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

Optimal. Leaf size=22 \[ \frac{1}{2 \left (x^2+1\right )}+\frac{1}{2} \log \left (x^2+1\right ) \]

[Out]

1/(2*(1 + x^2)) + Log[1 + x^2]/2

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Rubi [A]  time = 0.0108123, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {28, 266, 43} \[ \frac{1}{2 \left (x^2+1\right )}+\frac{1}{2} \log \left (x^2+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3/(1 + 2*x^2 + x^4),x]

[Out]

1/(2*(1 + x^2)) + Log[1 + x^2]/2

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0045505, size = 18, normalized size = 0.82 \[ \frac{1}{2} \left (\frac{1}{x^2+1}+\log \left (x^2+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(1 + 2*x^2 + x^4),x]

[Out]

((1 + x^2)^(-1) + Log[1 + x^2])/2

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Maple [A]  time = 0.047, size = 19, normalized size = 0.9 \begin{align*}{\frac{1}{2\,{x}^{2}+2}}+{\frac{\ln \left ({x}^{2}+1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.965903, size = 24, normalized size = 1.09 \begin{align*} \frac{1}{2 \,{\left (x^{2} + 1\right )}} + \frac{1}{2} \, \log \left (x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(x^2 + 1) + 1/2*log(x^2 + 1)

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Fricas [A]  time = 1.43392, size = 59, normalized size = 2.68 \begin{align*} \frac{{\left (x^{2} + 1\right )} \log \left (x^{2} + 1\right ) + 1}{2 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.083989, size = 15, normalized size = 0.68 \begin{align*} \frac{\log{\left (x^{2} + 1 \right )}}{2} + \frac{1}{2 x^{2} + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x**2 + 1)/2 + 1/(2*x**2 + 2)

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Giac [A]  time = 1.14005, size = 24, normalized size = 1.09 \begin{align*} \frac{1}{2 \,{\left (x^{2} + 1\right )}} + \frac{1}{2} \, \log \left (x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(x^2 + 1) + 1/2*log(x^2 + 1)