\(\int \frac {x^3}{16-8 x^2+x^4} \, dx\) [540]

Optimal result

Integrand size = 16, antiderivative size = 24 \[ \int \frac {x^3}{16-8 x^2+x^4} \, dx=\frac {2}{4-x^2}+\frac {1}{2} \log \left (4-x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {28, 272, 45} \[ \int \frac {x^3}{16-8 x^2+x^4} \, dx=\frac {2}{4-x^2}+\frac {1}{2} \log \left (4-x^2\right ) \]

[In]

Int[x^3/(16 - 8*x^2 + x^4),x]

[Out]

2/(4 - x^2) + Log[4 - 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 45

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])

Rule 272

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]]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {x^3}{16-8 x^2+x^4} \, dx=-\frac {2}{-4+x^2}+\frac {1}{2} \log \left (-4+x^2\right ) \]

[In]

Integrate[x^3/(16 - 8*x^2 + x^4),x]

[Out]

-2/(-4 + x^2) + Log[-4 + x^2]/2

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79

method	result	size
default	\(-\frac {2}{x^{2}-4}+\frac {\ln \left (x^{2}-4\right )}{2}\)	\(19\)
risch	\(-\frac {2}{x^{2}-4}+\frac {\ln \left (x^{2}-4\right )}{2}\)	\(19\)
norman	\(-\frac {2}{x^{2}-4}+\frac {\ln \left (x -2\right )}{2}+\frac {\ln \left (x +2\right )}{2}\)	\(23\)
parallelrisch	\(\frac {\ln \left (x -2\right ) x^{2}+\ln \left (x +2\right ) x^{2}-4-4 \ln \left (x -2\right )-4 \ln \left (x +2\right )}{2 x^{2}-8}\)	\(40\)

[In]

int(x^3/(x^4-8*x^2+16),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {x^3}{16-8 x^2+x^4} \, dx=\frac {{\left (x^{2} - 4\right )} \log \left (x^{2} - 4\right ) - 4}{2 \, {\left (x^{2} - 4\right )}} \]

[In]

integrate(x^3/(x^4-8*x^2+16),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.58 \[ \int \frac {x^3}{16-8 x^2+x^4} \, dx=\frac {\log {\left (x^{2} - 4 \right )}}{2} - \frac {2}{x^{2} - 4} \]

[In]

integrate(x**3/(x**4-8*x**2+16),x)

[Out]

log(x**2 - 4)/2 - 2/(x**2 - 4)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {x^3}{16-8 x^2+x^4} \, dx=-\frac {2}{x^{2} - 4} + \frac {1}{2} \, \log \left (x^{2} - 4\right ) \]

[In]

integrate(x^3/(x^4-8*x^2+16),x, algorithm="maxima")

[Out]

-2/(x^2 - 4) + 1/2*log(x^2 - 4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {x^3}{16-8 x^2+x^4} \, dx=-\frac {2}{x^{2} - 4} + \frac {1}{2} \, \log \left ({\left | x^{2} - 4 \right |}\right ) \]

[In]

integrate(x^3/(x^4-8*x^2+16),x, algorithm="giac")

[Out]

-2/(x^2 - 4) + 1/2*log(abs(x^2 - 4))

Mupad [B] (verification not implemented)

Time = 13.45 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {x^3}{16-8 x^2+x^4} \, dx=\frac {\ln \left (x^2-4\right )}{2}-\frac {2}{x^2-4} \]

[In]

int(x^3/(x^4 - 8*x^2 + 16),x)

[Out]

log(x^2 - 4)/2 - 2/(x^2 - 4)