x^2+1)

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Maxima [A]  time = 0.965276, size = 51, normalized size = 1.21 \begin{align*} \frac{25 \, x^{2} + 24}{2 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} - \frac{3}{2} \, \log \left (x^{2} + 2\right ) + 4 \, \log \left (x^{2} + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(25*x^2 + 24)/(x^4 + 3*x^2 + 2) - 3/2*log(x^2 + 2) + 4*log(x^2 + 1)

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Fricas [A]  time = 2.00301, size = 144, normalized size = 3.43 \begin{align*} \frac{25 \, x^{2} - 3 \,{\left (x^{4} + 3 \, x^{2} + 2\right )} \log \left (x^{2} + 2\right ) + 8 \,{\left (x^{4} + 3 \, x^{2} + 2\right )} \log \left (x^{2} + 1\right ) + 24}{2 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(25*x^2 - 3*(x^4 + 3*x^2 + 2)*log(x^2 + 2) + 8*(x^4 + 3*x^2 + 2)*log(x^2 + 1) + 24)/(x^4 + 3*x^2 + 2)

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Sympy [A]  time = 0.150097, size = 36, normalized size = 0.86 \begin{align*} \frac{25 x^{2} + 24}{2 x^{4} + 6 x^{2} + 4} + 4 \log{\left (x^{2} + 1 \right )} - \frac{3 \log{\left (x^{2} + 2 \right )}}{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(25*x**2 + 24)/(2*x**4 + 6*x**2 + 4) + 4*log(x**2 + 1) - 3*log(x**2 + 2)/2

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Giac [A]  time = 1.12106, size = 54, normalized size = 1.29 \begin{align*} \frac{25 \, x^{2} + 24}{2 \,{\left (x^{2} + 2\right )}{\left (x^{2} + 1\right )}} - \frac{3}{2} \, \log \left (x^{2} + 2\right ) + 4 \, \log \left (x^{2} + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(25*x^2 + 24)/((x^2 + 2)*(x^2 + 1)) - 3/2*log(x^2 + 2) + 4*log(x^2 + 1)