3.171 \(\int \frac{x (1+x^2)^3}{(2+2 x^2+x^4)^2} \, dx\)

Optimal. Leaf size=32 \[ \frac{1}{4 \left (x^4+2 x^2+2\right )}+\frac{1}{4} \log \left (x^4+2 x^2+2\right ) \]

[Out]

1/(4*(2 + 2*x^2 + x^4)) + Log[2 + 2*x^2 + x^4]/4

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Rubi [A]  time = 0.0263224, antiderivative size = 39, normalized size of antiderivative = 1.22, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {1247, 686, 628} \[ \frac{1}{4} \log \left (x^4+2 x^2+2\right )-\frac{\left (x^2+1\right )^2}{4 \left (x^4+2 x^2+2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x*(1 + x^2)^3)/(2 + 2*x^2 + x^4)^2,x]

[Out]

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

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 686

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*(d + e*x)^(m - 1)*
(a + b*x + c*x^2)^(p + 1))/(b*(p + 1)), x] - Dist[(d*e*(m - 1))/(b*(p + 1)), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2
*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{x \left (1+x^2\right )^3}{\left (2+2 x^2+x^4\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(1+x)^3}{\left (2+2 x+x^2\right )^2} \, dx,x,x^2\right )\\ &=-\frac{\left (1+x^2\right )^2}{4 \left (2+2 x^2+x^4\right )}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1+x}{2+2 x+x^2} \, dx,x,x^2\right )\\ &=-\frac{\left (1+x^2\right )^2}{4 \left (2+2 x^2+x^4\right )}+\frac{1}{4} \log \left (2+2 x^2+x^4\right )\\ \end{align*}

Mathematica [A]  time = 0.0127078, size = 26, normalized size = 0.81 \[ \frac{1}{4} \left (\frac{1}{\left (x^2+1\right )^2+1}+\log \left (\left (x^2+1\right )^2+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(1 + x^2)^3)/(2 + 2*x^2 + x^4)^2,x]

[Out]

((1 + (1 + x^2)^2)^(-1) + Log[1 + (1 + x^2)^2])/4

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Maple [A]  time = 0.007, size = 29, normalized size = 0.9 \begin{align*}{\frac{1}{4\,{x}^{4}+8\,{x}^{2}+8}}+{\frac{\ln \left ({x}^{4}+2\,{x}^{2}+2 \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(x^2+1)^3/(x^4+2*x^2+2)^2,x)

[Out]

1/4/(x^4+2*x^2+2)+1/4*ln(x^4+2*x^2+2)

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Maxima [A]  time = 0.949625, size = 38, normalized size = 1.19 \begin{align*} \frac{1}{4 \,{\left (x^{4} + 2 \, x^{2} + 2\right )}} + \frac{1}{4} \, \log \left (x^{4} + 2 \, x^{2} + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^2+1)^3/(x^4+2*x^2+2)^2,x, algorithm="maxima")

[Out]

1/4/(x^4 + 2*x^2 + 2) + 1/4*log(x^4 + 2*x^2 + 2)

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Fricas [A]  time = 1.9647, size = 92, normalized size = 2.88 \begin{align*} \frac{{\left (x^{4} + 2 \, x^{2} + 2\right )} \log \left (x^{4} + 2 \, x^{2} + 2\right ) + 1}{4 \,{\left (x^{4} + 2 \, x^{2} + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^2+1)^3/(x^4+2*x^2+2)^2,x, algorithm="fricas")

[Out]

1/4*((x^4 + 2*x^2 + 2)*log(x^4 + 2*x^2 + 2) + 1)/(x^4 + 2*x^2 + 2)

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Sympy [A]  time = 0.128267, size = 26, normalized size = 0.81 \begin{align*} \frac{\log{\left (x^{4} + 2 x^{2} + 2 \right )}}{4} + \frac{1}{4 x^{4} + 8 x^{2} + 8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x**2+1)**3/(x**4+2*x**2+2)**2,x)

[Out]

log(x**4 + 2*x**2 + 2)/4 + 1/(4*x**4 + 8*x**2 + 8)

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Giac [A]  time = 1.05774, size = 38, normalized size = 1.19 \begin{align*} \frac{1}{4 \,{\left (x^{4} + 2 \, x^{2} + 2\right )}} + \frac{1}{4} \, \log \left (x^{4} + 2 \, x^{2} + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^2+1)^3/(x^4+2*x^2+2)^2,x, algorithm="giac")

[Out]

1/4/(x^4 + 2*x^2 + 2) + 1/4*log(x^4 + 2*x^2 + 2)